From 7480de927c1433d41f5a8494bcdc13d1319a7047 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Maximilian=20Ke=C3=9Fler?= Date: Wed, 16 Feb 2022 05:19:03 +0100 Subject: [PATCH] latexindent: use on everything (experimental) --- 2021_Algebra_I.bak0 | 43 + 2021_Algebra_I.tex | 4 +- inputs/finiteness_conditions.tex | 568 ++- .../nullstellensatz_and_zariski_topology.tex | 3064 +++++++++++++---- inputs/preamble.tex | 16 +- inputs/projective_spaces.tex | 666 +++- inputs/uebersicht.bak0 | 125 + inputs/uebersicht.tex | 172 +- inputs/varieties.tex | 1046 ++++-- 9 files changed, 4319 insertions(+), 1385 deletions(-) create mode 100644 2021_Algebra_I.bak0 create mode 100644 inputs/uebersicht.bak0 diff --git a/2021_Algebra_I.bak0 b/2021_Algebra_I.bak0 new file mode 100644 index 0000000..58fc42f --- /dev/null +++ b/2021_Algebra_I.bak0 @@ -0,0 +1,43 @@ +\documentclass[10pt,ngerman,a4paper, fancyfoot, git]{mkessler-script} + +\course{Algebra I} +\lecturer{Prof.~Dr.~Jens Franke} +\author{Josia Pietsch} + +\usepackage{algebra} + +\begin{document} + +\maketitle +\cleardoublepage + +\tableofcontents +\cleardoublepage + +\input{inputs/preamble} +\cleardoublepage + + +\section{Finiteness conditions} +\input{inputs/finiteness_conditions} + +\section{The Nullstellensatz and the Zariski topology} +\input{inputs/nullstellensatz_and_zariski_topology} + +% Lecture 11 +\section{Projective spaces} +\input{inputs/projective_spaces} + +% Lecture 13 +\section{Varieties} +\input{inputs/varieties} + +\iffalse +\section{Übersicht} +\input{inputs/uebersicht} +\fi + +\cleardoublepage +\printvocabindex + +\end{document} diff --git a/2021_Algebra_I.tex b/2021_Algebra_I.tex index 58fc42f..acf0b7e 100644 --- a/2021_Algebra_I.tex +++ b/2021_Algebra_I.tex @@ -33,8 +33,8 @@ \input{inputs/varieties} \iffalse -\section{Übersicht} -\input{inputs/uebersicht} + \section{Übersicht} + \input{inputs/uebersicht} \fi \cleardoublepage diff --git a/inputs/finiteness_conditions.tex b/inputs/finiteness_conditions.tex index 95fa208..15c0ca0 100644 --- a/inputs/finiteness_conditions.tex +++ b/inputs/finiteness_conditions.tex @@ -4,278 +4,562 @@ Let $R$ be a ring, $M$ an $R$-module, $S \subseteq M$. Then the following sets coincide \begin{enumerate} - \item $\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{finite}, r_s \in R, \right\}$ - \item $\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$ - \item The $\subseteq$-smallest submodule of $M$ containing $S$ + \item + $\left\{ \sum_{s \in + S'} r_{s} \cdot s ~ |~ S + \subseteq S' \text{finite}, r_s \in R, \right\}$ + \item + $\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$ + \item + The $\subseteq$-smallest submodule of $M$ containing $S$ \end{enumerate} - This subset of $N \subseteq M$ is called the \vocab[Module!Submodule]{submodule of $M $ generated by $S$}. If $N= M$ we say that \vocab[Module!generated by subset $S$]{$ M$ is generated by $S$}. - $M$ is finitely generated $:\iff \exists S \subseteq M$ finite such that $M$ is generated by $S$. + This subset of $N \subseteq M$ is called the \vocab[Module! + Submodule]{submodule of $M $ generated by $S$}. + If $N= M$ we say that \vocab[Module! + generated by subset $S$]{$ M$ is generated by $S$}. + $M$ is finitely generated $:\iff \exists S \subseteq M$ finite such that $M$ is + generated by $S$. \end{definition} \begin{definition}[Noetherian $R$-module] - $M$ is a \vocab{Noetherian} $R$-module if the following equivalent conditions hold: + $M$ is a \vocab{Noetherian} $R$-module if the + following equivalent conditions hold: \begin{enumerate} - \item Every submodule $N \subseteq M$ is finitely generated. - \item Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates - \item Every set $\mathfrak{M} \neq \emptyset$ of submodules of $M$ has a $\subseteq$-largest element. + \item + Every submodule $N \subseteq M$ is finitely generated. + \item + Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates + \item + Every set $\mathfrak{M} \neq \emptyset$ of submodules of $M$ has a + $\subseteq$-largest element. \end{enumerate} \end{definition} -\begin{proposition}[Hilbert's Basissatz]\label{basissatz} - If $R$ is a Noetherian ring, then the polynomial rings $R[X_1,\ldots, X_n]$ in finitely many variables are Noetherian. +\begin{proposition}[Hilbert's Basissatz] + \label{basissatz} + If $R$ is a Noetherian ring, then the polynomial rings + $R[X_1,\ldots, X_n]$ in + finitely many variables are Noetherian. \end{proposition} \subsubsection{Properties of finite generation and Noetherianness} \begin{fact}[Properties of Noetherian modules] \begin{enumerate} - \item Every Noetherian module over an arbitrary ring is finitely generated. - \item If $R$ is a Noetherian ring, then an $R$-module is Noetherian iff it is finitely generated. - \item Every submodule of a Noetherian module is Noetherian. + \item + Every Noetherian module over an arbitrary ring is finitely generated. + \item + If $R$ is a Noetherian ring, then an $R$-module is Noetherian iff it is + finitely generated. + \item + Every submodule of a Noetherian module is Noetherian. \end{enumerate} \end{fact} \begin{proof} \begin{enumerate} - \item By definition, $M$ is a submodule of itself. Thus it is finitely generated. - \item Since $M$ is finitely generated, there exists a surjective homomorphism $R^n \to M$. As $R$ is Noetherian, $R^n$ is Noethrian as well. - \item trivial + \item + By definition, $M$ is a submodule of itself. + Thus it is finitely generated. + \item + Since $M$ is finitely generated, there exists a surjective homomorphism $R^n + \to M$. + As $R$ is Noetherian, $R^n$ is Noethrian as well. + \item + trivial \end{enumerate} \end{proof} \begin{fact} Let $M, M', M''$ be $R$-modules. \begin{enumerate} - \item Suppose $M \xrightarrow{p} M''$ is surjective. If $M$ is finitely generated (resp. Noetherian), then so is $M''$. - \item Let $M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ be exact. If $M'$ and $M ''$ are finitely generated (reps. Noetherian), so is $M$. + \item + Suppose $M \xrightarrow{p} + M''$ is surjective. + If $M$ is finitely generated (resp. + Noetherian), then so is $M''$. + \item + Let $M' \xrightarrow{f} + M \xrightarrow{p} M'' \to 0$ be exact. + If $M'$ and $M ''$ are finitely generated (reps. + Noetherian), so is $M$. \end{enumerate} \end{fact} \begin{proof} \begin{enumerate} - \item Consider a sequence $M_0'' \subset M_1'' \subset \ldots \subset M''$. Then $p^{-1} M_i''$ yields a strictly ascending sequence. - If $M$ is generated by $S, |S| < \omega$, then $M''$ is generated by $p(S)$. - \item Because of 1. we can replace $M'$ by $f(M')$ and assume $0 \to M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ to be exact. The fact about finite generation follows from EInführung in die Algebra. + \item + Consider a sequence $M_0'' \subset M_1'' \subset \ldots \subset M''$. + Then $p^{-1} M_i''$ yields a strictly ascending + sequence. + If $M$ is generated by $S, |S| < \omega$, then $M''$ is generated by $p(S)$. + \item + Because of 1. we can replace $M'$ by $f(M')$ and assume $0 \to M' + \xrightarrow{f} + M \xrightarrow{p} M'' \to 0$ to be exact. + The fact about finite generation follows from EInführung in die Algebra. - If $M', M''$ are Noetherian, $N \subseteq M$ a submodule, then $N' \coloneqq f^{-1}(N)$ and $N''\coloneqq p(N)$ are finitely generated. Since $0 \to N' \to N \to N'' \to 0$ is exact, $N$ is finitely generated. + If $M', M''$ are Noetherian, $N \subseteq M$ a submodule, then $N' \coloneqq + f^{-1}(N)$ and $N''\coloneqq p(N)$ are finitely + generated. + Since $0 \to N' \to N \to N'' \to 0$ is exact, $N$ is finitely generated. \end{enumerate} \end{proof} \subsection{Ring extensions of finite type} \begin{definition}[$R$-algebra] - Let $R$ be a ring. An $R$-algebra $(A, \alpha)$ is a ring $A$ with a ring homomorphism $R \xrightarrow{\alpha} A$. - $\alpha$ will usually be omitted. In general $\alpha$ is not assumed to be injective.\\ + Let $R$ be a ring. + An $R$-algebra $(A, \alpha)$ is a ring $A$ with a ring homomorphism $R + \xrightarrow{\alpha} A$. + $\alpha$ will usually be omitted. + In general $\alpha$ is not assumed to be injective. \\ - An $R$-subalgebra is a subring $\alpha(R) \subseteq A' \subseteq A$.\\ -A morphism of $R$-algebras $A \xrightarrow{f} \tilde{A}$ is a ring homomorphism with $\tilde{\alpha} = f \alpha$. + \\ + An $R$-subalgebra is a subring $\alpha(R) \subseteq A' \subseteq + A$.\\ + A morphism of $R$-algebras $A \xrightarrow{f} \tilde{A}$ is + a ring homomorphism with $\tilde{\alpha} = f \alpha$. \end{definition} \begin{definition}[Generated (sub)algebra, algebra of finite type] Let $(A, \alpha)$ be an $R$-algebra. \begin{align} - \alpha: R[X_1,\ldots,X_m] &\longrightarrow A[X_1,\ldots,X_m] \\ - P = \sum_{\beta \in \N^m} p_\beta X^{\beta} &\longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta) X^{\beta} + \alpha: R[X_1,\ldots,X_m] & \longrightarrow A[X_1,\ldots,X_m] \\ + P = \sum_{\beta \in \N^m} p_\beta X^{\beta} & \longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta) + X^{\beta} \end{align} - is a ring homomorphism. We will sometimes write $P(a_1,\ldots,a_m)$ instead of $(\alpha(P))(a_1,\ldots,a_m)$. + is a ring homomorphism. + We will sometimes write $P(a_1,\ldots,a_m)$ instead of + $(\alpha(P))(a_1,\ldots,a_m)$. - Fix $a_1,\ldots,a_m \in A^m$. Then we get a ring homomorphism $R[X_1,\ldots,X_m] \to A$. The image of this ring homomorphism is the $R$-subalgebra of $A$ \vocab[Algebra!generated subalgebra]{generated by the $a_i$}. - $A$ is \vocab[Algebra!of finite type]{of finite type} if it can be generated by finitely many $a_i \in I$. + Fix $a_1,\ldots,a_m \in A^m$. + Then we get a ring homomorphism $R[X_1,\ldots,X_m] \to A$. + The image of this ring homomorphism is the $R$-subalgebra of $A$ + \vocab[Algebra! + generated subalgebra]{generated by the $a_i$}. + $A$ is \vocab[Algebra!of finite type]{of finite type} if it can be generated by finitely many $a_i + \in I$. + + For arbitrary $S \subseteq A$ the subalgebra generated by $S$ is the + intersection of all subalgebras containing $S$ \\ $=$ the union of subalgebras + generated by finite $S' \subseteq S$\\ $= $ the image of + $R[X_s | s \in S]$ + under $P \mapsto (\alpha(P))(S)$. - For arbitrary $S \subseteq A$ the subalgebra generated by $S$ is the intersection of all subalgebras containing $S$ \\ - $=$ the union of subalgebras generated by finite $S' \subseteq S$\\ - $= $ the image of $R[X_s | s \in S]$ under $P \mapsto (\alpha(P))(S)$. - \end{definition} \subsection{Finite ring extensions} % LECTURE 2 \begin{definition}[Finite ring extension] - Let $R$ be a ring and $A$ an $R$-algebra. $A$ is a module over itself and the ringhomomorphism $R \to A$ allows us to derive an $R$-module structure on $A$. - $A$ \vocab[Algebra!finite over]{is finite over} $R$ / the $R$-algebra $A$ is finite / $A / R$ is finite if $A$ is finitely generated as an $R$-module. + Let $R$ be a ring and $A$ an $R$-algebra. $A$ is a + module over itself and the ringhomomorphism $R \to A$ allows us to derive an + $R$-module structure on $A$. + $A$ \vocab[Algebra!finite over]{is finite over} $R$ / the $R$-algebra $A$ is finite / $A / R$ is + finite if $A$ is finitely generated as an $R$-module. \end{definition} \begin{fact}[Basic properties of finiteness] \begin{enumerate}[A] - \item Every ring is finite over itself. - \item A field extension is finite as a ring extension iff it is finite as a field extension. - \item $A$ finite $\implies$ $A$ of finite type. - \item $A / R$ and $B / A$ finite $\implies$ $B / R$ finite. + \item + Every ring is finite over itself. + \item + A field extension is finite as a ring extension iff it is finite as a field + extension. + \item + $A$ finite $\implies$ $A$ of finite type. + \item + $A / R$ and $B / A$ finite $\implies$ $B / R$ finite. \end{enumerate} \end{fact} \begin{proof} \begin{enumerate}[A] - \item $1$ generates $R$ as a module - \item trivial - \item Let $A $ be generated by $a_1,\ldots,a_n$ as an $R$-module. Then $A$ is generated by $a_1,\ldots,a_n$ as an $R$-algebra. - \item Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module and $B$ by $b_1,\ldots,b_n$ as an $A$-module. - For every $b$ there exist $\alpha_j \in A$ such that $b = \sum_{j=1}^{n} \alpha_j b_j$. We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$ for some $\rho_{ij} \in R$ thus - $b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij} a_i b_j$ and the $a_ib_j$ generate $B$ as an $R$-module. -\end{enumerate} - + \item + $1$ generates $R$ as a module + \item + trivial + \item + Let $A $ be generated by $a_1,\ldots,a_n$ as an $R$-module. + Then $A$ is generated by $a_1,\ldots,a_n$ as an $R$-algebra. + \item + Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module and $B$ by + $b_1,\ldots,b_n$ as an $A$-module. + For every $b$ there exist $\alpha_j \in A$ such that $b = + \sum_{j=1}^{n} + \alpha_j b_j$. + We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$ for some + $\rho_{ij} \in R$ + thus $b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij} + a_i b_j$ and the $a_ib_j$ + generate $B$ as an $R$-module. + \end{enumerate} + \end{proof} \subsection{Determinants and Caley-Hamilton} %LECTURE 2 TODO: move to int. elements? -This generalizes some facts about matrices to matrices with elements from commutative rings with $1$. +This generalizes some facts about matrices to matrices with elements from +commutative rings with $1$. \footnote{Most of this even works in commutative rings without $ 1$, since $1$ simply can be adjoined.} \begin{definition}[Determinant] - Let $A = (a_{ij}) \Mat(n,n,R)$. We define the determinant by the Leibniz formula \[ - \det(A) \coloneqq \sum_{\pi \in S_n} \sgn(\pi) \prod_{i=1}^{n} a_{i, \pi(i)} + Let $A = (a_{ij}) + \Mat(n,n,R)$. + We define the determinant by the Leibniz formula + \[ + \det(A) \coloneqq \sum_{\pi + \in S_n} \sgn(\pi) + \prod_{i=1}^{n} a_{i, \pi(i)} \] - Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij} \coloneqq (-1)^{i+j} \cdot M_{ij}$, where $M_{ij}$ is the determinant of the matrix resulting from $A$ after deleting the $i^{\text{th}}$ row and the $j^{\text{th}}$ column. + Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij} + \coloneqq (-1)^{i+j} \cdot + M_{ij}$, where $M_{ij}$ is the + determinant of the matrix resulting from $A$ + after deleting the $i^{\text{th}}$ row and the + $j^{\text{th}}$ column. \end{definition} \begin{fact} \begin{enumerate} - \item $\det(AB) = \det(A)\det(B)$ - \item Development along a row or column works. - \item Cramer's rule: $A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot A = \det(A) \cdot \mathbf{1}_n$. $A$ is invertible iff $\det(A)$ is a unit. - \item Caley-Hamilton: If $P_A = \det(T \cdot \mathbf{1}_n - A)$ \footnote{$T \cdot \mathbf{1}_n -A \in \Mat(n,n,A[T])$}, then $P_A(A) = 0$. + \item + $\det(AB) = \det(A)\det(B)$ + \item + Development along a row or column works. + \item + Cramer's rule: $A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot + A = \det(A) \cdot \mathbf{1}_n$. $A$ is invertible + iff $\det(A)$ is a unit. + \item + Caley-Hamilton: If $P_A = \det(T \cdot \mathbf{1}_n - A)$ \footnote{$T \cdot \mathbf{1}_n -A \in \Mat(n,n,A[T])$}, then + $P_A(A) = 0$. \end{enumerate} - + \end{fact} \begin{proof} - All rules hold for the image of a matrix under a ring homomorphism if they hold for the original matrix. The converse holds in the case of injective ring homomorphisms. - Caley-Hamilton was shown for algebraically closed fields in LA2 using the Jordan normal form. - Fields can be embedded into their algebraic closure, thus Caley-Hamilton holds for fields. Every domain can be embedded in its field of quotients $\implies$ Caley-Hamilton holds for domains. - - In general, $A$ is the image of $(X_{i,j})_{i,j = 1}^{n} \in \Mat(n,n,S)$ where $S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain) under the morphism $S \to A$ of evaluation defined by $X_{i,j} \mapsto a_{i,j}$. Thus Caley-Hamilton holds in general. -\end{proof} %TODO: lernen + All rules hold for the image of a matrix under a ring homomorphism if they hold + for the original matrix. + The converse holds in the case of injective ring homomorphisms. + Caley-Hamilton was shown for algebraically closed fields in LA2 using the + Jordan normal form. + Fields can be embedded into their algebraic closure, thus Caley-Hamilton holds + for fields. + Every domain can be embedded in its field of quotients $\implies$ + Caley-Hamilton holds for domains. + + In general, $A$ is the image of + $(X_{i,j})_{i,j = 1}^{n} \in + \Mat(n,n,S)$ where + $S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain) under the + morphism $S \to A$ of evaluation defined by $X_{i,j} + \mapsto a_{i,j}$. + Thus Caley-Hamilton holds in general. +\end{proof} +%TODO: lernen \subsection{Integral elements and integral ring extensions} %LECTURE 2 -\begin{proposition}[on integral elements]\label{propinte} - Let $A$ be an $R$-algebra, $a \in A$. Then the following are equivalent: +\begin{proposition}[on integral elements] + \label{propinte} + Let $A$ be an $R$-algebra, $a \in A$. + Then the following are equivalent: \begin{enumerate}[A] - \item $\exists n \in \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = \sum_{i=0}^{n-1} r_i a^i$ - \item There exists a subalgebra $B \subseteq A$ finite over $R$ and containing $a$. + \item + $\exists n \in + \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = + \sum_{i=0}^{n-1} r_i a^i$ + \item + There + exists a subalgebra $B \subseteq A$ finite over $R$ and containing $a$. \end{enumerate} - If $a_1, \ldots, a_k \in A$ satisfy these conditions, there is a subalgebra of $A$ finite over $R$ and containing all $a_i$. + If $a_1, \ldots, a_k \in A$ satisfy these conditions, there is a subalgebra of + $A$ finite over $R$ and containing all $a_i$. \end{proposition} -\begin{definition}\label{intclosure} - Elements that satisfy the conditions from \ref{propinte} are called \vocab{integral over} $R$. +\begin{definition} + \label{intclosure} + Elements that satisfy the conditions from + \ref{propinte} are called + \vocab{integral over} $R$. $A / R$ is \vocab[Algebra!integral]{integral}, if all $a \in A$ are integral over $R$. - The set of elements of $A$ integral over $R$ is called the \vocab{integral closure} of $R$ in $A$. + The set of elements of $A$ integral over $R$ is called the + \vocab{integral + closure} of $R$ in $A$. \end{definition} \begin{proof} \hskip 10pt \begin{enumerate} - {\color{gray} \item[B $\implies$ A] Let $a \in A$ such that there is a subalgebra $B \subseteq A$ containing $a$ and finite over $R$. - Let $(b_i)_{i=1}^{n}$ generate $B$ as an $R$-module. + {\color{gray} + \item[B $\implies$ A] + Let $a \in A$ such that there is a subalgebra $B \subseteq A$ containing $a$ + and finite over $R$. + Let $(b_i)_{i=1}^{n}$ generate $B$ as an + $R$-module. \begin{align} - q: R^n &\longrightarrow B \\ - (r_1,\ldots,r_n) &\longmapsto \sum_{i=1}^{n} r_i b_i + q: R^n & \longrightarrow B \\ + (r_1,\ldots,r_n) & \longmapsto \sum_{i=1}^{n} r_i b_i \end{align} - is surjective. Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in R^n$ such that $a b_i = q(\rho_i)$. Let $\mathfrak{A}$ be the matrix with the $\rho_i$ as columns. - Then for all $v \in R^n: q(\mathfrak{A} \cdot v) = a \cdot q(v)$. By induction it follows that $q(P(\mathfrak{A}) \cdot v) = P(a)q(v)$ for all $P \in R[T]$. Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \mathfrak{A})$ and using Caley-Hamilton, we obtain $P(a) \cdot q(v) = 0$. $P$ is monic. Since $q$ is surjective, we find $v \in R^{n} : q(v) = 1$. Thus $P(a) = 0$ and $a$ satisfies A. - } - \item[B $\implies$ A] if $R$ is Noetherian.\footnote{This suffices in the exam.} - Let $a \in A$ satisfy B. Let $B$ be a subalgebra of $A$ containing $b$ and finite over $R$. Let $M_n \subseteq B$ be the $R$-submodule generated by the $a^i$ with $0 \le i < n$. As a finitely generated module over the Noetherian ring $R$, $B$ is a Noetherian $R$-module. Thus the ascending sequence $M_n$ stabilizes at some step $d$ and $a^d \in M_d$. Thus there are $(r_i)_{i=0}^{d-1} \in R^d$ such that $a^d = \sum_{i=0}^{d-1} r_ia^i$. - \item[A $\implies$ B] Let $a = (a_i)_{i=1}^n$ where all $a_i$ satisfy A, i.e. $a_i^{d_i} = \sum_{j=0}^{d_i - 1} r_{i,j}a_i^j$ with $r_{i,j} \in R$. Let $B \subseteq A$ be the sub-$R$-module generated by $a^\alpha = \prod_{i=1}^n a_i^{\alpha_i}$ with $0 \le \alpha_i < d_i$. - $B$ is closed under $a_1 \cdot $ since \[a_1a^{\alpha} = \begin{cases} - a^{(\alpha_1 + 1, \alpha')} &\text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1\\ - \sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} &\text{if } \alpha_1 = d_1 - 1 - \end{cases}\] - By symmetry, this hold for all $a_i$. By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$, $B$ is invariant under $a^{\alpha}\cdot $. Since these generate $B$ as an $R$-module, $B$ is multiplicatively closed. Thus A holds. Furthermore we have shown the final assertion of the proposition. + is surjective. + Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in R^n$ + such that + $a b_i = q(\rho_i)$. + Let $\mathfrak{A}$ be the matrix with the $\rho_i$ as columns. + Then for all $v \in R^n: q(\mathfrak{A} \cdot v) = a \cdot q(v)$. + By induction it follows that $q(P(\mathfrak{A}) \cdot v) = P(a)q(v)$ + for all $P + \in R[T]$. + Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \mathfrak{A})$ and using + Caley-Hamilton, we obtain $P(a) \cdot q(v) = 0$. + $P$ is monic. + Since $q$ is surjective, we find $v \in R^{n} : q(v) = + 1$. + Thus $P(a) = 0$ and $a$ satisfies A. + } + \item[B $\implies$ A] + if $R$ is Noetherian.\footnote{This suffices in the exam.} + Let $a \in A$ satisfy B. + Let $B$ be a subalgebra of $A$ containing $b$ and finite over $R$. + Let $M_n \subseteq B$ be the $R$-submodule generated by the $a^i$ with $0 \le + i < n$. + As a finitely generated module over the Noetherian ring $R$, $B$ is a + Noetherian $R$-module. + Thus the ascending sequence $M_n$ stabilizes at some step $d$ and $a^d \in + M_d$. + Thus there are $(r_i)_{i=0}^{d-1} \in R^d$ such + that $a^d = \sum_{i=0}^{d-1} + r_ia^i$. + \item[A $\implies$ B] + Let $a = (a_i)_{i=1}^n$ where all $a_i$ + satisfy A, i.e. $a_i^{d_i} = \sum_{j=0}^{d_i - 1} + r_{i,j}a_i^j$ with $r_{i,j} \in + R$. + Let $B \subseteq A$ be the sub-$R$-module generated by $a^\alpha = + \prod_{i=1}^n a_i^{\alpha_i}$ with $0 \le \alpha_i < d_i$. + $B$ is closed under $a_1 \cdot $ since + \[ + a_1a^{\alpha} = + \begin{cases} + a^{(\alpha_1 + 1, \alpha')} & \text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1 \\ + \sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} & \text{if } \alpha_1 = d_1 - 1 + \end{cases} + \] + By symmetry, this hold for all $a_i$. + By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$, $B$ is invariant + under + $a^{\alpha}\cdot $. + Since these generate $B$ as an $R$-module, $B$ is multiplicatively closed. + Thus A holds. + Furthermore we have shown the final assertion of the proposition. \end{enumerate} \end{proof} -\begin{corollary}\label{cintclosure} +\begin{corollary} + \label{cintclosure} \begin{enumerate} - \item[Q] Every finite $R$-algebra $A$ is integral. - \item[R] The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$ - \item[S] If $A$ is an $R$-algebra, $B$ an $A$-algebra and $b \in B$ integral over $R$, then it is integral over $A$. - \item[T] If $A$ is an integral $R$-algebra and $B$ any $A$-algebra, $b \in B$ integral over $A$, then $b$ is integral over $R$. + \item[Q] + Every finite $R$-algebra $A$ is integral. + \item[R] + The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$ + \item[S] + If $A$ is an $R$-algebra, $B$ an $A$-algebra and $b \in B$ integral over $R$, + then it is integral over $A$. + \item[T] + If $A$ is an integral $R$-algebra and $B$ any $A$-algebra, $b \in B$ integral + over $A$, then $b$ is integral over $R$. \end{enumerate} \end{corollary} \begin{proof} \begin{enumerate} - \item[Q] Put $ B = A $ in B. - \item[R] For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$, hence integral over $R$. + \item[Q] + Put $ B = A $ in B. + \item[R] + For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$, hence integral + over $R$. From B it follows, that the integral closure is closed under ring operations. - \item[S] trivial - \item[T] Let $b \in B$ such that $b^n = \sum_{i=0}^{n-1} a_ib^{i}$. Then there is a subalgebra $\tilde{A} \subseteq A$ finite over $R$, such that all $a_i \in \tilde{A}$. - $b$ is integral over $\tilde{A} \implies \exists \tilde{B} \subseteq B$ finite over $\tilde{A}$ and $b \in \tilde{B}$. Since $\tilde{B} / \tilde{A} $ and $\tilde{A} / R$ are finite, $\tilde{B} / R$ is finite and $b$ satisfies B. + \item[S] + trivial + \item[T] + Let $b \in B$ such that $b^n = \sum_{i=0}^{n-1} a_ib^{i}$. + Then there is a subalgebra $\tilde{A} \subseteq A$ finite over + $R$, such that + all $a_i \in \tilde{A}$. + $b$ is integral over $\tilde{A} \implies \exists + \tilde{B} \subseteq B$ finite over $\tilde{A}$ and + $b \in \tilde{B}$. + Since $\tilde{B} / \tilde{A} $ and + $\tilde{A} / R$ are finite, $\tilde{B} / R$ + is finite and $b$ satisfies B. \end{enumerate} \end{proof} \subsection{Finiteness, finite generation and integrality} %some more remarks on finiteness, finite generation and integrality -\begin{fact}[Finite type and integral $\implies$ finite]\label{ftaiimplf} - If $A$ is an integral $R$-algebra of finite type, then it is a finite $R$-algebra. +\begin{fact}[Finite type and integral $\implies$ finite] + \label{ftaiimplf} + If $A$ is an integral $R$-algebra of finite type, then it is a finite + $R$-algebra. \end{fact} \begin{proof} - Let $A $ be generated by $\left( a_i \right) _{i=1}^{n}$ as an $R$- algebra. By the proposition on integral elements (\ref{propinte}), there is a finite $R$-algebra $B \subseteq A$ such that all $a_i \in B$. + Let $A $ be generated by $\left( a_i \right) _{i=1}^{n}$ as an $R$- algebra. + By the proposition on integral elements ( + \ref{propinte}), there is a + finite + $R$-algebra $B \subseteq A$ such that all $a_i \in B$. We have $B = A$, as $A$ is generated by the $a_i$ as an $R$-algebra. \end{proof} \begin{fact}[Finite type in tower] - If $A$ is an $R$-algebra of finite type and $B$ an $A$-algebra of finite type, then $B$ is an $R$-algebra of finite type. + If $A$ is an $R$-algebra of finite type and $B$ an + $A$-algebra of finite type, then $B$ is an $R$-algebra of finite type. \end{fact} \begin{proof} - If $A / R$ is generated by $(a_i)_{i=1}^m$ and $B / A$ by $(b_j)_{j=1}^{n}$, then $B /R$ is generated by the $b_j$ and the images of the $a_i$ in $B$. + If $A / R$ is generated by $(a_i)_{i=1}^m$ and + $B / A$ by $(b_j)_{j=1}^{n}$, + then $B /R$ is generated by the $b_j$ and the images of the $a_i$ in $B$. \end{proof} {\color{red} - \begin{fact}[About integrality and fields] \label{fintaf} - Let $B$ be a domain integral over its subring $A$. Then $B$ is a field iff $A$ is a field. +\begin{fact}[About integrality and fields] + \label{fintaf} + Let $B$ be a domain integral over its subring $A$. + Then $B$ is a field iff $A$ is a field. \end{fact} } \begin{proof} - Let $B$ be a field and $a \in A \setminus \{0\} $. Then $a^{-1} \in B$ is integral over $A$, hence $a^{-d} = \sum_{i=0}^{d-1} \alpha_i a^{-i}$ for some $\alpha_i \in A$. Multiplication by $a^{d-1}$ yields - $a^{-1} = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$. + Let $B$ be a field and $a \in A \setminus \{0\} $. + Then $a^{-1} \in B$ is integral over $A$, hence + $a^{-d} = \sum_{i=0}^{d-1} + \alpha_i a^{-i}$ for some $\alpha_i \in A$. + Multiplication by $a^{d-1}$ yields + $a^{-1} = \sum_{i=0}^{d-1} \alpha_i + a^{d-1-i} \in A$. - On the other hand, let $B$ be integral over the field $A$. Let $b \in B \setminus \{0\}$. As $B$ is integral over $A$, there is a sub-$A$-algebra $\tilde{B} \subseteq B, b \in \tilde{B}$ finitely generated as an $A$-module, i.e. a finite-dimensional $A$-vector space. Since $B$ is a domain, $\tilde{B} \xrightarrow{b\cdot } \tilde{B}$ is injective, hence surjective, thus $\exists x \in \tilde{B} : b \cdot x \cdot 1$. + On the other hand, let $B$ be integral over the field $A$. + Let $b \in B \setminus \{0\}$. + As $B$ is integral over $A$, there is a sub-$A$-algebra $\tilde{B} + \subseteq B, + b \in \tilde{B}$ finitely generated as an $A$-module, i.e. a + finite-dimensional + $A$-vector space. + Since $B$ is a domain, $\tilde{B} \xrightarrow{b\cdot } + \tilde{B}$ is + injective, hence surjective, thus $\exists x \in \tilde{B} : b + \cdot x \cdot + 1$. \end{proof} \subsection{Noether normalization theorem} -\begin{lemma}\label{nntechlemma} - Let $S \subseteq \N^n$ be finite. Then there exists $\vec k \in \N^n$ such that $k_1 =1$ and $w_{\vec k}(\alpha) \neq w_{\vec k}(\beta)$ for $\alpha \neq \beta \in S$, - where $w_{\vec k}(\alpha) = \sum_{i=1}^{n} k_i \alpha_i$. +\begin{lemma} + \label{nntechlemma} + Let $S \subseteq \N^n$ be finite. + Then there exists $\vec k \in \N^n$ such that $k_1 =1$ and + $w_{\vec k}(\alpha) + \neq w_{\vec k}(\beta)$ for $\alpha \neq \beta \in S$, + where $w_{\vec + k}(\alpha) = \sum_{i=1}^{n} k_i + \alpha_i$. \end{lemma} \begin{proof} - Intuitive: - For $\alpha \neq \beta$ the equation $w_{(1, \vec \kappa)}(\alpha) = w_{(1, \vec \kappa)}(\beta)$ ($\kappa \in \R^{n-1}$) - defines a codimension $1$ affine hyperplane in $\R^{n-1}$. It is possible to choose $\kappa$ such that all $\kappa_i$ are $> \frac{1}{2}$ and with Euclidean distance $> \frac{\sqrt{n-1} }{2}$ from the union of these hyperplanes. By choosing the closest $\kappa'$ with integral coordinates, each coordinate will be disturbed by at most $\frac{1}{2}$, thus at Euclidean distance $\le \frac{\sqrt{n-1} }{2}$. + Intuitive: For $\alpha \neq \beta$ the equation + $w_{(1, \vec \kappa)}(\alpha) = + w_{(1, \vec \kappa)}(\beta)$ ($\kappa \in \R^{n-1}$) + defines a codimension $1$ + affine hyperplane in $\R^{n-1}$. + It is possible to choose $\kappa$ such that all $\kappa_i$ are $> + \frac{1}{2}$ + and with Euclidean distance $> \frac{\sqrt{n-1} }{2}$ from the union of these + hyperplanes. + By choosing the closest $\kappa'$ with integral coordinates, each coordinate + will be disturbed by at most $\frac{1}{2}$, thus at Euclidean + distance $\le + \frac{\sqrt{n-1} }{2}$. - More formally:\footnote{The intuitive version suffices in the exam.} - Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le i \le n\} $. We can choose $k$ such that $k_i > (i-1) M k_{i-1}$. - Suppose $\alpha \neq \beta$. Let $i$ be the maximal index such that $\alpha_i \neq \beta_i$. Then the contributions of $\alpha_j$ (resp. $\beta_j$) with $1 \le j < i$ to $w_{\vec k}(\alpha)$ (resp. $w_{\vec k}(\beta)$) cannot undo the difference $k_i(\alpha_i - \beta_i)$. + More formally:\footnote{The intuitive version suffices in the exam. + } + Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le i \le n\} $. + We can choose $k$ such that $k_i > (i-1) M k_{i-1}$. + Suppose $\alpha \neq \beta$. + Let $i$ be the maximal index such that $\alpha_i \neq \beta_i$. + Then the contributions of $\alpha_j$ (resp. + $\beta_j$) with $1 \le j < i$ to $w_{\vec k}(\alpha)$ + (resp. $w_{\vec k}(\beta)$) cannot undo the difference + $k_i(\alpha_i - \beta_i)$. \end{proof} -\begin{theorem}[Noether normalization] \label{noenort} - Let $K$ be a field and $A$ a $K$-algebra of finite type. Then there are $a = (a_i)_{i=1}^{n} \in A$ which are algebraically independent over $K$, i.e. the ring homomorphism \begin{align} - \ev_a: K[X_1,\ldots,X_n] &\longrightarrow A \\ - P &\longmapsto P(a_1,\ldots,a_n) -\end{align} -is injective. $n$ and the $a_i$ can be chosen such that $A$ is finite over the image of $\ev_a$. +\begin{theorem}[Noether normalization] + \label{noenort} + Let $K$ be a field and $A$ a $K$-algebra of finite type. + Then there are $a = (a_i)_{i=1}^{n} \in A$ which + are algebraically independent + over $K$, i.e. the ring homomorphism + \begin{align} + \ev_a: K[X_1,\ldots,X_n] & + \longrightarrow A \\ P & \longmapsto P(a_1,\ldots,a_n) + \end{align} + is + injective. + $n$ and the $a_i$ can be chosen such that $A$ is finite over the image of + $\ev_a$. \end{theorem} \begin{proof} - Let $(a_i)_{i=1}^n$ be a minimal number of elements such that $A$ is integral over its $K$-subalgebra generated by $a_1, \ldots, a_n$. (Such $a_i$ exist, since $A$ is of finite type). + Let $(a_i)_{i=1}^n$ be a minimal number of + elements such that $A$ is integral + over its $K$-subalgebra generated by $a_1, \ldots, a_n$. + (Such $a_i$ exist, since $A$ is of finite type). Let $\tilde{A}$ be the $K$-subalgebra generated by the $a_i$. If suffices to show that the $a_i$ are algebraically independent. - Since $A$ is of finite type over $K$ and thus over $\tilde{A}$, by fact \ref{ftaiimplf} (integral and finite type $\implies$ finite) $A$ is finite over $\tilde{A}$. - Thus we only need to show that the $a_i$ are algebraically independent over $K$. - Assume there is $P \in K[X_1,\ldots,X_n] \setminus \{0\} $ such that $P(a_1,\ldots,a_n) = 0$. Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and $S = \{ \alpha \in \N^n | p_\alpha \neq 0\}$. For $\vec{k} = (k_i)_{i=1}^{n} \in \N^n$ and $\alpha \in \N^n$ we define $w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$. + Since $A$ is of finite type over $K$ and thus over $\tilde{A}$, + by fact + \ref{ftaiimplf} (integral and finite type $\implies$ finite) $A$ is finite over + $\tilde{A}$. + Thus we only need to show that the $a_i$ are algebraically independent over + $K$. + Assume there is $P \in K[X_1,\ldots,X_n] \setminus \{0\} $ such + that + $P(a_1,\ldots,a_n) = 0$. + Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and + $S = \{ \alpha \in + \N^n | p_\alpha \neq 0\}$. + For $\vec{k} = (k_i)_{i=1}^{n} + \in \N^n$ and $\alpha \in \N^n$ we define + $w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} + k_i\alpha_i$. - By \ref{nntechlemma} it is possible to choose $\vec{k} \in \N^n$ such that - $k_1 = 1$ and for $\alpha \neq \beta \in S$ we have $w_{\vec{k}}(\alpha) \neq w_{\vec{k}}(\beta)$. + By + \ref{nntechlemma} it is possible to choose $\vec{k} + \in \N^n$ such that $k_1 + = 1$ and for $\alpha \neq \beta \in S$ we have + $w_{\vec{k}}(\alpha) \neq + w_{\vec{k}}(\beta)$. - Define $b_i \coloneqq a_{i+1} - a^{k_{i+1}}_1$ for $1 \le i < n$. + Define $b_i \coloneqq a_{i+1} - + a^{k_{i+1}}_1$ for $1 \le i < n$. \begin{claim} $A$ is integral over the subalgebra $B$ generated by the $b_i$. \end{claim} \begin{subproof} - By the transitivity of integrality, it is sufficient to show that the $a_i$ are integral over $B$. - For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$. Thus it suffices to show this for $a_1$. - Define $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n}) \in B[T]$. - We have $0 = P(a_1,\ldots, a_n) = Q(a_1)$. Hence it suffices to show that the leading coefficient of $Q$ is a unit. + By the transitivity of integrality, it is sufficient to show that the $a_i$ are + integral over $B$. + For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$. + Thus it suffices to show this for $a_1$. + Define $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, + b_{n-1} + T^{k_n}) \in + B[T]$. + We have $0 = P(a_1,\ldots, a_n) = Q(a_1)$. + Hence it suffices to show that the leading coefficient of $Q$ is a unit. - We have + We have \[ - T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} = T^{w_{\vec k}(\alpha)} + \sum_{l = 0}^{w_{\vec k}(\alpha) - 1} \beta_{\alpha, l} T^l - \] + T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} + = + T^{w_{\vec k}(\alpha)} + + \sum_{l = 0}^{w_{\vec k}(\alpha) - 1} + \beta_{\alpha, + l} T^l + \] with suitable $\beta_{\alpha, l} \in B$. - By the choice of $\vec k$, we have \[ - Q(T) = p_{\alpha} T^{w_{\vec k}(\alpha)} + \sum_{j=0}^{w_{\vec k}(\alpha) - 1} q_j T^j - \] - with $q_j \in B$ and $\alpha$ such that $w_{\vec k }(\alpha)$ is maximal subject to the condition $p_\alpha \neq 0$. + By the choice of $\vec k$, we have + \[ + Q(T) = p_{\alpha} T^{w_{\vec k}(\alpha)} + + \sum_{j=0}^{w_{\vec k}(\alpha) - 1} q_j T^j + \] + with $q_j \in B$ and + $\alpha$ such that $w_{\vec k }(\alpha)$ is maximal subject + to the condition + $p_\alpha \neq 0$. Thus the leading coefficient of $Q$ is a unit. \end{subproof} - This contradicts the minimality of $n$, as $B$ can be generated by $< n$ elements $b_i$. + This contradicts the minimality of $n$, as $B$ can be generated by $< n$ + elements $b_i$. \end{proof} diff --git a/inputs/nullstellensatz_and_zariski_topology.tex b/inputs/nullstellensatz_and_zariski_topology.tex index 7b0220a..55299a1 100644 --- a/inputs/nullstellensatz_and_zariski_topology.tex +++ b/inputs/nullstellensatz_and_zariski_topology.tex @@ -1,238 +1,453 @@ \subsection{The Nullstellensatz} %LECTURE 1 -Let $\mathfrak{k}$ be a field, $R \coloneqq \mathfrak{k}[X_1,\ldots,X_n], I \subseteq R$ an ideal. +Let $\mathfrak{k}$ be a field, $R \coloneqq +\mathfrak{k}[X_1,\ldots,X_n], I +\subseteq R$ an ideal. \begin{definition}[zero] - $x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$} if $\forall x \in I: P(x) = 0$. Let $\Va(I)$ denote the set of zeros if $I$ in $\mathfrak{k}^n$. + $x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$} + if $\forall x \in I: P(x) = 0$. + Let $\Va(I)$ denote the set of zeros if $I$ in + $\mathfrak{k}^n$. - The \vocab[Ideal!zero]{zero in a field extension $\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly. + The \vocab[Ideal! + zero]{zero in a field extension $\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly. \end{definition} \begin{remark}[Set of zeros and generators] - Let $I$ be generated by $S$. Then $\{x \in R | \forall s \in S: s(x) = 0\} = \Va(I)$. Thus zero sets of ideals correspond to solutions sets to systems of polynomial equations. - If $S, \tilde{S}$ generate the same ideal $I$ they have the same set of solutions. Therefore we only consider zero sets of ideals. + Let $I$ be generated by $S$. + Then $\{x \in R | \forall s \in S: s(x) = 0\} = \Va(I)$. + Thus zero sets of ideals correspond to solutions sets to systems of polynomial + equations. + If $S, \tilde{S}$ generate the same ideal $I$ they have the same + set of + solutions. + Therefore we only consider zero sets of ideals. \end{remark} -\begin{theorem}[Hilbert's Nullstellensatz (1)]\label{hns1} - If $\mathfrak{k}$ is algebraically closed and $I \subsetneq R$ a proper ideal, then $I$ has a zero in $\mathfrak{k}^n$. +\begin{theorem}[Hilbert's Nullstellensatz (1)] + \label{hns1} + If $\mathfrak{k}$ is algebraically closed and $I \subsetneq R$ a + proper ideal, + then $I$ has a zero in $\mathfrak{k}^n$. \end{theorem} \begin{remark} - Will be shown later (see proof of \ref{hns1b}). - Trivial if $n = 1$: $R$ is a PID, thus $I = pR$ for some $p \in R$. Since $I \neq R$ $p = 0$ or $P$ is non-constant. $\mathfrak{k}$ algebraically closed $\leadsto$ there exists a zero of $p$.\\ + Will be shown later (see proof of + \ref{hns1b}). + Trivial if $n = 1$: $R$ is a PID, thus $I = pR$ for some $p \in R$. + Since $I \neq R$ $p = 0$ or $P$ is non-constant. + $\mathfrak{k}$ algebraically closed $\leadsto$ there exists a zero of + $p$.\\ - If $\mathfrak{k}$ is not algebraically closed and $n > 0$, the theorem fails (consider $I = p(X_1) R$). + If $\mathfrak{k}$ is not algebraically closed and $n > 0$, the + theorem fails + (consider $I = p(X_1) R$). \end{remark} Equivalent\footnote{used in a vague sense here} formulation: -\begin{theorem}[Hilbert's Nullstellensatz (2)] \label{hns2} - Let $L / K$ be an arbitrary field extension. Then $L / K$ is a finite field extension ($\dim_K L < \infty$) iff $L $ is a $K$-algebra of finite type. +\begin{theorem}[Hilbert's Nullstellensatz (2)] + \label{hns2} Let $L / K$ be an + arbitrary field extension. + Then $L / K$ is a finite field extension ($\dim_K L < \infty$) iff $L $ is a + $K$-algebra of finite type. \end{theorem} \begin{proof} \begin{itemize} - \item[$\implies$] If $(l_i)_{i=1}^{m}$ is a base of $L$ as a $K$-vector space, then $L$ is generated by the $l_i$ as a $K$-algebra. - \item[$\impliedby$ ] Apply the Noether normalization theorem (\ref{noenort}) to $A = L$. This yields an injective ring homomorphism $\ev_a: K[X_1,\ldots,X_n] \to A$ such that $A$ is finite over the image of $\ev_a$. - By the fact about integrality and fields (\ref{fintaf}), the isomorphic image of $\ev_a$ is a field. Thus $K[X_1,\ldots, X_n]$ is a field $\implies n = 0$. Thus $L / K$ is a finite ring extension, hence a finite field extension. + \item[$\implies$] + If $(l_i)_{i=1}^{m}$ is a base of $L$ as a + $K$-vector space, then $L$ is generated by the $l_i$ as a $K$-algebra. + \item[$\impliedby$ ] + Apply the Noether normalization theorem ( + \ref{noenort}) to $A = L$. + This yields an injective ring homomorphism $\ev_a: + K[X_1,\ldots,X_n] \to A$ + such that $A$ is finite over the image of $\ev_a$. + By the fact about integrality and fields ( + \ref{fintaf}), the + isomorphic image + of $\ev_a$ is a field. + Thus $K[X_1,\ldots, X_n]$ is a field $\implies n = 0$. + Thus $L / K$ is a finite ring extension, hence a finite field extension. \end{itemize} \end{proof} \begin{remark} - We will see several additional proofs of this theorem. See \ref{hns2unc} and \ref{rfuncnft}. + We will see several additional proofs of this theorem. + See + \ref{hns2unc} and + \ref{rfuncnft}. All will be accepted in the exam. - \ref{hns3} and \ref{hnsp} are closely related. + \ref{hns3} and + \ref{hnsp} are closely related. \end{remark} -\begin{theorem}[Hilbert's Nullstellensatz (1b)] \label{hns1b} - Let $\mathfrak{l}$ be a field and $I \subset R = \mathfrak{l}[X_1,\ldots,X_m]$ a proper ideal. Then there are a finite field extension $\mathfrak{i}$ of $\mathfrak{l}$ and a zero of $I$ in $\mathfrak{i}^m$. +\begin{theorem}[Hilbert's Nullstellensatz (1b)] + \label{hns1b} + Let $\mathfrak{l}$ be a field and $I \subset R = + \mathfrak{l}[X_1,\ldots,X_m]$ + a proper ideal. + Then there are a finite field extension $\mathfrak{i}$ of + $\mathfrak{l}$ and a + zero of $I$ in $\mathfrak{i}^m$. \end{theorem} -\begin{proof} (HNS2 (\ref{hns2}) $\implies$ HNS1b (\ref{hns1b})) - $I \subseteq \mathfrak{m}$ for some maximal ideal. $R / \mathfrak{m}$ is a field, since $\mathfrak{m}$ is maximal. - $R / \mathfrak{m}$ is of finite type, since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra. - There are thus a field extension $\mathfrak{i} / \mathfrak{l}$ and an isomorphism $R / \mathfrak{m} \xrightarrow{\iota} \mathfrak{i}$ of $\mathfrak{l}$-algebras. - By HNS2 (\ref{hns2}), $\mathfrak{i} / \mathfrak{l}$ is a finite field extension. +\begin{proof} + (HNS2 ( + \ref{hns2}) $\implies$ HNS1b ( + \ref{hns1b})) + $I \subseteq \mathfrak{m}$ for some maximal ideal. $R / + \mathfrak{m}$ is a field, since $\mathfrak{m}$ is maximal. + $R / \mathfrak{m}$ is of finite type, since the images of the $X_i$ + generate it as a $\mathfrak{l}$-algebra. + There are thus a field extension $\mathfrak{i} / + \mathfrak{l}$ and an + isomorphism $R / \mathfrak{m} \xrightarrow{\iota} + \mathfrak{i}$ of + $\mathfrak{l}$-algebras. + By HNS2 ( + \ref{hns2}), $\mathfrak{i} / + \mathfrak{l}$ is a finite field + extension. Let $x_i \coloneqq \iota (X_i \mod \mathfrak{m})$. \[ P(x_1,\ldots,x_m) = \iota(P \mod \mathfrak{m}) - \] - Both sides are morphisms $R \to \mathfrak{i}$ of $\mathfrak{l}$-algebras. For for $P = X_i$ the equality is trivial. It follows in general, since the $X_i$ generate $R$ as a $\mathfrak{l}$-algebra. + \] + Both sides are morphisms $R \to \mathfrak{i}$ of + $\mathfrak{l}$-algebras. + For for $P = X_i$ the equality is trivial. + It follows in general, since the $X_i$ generate $R$ as a + $\mathfrak{l}$-algebra. - Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \mathfrak{m} = 0$ for $P \in I \subseteq \mathfrak{m}$). - HNS1 (\ref{hns1}) can easily be derived from HNS1b. + Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \mathfrak{m} + = 0$ for + $P \in I \subseteq \mathfrak{m}$). + HNS1 ( + \ref{hns1}) can easily be derived from HNS1b. \end{proof} \subsubsection{Nullstellensatz for uncountable fields} % from lecture 5 Yet another proof of the Nullstellensatz -The following proof of the Nullstellensatz only works for uncountable fields, but will be accepted in the exam. +The following proof of the Nullstellensatz only works for uncountable fields, +but will be accepted in the exam. -\begin{lemma}\label{dimrfunc} +\begin{lemma} + \label{dimrfunc} If $K$ is an uncountable field, then $\dim_K K(T)$ is uncountable. \end{lemma} \begin{proof} - We will show, that $S \coloneqq \left\{ \frac{1}{T - \kappa} | \kappa \in K\right\} $ is $K$-linearly independent. It follows that $\dim_K K(T) \ge \#S > \aleph_0$. + We will show, that $S \coloneqq \left\{ \frac{1}{T - \kappa} | \kappa \in + K\right\} $ is $K$-linearly independent. + It follows that $\dim_K K(T) \ge \#S > \aleph_0$. - Suppose $(x_{\kappa})_{\kappa \in K}$ is a selection of coefficients from $K$ such that $I \coloneqq \{\kappa \in K | x_{\kappa} \neq 0\} $ is finite and + Suppose + $(x_{\kappa})_{\kappa \in K}$ is a + selection of coefficients from $K$ + such that $I \coloneqq \{\kappa \in K | x_{\kappa} \neq 0\} + $ is finite and \[ g \coloneqq \sum_{\kappa \in K} \frac{x_\kappa}{T-\kappa} = 0 \] - Let $d \coloneqq \prod_{\kappa \in I} (T - \kappa) $. Then for $\lambda \in I$ we have + Let $d + \coloneqq \prod_{\kappa \in I} (T - \kappa) $. + Then for $\lambda \in I$ we have \[ - 0 = (dg)(\lambda) = x_\lambda \prod_{\kappa \in I \setminus \{\lambda\} } (\lambda - \kappa) - \] - This is a contradiction as $x_\lambda \neq 0$. + 0 = (dg)(\lambda) = x_\lambda \prod_{\kappa + \in I \setminus \{\lambda\} } (\lambda - \kappa) + \] + This is a contradiction as + $x_\lambda \neq 0$. \end{proof} -\begin{theorem}[Hilbert's Nullstellensatz for uncountable fields]\label{hns2unc} - If $K$ is an uncountable field and $L / K$ a field extension and $L$ of finite type as a $K$-algebra, then this field extension is finite. +\begin{theorem}[Hilbert's Nullstellensatz for uncountable fields] + \label{hns2unc} + If $K$ is an uncountable field and $L / K$ a field extension and $L$ of finite + type as a $K$-algebra, then this field extension is finite. \end{theorem} \begin{proof} - If $(x_i)_{i=1}^{n}$ generate $L$ as an $K$-algebra, then the countably many monomials $x^{\alpha} = \prod_{i = 1}^{n} x_i^{\alpha_i} $ in the $x_i$ with $\alpha \in \N^n$ generate $L$ as a $K$-vector space. - Thus $\dim_K L \le \aleph_0$ and the same holds for any intermediate field $K \subseteq M \subseteq L$ . If $l \in L$ is transcendent over $K$ and $M = K(l)$, then $M \cong K(T)$ has uncountable dimension by \ref{dimrfunc}. Thus $L / K$ is algebraic, hence integral, hence finite (\ref{ftaiimplf}). + If $(x_i)_{i=1}^{n}$ generate $L$ as an + $K$-algebra, then the countably many + monomials $x^{\alpha} = \prod_{i = 1}^{n} + x_i^{\alpha_i} $ in the $x_i$ with + $\alpha \in \N^n$ generate $L$ as a $K$-vector space. + Thus $\dim_K L \le \aleph_0$ and the same holds for any intermediate field $K + \subseteq M \subseteq L$ . + If $l \in L$ is transcendent over $K$ and $M = K(l)$, then $M \cong K(T)$ has + uncountable dimension by + \ref{dimrfunc}. + Thus $L / K$ is algebraic, hence integral, hence finite + ( + \ref{ftaiimplf}). \end{proof} \subsection{The Zariski topology} \subsubsection{Operations on ideals and \texorpdfstring{$\Va\left( I \right) $}{V(I)}} -Let $R$ be a ring and $I,J, I_\lambda \subseteq R$ ideals, $\lambda \in \Lambda$. +Let $R$ be a ring and $I,J, I_\lambda \subseteq R$ ideals, $\lambda \in +\Lambda$. \begin{definition}[Radical, product and sum of ideals] \[ - \sqrt{I} \coloneqq \bigcap_{n=0} ^{\infty} \{ f \in R | f^n \in I\} - \] + \sqrt{I} \coloneqq \bigcap_{n=0} ^{\infty} \{ f \in R | f^n \in + I\} + \] \[ I \cdot J \coloneqq \langle\{ i \cdot j | i \in I , j \in J\}\rangle_R - \] + \] \[ - \sum_{\lambda \in \Lambda} I_\lambda \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda | \Lambda' \subseteq \Lambda \text{ finite}\right\} - \] + \sum_{\lambda \in \Lambda} + I_\lambda \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda | \Lambda' + \subseteq \Lambda \text{ finite}\right\} + \] \end{definition} \begin{fact} - The radical is an ideal in $R$ and $\sqrt{\sqrt{I} } = \sqrt{I}$.\\ + The + radical is an ideal in $R$ and $\sqrt{\sqrt{I} } = + \sqrt{I}$. + \\ $I \cdot J$ is an ideal.\\ - $\sum_{\lambda \in \Lambda} I_\lambda$ coincides with the ideal generated by $\bigcap_{\lambda \in \Lambda} I_\lambda$ in $R$.\\ - $\bigcap_{\lambda \in \Lambda} I_\lambda$ is an ideal. + $\sum_{\lambda \in \Lambda} + I_\lambda$ coincides with the ideal generated by $\bigcap_{\lambda \in + \Lambda} + I_\lambda$ in $R$. + \\ + $\bigcap_{\lambda \in \Lambda} + I_\lambda$ is an ideal. \end{fact} -Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an algebraically closed field. +Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an +algebraically +closed field. -\begin{fact} \label{fvop} - Let $I, J, (I_{\lambda})_{\lambda \in \Lambda}$ be ideals in $R$. $\Lambda$ may be infinite. +\begin{fact} + \label{fvop} + Let $I, J, + (I_{\lambda})_{\lambda \in \Lambda}$ be + ideals in $R$. + $\Lambda$ may be infinite. \begin{enumerate}[A] - \item $\Va(I) = \Va(\sqrt{I})$ - \item $\sqrt{J} \subseteq \sqrt{I} \implies \Va(I) \subseteq \Va(J)$ - \item $\Va(R) = \emptyset, \Va(\{0\} =\mathfrak{k}^n$ - \item $\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$ - \item $\Va(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ + \item + $\Va(I) = \Va(\sqrt{I})$ + \item + $\sqrt{J} \subseteq \sqrt{I} \implies + \Va(I) \subseteq \Va(J)$ + \item + $\Va(R) = \emptyset, \Va(\{0\} =\mathfrak{k}^n$ + \item + $\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$ + \item + $\Va(\sum_{\lambda \in \Lambda} + I_\lambda) = \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ \end{enumerate} \end{fact} \begin{proof} - \begin{enumerate} - \item[A-C] trivial - \item[D] $I \cdot J \subseteq I \cap J \subseteq I$. Thus $\Va(I) \subseteq \Va(I \cap J) \subseteq \Va(I \cdot J)$. By symmetry we have $\Va(I) \cup \Va(J) \subseteq \Va(I \cap J) \subseteq \Va(I \cdot J)$. - Let $x \not\in \Va(I) \cup \Va(J)$. Then there are $f \in I, g \in J$ such that $f(x) \neq 0, g(x) \neq 0$ thus $(f \cdot g)(x) \neq 0 \implies x \not\in \Va(I\cdot J)$. - Therefore \[ - \Va(I) \cup \Va(J) \subseteq \Va(I \cap J) \subseteq \Va(I \cdot J) \subseteq \Va(I) \cup \Va(J) - \] - \item[E] $I_\lambda \subseteq \sum_{\lambda \in \Lambda} I_\lambda \implies \Va(\sum_{\lambda \in \Lambda} I_\lambda) \subseteq \Va(I_\lambda)$. - Thus $\Va(\sum_{\lambda \in \Lambda} I_\lambda) \subseteq \bigcap_{\lambda \in \Lambda} \Va(I_\lambda)$. - On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$ we have $f = \sum_{\lambda \in \Lambda} f_\lambda$. Thus $f$ vanishes on $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ and we have $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda) \subseteq \Va(\sum_{\lambda \in \Lambda} I_\lambda)$. - \end{enumerate} + \begin{enumerate} + \item[A-C] + trivial + \item[D] + $I \cdot + J \subseteq I \cap J \subseteq I$. + Thus $\Va(I) \subseteq \Va(I \cap J) \subseteq + \Va(I \cdot J)$. + By symmetry we have $\Va(I) \cup \Va(J) + \subseteq \Va(I \cap J) \subseteq \Va(I + \cdot J)$. + Let $x \not\in \Va(I) \cup \Va(J)$. + Then there are $f \in I, g \in J$ such that $f(x) \neq 0, g(x) \neq 0$ thus + $(f \cdot g)(x) \neq 0 \implies x \not\in \Va(I\cdot J)$. + Therefore + \[ + \Va(I) \cup \Va(J) \subseteq + \Va(I \cap J) \subseteq \Va(I \cdot + J) \subseteq + \Va(I) \cup \Va(J) + \] + \item[E] + $I_\lambda \subseteq \sum_{\lambda + \in \Lambda} I_\lambda \implies + \Va(\sum_{\lambda \in \Lambda} I_\lambda) + \subseteq \Va(I_\lambda)$. + Thus $\Va(\sum_{\lambda \in \Lambda} I_\lambda) \subseteq \bigcap_{\lambda \in + \Lambda} + \Va(I_\lambda)$. + On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$ we have $f = + \sum_{\lambda \in \Lambda} f_\lambda$. + Thus $f$ vanishes on $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ and we + have $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda) \subseteq + \Va(\sum_{\lambda + \in \Lambda} I_\lambda)$. + \end{enumerate} \end{proof} \begin{remark} - There is no similar way to describe $\Va(\bigcap_{\lambda \in \Lambda} I_\lambda)$ in terms of the $\Va(I_{\lambda})$ when $\Lambda$ is infinite. - For instance if $n = 1, I_k \coloneqq X_1^k R$ then $\bigcap_{k=0}^\infty I_k = \{0\} $ but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$. + There is no similar way to describe $\Va(\bigcap_{\lambda \in \Lambda} + I_\lambda)$ in terms of the + $\Va(I_{\lambda})$ when $\Lambda$ is infinite. + For instance if $n = 1, I_k \coloneqq X_1^k R$ then + $\bigcap_{k=0}^\infty I_k = + \{0\} $ but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$. \end{remark} \subsubsection{Definition of the Zariski topology} -Let $\mathfrak{k}$ be algebraically closed, $R = \mathfrak{k}[X_1,\ldots,X_n]$. -\begin{corollary} (of \ref{fvop}) - There is a topology on $\mathfrak{k}^n$ for which the set of closed sets coincides with the set $\mathfrak{A}$ of subsets of the form $\Va\left(I \right) $ for ideals $I \subseteq R$. - This topology is called the \vocab{Zariski-Topology} +Let $\mathfrak{k}$ be algebraically closed, $R = +\mathfrak{k}[X_1,\ldots,X_n]$. +\begin{corollary} + (of + \ref{fvop}) + There is a topology on $\mathfrak{k}^n$ for which the set of closed + sets + coincides with the set $\mathfrak{A}$ of subsets of the form + $\Va\left(I + \right) $ for ideals $I \subseteq R$. + This topology is called the \vocab{Zariski-Topology} \end{corollary} -\begin{example}\label{zariskinothd} - Let $n = 1$. Then $R$ is a PID. Hence every ideal is a principal ideal and the Zariski-closed subsets of $\mathfrak{k}$ are the subsets of the form $\Va(P)$ for $P \in R$. -As $\Va(0) = \mathfrak{k}$ and $\Va(P)$ finite for $P \neq 0$ and $\{x_1,\ldots,x_n\} = \Va(\prod_{i=1}^{n}(T-x_i))$ the Zariski-closed subsets of $\mathfrak{k}$ are $\mathfrak{k}$ and the finite subsets. +\begin{example} + \label{zariskinothd} Let $n = 1$. + Then $R$ is a PID. + Hence every ideal is a principal ideal and the Zariski-closed subsets of + $\mathfrak{k}$ are the subsets of the form $\Va(P)$ + for $P \in R$. + As $\Va(0) = \mathfrak{k}$ and + $\Va(P)$ finite for $P \neq 0$ and + $\{x_1,\ldots,x_n\} = \Va(\prod_{i=1}^{n}(T-x_i))$ the Zariski-closed subsets + of $\mathfrak{k}$ are $\mathfrak{k}$ and the finite + subsets. Because $\mathfrak{k}$ is infinite, this topology is not Hausdorff. \end{example} \subsubsection{Separation properties of topological spaces} \begin{definition} - Let $X$ be a topological space. $X$ satisfies the separation properties $T_{0-2}$ if for any $x \neq y \in X$ + Let $X$ be a topological space. + $X$ satisfies the separation properties $T_{0-2}$ if for + any $x \neq y \in X$ \begin{enumerate} - \item[$T_0$ ] $\exists U \subseteq X$ open such that $|U \cap \{x,y\}| = 1$ - \item[$T_1$ ] $\exists U \subseteq X$ open such that $x \in U, y \not\in U$. - \item[$T_2$ ] There are disjoined open sets $U, V \subseteq X$ such that $x \in U, y \in V$. (Hausdorff) + \item[$T_0$ ] + $\exists U \subseteq X$ open such that $|U \cap \{x,y\}| = 1$ + \item[$T_1$ ] + $\exists U \subseteq X$ open such that $x \in U, y \not\in U$. + \item[$T_2$ ] + There are disjoined open sets $U, V \subseteq X$ such that $x \in U, y \in V$. + (Hausdorff) \end{enumerate} \end{definition} \begin{remark} - Let $x \sim y :\iff$ the open subsets of $X$ containing $x$ are precisely the open subsets of $X$ containing $y$. Then $T_0$ holds iff $x \sim y \implies x =y$. + Let $x \sim y :\iff$ the open subsets of $X$ containing $x$ are precisely the + open subsets of $X$ containing $y$. + Then $T_0$ holds iff $x \sim y \implies x =y$. \end{remark} \begin{fact} $T_0 \iff$ every point is closed. \end{fact} \begin{fact} - The Zariski topology on $\mathfrak{k}^n$ is $T_1$ but for $n \ge 1$ not Hausdorff. For $n \ge 1$ the intersection of two non-empty open subsets of $\mathfrak{k}^n$ is always non-empty. + The Zariski topology on $\mathfrak{k}^n$ is $T_1$ but for $n \ge 1$ + not + Hausdorff. + For $n \ge 1$ the intersection of two non-empty open subsets of + $\mathfrak{k}^n$ is always non-empty. \end{fact} \begin{proof} - $\{x\} $ is closed, as $\{x\} = V(\Span{X_1 - x_1, \ldots, X_n - x_n}_R)$. If $A = V(I), B = V(J)$ are two proper closed subsets of $\mathfrak{k}^n$ then $I \neq \{0\} , J \neq \{0\} $ and thus $IJ \neq \{0\} $. Therefore $A \cup B = V(IJ)$ is a proper closed subset of $\mathfrak{k}^n$. + $\{x\} $ is closed, as $\{x\} = V(\Span{X_1 - x_1, \ldots, X_n - x_n}_R)$. + If $A = V(I), B = V(J)$ are two proper closed subsets of + $\mathfrak{k}^n$ then + $I \neq \{0\} , J \neq \{0\} $ and thus $IJ \neq \{0\} $. + Therefore $A \cup B = V(IJ)$ is a proper closed subset of + $\mathfrak{k}^n$. \end{proof} \subsubsection{Compactness properties of topological spaces} Let $X$ be a topological space. \begin{definition}[Compact, quasi-compact] - $X$ is called \vocab[Topological space!quasi-compact]{quasi-compact} if every open covering of $X$ has a finite subcovering. - It is called \vocab[Topological space!compact]{compact}, if it is quasi-compact and Hausdorff. + $X$ is called \vocab[Topological space!quasi-compact]{quasi-compact} if every open + covering of $X$ has a finite subcovering. + It is called \vocab[Topological space! + compact]{compact}, if it is quasi-compact and Hausdorff. \end{definition} \begin{definition}[Noetherian topological spaces] - $X$ is called \vocab{Noetherian}, if the following equivalent conditions hold: + $X$ is called \vocab{Noetherian}, if the + following equivalent conditions hold: \begin{enumerate}[A] - \item Every open subset of $X$ is quasi-compact. - \item Every descending sequence $A_0 \supseteq A_1 \supseteq \ldots$ of closed subsets of $X$ stabilizes. - \item Every non-empty set $\mathcal{M}$ of closed subsets of $X$ has a $\subseteq$-minimal element. + \item + Every open subset of $X$ is quasi-compact. + \item + Every descending sequence $A_0 \supseteq A_1 \supseteq \ldots$ of closed + subsets of $X$ stabilizes. + \item + Every non-empty set $\mathcal{M}$ of closed subsets of $X$ has a + $\subseteq$-minimal element. \end{enumerate} \end{definition} -\begin{proof}\, +\begin{proof} + \, \begin{enumerate} - \item[A $\implies$ B] Let $A_j$ be a descending chain of closed subsets. Define $A \coloneqq \bigcap_{j = 0}^{\infty} A_j$. If A holds, the covering $X \setminus A = \bigcup_{j = 0}^{\infty} (X \setminus A_j)$ has a finite subcovering. - \item[B $\implies$ C] Suppose $\mathcal{M}$ does not have a $\subseteq$-minimal element. Using DC, one can construct a counterexample $A_1 \subsetneq A_2 \supsetneq \ldots$ to B. - \item[C $\implies$ A] Let $\bigcup_{i \in I} V_i$ be an open covering of an open subset $U \subseteq X$. - By C, the set $\mathcal{M} \coloneqq \{X \setminus \bigcup_{i \in F} V_i | F \subseteq I \text{ finite} \}$ has a $\subseteq$-minimal element. + \item[A $\implies$ B] + Let $A_j$ be a descending chain of closed subsets. + Define $A \coloneqq \bigcap_{j = 0}^{\infty} A_j$. + If A holds, the covering $X \setminus A = \bigcup_{j = 0}^{\infty} (X \setminus + A_j)$ has a finite + subcovering. + \item[B $\implies$ C] + Suppose $\mathcal{M}$ does not have a $\subseteq$-minimal element. + Using DC, one can construct a counterexample $A_1 \subsetneq A_2 \supsetneq + \ldots$ to B. + \item[C $\implies$ A] + Let $\bigcup_{i \in I} + V_i$ be an open covering of an open subset $U \subseteq X$. + By C, the set $\mathcal{M} \coloneqq \{X \setminus + \bigcup_{i \in F} V_i | F + \subseteq I \text{ finite} \}$ has a $\subseteq$-minimal element. \end{enumerate} \end{proof} -\subsection{Another form of the Nullstellensatz and Noetherianness of \texorpdfstring{$\mathfrak{k}^n$}{kn}} -Let $\mathfrak{k}$ be algebraically closed, $R = \mathfrak{k}[X_1,\ldots,X_n]$. +\subsection{Another form of the Nullstellensatz and Noetherianness of + \texorpdfstring{$\mathfrak{k}^n$}{kn}} +Let $\mathfrak{k}$ be algebraically closed, $R = +\mathfrak{k}[X_1,\ldots,X_n]$. For $f \in R$ let $V(f) = V(fR)$. -\begin{theorem}[Hilbert's Nullstellensatz (3)] \label{hns3} - Let $I \subseteq R$ be an ideal. Then $V(I) \subseteq V(f)$ iff $f \in \sqrt{I}$. +\begin{theorem}[Hilbert's Nullstellensatz (3)] + \label{hns3} + Let $I \subseteq R$ be an ideal. + Then $V(I) \subseteq V(f)$ iff $f \in \sqrt{I}$. \end{theorem} \begin{proof} - Suppose $f$ vanishes on all zeros of $I$. Let $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n,T]$, - $g(X_1,\ldots,X_n,T) \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)$ - and $J \subseteq R'$ the ideal generated by $g$ and the elements of $I$ (viewed as elements of $R'$ which are constant in the $T$-direction). + Suppose $f$ vanishes on all zeros of $I$. + Let $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n,T]$, $g(X_1,\ldots,X_n,T) + \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)$ and $J \subseteq R'$ the ideal + generated by $g$ and the elements of $I$ (viewed as elements of $R'$ which are + constant in the $T$-direction). - If $f$ vanishes on all zeros of $I$, then $J$ has no zeros in $\mathfrak{k}^{n+1}$. + If $f$ vanishes on all zeros of $I$, then $J$ has no zeros in + $\mathfrak{k}^{n+1}$. - Thus there exist $p_i \in I, i=1,\ldots,n, q_i \in \mathfrak{k}[X_1,\ldots,X_n,T], i = 1,\ldots,n$ and $q \in \mathfrak{k}[X_1,\ldots,X_n,T]$ such that + Thus there exist $p_i \in I, i=1,\ldots,n, q_i \in + \mathfrak{k}[X_1,\ldots,X_n,T], i = 1,\ldots,n$ and $q \in + \mathfrak{k}[X_1,\ldots,X_n,T]$ such that \[ - 1 = g \cdot q + \sum_{i=1}^{n} p_{i}q_i + 1 = g \cdot q + \sum_{i=1}^{n} + p_{i}q_i \] - Formally substituting $\frac{1}{f(x_1,\ldots,x_n)}$ for $Y$, one obtains: + Formally substituting $\frac{1}{f(x_1,\ldots,x_n)}$ for $Y$, one + obtains: \[ - 1 = \sum_{i=1}^{n} p_{i}\left(x_1,\ldots,x_n\right) q_i\left( x_1,\ldots,x_n, \frac{1}{f(x_1,\ldots,x_n)} \right) - \] - Multiplying by a sufficient power of $f$, this yields an equation in $R$ : + 1 = \sum_{i=1}^{n} + p_{i}\left(x_1,\ldots,x_n\right) q_i\left( + x_1,\ldots,x_n, \frac{1}{f(x_1,\ldots,x_n)} \right) + \] + Multiplying by a + sufficient power of $f$, this yields an equation in $R$ : \[ - f^d = \sum_{i=1}^{n} p_{i}(x_1,\ldots,_n) \cdot q_i'(x_1,\ldots,x_n) \in I - \] - Thus $f \in \sqrt{I}$. + f^d = + \sum_{i=1}^{n} p_{i}(x_1,\ldots,_n) \cdot + q_i'(x_1,\ldots,x_n) \in I + \] + Thus $f + \in \sqrt{I}$. \end{proof} -\begin{corollary}\label{antimonbij} +\begin{corollary} + \label{antimonbij} \begin{align} - f: \{I \subseteq R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\} \\ - I &\longmapsto V(I)\\ - \{f \in R | A \subseteq V(f)\} &\longmapsfrom A + f: \{I \subseteq R | I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\} \\ + I & \longmapsto V(I) \\ + \{f \in R | A \subseteq V(f)\} & \longmapsfrom A \end{align} is a $\subseteq$-antimonotonic bijection. \end{corollary} @@ -240,8 +455,14 @@ For $f \in R$ let $V(f) = V(fR)$. The topological space $\mathfrak{k}^n$ is Noetherian. \end{corollary} \begin{proof} - Because the map from \ref{antimonbij} is antimonotonic, strictly decreasing chains of closed subsets of $\mathfrak{k}^n$ are mapped to strictly increasing chains of ideals in $R$. - By the Basissatz (\ref{basissatz}), $R$ is Noetherian. + Because the map from + \ref{antimonbij} is antimonotonic, strictly + decreasing + chains of closed subsets of $\mathfrak{k}^n$ are mapped to strictly + increasing + chains of ideals in $R$. + By the Basissatz ( + \ref{basissatz}), $R$ is Noetherian. \end{proof} % Lecture 04 @@ -252,64 +473,99 @@ For $f \in R$ let $V(f) = V(fR)$. Let $X$ be a topological space. \begin{definition} - $X$ is called \vocab[Topological space!irreducible]{irreducible}, if $X \neq \emptyset$ and the following equivalent conditions hold: + $X$ is called \vocab[Topological space!irreducible]{irreducible}, if $X \neq \emptyset$ and the following + equivalent conditions hold: \begin{enumerate}[A] - \item Every open $\emptyset \neq U \subseteq X$ is dense. - \item The intersection of non-empty, open subsets $U, V \subseteq X$ is non-empty. - \item If $A, B \subseteq X$ are closed, $X = A \cup B$ then $X = A$ or $X = B$. - \item Every open subset of $X$ is connected. + \item + Every open $\emptyset \neq U \subseteq X$ is dense. + \item + The intersection of non-empty, open subsets $U, V \subseteq X$ is non-empty. + \item + If $A, B \subseteq X$ are closed, $X = A \cup B$ then $X = A$ or $X = B$. + \item + Every open subset of $X$ is connected. \end{enumerate} \end{definition} -\begin{proof}\, +\begin{proof} + \, \begin{itemize} - \item[$A \iff B$] by definition of denseness. - \item[B $\iff$ C] Let $U \coloneqq X \setminus A, V \coloneqq X \setminus B$. - \item[B $\implies$ D] Suppose $W$ is a non-connected open subset. Then there exists a decomposition $W = U \cup V$ into disjoint open subsets. - \item[D $\implies$ B] If $U,V \neq \emptyset$ are disjoint open subsets, then $U \cup V$ is non-connected. + \item[$A \iff B$] + by definition of denseness. + \item[B $\iff$ C] + Let $U \coloneqq X \setminus A, V \coloneqq X \setminus B$. + \item[B $\implies$ D] + Suppose $W$ is a non-connected open subset. + Then there exists a decomposition $W = U \cup V$ into disjoint open subsets. + \item[D $\implies$ B] + If $U,V \neq \emptyset$ are disjoint open subsets, then $U \cup V$ is + non-connected. \end{itemize} \end{proof} \begin{corollary} Every irreducible topological space is connected. \end{corollary} \begin{example} - $\mathfrak{k}^n$ is irreducible as shown in \ref{zariskinothd}. + $\mathfrak{k}^n$ is irreducible as shown in + \ref{zariskinothd}. \end{example} \begin{fact} \begin{enumerate}[A] - \item A single point is always irreducible. - \item If $X$ is Hausdorff then it is irreducible iff it has precisely one point. - \item $X$ is irreducible iff it cannot be written as a finite union of proper closed subsets. - \item $X$ is irreducible iff any finite intersection of non-empty open subsets is non-empty. ($\bigcap \emptyset \coloneqq X$) + \item + A single point is always irreducible. + \item + If $X$ is Hausdorff then it is irreducible iff it has precisely one point. + \item + $X$ is irreducible iff it cannot be written as a finite union of proper closed + subsets. + \item + $X$ is irreducible iff any finite intersection of non-empty open subsets is + non-empty. ($\bigcap \emptyset \coloneqq X$) \end{enumerate} \end{fact} \begin{proof} \begin{enumerate} - \item[A,B] trivial - \item[C] $\implies$ : Induction on the cardinality of the union. $\impliedby $: $\bigcap \emptyset = X$ is non-empty and any intersection of two non-empty open subsets is non-empty. - \item[D] Follows from C. + \item[A,B] + trivial + \item[C] + $\implies$ : Induction on the cardinality of the union. $\impliedby $: $\bigcap + \emptyset = X$ is non-empty and any intersection of two non-empty open subsets + is non-empty. + \item[D] + Follows from C. \end{enumerate} \end{proof} \subsubsection{Irreducible components} \begin{fact} - If $D \subseteq X$ is dense, then $X$ is irreducible iff $D$ is irreducible with its induced topology. + If $D \subseteq X$ is dense, then $X$ is irreducible iff $D$ is irreducible + with its induced topology. \end{fact} \begin{proof} $X = \emptyset$ iff $D = \emptyset$. - Suppose $B$ is the union of its proper closed subsets $A,B$. Then $X = \overline{A} \cup \overline{B}$. These are proper closed subsets of $X$, as $\overline{A} \cap D = A \cap D$ (by closedness of $D$) and thus $\overline{A} \cap D \neq D$. + Suppose $B$ is the union of its proper closed subsets $A,B$. + Then $X = \overline{A} \cup \overline{B}$. + These are proper closed subsets of $X$, as $\overline{A} \cap D = A + \cap D$ (by + closedness of $D$) and thus $\overline{A} \cap D \neq D$. - On the other hand, if $U$ and $V$ are disjoint non-empty open subsets of $X$, then $U \cap D$ and $V \cap D$ are disjoint non-empty open subsets of $D$. + On the other hand, if $U$ and $V$ are disjoint non-empty open subsets of $X$, + then $U \cap D$ and $V \cap D$ are disjoint non-empty open subsets of $D$. \end{proof} \begin{definition}[Irreducible subsets] - A subset $Z \subseteq X$ is called \vocab[Topological space!irreducible]{irreducible} if it is irreducible with its induced topology. - $Z$ is called an \vocab{irreducible component} of $X$, if it is irreducible and if every irreducible subset $Z \subseteq Y \subseteq X$ coincides with $Z$. + A subset $Z \subseteq X$ is called + \vocab[Topological space!irreducible]{irreducible} if it is irreducible with its induced topology. + $Z$ is called an \vocab{irreducible component} of $X$, if it is irreducible and if + every irreducible subset $Z \subseteq Y \subseteq X$ coincides with $Z$. \end{definition} \begin{corollary} \begin{enumerate} - \item $Z \subseteq X$ is irreducible iff $\overline{Z} \subseteq X$ is irreducible. - \item Every irreducible component of $X$ is a closed subset of $X$. + \item + $Z \subseteq X$ is irreducible iff $\overline{Z} \subseteq X$ is + irreducible. + \item + Every irreducible component of $X$ is a closed subset of $X$. \end{enumerate} \end{corollary} \begin{notation} @@ -318,280 +574,538 @@ Let $X$ be a topological space. \subsubsection{Decomposition into irreducible subsets} \begin{proposition} - Let $X$ be a Noetherian topological space. Then $X$ can be written as a finite union $X = \bigcup_{i = 1}^n Z_i$ of irreducible closed subsets of $X$. - One may additionally assume that $i \neq j \implies Z_i \not\subseteq Z_i$. With this minimality condition, $n$ and the $Z_i$ are unique (up to permutation) and $\{Z_1,\ldots,Z_n\}$ is the set of irreducible components of $X$. + Let $X$ be a Noetherian topological space. + Then $X$ can be written as a finite union $X = \bigcup_{i = 1}^n Z_i$ + of + irreducible closed subsets of $X$. + One may additionally assume that $i \neq j \implies Z_i \not\subseteq Z_i$. + With this minimality condition, $n$ and the $Z_i$ are unique (up to + permutation) and $\{Z_1,\ldots,Z_n\}$ is the set of irreducible components of + $X$. \end{proposition} \begin{proof} % i = ic - Let $\mathfrak{M}$ be the set of closed subsets of $X$ which cannot be decomposed as a union of finitely many irreducible subsets. - Suppose $\mathfrak{M} \neq \emptyset$. Then there exists a $\subseteq$-minimal $Y \in \mathfrak{M}$. $Y$ cannot be empty or irreducible. Hence $Y = A \cup B$ where $A,B$ are proper closed subsets of $ Y$. By the minimality of $Y$, $A$ and $B$ can be written as a union of proper closed subsets $\lightning$. + Let $\mathfrak{M}$ be the set of closed subsets of $X$ which cannot + be + decomposed as a union of finitely many irreducible subsets. + Suppose $\mathfrak{M} \neq \emptyset$. + Then there exists a $\subseteq$-minimal $Y \in \mathfrak{M}$. + $Y$ cannot be empty or irreducible. + Hence $Y = A \cup B$ where $A,B$ are proper closed subsets of $ Y$. + By the minimality of $Y$, $A$ and $B$ can be written as a union of proper + closed subsets $\lightning$. - Let $X = \bigcup_{i = 1}^n Z_i$, where there are no inclusions between the $Z_i$. If $Y$ is an irreducible subsets of $X$, $Y = \bigcup_{i = 1}^n (Y \cap Z_i)$ and there exists $1 \le i \le n$ such that $Y = Y \cap Z_i$. - Hence $Y \subseteq Z_i$. Thus the $Z_i$ are irreducible components. Conversely, if $Y$ is an irreducible component of $X$, $Y \subseteq Z_i$ for some $i$ and $Y = Z_i$ by the definition of irreducible component. + Let $X = \bigcup_{i = 1}^n Z_i$, where there are no inclusions between + the + $Z_i$. + If $Y$ is an irreducible subsets of $X$, $Y = \bigcup_{i = 1}^n (Y \cap + Z_i)$ + and there exists $1 \le i \le n$ such that $Y = Y \cap Z_i$. + Hence $Y \subseteq Z_i$. + Thus the $Z_i$ are irreducible components. + Conversely, if $Y$ is an irreducible component of $X$, $Y \subseteq Z_i$ for + some $i$ and $Y = Z_i$ by the definition of irreducible component. \end{proof} \begin{remark} - The proof of existence was an example of \vocab{Noetherian induction} : If $E$ is an assertion about closed subsets of a Noetherian topological space $X$ and $E$ holds for $A$ if it holds for all proper subsets of $A$, then $E(A)$ holds for every closed subset $A \subseteq X$. + The proof of existence was an example of \vocab{Noetherian induction} : If $E$ + is an assertion about closed subsets of a Noetherian topological space $X$ and + $E$ holds for $A$ if it holds for all proper subsets of $A$, then $E(A)$ holds + for every closed subset $A \subseteq X$. \end{remark} -\begin{proposition}\label{bijiredprim} - By \ref{antimonbij} there exists a bijection +\begin{proposition} + \label{bijiredprim} + By + \ref{antimonbij} there exists a bijection \begin{align} - f: \{I \subseteq R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\} \\ - I &\longmapsto V(I)\\ - \{f \in R | A \subseteq V(f)\} &\longmapsfrom A + f: \{I \subseteq R | + I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n + | A \text{ Zariski-closed}\} \\ I & \longmapsto V(I)\\ \{f \in R | A + \subseteq V(f)\} & \longmapsfrom A \end{align} - Under this correspondence $A \subseteq \mathfrak{k}^n$ is irreducible iff $I \coloneqq f^{-1}(A)$ is a prime ideal. + Under this correspondence $A \subseteq \mathfrak{k}^n$ is + irreducible iff $I + \coloneqq f^{-1}(A)$ is a prime ideal. Moreover, $\#A = 1$ iff $I$ is a maximal ideal. \end{proposition} \begin{proof} - By the Nullstellensatz (\ref{hns1}), $A = \emptyset \iff I = R$. Suppose $A = B \cup C$ is a decomposition into proper closed subsets $A = V(J), B = V(K)$ where $J = \sqrt{J}. K = \sqrt{K}$. - Since $A \neq B$ and $A \neq C$, there are $f \in J \setminus I, g \in K \setminus I$. $fg$ vanishes on $A = B \cup C$. By the Nullstellensatz (\ref{hns3}) $fg \in \sqrt{I} = I$ and $I$ fails to be prime. + By the Nullstellensatz ( + \ref{hns1}), $A = \emptyset \iff I = R$. + Suppose $A = B \cup C$ is a decomposition into proper closed subsets $A = V(J), + B = V(K)$ where $J = \sqrt{J}. + K = \sqrt{K}$. + Since $A \neq B$ and $A \neq C$, there are $f \in J \setminus I, g \in K + \setminus I$. + $fg$ vanishes on $A = B \cup C$. + By the Nullstellensatz ( + \ref{hns3}) $fg \in + \sqrt{I} = I$ and $I$ fails to be + prime. - On the other hand suppose that $fg \in I, f \notin I, g \not\in I$. By the Nullstellensatz (\ref{hns3}) and $I = \sqrt{I} $ neither $f$ nor $g$ vanishes on all of $A$. Thus $(A \cap V(f)) \cup (A \cap V(g))$ is a decomposition and $A$ fails to be irreducible. + On the other hand suppose that $fg \in I, f \notin I, g \not\in I$. + By the Nullstellensatz ( + \ref{hns3}) and $I = + \sqrt{I} $ neither $f$ nor $g$ + vanishes on all of $A$. + Thus $(A \cap V(f)) \cup (A \cap V(g))$ is a decomposition and $A$ fails to be + irreducible. - The remaining assertion follows from the fact, that the bijection is $\subseteq$-antimonotonic and thus maximal ideals correspond to minimal irreducible closed subsets, which are the one-point subsets as $\mathfrak{k}^n$ is T${}_1$. + The remaining assertion follows from the fact, that the bijection is + $\subseteq$-antimonotonic and thus maximal ideals correspond to minimal + irreducible closed subsets, which are the one-point subsets as + $\mathfrak{k}^n$ + is T${}_1$. \end{proof} \subsection{Krull dimension} \begin{definition} - Let $Z $ be an irreducible subset of the topological space $X$. Let $\codim(Z,X)$ be the maximum of the length $n$ of strictly increasing chains $Z \subseteq Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n$ of irreducible closed subsets of $X$ containing $Z$ or $\infty$ if such chains can be found for arbitrary $n$. - Let + Let $Z $ be an irreducible subset of the topological space $X$. + Let $\codim(Z,X)$ be the maximum of the length $n$ of strictly + increasing + chains $Z \subseteq Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n$ of + irreducible closed subsets of $X$ containing $Z$ or $\infty$ if such chains can + be found for arbitrary $n$. + Let \[ - \dim X \coloneqq \begin{cases} - - \infty &\text{if } X = \emptyset\\ - \sup_{\substack{Z \subseteq X\\ Z \text{ irreducible}}} \codim(Z,X) & \text{otherwise} - \end{cases} - \] + \dim X \coloneqq + \begin{cases} + - \infty & \text{if } X = \emptyset \\ + \sup_{\substack{Z \subseteq X \\ Z \text{ irreducible}}} \codim(Z,X) & + \text{otherwise} + \end{cases} + \] \end{definition} \begin{remark} \begin{itemize} - \item In the situation of the definition $\overline{Z}$ is irreducible. Hence $\codim(Z,X)$ is well-defined and one may assume without losing much generality that $Z$ is closed. - \item Because a point is always irreducible, every non-empty topological space has an irreducible subset and for $X \neq \emptyset$, $\dim X$ is $\infty$ or $\max_{x \in X} \codim(\{x\}, X)$. - \item Even for Noetherian $X$, it may happen that $\codim(Z,X) = \infty$. - \item Even for if $X$ is Noetherian and $\codim(Z,X)$ is finite for all irreducible subsets $Z$ of $X$, $\dim X$ may be infinite. + \item + In the situation of the definition $\overline{Z}$ is irreducible. + Hence $\codim(Z,X)$ is well-defined and one may assume without + losing much + generality that $Z$ is closed. + \item + Because a point is always irreducible, every non-empty topological space has an + irreducible subset and for $X \neq \emptyset$, $\dim X$ is $\infty$ or + $\max_{x \in X} \codim(\{x\}, X)$. + \item + Even for Noetherian $X$, it may happen that $\codim(Z,X) = \infty$. + \item + Even for if $X$ is Noetherian and $\codim(Z,X)$ is finite for all irreducible + subsets $Z$ of $X$, $\dim X$ may be infinite. \end{itemize} \end{remark} \begin{fact} If $X = \{x\}$, then $\dim X = 0$. \end{fact} \begin{fact} - For every $x \in \mathfrak{k}$, $\codim( \{x\} ,\mathfrak{k}) = 1$. The only other irreducible closed subset of $\mathfrak{k}$ is $\mathfrak{k}$ itself, which has codimension zero. Thus $\dim \mathfrak{k} = 1$. + For every $x \in \mathfrak{k}$, $\codim( \{x\} ,\mathfrak{k}) = 1$. + The only other irreducible closed subset of $\mathfrak{k}$ is + $\mathfrak{k}$ + itself, which has codimension zero. + Thus $\dim \mathfrak{k} = 1$. \end{fact} \begin{fact} - Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that $U \cap Y \neq \emptyset$. Then we have a bijection + Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that + $U \cap Y \neq \emptyset$. + Then we have a bijection \begin{align} - f: \{A \subseteq X | A \text{ irreducible, closed and } Y \subseteq A\} &\longrightarrow \{B \subseteq U | B \text{ irreducible, closed and } Y \cap U \subseteq B\} \\ - A&\longmapsto A \cap U\\ - \overline{B}&\longmapsfrom B + f: \{A \subseteq X | A \text{ + irreducible, closed and } Y \subseteq A\} & \longrightarrow \{B \subseteq U | + B \text{ irreducible, closed and } Y \cap U \subseteq B\} \\ A & \longmapsto A + \cap U \\ \overline{B} & \longmapsfrom B \end{align} - where $\overline{B}$ denotes the closure in $X$. + where $\overline{B}$ denotes + the closure in $X$. \end{fact} \begin{proof} - If $A$ is given and $B = A \cap U$, then $B \neq \emptyset$ and B is open hence (irreducibility of $A$) dense in $A$, hence $A = \overline{B}$. The fact that $B = \overline{B} \cap U$ is a general property of the closure operator. + If $A$ is given and $B = A \cap U$, then $B \neq \emptyset$ and B is open + hence (irreducibility of $A$) dense in $A$, hence $A = + \overline{B}$. + The fact that $B = \overline{B} \cap U$ is a general property of the + closure + operator. \end{proof} -\begin{corollary}[Locality of Krull codimension] \label{lockrullcodim} - Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that $U \cap Y \neq \emptyset$. +\begin{corollary}[Locality of Krull codimension] + \label{lockrullcodim} + Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that + $U \cap Y \neq \emptyset$. Then $\codim(Y,X) = \codim(Y \cap U, U)$. \end{corollary} \begin{fact} - Let $Z \subseteq Y \subseteq X$ be irreducible closed subsets of the topological space $X$. Then + Let $Z \subseteq Y \subseteq X$ be irreducible closed subsets of the + topological space $X$. + Then \[ - \codim(Z,Y) + \codim(Y,X) \le \codim(Z,X) \tag{CD+}\label{eq:cdp} + \codim(Z,Y) + \codim(Y,X) \le \codim(Z,X) + \tag{CD+} + \label{eq:cdp} \] \end{fact} \begin{proof} - A chain of irreducible closed subsets between $Z$ and $Y$ and a chain of irreducible closed between $Y$ and $X$ can be spliced together. + A chain of irreducible closed subsets between $Z$ and + $Y$ and a chain of irreducible closed between $Y$ and $X$ can be spliced + together. \end{proof} Taking the supremum over all $Z$ we obtain: \begin{fact} - If $Y$ is an irreducible closed subset of the topological space $X$, then + If $Y$ is an + irreducible closed subset of the topological space $X$, then \[ - \dim(Y) + \codim(Y,X) \le \dim(X) \tag{D+}\label{eq:dp} - \] + \dim(Y) + + \codim(Y,X) \le \dim(X) + \tag{D+} + \label{eq:dp} + \] \end{fact} -In general, these inequalities may be strict. +In general, these +inequalities may be strict. \begin{definition}[Catenary topological spaces] - A topological space $T$ is called \vocab[Topological space!catenary]{catenary} if equality holds in \eqref{eq:cdp} whenever $X$ is an irreducible closed subset of $T$. + A topological space $T$ is called + \vocab[Topological space!catenary]{catenary} if equality holds in \eqref{eq:cdp} whenever + $X$ is an irreducible closed subset of $T$. \end{definition} \subsubsection{Krull dimension of \texorpdfstring{$\mathfrak{k}^n$}{kn}} % from lecture 04 -\begin{theorem}\label{kdimkn} - $\dim \mathfrak{k}^n = n$ and $\mathfrak{k}^n$ is catenary. Moreover, if $X$ is an irreducible closed subset of $\mathfrak{k}^n$, then equality occurs in \eqref{eq:dp}. +\begin{theorem} + \label{kdimkn} + $\dim \mathfrak{k}^n = n$ and $\mathfrak{k}^n$ is + catenary. + Moreover, if $X$ is an irreducible closed subset of + $\mathfrak{k}^n$, then + equality occurs in \eqref{eq:dp}. \end{theorem} \begin{proof} Considering \[ - \{0\} \subsetneq \mathfrak{k} \times \{0\} \subsetneq \mathfrak{k}^2 \times \{0\} \subsetneq \ldots \subsetneq \mathfrak{k}^n + \{0\} \subsetneq \mathfrak{k} \times \{0\} \subsetneq + \mathfrak{k}^2 \times \{0\} \subsetneq \ldots \subsetneq + \mathfrak{k}^n \] - it is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$.Translation by $x \in \mathfrak{k}^n$ gives us $\codim(\{x\} , \mathfrak{k}^n) \ge n$. + it + is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$.Translation by $x \in + \mathfrak{k}^n$ gives us $\codim(\{x\} , \mathfrak{k}^n) \ge n$. - The opposite inequality follows from \ref{upperbounddim} ($Z = \mathfrak{k}^n$ $\dim \mathfrak{k}^n \le \trdeg(\mathfrak{K}(Z) / \mathfrak{k}) = \trdeg(Q(\mathfrak{k}[X_1,\ldots,X_n]) / \mathfrak{k}) = n$). + The opposite inequality follows from \ref{upperbounddim} ($Z = + \mathfrak{k}^n$ + $\dim \mathfrak{k}^n \le \trdeg(\mathfrak{K}(Z) / \mathfrak{k}) = + \trdeg(Q(\mathfrak{k}[X_1,\ldots,X_n]) / \mathfrak{k}) = n$). - The theorem is a special case of \ref{htandtrdeg}. - % DIMT + The theorem is a special case of + \ref{htandtrdeg}. + % DIMT \end{proof} -\begin{lemma}\label{ufdprimeideal} +\begin{lemma} + \label{ufdprimeideal} Every non-zero prime ideal $\fp$ of a UFD $R$ contains a prime element. \end{lemma} \begin{proof} - Let $p \in \fp \setminus \{0\} $ with the minimal number of prime factors, counted by multiplicity. - If $p $ was a unit, then $\fp \supseteq pR = R$. If $p = ab$ with non-units $a,b$, it follows that $a \in \fp$ or $b \in \fp$ contradicting the minimality assumption. + Let $p \in \fp \setminus \{0\} $ with the minimal number of prime factors, + counted by multiplicity. + If $p $ was a unit, then $\fp \supseteq pR = R$. + If $p = ab$ with non-units $a,b$, it follows that $a \in \fp$ or $b \in \fp$ + contradicting the minimality assumption. Thus $p$ is a prime element of $R$. \end{proof} -\begin{proposition}[Irreducible subsets of codimension one]\label{irredcodimone} - Let $p \in R = \mathfrak{k}[X_1,\ldots, X_n]$ be a prime element. Then the irreducible subset $X = V(p) \subseteq \mathfrak{k}^n$ has codimension one, and every codimension one subset of $\mathfrak{k}^n$ has this form. +\begin{proposition}[Irreducible subsets of codimension one] + \label{irredcodimone} + Let $p \in R = \mathfrak{k}[X_1,\ldots, X_n]$ be a prime element. + Then the irreducible subset $X = V(p) \subseteq \mathfrak{k}^n$ has + codimension + one, and every codimension one subset of $\mathfrak{k}^n$ has this + form. \end{proposition} \begin{proof} - Since $pR$ is a prime ideal, $X = V(p)$ is irreducible. Since $p \neq 0$, $X$ is a proper subset of $\mathfrak{k}^n$. - If $X \subseteq Y \subseteq \mathfrak{k}^n$ is irreducible and closed, then $Y = V(\fq)$ for some prime ideal $\fp \subseteq pR$. - If $Y \neq \mathfrak{k}^n$, then $\fp \neq \{0\}$. By \ref{ufdprimeideal} there exists a prime element $q \in \fq$. As $\fq \subseteq pR$ we have $p \divides q$. - By the irreducibility of $p$ and $q$ it follows that $p \sim q$. Hence $\fq = pR$ and $X = Y$. + Since $pR$ is a prime ideal, $X = V(p)$ is irreducible. + Since $p \neq 0$, $X$ is a proper subset of $\mathfrak{k}^n$. + If $X \subseteq Y \subseteq \mathfrak{k}^n$ is irreducible and + closed, then $Y + = V(\fq)$ for some prime ideal $\fp \subseteq pR$. + If $Y \neq \mathfrak{k}^n$, then $\fp \neq \{0\}$. + By + \ref{ufdprimeideal} there exists a prime element $q \in \fq$. + As $\fq \subseteq pR$ we have $p \divides q$. + By the irreducibility of $p$ and $q$ it follows that $p \sim q$. + Hence $\fq = pR$ and $X = Y$. - Suppose $X = V(\fp) \subseteq \mathfrak{k}^n$ is closed, irreducible and of codimension one. - Then $\fp \neq \{0\}$, hence $X \neq \mathfrak{k}^n$. By \ref{ufdprimeideal} there is a prime element $p \in \fp$. If $\fp \neq pR$, then - $X \subsetneq V(p) \subsetneq \mathfrak{k}^n$ contradicts $\codim(X, \mathfrak{k}^n) = 1$. + Suppose $X = V(\fp) \subseteq \mathfrak{k}^n$ is closed, irreducible + and of + codimension one. + Then $\fp \neq \{0\}$, hence $X \neq \mathfrak{k}^n$. + By + \ref{ufdprimeideal} there is a prime element $p \in \fp$. + If $\fp \neq pR$, then $X \subsetneq V(p) \subsetneq + \mathfrak{k}^n$ + contradicts $\codim(X, \mathfrak{k}^n) = 1$. \end{proof} % Lecture 05 \subsection{Transcendence degree} \subsubsection{Matroids} \begin{definition}[Hull operator] - - Let $X$ be a set, $\mathcal{P}(X)$ the power set of $X$. A \vocab{Hull operator} on $X$ is a map $\mathcal{P}(X) \xrightarrow{\mathcal{H}} \mathcal{P}(X)$ such that + + Let $X$ be a set, $\mathcal{P}(X)$ the power set of $X$. + A \vocab{Hull operator} on $X$ is a map $\mathcal{P}(X) + \xrightarrow{\mathcal{H}} \mathcal{P}(X)$ such that \begin{enumerate} - \item[H1] $\forall A \in \mathcal{P}(X) ~ A \subseteq \mathcal{H}(A)$. - \item[H2] $A \subseteq B \subseteq X \implies \mathcal{H}(A) \subseteq \mathcal{H}(B)$. - \item[H3] $\mathcal{H}(\mathcal{H}(X)) = \mathcal{H}(X)$. + \item[H1] + $\forall A \in \mathcal{P}(X) ~ A \subseteq \mathcal{H}(A)$. + \item[H2] + $A \subseteq B \subseteq X \implies \mathcal{H}(A) \subseteq + \mathcal{H}(B)$. + \item[H3] + $\mathcal{H}(\mathcal{H}(X)) = \mathcal{H}(X)$. \end{enumerate} - We call $\mathcal{H}$ \vocab{matroidal} if in addition the following conditions hold: + We call $\mathcal{H}$ \vocab{matroidal} if in addition the + following + conditions hold: \begin{enumerate} - \item[M] If $m,n \in X$ and $A \subseteq X$ then $m \in \mathcal{H}( \{n\} \cup A) \setminus \mathcal{H}(A) \iff n \in \mathcal{H}(\{m\} \cup A) \setminus \mathcal{H}(A).$ - \item[F] $\mathcal{H}(A) = \bigcup_{F \subseteq A \text{ finite}} \mathcal{H}(F)$. + \item[M] + If $m,n \in X$ and $A \subseteq X$ + then $m \in \mathcal{H}( \{n\} \cup A) \setminus \mathcal{H}(A) \iff n + \in + \mathcal{H}(\{m\} \cup A) \setminus \mathcal{H}(A). + $ + \item[F] + $\mathcal{H}(A) = \bigcup_{F \subseteq A \text{ finite}} \mathcal{H}(F)$. \end{enumerate} - In this case, $S \subseteq X$ is called \vocab{Independent subset}, if $s \not\in \mathcal{H}(S \setminus \{s\})$ for all $s \in S$ and + In this case, $S \subseteq X$ is called \vocab{Independent subset}, if $s + \not\in \mathcal{H}(S \setminus \{s\})$ for all $s \in S$ and \vocab[Generating subset]{generating} if $X = \mathcal{H}(S)$. - $S$ is called a \vocab{base}, if it is both generating and independent. + $S$ is called a \vocab{base}, if it is both generating and + independent. \end{definition} \begin{theorem} - If $\mathcal{H}$ is a matroidal hull operator on $X$, then a basis exists, every independent set is contained in a base and two arbitrary bases have the same cardinality. + If $\mathcal{H}$ is a matroidal hull operator on $X$, then a basis + exists, + every independent set is contained in a base and two arbitrary bases have the + same cardinality. \end{theorem} \begin{example} - Let $K$ be a field, $V$ a $K$-vector space and $\mathcal{L}(T)$ the $K$-linear hull of $T$ for $T \subseteq V$. + Let $K$ be a field, $V$ a $K$-vector space and $\mathcal{L}(T)$ the + $K$-linear + hull of $T$ for $T \subseteq V$. Then $\mathcal{L}$ is a matroidal hull operator on $V$. \end{example} \subsubsection{Transcendence degree} \begin{lemma} - Let $L / K$ be a field extension and let $\mathcal{H}(T)$ be the algebraic closure in $L$ of the subfield of $L$ generated by $K$ and $T$.\footnote{This is the intersection of all subfields of $L$ containing $K \cup T$, or the field of quotients of the sub-$K$-algebra of $L$ generated by $T$.} + Let $L / K$ be a field extension and let $\mathcal{H}(T)$ be the + algebraic + closure in $L$ of the subfield of $L$ generated by $K$ and $T$. + \footnote{This is the intersection of all subfields of $L$ containing $K \cup + T$, or the field of quotients of the sub-$K$-algebra of $L$ generated by $T$.} Then $\mathcal{H}$ is a matroidal hull operator. \end{lemma} \begin{proof} - H1, H2 and F are trivial. For an algebraically closed subfield $K \subseteq M \subseteq L$ we have $\mathcal{H}(M) = M$. Thus $\mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T)$ (H3). + H1, H2 and F are trivial. + For an algebraically closed subfield $K \subseteq M \subseteq L$ we have + $\mathcal{H}(M) = M$. + Thus $\mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T)$ (H3). - Let $x,y \in L$, $T \subseteq L$ and $x \in \mathcal{H}(T \cup \{y\}) \setminus \mathcal{H}(T)$. We have to show that $y \in \mathcal{H}(T \cup \{x\}) \setminus \mathcal{H}(T)$. - If $y \in \mathcal{H}(T)$ we have $\mathcal{H}(T \cup \{y\}) \subseteq \mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T) \implies x \in \mathcal{H}(T) \setminus \mathcal{H}(T) \lightning$. - Hence it is sufficient to show $y \in \mathcal{H}(T \cup \{x\})$. Without loss of generality loss of generality $T = \emptyset$ (replace $K$ be the subfield generated by $K \cup T$). - Then $x$ is algebraic over the subfield $M$ of $L$ generated by $K \cup \{y\}$. Thus there exists $0 \neq P \in M[T]$ with $P(x) = 0$. - The coefficients $p_i$ of $P$ belong to the field of quotients of the $K$-subalgebra of $L$ generated by $y$. There are thus polynomials $Q_i, R \in K[Y]$ such that $p_i = \frac{Q_i(y)}{R(y)}$, $R(y) \neq 0$. + Let $x,y \in L$, $T \subseteq L$ and $x \in \mathcal{H}(T \cup \{y\}) + \setminus \mathcal{H}(T)$. + We have to show that $y \in \mathcal{H}(T \cup \{x\}) \setminus + \mathcal{H}(T)$. + If $y \in \mathcal{H}(T)$ we have $\mathcal{H}(T \cup \{y\}) \subseteq + \mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T) \implies x \in + \mathcal{H}(T) + \setminus \mathcal{H}(T) \lightning$. + Hence it is sufficient to show $y \in \mathcal{H}(T \cup \{x\})$. + Without loss of generality loss of generality $T = \emptyset$ (replace $K$ be + the subfield generated by $K \cup T$). + Then $x$ is algebraic over the subfield $M$ of $L$ generated by $K \cup \{y\}$. + Thus there exists $0 \neq P \in M[T]$ with $P(x) = + 0$. + The coefficients $p_i$ of $P$ belong to the field of quotients of the + $K$-subalgebra of $L$ generated by $y$. + There are thus polynomials $Q_i, R \in K[Y]$ such + that $p_i = + \frac{Q_i(y)}{R(y)}$, $R(y) \neq 0$. Let \[ - Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty} q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j \hat{Q_j}(X) \in K[X,Y] - \]. + Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty} + q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j + \hat{Q_j}(X) \in K[X,Y] + \] + . Then $Q(x,y) = 0$. - Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$. Then $\hat{P}(y) = 0$. As $Q \neq 0$ there is $(i,j) \in \N^2$ such that $q_{i,j} \neq 0$ and then $\hat{p_j} \neq 0$ as $x \not\in \mathcal{H}(\emptyset)$. Thus $\hat{P} \in \hat{M}[X] \setminus \{0\} $, where $\hat{M}$ is the subfield of $L$ generated by $K$ and $x$. Thus $y$ is algebraic over $\hat{M}$ and $y \in \mathcal{H}(\{x\})$, + Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$. + Then $\hat{P}(y) = 0$. + As $Q \neq 0$ there is $(i,j) \in \N^2$ such that + $q_{i,j} \neq 0$ and then + $\hat{p_j} \neq 0$ as $x \not\in \mathcal{H}(\emptyset)$. + Thus $\hat{P} \in \hat{M}[X] \setminus \{0\} $, + where $\hat{M}$ is the + subfield of $L$ generated by $K$ and $x$. + Thus $y$ is algebraic over $\hat{M}$ and $y \in + \mathcal{H}(\{x\})$, \end{proof} -\begin{definition}[Transcendence Base] - Let $L / K$ be a field extension and $\mathcal{H}(T)$ the algebraic closure in $L$ of the subfield generated by $K$ and $T$. A base for $(L, \mathcal{H})$ is called a \vocab{transcendence base} and the \vocab{transcendence degree} $\trdeg(L / K)$ is defined as the cardinality of any transcendence base of $L / K$. +\begin{definition}[Transcendence Base] Let $L / K$ be a field + extension and $\mathcal{H}(T)$ the algebraic closure in $L$ of the + subfield + generated by $K$ and $T$. + A base for $(L, \mathcal{H})$ is called a \vocab{transcendence base} + and the + \vocab{transcendence degree} $\trdeg(L / K)$ is defined as the + cardinality of + any transcendence base of $L / K$. \end{definition} \begin{remark} $L / K$ is algebraic iff $\trdeg(L / K) = 0$. \end{remark} -\subsection{Inheritance of Noetherianness and of finite type by subrings and subalgebras / Artin-Tate} -The following will lead to another proof of the Nullstellensatz, which uses the transcendence degree. +\subsection{Inheritance of Noetherianness and of finite type by subrings and subalgebras / + Artin-Tate} +The following will lead to another proof of the Nullstellensatz, which uses the +transcendence degree. \begin{remark} - There exist non-Noetherian domains, which are subrings of Noetherian domains (namely the field of quotients is Noetherian). + There exist non-Noetherian domains, which are subrings of Noetherian domains + (namely the field of quotients is Noetherian). \end{remark} \begin{theorem}[Eakin-Nagata] - Let $A$ be a subring of the Noetherian ring $B$. If the ring extension $B / A$ is finite (i.e. $B$ finitely generated as an $A$-module) then $A$ is Noetherian. + Let $A$ be a subring of the Noetherian ring $B$. + If the ring extension $B / A$ is finite (i.e. $B$ finitely generated as an + $A$-module) then $A$ is Noetherian. \end{theorem} -\begin{fact}+\label{noethersubalg} - Let $R$ be Noetherian and let $B$ be a finite $R$-algebra. Then every $R$-subalgebra $A \subseteq B$ is finite over $R$. +\begin{fact} + + + \label{noethersubalg} + Let $R$ be Noetherian and let $B$ be a finite $R$-algebra. + Then every $R$-subalgebra $A \subseteq B$ is finite over $R$. \end{fact} \begin{proof} - Since $B$ a finitely generated $R$-module and $R$ a Noetherian ring, $B$ is a Noetherian $R$-module (this is a stronger assertion than Noetherian algebra). + Since $B$ a finitely generated $R$-module and $R$ a Noetherian ring, $B$ is a + Noetherian $R$-module (this is a stronger assertion than Noetherian algebra). Thus the sub- $R$-module $A$ is finitely generated. \end{proof} \begin{proposition}[Artin-Tate] \label{artintate} - Let $A$ be a subalgebra of the $R$-algebra $B$, where $R$ is Noetherian. If $ B / R$ is of finite type and $B / A$ is finite, then $A / R$ is also of finite type. + Let $A$ be a subalgebra of the $R$-algebra $B$, where $R$ is Noetherian. + If $ B / R$ is of finite type and $B / A$ is finite, then $A / R$ is also of + finite type. - \[ + \[ \begin{tikzcd} - A \arrow[hookrightarrow]{rr}{\subseteq}& & B \\ - &R \arrow{ul}{\alpha} \arrow{ur}{\alpha} \text{~(Noeth.)} + A \arrow[hookrightarrow]{rr}{\subseteq}& & B \\ &R \arrow{ul}{\alpha} + \arrow{ur}{\alpha} \text{~(Noeth.) + } \end{tikzcd} - \] + \] \end{proposition} \begin{proof} - Let $(b_i)_{i=1}^{m}$ generate $B$ as an $A$-module and $(\beta_j)_{j=1}^m$ as an $R$-algebra. - There are $a_{ijk} \in A$ such that $b_i b_j = \sum_{k=1}^{m} a_{ijk}b_k$. And $\alpha_{ij} \in A$ such that $\beta_i = \sum_{j=1}^{m} \alpha_{ij}b_j$. Let $\tilde{A}$ be the sub- $R$-algebra of $A$ generated by the $a_{ijk}$ and $\alpha_{ij}$. $\tilde{A}$ is of finite type over $ R$, hence Noetherian. The $\tilde{A}$-submodule generated by $1$ and the $b_i$ is a sub-$R$-algebra containing the $\beta_i$ and thus coincides with $B$. - Hence $B / \tilde{A}$ is finite. Since $A \subseteq B, A / \tilde{A}$ is finite (\ref{noethersubalg}). - Hence $A / \tilde{A}$ is of finite type. By the transitivity of ``of finite type'', it follows that $A / R$ is of finite type. + Let $(b_i)_{i=1}^{m}$ generate $B$ as an + $A$-module and $(\beta_j)_{j=1}^m$ as + an $R$-algebra. + There are $a_{ijk} \in A$ such that $b_i b_j = + \sum_{k=1}^{m} a_{ijk}b_k$. + And $\alpha_{ij} \in A$ such that $\beta_i = + \sum_{j=1}^{m} \alpha_{ij}b_j$. + Let $\tilde{A}$ be the sub- $R$-algebra of $A$ generated by the + $a_{ijk}$ and + $\alpha_{ij}$. + $\tilde{A}$ is of finite type over $ R$, hence Noetherian. + The $\tilde{A}$-submodule generated by $1$ and the $b_i$ is a + sub-$R$-algebra + containing the $\beta_i$ and thus coincides with $B$. + Hence $B / \tilde{A}$ is finite. + Since $A \subseteq B, A / \tilde{A}$ is finite + ( + \ref{noethersubalg}). + Hence $A / \tilde{A}$ is of finite type. + By the transitivity of ``of finite type'', it follows that $A / R$ is of finite + type. \[ \begin{tikzcd} \tilde A \arrow[hookrightarrow]{r}{\subseteq}& A \arrow[hookrightarrow]{r}{\subseteq} & B \\ - &R \arrow[bend left, swap]{ul}{\alpha} \arrow{u}{\alpha} \arrow[bend right]{ur}{\alpha} + &R \arrow[bend left, swap]{ul}{\alpha} \arrow{u}{\alpha} \arrow[bend right]{ur}{\alpha} \end{tikzcd} - \] - + \] + \end{proof} \subsubsection{Artin-Tate proof of the Nullstellensatz} -Let $K$ be a field and $R = K[X_1,\ldots,X_n]$. +Let $K$ be a field and $R = K[X_1,\ldots,X_n]$. \begin{definition}[Rational functions] - Let $K(X_1,\ldots,X_n) \coloneqq Q(R)$ be the field of quotients of $R$. + Let $K(X_1,\ldots,X_n) \coloneqq Q(R)$ be the field of + quotients of $R$. - $K(X_1,\ldots,X_n)$ is called the \vocab{field of rational functions} in $n$ variables over $K$. + $K(X_1,\ldots,X_n)$ is called the \vocab{field of rational functions} in $n$ variables + over $K$. \end{definition} \begin{lemma}[Infinitely many prime elements] - There are infinitely many multiplicative equivalence classes of prime elements in $R$. + There are infinitely many multiplicative equivalence + classes of prime elements in $R$. \end{lemma} \begin{proof} - Suppose $(P_i)_{i =1}^m$ is a complete (up to multiplicative equvialence) lsit of prime elements of $R$. - $m > 0$, as $X_1$ is prime. The polynomial $f \coloneqq 1 + \prod_{i=1}^{m} P_i $ is non-constant, hence not a unit in $R$. Hence there exists a prime divisor $P \in R$. As no $P_i$ divides $f$, $P$ cannot be multiplicatively equivalent to any $P_i \lightning$. + Suppose $(P_i)_{i =1}^m$ is a complete (up to + multiplicative equvialence) lsit + of prime elements of $R$. + $m > 0$, as $X_1$ is prime. + The polynomial $f \coloneqq 1 + \prod_{i=1}^{m} P_i $ is non-constant, + hence + not a unit in $R$. + Hence there exists a prime divisor $P \in R$. + As no $P_i$ divides $f$, $P$ cannot be multiplicatively equivalent to any $P_i + \lightning$. \end{proof} -\begin{lemma}[Ring of rational functions not of finite type]\label{rfuncnft} +\begin{lemma}[Ring of rational functions not of finite type] + \label{rfuncnft} If $n > 0$, then $K(X_1,\ldots,X_n) / K$ is not of finite type. \end{lemma} \begin{proof} - Suppose $(f_i)_{i=1}^m$ generate $K(X_1,\ldots,X_n)$ as a $K$-algebra. Let $f_i = \frac{a_i}{b}, a_i \in R, b \in R \setminus \{0\}$. - Then $bf_i \in R$, and as the $f_i$ generate $K(X_1,\ldots,X_n)$ as a $K$-algebra, for every $g \in K(X_1,\ldots,X_n)$ there is $N \in \N$ with + Suppose $(f_i)_{i=1}^m$ generate + $K(X_1,\ldots,X_n)$ as a $K$-algebra. + Let $f_i = \frac{a_i}{b}, a_i \in R, b \in R \setminus \{0\}$. + Then $bf_i \in R$, and as the $f_i$ generate $K(X_1,\ldots,X_n)$ as a + $K$-algebra, for every $g \in K(X_1,\ldots,X_n)$ there is $N \in \N$ with \[ - b^Ng \in R \tag{+} \label{bNginR} + b^Ng \in R \tag{+} + \label{bNginR} \] - However, if $b = \varepsilon \prod_{i=1}^{l} P_i$ is a decomposition of $b$ into prime factors $P_i$ and a unit $\varepsilon$ in $R$ and $g = \frac{1}{P}$, wehere $P \in R$ is a prime element not multiplicatively equvalent to any $P_i$, - then \eqref{bNginR} fails for any $N \in \N$. + However, if $b = \varepsilon + \prod_{i=1}^{l} P_i$ is a decomposition of $b$ into prime factors $P_i$ + and a + unit $\varepsilon$ in $R$ and $g = \frac{1}{P}$, wehere $P \in R$ + is a prime + element not multiplicatively equvalent to any $P_i$, then + \eqref{bNginR} fails + for any $N \in \N$. \end{proof} -The Nullstellensatz (\ref{hns2}) can be reduced to the case of \ref{rfuncnft}: +The Nullstellensatz ( +\ref{hns2}) can be reduced to the case of +\ref{rfuncnft}: -\begin{proof}(Artin-Tate proof of HNS) - Let $(l_i)_{i=1}^n$ be a transcendence base of $L / K$. If $n = 0$ then $L / K$ is algebraic, hence an integral ring extension, hence a finite ring extension (\ref{ftaiimplf}). +\begin{proof} + (Artin-Tate proof of HNS) Let $(l_i)_{i=1}^n$ + be a transcendence + base of $L / K$. + If $n = 0$ then $L / K$ is algebraic, hence an integral ring extension, hence a + finite ring extension ( + \ref{ftaiimplf}). - Suppose $n > 0$. Let $\tilde R \subseteq L$ be the $K$-subalgebra generated by the $l_i$. $\tilde R \cong R \coloneqq K[X_1,\ldots,X_n]$, as the $l_i$ are algebraically independent. - As they are a transcendence base, $L$ is algebraic over the field of quotients $Q(\tilde R)$, hence integral over $Q(\tilde R)$. + Suppose $n > 0$. + Let $\tilde R \subseteq L$ be the $K$-subalgebra generated by the $l_i$. + $\tilde R \cong R \coloneqq K[X_1,\ldots,X_n]$, as the $l_i$ are + algebraically independent. + As they are a transcendence base, $L$ is algebraic over the field of quotients + $Q(\tilde R)$, hence integral over $Q(\tilde R)$. - As $L / K$ is of finite type, so is $L / Q(\tilde R)$ and it follows that $L / Q(\tilde R)$ is a finite ring extension. - By Artin-Tate (\ref{artintate}), $Q(\tilde K)$ is of finite type over $K$. This contradicts \ref{rfuncnft}, as $R \cong \tilde R \implies K(X_1,\ldots,X_n) \cong Q(\tilde R)$. + As $L / K$ is of finite type, so is $L / Q(\tilde R)$ and it follows that $L / + Q(\tilde R)$ is a finite ring extension. + By Artin-Tate ( + \ref{artintate}), $Q(\tilde K)$ is of finite type over + $K$. + This contradicts + \ref{rfuncnft}, as $R \cong \tilde R \implies + K(X_1,\ldots,X_n) \cong Q(\tilde R)$. \end{proof} @@ -599,25 +1113,44 @@ The Nullstellensatz (\ref{hns2}) can be reduced to the case of \ref{rfuncnft}: Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. %i = ic \begin{notation} - Let $X \subseteq \mathfrak{k}^n$ be an irreducible closed subset. Then $X = V(\fp)$ for a unique prime ideal $\fp \subseteq R$. - Let $\mathfrak{K}(X) \coloneqq Q(R / \fp)$ denote the field of quotients of $R / \fp$. + Let $X \subseteq \mathfrak{k}^n$ be an irreducible closed subset. + Then $X = V(\fp)$ for a unique prime ideal $\fp \subseteq R$. + Let $\mathfrak{K}(X) \coloneqq Q(R / \fp)$ denote the field of + quotients of $R + / \fp$. \end{notation} \begin{remark} - As the elements of $\fp$ vanish on $X$, $R / \fp$ may be viewed as the ring of polynomials and $\mathfrak{K}(X)$ as the field of rational functions on $X$. + As the elements of $\fp$ vanish on $X$, $R / \fp$ may be viewed as the ring of + polynomials and $\mathfrak{K}(X)$ as the field of rational functions + on $X$. \end{remark} -\begin{theorem}\label{trdegandkdim} - If $X \subseteq \mathfrak{k}^n$ is irreducible, then $\dim X = \trdeg (\mathfrak{k}(X) / \mathfrak{k})$ and $\codim(X, \mathfrak{k}^n) = n - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$. - More generally if $Y \subseteq \mathfrak{k}^n$ is irreducible and $X \subseteq Y$, then $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$. +\begin{theorem} + \label{trdegandkdim} + If $X \subseteq \mathfrak{k}^n$ is irreducible, then $\dim X = + \trdeg + (\mathfrak{k}(X) / \mathfrak{k})$ and $\codim(X, \mathfrak{k}^n) = n - + \trdeg(\mathfrak{K}(X) / \mathfrak{k})$. + More generally if $Y \subseteq \mathfrak{k}^n$ is irreducible and $X + \subseteq + Y$, then $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - + \trdeg(\mathfrak{K}(X) / \mathfrak{k})$. \end{theorem} \begin{proof} % DIMT - One part will be shown in "A first result on dimension theory" (\ref{upperboundcodim}) - and other one in "Aplication to dimension theory: Proof of $\dim Y = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$" (\ref{lowerbounddimy}). - The theorem is a special case of \ref{htandtrdeg}. + One part will be shown in "A first result on dimension theory" + ( + \ref{upperboundcodim}) and other one in "Aplication to dimension theory: + Proof + of $\dim Y = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$" ( + \ref{lowerbounddimy}). + The theorem is a special case of + \ref{htandtrdeg}. \end{proof} \begin{remark} - Loosely speaking, the Krull dimension of $X$ is equal to the maximal number of $\mathfrak{k}$-algebraically independent rational functions on $X$. - This is yet another indication that the notion of dimension is the ``correct'' one. + Loosely speaking, the Krull dimension of $X$ is equal to the maximal number of + $\mathfrak{k}$-algebraically independent rational functions on $X$. + This is yet another indication that the notion of dimension is the ``correct'' + one. \end{remark} \begin{remark} \ref{kdimkn} follows. @@ -633,102 +1166,185 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. \begin{definition}[Spectrum] Let $R$ be a commutative ring. \begin{itemize} - \item Let $\Spec R$ denote the set of prime ideals and $\MaxSpec R \subseteq \Spec R$ the set of maximal ideals of $R$. - \item For an ideal $I \subseteq R$ let $V(I) \coloneqq \{\fp \in \Spec R | I \subseteq \fp\}$ - \item We equip $\Spec R$ with the \vocab{Zariski-Topology} for which the closed subsets are the subsets of the form $V(I)$, where $I$ runs over the set of ideals in $R$. + \item + Let $\Spec R$ denote the set of prime ideals and $\MaxSpec R \subseteq \Spec R$ + the set of maximal ideals of $R$. + \item + For an ideal $I \subseteq R$ let $V(I) \coloneqq \{\fp \in \Spec R | I + \subseteq \fp\}$ + \item + We equip $\Spec R$ with the \vocab{Zariski-Topology} for which the closed + subsets are the subsets of the form $V(I)$, where $I$ runs over the set of + ideals in $R$. \end{itemize} \end{definition} \begin{remark} - When $R = \mathfrak{k}[X_1,\ldots,X_n]$, the notation $V(I)$ clashes with the previous notation. When several types of $V(I)$ will be in use, they will be distinguished using indices. + When $R = \mathfrak{k}[X_1,\ldots,X_n]$, the notation $V(I)$ clashes with the + previous notation. + When several types of $V(I)$ will be in use, they will be distinguished using + indices. \end{remark} \begin{remark} - Let $(I_{\lambda})_{\lambda \in \Lambda}$ and $(l_j)_{j=1}^n$ be ideals in $R$, where $\Lambda$ may be infinite. We have $V(\sum_{\lambda \in \Lambda} I_\lambda ) = \bigcap_{\lambda \in \Lambda} V(I_\lambda)$ and $V(\bigcap_{j=1}^n I_j) = V(\prod_{j=1}^{n} I_j) = \bigcup_{j = 1}^n V(I_j)$. + Let $(I_{\lambda})_{\lambda \in \Lambda}$ + and $(l_j)_{j=1}^n$ be ideals in $R$, + where $\Lambda$ may be infinite. + We have $V(\sum_{\lambda \in \Lambda} I_\lambda ) = \bigcap_{\lambda \in + \Lambda} + V(I_\lambda)$ and $V(\bigcap_{j=1}^n I_j) = V(\prod_{j=1}^{n} + I_j) = + \bigcup_{j = 1}^n V(I_j)$. Thus, the Zariski topology on $\Spec R$ is a topology. \end{remark} \begin{remark} - Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. Then there exists a bijection (\ref{antimonbij}, \ref{bijiredprim}) between $\Spec R$ and the set of irreducible closed subsets of $\mathfrak{k}^n$ sending $\fp \in \Spec R$ to $V_{\mathfrak{k}^n}(\fp)$ and identifying the one-point subsets with $\MaxSpec R$. - This defines a bijection $\mathfrak{k}^n \cong \MaxSpec R$ which is a homeomorphism if $\MaxSpec R$ is equipped with the induced topology from the Zariski topology on $\Spec R$. + Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. + Then there exists a bijection ( + \ref{antimonbij}, + \ref{bijiredprim}) between + $\Spec R$ and the set of irreducible closed subsets of + $\mathfrak{k}^n$ sending + $\fp \in \Spec R$ to $V_{\mathfrak{k}^n}(\fp)$ and identifying the + one-point + subsets with $\MaxSpec R$. + This defines a bijection $\mathfrak{k}^n \cong \MaxSpec R$ which is + a + homeomorphism if $\MaxSpec R$ is equipped with the induced topology from the + Zariski topology on $\Spec R$. \end{remark} \subsection{Localization of rings} \begin{definition}[Multiplicative subset] - A \vocab{multiplicative subset} of a ring $R$ is a subset $S \subseteq R$ such that $\prod_{i=1}^{n} f_i \in S $ when $n \in \N$ and all $f_i \in S$. + A \vocab{multiplicative subset} of a ring $R$ is a subset $S + \subseteq R$ such that $\prod_{i=1}^{n} f_i \in S $ when $n \in \N$ and + all $f_i \in S$. \end{definition} \begin{proposition} - Let $S \subseteq R$ be a multiplicative subset. Then there is a ring homomorphism $R \xrightarrow{i} R_S$ such that $i(S) \subseteq R_S^{\times }$ and $i$ has the \vocab{universal property} for such ring homomorphisms: - If $R \xrightarrow{j} T$ is a ring homomorphism with $j(S) \subseteq T^{\times }$, then there is a unique ring homomorphism $R_S \xrightarrow{\iota} T$ with $j = \iota i$. + Let $S \subseteq R$ be a multiplicative subset. + Then there is a ring homomorphism $R \xrightarrow{i} R_S$ such that + $i(S) + \subseteq R_S^{\times }$ and $i$ has the \vocab{universal property} for such + ring homomorphisms: If $R \xrightarrow{j} T$ is a ring homomorphism + with $j(S) + \subseteq T^{\times }$, then there is a unique ring + homomorphism $R_S + \xrightarrow{\iota} T$ with $j = \iota i$. \[ \begin{tikzcd} - R \arrow{r}{i}\arrow{d}{j}& R_S \arrow[dotted]{ld}{\existsone \iota}\\ - T + R \arrow{r}{i}\arrow{d}{j}& R_S \arrow[dotted]{ld}{\existsone \iota}\\ T \end{tikzcd} - \] - + \] + \end{proposition} \begin{proof} - The construction is similar to the construction of the field of quotients: + The construction is similar to the construction of the field of + quotients: - Let $R_S \coloneqq (R \times S) / \sim $, where $(r,s) \sim (\rho, \sigma) : \iff \exists t \in S ~ t \sigma r = ts\rho$.\footnote{$t$ does not appear in the construction of the field of quotients, but is important if $S$ contains zero divisors.} - $[r,s] + [\rho, \sigma] \coloneqq [r\sigma + \rho s, s \sigma]$, $[r,s] \cdot [\rho, \sigma] \coloneqq [r \cdot \rho, s \cdot \sigma]$. - - In order proof the universal property define $\iota([r,s]) \coloneqq \frac{j(r)}{j(s)}$. + Let $R_S \coloneqq (R \times S) / \sim $, where $(r,s) \sim (\rho, \sigma) + : + \iff \exists t \in S ~ t \sigma r = ts\rho$. + \footnote{$t$ does not appear in the construction of the field of + quotients, but is important if $S$ contains zero divisors.} + $[r,s] + [\rho, \sigma] + \coloneqq [r\sigma + \rho s, s \sigma]$, $[r,s] + \cdot [\rho, \sigma] \coloneqq [r \cdot \rho, s \cdot \sigma]$. + + In order proof the universal property define $\iota([r,s]) \coloneqq + \frac{j(r)}{j(s)}$. The universal property characterizes $R_S$ up to unique isomorphism. \end{proof} \begin{remark} - $i$ is often not injective and $\ker(i) = \{r \in R | \exists s \in S ~ s \cdot r = 0\} $. + $i$ is often not injective and $\ker(i) = \{r \in R | \exists s + \in S ~ s \cdot r = 0\} $. In particular $(r = 1)$, $R_S$ is the null ring iff $0 \in S$. \end{remark} \begin{notation} - Let $S \subseteq R$ be a multiplicative subset of $R$. We write $\frac{r}{s}$ for $[r,s]$. - The ring homomorphism $R \xrightarrow{i} R_S$ i given by $i(r) = \frac{r}{1}$. - For $X \subseteq R_S$ let $X \sqcap R$ denote $i^{-1}(X)$. + Let $S \subseteq R$ be a multiplicative subset of $R$. + We write $\frac{r}{s}$ for + $[r,s]$. + The ring homomorphism $R \xrightarrow{i} R_S$ i given by $i(r) = + \frac{r}{1}$. + For $X \subseteq R_S$ let $X \sqcap R$ denote + $i^{-1}(X)$. \end{notation} \begin{definition}[$S$-saturated ideal] - An ideal $I \subseteq R$ is called \vocab[Ideal!S-saturated]{$S$-saturated} if for all $s \in S, r \in R$ - $rs \in I \implies r \in I$. + An ideal $I \subseteq R$ is called + \vocab[Ideal! + S-saturated]{$S$-saturated} if for all $s \in S, r \in R$ $rs \in I \implies r + \in I$. \end{definition} -\begin{fact}\label{primeidealssat} - A prime ideal $\fp \subseteq \Spec R$ is $S$-saturated iff $\fp \cap S = \emptyset$. +\begin{fact} + \label{primeidealssat} + A prime ideal $\fp \subseteq \Spec R$ is $S$-saturated iff $\fp \cap S = + \emptyset$. \end{fact} -Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$. +Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an +$S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$. -\begin{fact}\label{ssatiis} - Let $I \subseteq R$ be an $S$-saturated ideal and let $I_S$ denote the ideal $\{\frac{r}{s} | r \in R, s \in S\} \subseteq R_S$. - Then for all $r \in R, s \in S$ - we have $\frac{r}{s} \in I_S \iff r \in I$. +\begin{fact} + \label{ssatiis} + Let $I \subseteq R$ be an $S$-saturated ideal and let $I_S$ denote the ideal + $\{\frac{r}{s} | r \in R, s \in S\} \subseteq R_S$. + Then for all $r \in R, s \in S$ we have $\frac{r}{s} \in I_S \iff r + \in I$. \end{fact} \begin{proof} - Clearly $i \in I \implies \frac{i}{s} \in I_S$. If $\frac{i}{s} \in J$ there are $\iota \in I$, $\sigma \in S$ such that $\frac{i}{s} = \frac{\iota}{\sigma}$ in $R_S$. - This equation holds iff there exists $t \in S$ such that $ts\iota = t \sigma i$. But $ts \iota \in I$ hence $i \in I$, as $I$ is $S $-saturated. + Clearly $i \in I \implies \frac{i}{s} \in I_S$. + If $\frac{i}{s} \in J$ there are $\iota \in I$, $\sigma \in S$ such + that + $\frac{i}{s} = \frac{\iota}{\sigma}$ in $R_S$. + This equation holds iff there exists $t \in S$ such that $ts\iota = t \sigma + i$. + But $ts \iota \in I$ hence $i \in I$, as $I$ is $S $-saturated. \end{proof} -\begin{fact}\label{invimgprimeideal} - The inverse image of a prime ideal under any ring homomorphism is a prime ideal. +\begin{fact} + \label{invimgprimeideal} + The inverse image of a prime ideal under any ring homomorphism is a prime + ideal. \end{fact} -\begin{proposition}\label{idealslocbij} +\begin{proposition} + \label{idealslocbij} \begin{align} - f: \{I \subseteq R | I \text{ $S$-saturated ideal}\} &\longrightarrow \left\{J \subseteq R_S | J \text{ ideal}\right\} \\ - I &\longmapsto I_S \coloneqq \left\{\frac{i}{s} | i \in I, s \in S\right\}\\ - J \sqcap R &\longmapsfrom J\\ + f: \{I \subseteq R | I \text{ $S$-saturated ideal}\} & \longrightarrow \left\{J \subseteq R_S | J \text{ ideal}\right\} \\ + I & \longmapsto I_S \coloneqq \left\{\frac{i}{s} | i \in I, s \in S\right\} \\ + J \sqcap R & \longmapsfrom J \\ \end{align} - is a bijection. Under this bijection $I$ is a prime ideal iff $f(I)$ is. + is a bijection. + Under this bijection $I$ is a prime ideal iff $f(I)$ is. \end{proposition} \begin{proof} - Applying \ref{ssatiis} to $s = 1$ gives $I_S \sqcap R = I$, when $I$ is $S$-saturated. + Applying + \ref{ssatiis} to $s = 1$ gives $I_S \sqcap R = I$, when $I$ + is + $S$-saturated. - Conversely, if $J$ is given and $I = J \sqcap R, \frac{r}{s} \in R_S$, then by \ref{ssatiis} $\frac{r}{s} \in IR_S \iff r \in I$. - But as $\frac{r}{1} = s \cdot \frac{r}{s}$ and $s \in R_S^{\times }$, we have $r \in I \iff \frac{r}{1} \in J \iff \frac{r}{s} \in J$ . - We have thus shown that the two maps between sets of ideals are well-defined and inverse to each other. + Conversely, if $J$ is given and $I = J \sqcap R, \frac{r}{s} \in + R_S$, then by + \ref{ssatiis} $\frac{r}{s} \in IR_S \iff r \in I$. + But as $\frac{r}{1} = s \cdot \frac{r}{s}$ and $s \in + R_S^{\times }$, we have + $r \in I \iff \frac{r}{1} \in J \iff \frac{r}{s} \in J$ + . + We have thus shown that the two maps between sets of ideals are well-defined + and inverse to each other. - By \ref{invimgprimeideal}, $J \in \Spec R_S \implies f^{-1}(J) = J \cap R \in \Spec R_S$. - Suppose $I \in \Spec R$, $\frac{a}{s} \cdot \frac{b}{t} \in I_S$ for some $a,b \in R, s,t \in S$. - By \ref{ssatiis} $ab \in I$. Thus $a \in I \lor b \in I$, hence $\frac{a}{s} \in I_S \lor \frac{b}{t} \in I_S$ and we have $I_S \in \Spec R_S$. + By + \ref{invimgprimeideal}, $J \in \Spec R_S \implies + f^{-1}(J) = J \cap R \in + \Spec R_S$. + Suppose $I \in \Spec R$, $\frac{a}{s} \cdot \frac{b}{t} + \in I_S$ for some $a,b + \in R, s,t \in S$. + By + \ref{ssatiis} $ab \in I$. + Thus $a \in I \lor b \in I$, hence $\frac{a}{s} \in I_S \lor + \frac{b}{t} \in + I_S$ and we have $I_S \in \Spec R_S$. @@ -736,172 +1352,359 @@ Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-satura % Some more remarks on localization -\begin{remark}\label{locandquot} - Let $R$ be a domain. If $S = R \setminus \{0\}$, then $R_S$ is the field of quotients $Q(R)$. +\begin{remark} + \label{locandquot} + Let $R$ be a domain. + If $S = R \setminus \{0\}$, then $R_S$ is the field of quotients $Q(R)$. If $S \subseteq R \setminus \{0\} $, then - \[ - R_S \cong \left\{ \frac{a}{s} \in K | a \in R, s \in S\right\} - \] + \[ + R_S \cong \left\{ \frac{a}{s} \in + K | a \in R, s \in S\right\} + \] In particular $Q(R) \cong Q(R_S)$. \end{remark} -\begin{definition}[$S$-saturation]\label{ssaturation} - Let $R$ be any ring, $I \subseteq R$ an ideal. Even if $I$ is not $S$-saturated, $J = I_S \coloneqq \{\frac{i}{s} | i \in I, s \in S\}$ is an ideal in $R_S$, and $I_S \sqcap R = \{r \in R | s\cdot r \in I, s \in S\}$ is called the \vocab[Ideal!$S$-saturation]{$S$-saturation of $I$ } which is the smallest $S$-saturated ideal containing $I$. +\begin{definition}[$S$-saturation] + \label{ssaturation} + Let $R$ be any ring, $I \subseteq R$ an ideal. + Even if $I$ is not $S$-saturated, $J = I_S \coloneqq \{\frac{i}{s} + | i \in I, s + \in S\}$ is an ideal in $R_S$, and $I_S \sqcap R = \{r \in R | s\cdot r \in + I, s \in S\}$ is called the \vocab[Ideal! + $S$-saturation]{$S$-saturation of $I$ } which is the smallest + $S$-saturated ideal containing $I$. \end{definition} -\begin{lemma}\label{locandfactor} - In the situation of \ref{ssaturation}, if $\overline{S}$ denotes the image of $S$ in $R / I$, there is a canonical isomorphism $R_S / I_S \cong (R / I)_{\overline{S}}$. +\begin{lemma} + \label{locandfactor} + In the situation of + \ref{ssaturation}, if $\overline{S}$ + denotes the image of + $S$ in $R / I$, there is a canonical isomorphism $R_S / I_S + \cong (R / + I)_{\overline{S}}$. \end{lemma} \begin{proof} - We show that both rings have the universal property for ring homomorphisms $R \xrightarrow{\tau} T$ with $\tau(I) = \{0\} $ and $\tau(S) \subseteq T^{\times }$. - For such $\tau$, by the fundamental theorem on homomorphisms (Homomorphiesatz) there is a unique $R/I \xrightarrow{\tau_1} T$ such that $\tau = \tau_1 \pi_{R,I}$. - We have $\tau_1(\overline{S}) = \tau(S) \subseteq T^{\times }$, hence there is a unique $(R / I)_{\overline{S}} \xrightarrow{\tau_2} T$ such that the composition $R / I \to (R / I)_{\overline{S}} \xrightarrow{\tau_2} T $ equals $\tau_1$. It is easy to see that this is the only one for which $R \to R / I \to (R / I)_{\overline{S}} \xrightarrow{\tau_2} T$ equals $\tau$. + We show that both rings have the universal property for ring homomorphisms $R + \xrightarrow{\tau} T$ with $\tau(I) = \{0\} $ and + $\tau(S) \subseteq + T^{\times }$. + For such $\tau$, by the fundamental theorem on homomorphisms (Homomorphiesatz) + there is a unique $R/I \xrightarrow{\tau_1} T$ such that $\tau = \tau_1 + \pi_{R,I}$. + We have $\tau_1(\overline{S}) = \tau(S) \subseteq + T^{\times }$, hence there is + a unique $(R / I)_{\overline{S}} + \xrightarrow{\tau_2} T$ such that the + composition $R / I \to (R / I)_{\overline{S}} + \xrightarrow{\tau_2} T $ equals + $\tau_1$. + It is easy to see that this is the only one for which $R \to R / I + \to (R / + I)_{\overline{S}} \xrightarrow{\tau_2} T$ + equals $\tau$. - Similarly, by the universal property of $R_S$ there is a unique $R_S \xrightarrow{\tau_3} T$ whose composition with $R \to R_S$ equals $\tau$. - $\tau_3(I_{S}) = 0$, hence a unique $R_S / I_S \xrightarrow{\tau_4} T$ whose composition with $\pi_{R_S, I_S}$ equals $\tau_3$ exists. - This is the only one for which the composition $R \to R_S \to R_S / I_S \xrightarrow{\tau_4} T$ equals $\tau$. + Similarly, by the universal property of $R_S$ there is a unique $R_S + \xrightarrow{\tau_3} T$ whose composition with $R \to R_S$ equals $\tau$. + $\tau_3(I_{S}) = 0$, hence a unique $R_S / I_S + \xrightarrow{\tau_4} + T$ whose composition with $\pi_{R_S, I_S}$ equals $\tau_3$ exists. + This is the only one for which the composition $R \to R_S \to R_S / I_S + \xrightarrow{\tau_4} T$ equals $\tau$. \[ - \begin{tikzcd} - R \arrow{r}{\tau}\arrow[swap]{d}{\pi_{R,I}}& T & R\arrow[swap]{l}{\tau}\arrow{d}{}\\ - R / I \arrow[dotted]{ru}{\existsone \tau_1}\arrow{d}{} & & R_S \arrow[dotted, swap]{lu}{\existsone \tau_3}\arrow{d}{\pi_{R_S, I_S}}\\ - (R / I)_{\overline{S}} \arrow[dotted,bend right]{ruu}{\existsone \tau_2} & & R_S / I_S \arrow[dotted, bend left, swap]{luu}{\existsone \tau_4}\\ - \end{tikzcd} - \] + \begin{tikzcd} + R \arrow{r}{\tau}\arrow[swap]{d}{\pi_{R,I}}& T & + R\arrow[swap]{l}{\tau}\arrow{d}{}\\ R / I \arrow[dotted]{ru}{\existsone + \tau_1}\arrow{d}{} & & R_S \arrow[dotted, swap]{lu}{\existsone + \tau_3}\arrow{d}{\pi_{R_S, I_S}}\\ (R / I)_{\overline{S}} \arrow[dotted,bend + right]{ruu}{\existsone \tau_2} & & R_S / I_S \arrow[dotted, bend left, + swap]{luu}{\existsone \tau_4}\\ + \end{tikzcd} + \] \end{proof} -\subsection{A first result of dimension theory} +\subsection{A first result of dimension theory} \begin{notation} - Let $R$ be a ring, $\fp \in \Spec R$. Let $\mathfrak{k}(\fp)$ denote the field of quotients of the domain $R / \fp$. This is called the \vocab{residue field} of $\fp$. + Let $R$ be a ring, $\fp \in \Spec R$. + Let $\mathfrak{k}(\fp)$ denote the field of quotients of the domain $R / + \fp$. + This is called the \vocab{residue field} of $\fp$. \end{notation} % i = ic -\begin{proposition}\label{trdegresfield} - Let $\mathfrak{l}$ be a %% ?? -field, $A$ a $\mathfrak{l}$-algebra of finite type and $\fp, \fq \in \Spec A$ with $\fp \subsetneq \fq$. -Then \[ - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) -\] +\begin{proposition} + \label{trdegresfield} + Let $\mathfrak{l}$ be a field, $A$ a + $\mathfrak{l}$-algebra of finite type and + $\fp, \fq \in \Spec A$ with $\fp \subsetneq \fq$. + %% ?? + Then + \[ + \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) / + \mathfrak{l}) + \] \end{proposition} \begin{proof} - Replacing $A$ by $A / \fp$, we may assume $\fp = \{0\} $ and $A$ to be a domain. Then $\mathfrak{k}(\fp) = Q(A / \fp) = Q(A)$. + Replacing $A$ by $A / \fp$, we + may assume $\fp = \{0\} $ and $A$ to be a domain. + Then $\mathfrak{k}(\fp) = Q(A / \fp) = Q(A)$. - If $\fq$ is a maximal ideal, $\mathfrak{k}(\fq) = A / \fq$ is of finite type over $\mathfrak{l}$, hence a finite field extension of $\mathfrak{l}$ by the Nullstellensatz (\ref{hns2}). + If $\fq$ is a maximal ideal, $\mathfrak{k}(\fq) = A / \fq$ is of finite + type + over $\mathfrak{l}$, hence a finite field extension of + $\mathfrak{l}$ by the + Nullstellensatz ( + \ref{hns2}). Thus, $\trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$. - If $\trdeg(Q(A) / \mathfrak{l}) = 0$, $A$ would be integral over $\mathfrak{l}$, hence a field (fact about integrality and fields, \ref{fintaf}). But if $A$ is a field, $\fp = \{0\}$ is a maximal ideal of $A$, hence $\fq = \fp \lightning$. + If $\trdeg(Q(A) / \mathfrak{l}) = 0$, $A$ would be integral over + $\mathfrak{l}$, hence a field (fact about integrality and fields, + \ref{fintaf}). + But if $A$ is a field, $\fp = \{0\}$ is a maximal ideal of $A$, hence $\fq = + \fp \lightning$. This finishes the proof for $\fq \in \MaxSpec A$. We will use the following lemma to reduce the general case to this case: - \begin{lemma}\label{ltrdegresfieldtrbase} - There are algebraically independent $a_1,\ldots,a_n \in A$ whose images in $A / \fq$ form a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{l}$. + \begin{lemma} + \label{ltrdegresfieldtrbase} There are algebraically independent + $a_1,\ldots,a_n \in A$ whose images in $A / \fq$ form a transcendence base for + $\mathfrak{k}(\fq) / \mathfrak{l}$. \end{lemma} \begin{subproof} - There exist $a_1,\ldots,a_n \in A$ such that $\mathfrak{k}(\fq)$ is algebraic over the subfield generated by $\mathfrak{l}$ and their images $\overline{a_i}$ (for instance generators of $A$ as a $\mathfrak{l}$-algebra). - We may assume that $n$ is minimal. If the $a_i$ are $\mathfrak{l}$-algebraically dependent, then w.l.o.g. $\overline{a_n}$ can be assumed to be algebraic over the subfield generated by $\mathfrak{l}$ and the $\overline{a_i}, 1\le i 0$, then $L^{\Aut(L / K)} = \{l \in L | \exists n \in \N ~ l^{p^n} \in K\}$. +\begin{proposition} + \label{characfixnormalfe} Let $L / K$ be a normal field + extension. + If the characteristic of the fields is $O$, then + $L^{\Aut( L / K)} = K$. + If the characteristic is $p > 0$, then $L^{\Aut(L / K)} = \{l + \in L | \exists n + \in \N ~ l^{p^n} \in K\}$. \end{proposition} \begin{proof} In both cases $L^G \supseteq$ is easy to see. - If $K \subseteq M \subseteq L$ is an intermediate field, then $L$ is normal over $M$. If $\sigma \in \Aut(M /K)$, an application of Zorn's lemma to the set of all $(N, \vartheta)$ where $N$ is an intermediate field $M \subseteq N \subseteq L$ and $N \xrightarrow{\vartheta} L$ a ring homomorphism such that $\vartheta\defon{M} = \sigma$ shows that $\sigma$ has an extension to an element of $\Aut(L / K)$. % TODO make this rigorous - If $M$ is normal over $K$, it is easily seen to be $\Aut(L / K)$ invariant. - Thus $L^G$ is the union of $M^{\Aut(M / K)}$ over all intermediate fields which are finite and normal over $K$, and it is sufficient to show the proposition for finite normal extensions $L / K$. + If $K \subseteq M \subseteq L$ is an intermediate field, then $L$ is normal + over $M$. + If $\sigma \in \Aut(M /K)$, an application of Zorn's lemma to + the set of all + $(N, \vartheta)$ where $N$ is an intermediate field $M \subseteq N \subseteq L$ + and $N \xrightarrow{\vartheta} L$ a ring homomorphism such that + $\vartheta\defon{M} = \sigma$ shows that $\sigma$ has an + extension to an + element of $\Aut(L / K)$. + % TODO make this rigorous + If $M$ is normal over $K$, it is easily seen to be $\Aut(L / K)$ + invariant. + Thus $L^G$ is the union of $M^{\Aut(M / K)}$ over all + intermediate fields which + are finite and normal over $K$, and it is sufficient to show the proposition + for finite normal extensions $L / K$. \begin{itemize} - \item Characteristic $0$: The extension is normal, hence Galois, and the assertion follows from Galois theory. - \item Characteristic $p > 0$: Let $l \in L^G$ and $P \in K[T]$ be the minimal polynomial of $l$ over $K$. - We show that $l^{p^n} \in K$ for some $n \in \N$ by induction on $\deg(l / K) \coloneqq \deg(P)$. + \item + Characteristic $0$: The extension is normal, hence Galois, and the assertion + follows from Galois theory. + \item + Characteristic $p > 0$: Let $l \in L^G$ and $P \in + K[T]$ be the minimal polynomial of $l$ over $K$. + We show that $l^{p^n} \in K$ for some $n \in \N$ by + induction on $\deg(l / K) + \coloneqq \deg(P)$. - If $\deg(l / K) = 1$, we have $l \in K$. - Otherwise, assume that the assertion has been shown for elements of $L^G$ whose degree over $K$ is smaller than $\deg( l / K)$. - Let $\overline{L}$ be an algebraic closure of $L$ and $\lambda$ a zero of $P$ in $\overline{L}$. - If $M = K(l) \subseteq L$, then there is a ring homomorphism $M - \overline{L}$ sending $l$ to $\lambda$. This can be extended to a ring homomorphism $L \xrightarrow{\sigma} \overline{L}$. We have $\sigma \in G$ because $L / K$ is normal. Hence $\lambda = \sigma(l) = l$, as $l \in L^G$. Thus $l$ is the only zero of $P$ in $\overline{L}$ and because $\deg P >1$ it is a multiple zero. - It is shown in the Galois theory lecture % TODO: link to EinfAlg - that this is possible only when $P(T) = Q(T^p)$ for some $Q \in K[T]$. Then $Q(l^p) = 0$ and the induction assumption can be applied to $x = l^p$ showing $x^{p^m} \in K$ hence $l^{p^{m+1}} \in K$ for some $m \in \N$. - \end{itemize} + If $\deg(l / K) = 1$, we have $l \in K$. + Otherwise, assume that the assertion has been shown for elements of $L^G$ whose + degree over $K$ is smaller than $\deg( l / K)$. + Let $\overline{L}$ be an algebraic closure of $L$ and $\lambda$ a + zero of $P$ + in $\overline{L}$. + If $M = K(l) \subseteq L$, then there is a ring homomorphism $M - + \overline{L}$ + sending $l$ to $\lambda$. + This can be extended to a ring homomorphism $L \xrightarrow{\sigma} + \overline{L}$. + We have $\sigma \in G$ because $L / K$ is normal. + Hence $\lambda = \sigma(l) = l$, as $l \in L^G$. + Thus $l$ is the only zero of $P$ in $\overline{L}$ and because $\deg + P >1$ it + is a multiple zero. + It is shown in the Galois theory lecture that this is possible only when $P(T) + = Q(T^p)$ for some $Q \in K[T]$. + % TODO: link to EinfAlg + Then $Q(l^p) = 0$ and the induction assumption can be applied to $x = l^p$ + showing $x^{p^m} \in K$ hence + $l^{p^{m+1}} \in K$ for some $m \in \N$. + \end{itemize} \end{proof} \subsubsection{Integral closure and normal domains} \begin{definition}[Integral closure, normal domains] - Let $A$ be a domain with field of quotients $Q(A)$ and let $L$ be a field extension of $Q(A)$. - By \ref{intclosure} the set of elements of $L$ integral over $A$ is a subring of $L$, the \vocab{integral closure} of $A$ in $L$. - $A$ is \vocab{Domain!integrally closed} in $L$ if the integral closure of $A$ in $L$ equals $A$. + Let $A$ be a domain with field of quotients $Q(A)$ and + let $L$ be a field extension of $Q(A)$. + By + \ref{intclosure} the set of elements of $L$ integral over $A$ is a + subring + of $L$, the \vocab{integral closure} of $A$ in $L$. + $A$ is \vocab{Domain!integrally closed} in $L$ if the integral closure of $A$ in $L$ + equals $A$. $A$ is \vocab{Domain!normal} if it is integrally closed in $Q(A)$. \end{definition} -\begin{proposition}\label{ufdnormal} +\begin{proposition} + \label{ufdnormal} Any factorial domain (UFD) is normal. \end{proposition} \begin{proof} - Let $x \in Q(A)$ be integral over $A$. Then there is a normed polynomial $P \in A[T]$ with $P(x) = 0$. - In EInführung in die Algebra it was shown that $A[T]$ is a UFD and that the prime elements of $A[T]$ are the elements which are irreducible in $Q(A)[T]$ and for which the $\gcd$ of the coefficients is $\sim 1$. % TODO reference - The prime factors of a normed polynomial are all normed up to multiplicative equivalence. We may thus assume $P$ to be irreducible in $Q(A)[T]$. - But then $\deg P = 1$ as $x$ is a zero of $P$ in $Q(A)$, hence $P(T) = T - x$ and $x \in A$ as $P \in A[T]$. - + Let $x \in Q(A)$ be integral over $A$. + Then there is a normed polynomial $P \in A[T]$ with + $P(x) = 0$. + In EInführung in die Algebra it was shown that $A[T]$ + is a UFD and that the + prime elements of $A[T]$ are the elements which are + irreducible in $Q(A)[T]$ + and for which the $\gcd$ of the coefficients is $\sim 1$. + % TODO reference + The prime factors of a normed polynomial are all normed up to multiplicative + equivalence. + We may thus assume $P$ to be irreducible in + $Q(A)[T]$. + But then $\deg P = 1$ as $x$ is a zero of $P$ in $Q(A)$, hence $P(T) = T - x$ + and $x \in A$ as $P \in A[T]$. - Alternative proof\footnote{\url{http://www.math.lsa.umich.edu/~tfylam/Math221/2.pdf}}: - Let $x = \frac{a}{b} \in Q(A)$ be integral over $A$. Without loss of generality loss of generality $\gcd(a,b) = 1$. Then $x^n + c_{n-1} x^{n-1} + \ldots + c_0 = 0$ for some $c_i \in A$. - Multiplication with $b^n$ yields $a^n + c_{n-1} b a^{n-1} + \ldots +c_0 b^n = 0$. Thus $b | a^n$. Since $\gcd(a,b) = 1$ it follows that $b$ is a unit, hence $x \in A$. + + Alternative + proof\footnote{\url{http://www.math.lsa.umich.edu/~tfylam/Math221/2.pdf}}: Let + $x = \frac{a}{b} \in Q(A)$ be integral over $A$. + Without loss of generality loss of generality $\gcd(a,b) = 1$. + Then $x^n + c_{n-1} x^{n-1} + + \ldots + c_0 = 0$ for some $c_i \in A$. + Multiplication with $b^n$ yields $a^n + c_{n-1} b + a^{n-1} + \ldots +c_0 b^n = + 0$. + Thus $b | a^n$. + Since $\gcd(a,b) = 1$ it follows that $b$ is a unit, hence $x \in + A$. \end{proof} \begin{remark} - It follows from \ref{cintclosure} and \ref{locandquot} that the integral closure of $A$ in some field extension $L$ of $Q(A)$ is always normal. + It follows from + \ref{cintclosure} and + \ref{locandquot} that + the integral + closure of $A$ in some field extension $L$ of $Q(A)$ is always normal. \end{remark} \begin{remark} - A finite field extension of $\Q$ is called an \vocab{algebraic number field} (ANF). If $K$ is an ANF, let $\mathcal{O}_K$ (the \vocab[Ring of integers in an ANF]{ring of integers in $K$}) be the integral closure of $\Z$ in $K$. - One can show that this is a finitely generated (hence free, by results of EInführung in die Algebra % EINFALG - ) abelian group. + A finite field extension of $\Q$ is called an \vocab{algebraic number field} + (ANF). + If $K$ is an ANF, let $\mathcal{O}_K$ (the + \vocab[Ring of integers in an + ANF]{ring of integers in $K$}) be the integral closure of $\Z$ in $K$. + One can show that this is a finitely generated (hence free, by results of + EInführung in die Algebra ) abelian group. + % EINFALG We have $\mathcal{O}_{\Q} = \Z$ by the proposiiton. \end{remark} \subsubsection{Action of \texorpdfstring{$\Aut(L / K)$}{Aut(L / K)} on prime ideals of a normal ring extension} -\begin{theorem}\label{autonprime} - Let $A$ be a normal domain, $L$ a normal field extension of $K \coloneqq Q(A)$, $B$ the integral closure of $A$ in $L$ and $\fp \in \Spec A$. - Then $G \coloneqq \Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$. +\begin{theorem} + \label{autonprime} + Let $A$ be a normal domain, $L$ a normal field extension of $K \coloneqq Q(A)$, + $B$ the integral closure of $A$ in $L$ and $\fp \in \Spec A$. + Then $G \coloneqq \Aut(L / K)$ transitively acts on $\{\fq \in + \Spec B | \fq + \cap A = \fp\}$. \end{theorem} \begin{proof} Let $\fq, \fr$ be prime ideals of $B$ above the given $\fp \in \Spec A$. - We must show that there exists $\sigma \in G$ such that $\fq = \sigma(\fr)$. - This is equivalent to $\fq \subseteq \sigma(\fr)$, since the Krull going-up theorem (\ref{cohenseidenberg}) applies to the integral ring extension $B / A$, showing that there are no inclusions between different elements of $\Spec B$ lying above $\fp \in \Spec A$. + We must show that there exists $\sigma \in G$ such that $\fq = + \sigma(\fr)$. + This is equivalent to $\fq \subseteq \sigma(\fr)$, since the Krull + going-up + theorem ( + \ref{cohenseidenberg}) applies to the integral ring extension $B / + A$, + showing that there are no inclusions between different elements of $\Spec B$ + lying above $\fp \in \Spec A$. - If $L / K$ is finite and there is no such $\sigma$, then by prime avoidance (\ref{primeavoidance}) there is $ x \in \fq \setminus \bigcup_{\sigma \in G} \sigma(\fr)$. - As $\fr$ is a prime ideal, $y = \prod_{\sigma \in G} \sigma(x) \in \fq \setminus \fr$.\footnote{$\prod_{\sigma \in G} \sigma(x) = \prod_{\sigma \in G} \sigma^{-1}(x)$} - By the characterization of $L^G$ for normal field extensions (\ref{characfixnormalfe}), there is a positive integer $k$ with $y^k \in K$. + If $L / K$ is finite and there is no such $\sigma$, then by prime avoidance + ( + \ref{primeavoidance}) there is $ x \in \fq \setminus + \bigcup_{\sigma \in G} + \sigma(\fr)$. + As $\fr$ is a prime ideal, $y = \prod_{\sigma \in G} \sigma(x) + \in \fq + \setminus \fr$. + \footnote{$\prod_{\sigma \in G} \sigma(x) = \prod_{\sigma \in G} + \sigma^{-1}(x)$} + By the characterization of $L^G$ for normal field extensions + ( + \ref{characfixnormalfe}), there is a positive integer $k$ with $y^k \in K$. As $A$ is normal, we have $y^k \in K \cap B = A$. - Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp = \emptyset \lightning$. + Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp = + \emptyset \lightning$. - If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs $(M, \sigma)$ where $M$ is an intermediate field and $\sigma \in \Aut(M / K)$ such that $\sigma(\fr \cap M) = \fq \cap M$. + If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs $(M, + \sigma)$ where $M$ is an intermediate field and $\sigma \in + \Aut(M / K)$ such + that $\sigma(\fr \cap M) = \fq \cap M$. \end{proof} \begin{remark} - The theorem is very important for its own sake. For instance, if $K$ is an ANF which is a Galois extension of $\Q$ it shows that $\Gal(K / \Q)$ transitively acts on the set of prime ideals of $\mathcal{O}_K$ over a given prime number $p$. More generally, if $L / K$ is a Galois extension of ANF then $\Gal(L / K)$ transitively acts on the set of $\fq \in \Spec \mathcal{O}_L$ for which $\fq \cap K$ is a given $\fp \in \Spec \mathcal{O}_K$. + The theorem is very important for its own sake. + For instance, if $K$ is an ANF which is a Galois extension of $\Q$ it shows + that $\Gal(K / \Q)$ transitively acts on the set of prime ideals of + $\mathcal{O}_K$ over a given prime number $p$. + More generally, if $L / K$ is a Galois extension of ANF then + $\Gal(L / K)$ + transitively acts on the set of $\fq \in \Spec \mathcal{O}_L$ for + which $\fq + \cap K$ is a given $\fp \in \Spec \mathcal{O}_K$. \end{remark} \subsubsection{A going-down theorem} -\begin{theorem}[Going-down for integral extensions of normal domains (Krull)]\label{gdkrull} - Let $B$ be a domain which is integral over its subring $A$. If $A$ is a normal domain, then going-down holds for $B / A$. +\begin{theorem}[Going-down for integral extensions of normal domains (Krull)] + \label{gdkrull} + Let $B$ be a domain which is integral over its subring $A$. + If $A$ is a normal domain, then going-down holds for $B / A$. \end{theorem} \begin{proof} - It follows from the assumptions that the field of quotients $Q(B)$ is an algebraic field extension of $Q(A)$. - There is an algebraic extension $L$ of $Q(B)$ such that $L / Q(A)$ is normal (for instance an algebraic closure of $Q(B)$). - Let $C$ be the integral closure of $A$ in $L$. Then $B \subseteq C$ and $C / B$ is integral. - \[ - \begin{tikzcd} - Q(A) \arrow[hookrightarrow]{r}{} & Q(B) \arrow[hookrightarrow]{r}{} & L \coloneqq \overline{Q(B)} \\ - A \arrow[hookrightarrow]{u}{}\arrow[hookrightarrow]{r}{} & B \arrow[hookrightarrow]{r}{}\arrow[hookrightarrow]{u}{}& C \arrow[hookrightarrow]{u}{}\\ - \end{tikzcd} - \] + It follows from the assumptions that the field of quotients $Q(B)$ is an + algebraic field extension of $Q(A)$. + There is an algebraic extension $L$ of $Q(B)$ such that $L / Q(A)$ is normal + (for instance an algebraic closure of $Q(B)$). + Let $C$ be the integral closure of $A$ in $L$. + Then $B \subseteq C$ and $C / B$ is integral. + \[ + \begin{tikzcd} + Q(A) \arrow[hookrightarrow]{r}{} & Q(B) \arrow[hookrightarrow]{r}{} & L + \coloneqq \overline{Q(B)} \\ A + \arrow[hookrightarrow]{u}{}\arrow[hookrightarrow]{r}{} & B + \arrow[hookrightarrow]{r}{}\arrow[hookrightarrow]{u}{}& C + \arrow[hookrightarrow]{u}{}\\ + \end{tikzcd} + \] \begin{claim} Going-down holds for $C / A$. \end{claim} \begin{subproof} - Let $\fp \subseteq \tilde \fp$ be an inclusion of prime ideals of $A$ and $\tilde \fr \in \Spec C$ with $\tilde \fr \cap A = \tilde \fp$. - By going-up for integral ring extensions (\ref{cohenseidenberg}), $\Spec C \xrightarrow{\cdot \cap A} \Spec A$ is surjectiv. Thus there is $\fr' \in \Spec C$ such that $\fr' \cap A = \fp$. By going up for $C / A$ there is $\tilde \fr' \in \Spec C$ with $\tilde \fr' \cap A = \tilde \fp, \fr' \subseteq \tilde \fr'$. - By the theorem about the action of the automorphism group on prime ideals of a normal ring extension (\ref{autonprime}) there exists a $\sigma \in \Aut(L / Q(A))$ with $\sigma(\tilde \fr') = \tilde \fr$. Then $\fr \coloneqq \sigma(\fr')$ satisfies $\fr \subseteq \tilde \fr$ and $\fr \cap A = \fp$. + Let $\fp \subseteq \tilde \fp$ be an inclusion of prime ideals of $A$ and + $\tilde \fr \in \Spec C$ with $\tilde \fr \cap A = \tilde \fp$. + By going-up for integral ring extensions ( + \ref{cohenseidenberg}), $\Spec C + \xrightarrow{\cdot \cap A} \Spec A$ is surjectiv. + Thus there is $\fr' \in \Spec C$ such that $\fr' \cap A = \fp$. + By going up for $C / A$ there is $\tilde \fr' \in \Spec C$ with $\tilde \fr' + \cap A = \tilde \fp, \fr' \subseteq \tilde \fr'$. + By the theorem about the action of the automorphism group on prime ideals of a + normal ring extension ( + \ref{autonprime}) there exists a $\sigma \in + \Aut(L / + Q(A))$ with $\sigma(\tilde \fr') = \tilde \fr$. + Then $\fr \coloneqq \sigma(\fr')$ satisfies $\fr \subseteq \tilde + \fr$ and $\fr + \cap A = \fp$. \end{subproof} - If $\fp \subseteq \tilde \fp$ is an inclusion of elements of $\Spec A$ and $\tilde \fq \in \Spec B$ with $\tilde \fp \cap A = \tilde \fp$, by the surjectivity of $\Spec C \xrightarrow{\cdot \cap B} \Spec B$ (\ref{cohenseidenberg}) there is $\tilde \fr \in \Spec C$ with $\tilde \fr \cap B = \fq$. - By going-down for $C / A$, there is $\fr \in \Spec C$ with $\fr \subseteq \tilde \fr$ and $\fr \cap A = \fp$. - Then $\fq \coloneqq \fr \cap B \in \Spec B, \fq \subseteq \tilde \fq$ and $\fq \cap A = \fp$. Thus going-down holds for $B / A$. + If $\fp \subseteq \tilde \fp$ is an inclusion of elements of $\Spec A$ and + $\tilde \fq \in \Spec B$ with $\tilde \fp \cap A = \tilde \fp$, by the + surjectivity of $\Spec C \xrightarrow{\cdot \cap B} \Spec B$ + ( + \ref{cohenseidenberg}) there is $\tilde \fr \in \Spec C$ with $\tilde \fr + \cap + B = \fq$. + By going-down for $C / A$, there is $\fr \in \Spec C$ with $\fr \subseteq + \tilde \fr$ and $\fr \cap A = \fp$. + Then $\fq \coloneqq \fr \cap B \in \Spec B, \fq \subseteq \tilde \fq$ and $\fq + \cap A = \fp$. + Thus going-down holds for $B / A$. \end{proof} \begin{remark}[Universally Japanese rings] - A Noetherian ring $A$ is called universally Japanese if for every $\fp \in \Spec A$ and every finite field extension $L$ of $\mathfrak{k}(\fp)$, the integral closure of $A / \fp$ in $L$ is a finitely generated $A$-module. This notion was coined by Grothendieck because the condition was extensively studied by the Japanese mathematician Nataga Masayoshji. - By a hard result of Nagata, algebras of finite type over a universally Japanese ring are universally Japanese. + A Noetherian ring $A$ is called universally Japanese + if for every $\fp \in \Spec A$ and every finite field extension $L$ of + $\mathfrak{k}(\fp)$, the integral closure of $A / \fp$ in $L$ is a + finitely generated $A$-module. + This notion was coined by Grothendieck because the condition was extensively + studied by the Japanese mathematician Nataga Masayoshji. + By a hard result of Nagata, algebras of finite type over a universally Japanese + ring are universally Japanese. Every field is universally Japanese, as is every PID of characteristic $0$. - There are, however, examples of Noetherian rings which fail to be universally Japanese. + There are, however, examples of Noetherian rings which fail to be universally + Japanese. \end{remark} -\begin{example}+[Counterexample to going down] - Let $R = \mathfrak{k}[X,Y]$ and $A = \mathfrak{k}[X,Y, \frac{X}{Y}]$. Then going down does not hold for $A / R$: +\begin{example} + +[Counterexample to going down] + Let $R = \mathfrak{k}[X,Y]$ and $A = \mathfrak{k}[X,Y, \frac{X}{Y}]$. + Then going down does not hold for $A / R$: - For any ideal $Y \in \fq \subseteq A$ we have $X = \frac{X}{Y} \cdot Y \in \fq$. - Consider $(Y)_R \subsetneq (X,Y)_R \subseteq \fq \cap R$. As $(X,Y)_R$ is maximal and the preimage of a prime ideal is prime and thus proper, we have $(X,Y)_R = \fq \cap R$. - The prime ideal $(\frac{X}{Y},Y)_A = (\frac{X}{Y}, X,Y)_A$ is lying over $(X,Y)_R$, so going down is violated. + For any ideal $Y \in \fq \subseteq A$ we have $X = \frac{X}{Y} + \cdot Y \in + \fq$. + Consider $(Y)_R \subsetneq (X,Y)_R \subseteq \fq \cap R$. + As $(X,Y)_R$ is maximal and the preimage of a prime ideal is prime and thus + proper, we have $(X,Y)_R = \fq \cap R$. + The prime ideal $(\frac{X}{Y},Y)_A = (\frac{X}{Y}, + X,Y)_A$ is lying over + $(X,Y)_R$, so going down is violated. \end{example} -\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}} +\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\mathfrak{K}(Y) / + \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}} \label{proofcodimletrdeg} -This is part of the proof of \ref{trdegandkdim}. %TODO: reorder +This is part of the proof of +\ref{trdegandkdim}. +%TODO: reorder \begin{proof} % DIMT - Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and $X \subseteq Y = V(\fp) \subseteq \mathfrak{k}^n$ irreducible closed subsets of $\mathfrak{k}^n$. - We want to show that $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$. - $\le $ was shown in \ref{upperboundcodim}. - $\dim Y \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$ was shown in \ref{lowerbounddimy} by + Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and $X \subseteq Y = V(\fp) \subseteq + \mathfrak{k}^n$ irreducible closed subsets of + $\mathfrak{k}^n$. + We want to show that $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - + \trdeg(\mathfrak{K}(X) / \mathfrak{k})$. + $\le $ was shown in + \ref{upperboundcodim}. + $\dim Y \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$ was shown in + \ref{lowerbounddimy} by - Applying Noether normalization to $A \coloneqq B / \fp$, giving us $(f_i)_{i=1}^d \in A^d$ such that the $f_i$ are algebraically independent and $A$ finite over the subalgebra generated by them. - We then used going-up to lift a chain of prime ideals corresponding to $\mathfrak{k}^d \supsetneq \{0\} \times \mathfrak{k}^{n-1} \supsetneq \ldots \supsetneq \{0\}$ under $Y \xrightarrow{F = (f_1,\ldots,f_d)} \mathfrak{k}^d$ to a chain of prime ideals in $A$. - This was done left-to-right as going-up was used to make prime ideals larger. In particular, when $\{0\} \in \mathfrak{k}^d$ has several preimages under $F$ we cannot control to which of them the maximal ideal terminating the lifted chain belongs. Thus, we can show that in the inequality + Applying Noether normalization to $A \coloneqq B / \fp$, giving us + $(f_i)_{i=1}^d \in A^d$ such that the $f_i$ + are algebraically independent and + $A$ finite over the subalgebra generated by them. + We then used going-up to lift a chain of prime ideals corresponding to + $\mathfrak{k}^d \supsetneq \{0\} \times \mathfrak{k}^{n-1} + \supsetneq \ldots + \supsetneq \{0\}$ under $Y \xrightarrow{F = (f_1,\ldots,f_d)} \mathfrak{k}^d$ + to a chain of prime ideals in $A$. + This was done left-to-right as going-up was used to make prime ideals larger. + In particular, when $\{0\} \in \mathfrak{k}^d$ has several preimages + under $F$ + we cannot control to which of them the maximal ideal terminating the lifted + chain belongs. + Thus, we can show that in the inequality \[ - \codim(\{y\}, Y) \le d = \trdeg(\mathfrak{K}(Y) \setminus \mathfrak{k}) - \] - (see \ref{upperboundcodim}) - equality holds for at least one pint $y \in F^{-1}(\{0\})$ but cannot rule out that there are other $y \in F^{-1}(\{0\})$ for which the inequality becomes strict. - However using going-down (\ref{gdkrull}) for $F$, we can use a similar argument, but start lifting of the chain at the right end for the point $y \in Y$ for which we would like to show equality. - From this $\codim(X,Y) \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$ can be derived similarly to \ref{upperboundcodim}. - Thus - \[ - \codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k}) - \] - follows (see \ref{htandcodim} and \ref{htandtrdeg}). + \codim(\{y\}, Y) \le d = + \trdeg(\mathfrak{K}(Y) \setminus \mathfrak{k}) + \] + (see + \ref{upperboundcodim}) + equality holds for at least one pint $y \in + F^{-1}(\{0\})$ but cannot rule out + that there are other $y \in F^{-1}(\{0\})$ for which + the inequality becomes + strict. + However using going-down ( + \ref{gdkrull}) for $F$, we can use a + similar + argument, but start lifting of the chain at the right end for the point $y \in + Y$ for which we would like to show equality. + From this $\codim(X,Y) \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - + \trdeg(\mathfrak{K}(X) / \mathfrak{k})$ can be derived similarly to + \ref{upperboundcodim}. + Thus + \[ + \codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - + \trdeg(\mathfrak{K}(X) / \mathfrak{k}) + \] + follows (see + \ref{htandcodim} and + \ref{htandtrdeg}). \end{proof} \begin{remark} - The going-down theorem used to prove this is somewhat more general, as it does not depend on $\mathfrak{k}$ being algebraically closed. + The going-down theorem used to prove this is somewhat more general, as it does + not depend on $\mathfrak{k}$ being algebraically closed. \end{remark} @@ -1256,123 +2403,300 @@ This is part of the proof of \ref{trdegandkdim}. %TODO: reorder % i = ic \subsection{The height of a prime ideal} -In order to complete the proof of \ref{proofcodimletrdeg} and show $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$, -we need to localize the $\mathfrak{k}$-algebra with respect to a multiplicative subset and replace the ground field by a larger subfield of that localization which is no longer algebraically closed. -To formulate a result which still applies in this context, we need the following: -\begin{definition}[Height of a prime ideal] - Let $A$ be a ring, $\fp \in \Spec A$. We define the \vocab[Height of a prime ideal]{height of the prime ideal $\fp$}, $\hght(\fp)$, to be the largest $k \in \N$ such that there is a strictly decreasing sequence $\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_k$ of prime ideals of $A$, or $\infty$ if there is no finite upper bound on the length of such sequences. +In order to complete the proof of +\ref{proofcodimletrdeg} and show +$\codim(X,Y) += \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / +\mathfrak{k})$, we need to localize +the $\mathfrak{k}$-algebra with respect to +a multiplicative subset and replace the ground field by a larger subfield of +that localization which is no longer algebraically closed. +To formulate a result which still applies in this context, we need the +following: +\begin{definition}[Height of a prime ideal] Let $A$ be a ring, $\fp + \in \Spec A$. + We define the \vocab[Height of a prime ideal]{height of the prime ideal $\fp$}, + $\hght(\fp)$, to be the largest $k \in \N$ such that there is a + strictly + decreasing sequence $\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq + \fp_k$ of prime ideals of $A$, or $\infty$ if there is no finite upper bound on + the length of such sequences. \end{definition} \begin{example} Let $A = \mathfrak{k}[X_1,\ldots,X_n]$, $X = V(\fp)$ for a prime ideal $\fp$. - By the correspondence between irreducible subsets of $\mathfrak{k}^n$ and prime ideals in $A$ (\ref{bijiredprim}), - the $\fp_i$ correspond to irreducible subsets $X_i \subseteq \mathfrak{k}^n$ containing $X$. Thus $\hght(\fp) = \codim(X, \mathfrak{k}^n)$. + By the correspondence between irreducible subsets of + $\mathfrak{k}^n$ and prime + ideals in $A$ ( + \ref{bijiredprim}), the $\fp_i$ correspond to irreducible + subsets $X_i \subseteq \mathfrak{k}^n$ containing $X$. + Thus $\hght(\fp) = \codim(X, \mathfrak{k}^n)$. \end{example} -\begin{example}\label{htandcodim} - Let $B = \mathfrak{k}[X_1,\ldots,X_n], \fq \in \Spec B$ and let $A \coloneqq B / \fp$. - Let $Y \coloneqq V(\fq) \subseteq \mathfrak{k}^n$, $\tilde \fp \coloneqq \pi_{B, \fq}^{-1}(\fp)$, where $B \xrightarrow{\pi_{B, \fp}} A $ is the projection to the ring of residue classes, and let $X = V(\tilde \fp)$. - By \ref{idealslocbij} we have a bijection between the prime ideals $\fr \subseteq \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime ideals $\tilde \fr \in \Spec B$ with $\fq \subseteq \tilde \fr \subseteq \tilde \fp$: +\begin{example} + \label{htandcodim} + Let $B = \mathfrak{k}[X_1,\ldots,X_n], \fq \in \Spec B$ and let $A \coloneqq B + / \fp$. + Let $Y \coloneqq V(\fq) \subseteq \mathfrak{k}^n$, $\tilde \fp + \coloneqq + \pi_{B, \fq}^{-1}(\fp)$, where $B \xrightarrow{\pi_{B, \fp}} A $ is the + projection to the ring of residue classes, and let $X = V(\tilde \fp)$. + By + \ref{idealslocbij} we have a bijection between the prime ideals $\fr + \subseteq \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime + ideals $\tilde \fr \in \Spec B$ with $\fq \subseteq \tilde \fr \subseteq \tilde + \fp$: \begin{align} - f: \{\fr \in \Spec A | \fr \subseteq \fp \} &\longrightarrow \{\tilde \fr \in \Spec B | \fq \subseteq \tilde \fr \subseteq \tilde \fp\} \\ - \fr &\longmapsto \pi_{B, \fq}^{-1}(\fr)\\ - \tilde \fr / \fq &\longmapsfrom \tilde \fr + f: \{\fr \in \Spec A | \fr \subseteq \fp \} + & \longrightarrow \{\tilde \fr \in \Spec B | \fq \subseteq \tilde \fr \subseteq + \tilde \fp\} \\ \fr & \longmapsto \pi_{B, \fq}^{-1}(\fr)\\ \tilde \fr / \fq + & \longmapsfrom \tilde \fr \end{align} - By \ref{bijiredprim}, the $\tilde \fr$ are in canonical bijection with the irreducible subsets $Z$ of $Y$ containing $X$. - Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in canonical bijection with the chains $X = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \subseteq Y$ of irreducible subsets and - $\hght(\fp) = \codim(X,Y)$. + By + \ref{bijiredprim}, the $\tilde \fr$ + are in canonical bijection with the irreducible subsets $Z$ of $Y$ containing + $X$. + Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in + canonical bijection with the chains $X = X_0 \subsetneq X_1 \subsetneq \ldots + \subsetneq X_k \subseteq Y$ of irreducible subsets and + $\hght(\fp) = + \codim(X,Y)$. \end{example} \begin{remark} - Let $A$ be an arbitrary ring. One can show that there is a bijection between $\Spec A$ and the set of irreducible subsets $Y \subseteq \Spec A$: - \begin{align} - f: \Spec A &\longrightarrow \{Y \subseteq \Spec A | Y\text{irreducible}\} \\ - \fp &\longmapsto \Vs(\fp)\\ - \bigcup_{\fp \in Y} \fp &\longmapsfrom Y + Let $A$ be an arbitrary ring. + One can show that there is a bijection between $\Spec A$ and the set of + irreducible subsets $Y \subseteq \Spec A$: + \begin{align} + f: \Spec A + & \longrightarrow \{Y \subseteq \Spec A | Y\text{irreducible}\} \\ \fp + & \longmapsto \Vs(\fp) \\ \bigcup_{\fp \in Y} \fp & \longmapsfrom Y \end{align} - Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in canonical bijection with the chains $V(\fp) = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \subseteq \Spec A$ of irreducible subsets, and $\hght(\fp) = \codim(V(\fp), \Spec A)$. + Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in + canonical bijection with the chains $V(\fp) = X_0 \subsetneq X_1 \subsetneq + \ldots \subsetneq X_k \subseteq \Spec A$ of irreducible subsets, and + $\hght(\fp) = \codim(V(\fp), \Spec A)$. \end{remark} \subsubsection{The relation between \texorpdfstring{$\hght(\fp)$}{ht(p)} and \texorpdfstring{$\trdeg$}{trdeg}} We will use the following -\begin{lemma}\label{extendtotrbase} - Let $\mathfrak{l}$ be an arbitrary field, $A$ a $\mathfrak{l}$-algebra of finite type which is a domain, $K \coloneqq Q(A)$ the field of quotients and let $(a_i)_{i=1}^n$ be $\mathfrak{l}$-algebraically independent elements of $A$. Then there exist a natural number $m \ge n$ and a transcendence base $(a_i)_{i = 1}^m$ for $K / \mathfrak{l}$ with $a_i \in A$ for $1 \le i \le m$. +\begin{lemma} + \label{extendtotrbase} Let + $\mathfrak{l}$ be an arbitrary field, $A$ a + $\mathfrak{l}$-algebra of finite + type which is a domain, $K \coloneqq Q(A)$ the field of quotients and let + $(a_i)_{i=1}^n$ be + $\mathfrak{l}$-algebraically independent elements of $A$. + Then there exist a natural number $m \ge n$ and a transcendence base + $(a_i)_{i + = 1}^m$ for $K / + \mathfrak{l}$ with $a_i \in A$ for $1 \le i \le m$. \end{lemma} \begin{proof} - The proof is similar to the proof of \ref{ltrdegresfieldtrbase}. - There are a natural number $m \ge n$ and elements $(a_i)_{i = n+1}^m \in A^{m-n}$ which generate $K$ in the sense of a matroid used in the definition of $\trdeg$. - For instance, one can use generators of the $\mathfrak{l}$-algebra $A$. We assume $m$ to be minimal and claim that $(a_i)_{i=1}^m$ are $\mathfrak{l}$-algebraically independent. - Otherwise there is $j \in \N$, $1 \le j \le m$ such that $a_j$ is algebraic over the subfield of $K$ generated by $\mathfrak{l}$ and the $(a_i)_{i=1}^{j-1}$. We have $j > n$ by the algebraic independence of $(a_i)_{i=1}^n$. - Exchanging $x_j$ and $x_m$, we may assume $j = m$. But then $K$ is algebraic over its subfield generated by $\mathfrak{l}$ and the $(a_i)_{i=1}^{m-1} $, contradicting the minimality of $m$. + The proof is similar to the proof of + \ref{ltrdegresfieldtrbase}. + There are a natural number $m \ge n$ and elements + $(a_i)_{i = n+1}^m \in + A^{m-n}$ which generate $K$ in the sense of a matroid + used in the definition of + $\trdeg$. + For instance, one can use generators of the $\mathfrak{l}$-algebra + $A$. + We assume $m$ to be minimal and claim that + $(a_i)_{i=1}^m$ are + $\mathfrak{l}$-algebraically independent. + Otherwise there is $j \in \N$, $1 \le j \le m$ such that $a_j$ is algebraic + over the subfield of $K$ generated by $\mathfrak{l}$ and the + $(a_i)_{i=1}^{j-1}$. + We have $j > n$ by the algebraic independence of + $(a_i)_{i=1}^n$. + Exchanging $x_j$ and $x_m$, we may assume $j = m$. + But then $K$ is algebraic over its subfield generated by + $\mathfrak{l}$ and the + $(a_i)_{i=1}^{m-1} $, contradicting the minimality + of $m$. \end{proof} -\begin{theorem}\label{htandtrdeg} - Let $\mathfrak{l}$ be an arbitrary field, $A$ a $\mathfrak{l}$-algebra of finite type which is a domain, and $\fp \in \Spec A$. - Let $K \coloneqq Q(A)$ be the field of quotients of $A$. Then +\begin{theorem} + \label{htandtrdeg} + Let $\mathfrak{l}$ be an arbitrary field, $A$ a + $\mathfrak{l}$-algebra of + finite type which is a domain, and $\fp \in \Spec A$. + Let $K \coloneqq Q(A)$ be the field of quotients of $A$. + Then \[ - \hght(\fp) = \trdeg(K /\mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) - \] + \hght(\fp) = \trdeg(K /\mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / + \mathfrak{l}) + \] \end{theorem} \begin{remark} - By example \ref{htandcodim}, theorem \ref{trdegandkdim} is a special case of this theorem. %(\ref{htandtrdeg}). + By example + \ref{htandcodim}, + theorem + \ref{trdegandkdim} is a special case of this theorem. + %(\ref{htandtrdeg}). \end{remark} \begin{proof} - If $\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_k$ is a chain of prime ideals in $A$, we have $\trdeg(\mathfrak{k}(\fp_i) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp_{i+1}) / \mathfrak{l})$ by \ref{trdegresfield} (``A first result of dimension theory''). -Thus -\[ - k \le \trdeg(\mathfrak{k}(\fp_k) / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) \le \trdeg(K / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) -\] -where the last inequality is another application of \ref{trdegresfield} (using $K = Q(A) = Q(A / \{0\}) = \mathfrak{k}(\{0\})$ and the fact that $\{0\} \subseteq \fp_k$ is a prime ideal). -Hence \[ - \hght(\fp) \le \trdeg( K / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) -\] -and it remains to show the opposite inequality. + If $\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_k$ is a chain + of prime ideals in $A$, we have $\trdeg(\mathfrak{k}(\fp_i) / \mathfrak{l}) < + \trdeg(\mathfrak{k}(\fp_{i+1}) / \mathfrak{l})$ by + \ref{trdegresfield} (``A + first result of dimension theory''). + Thus + \[ + k \le \trdeg(\mathfrak{k}(\fp_k) / \mathfrak{l}) - + \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) \le \trdeg(K / \mathfrak{l}) - + \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) + \] + where the last inequality is + another application of \ref{trdegresfield} (using $K = Q(A) = Q(A / \{0\}) = + \mathfrak{k}(\{0\})$ and the fact that $\{0\} \subseteq \fp_k$ is a prime + ideal). + Hence + \[ + \hght(\fp) \le \trdeg( K / \mathfrak{l}) - + \trdeg(\mathfrak{k}(\fp) / + \mathfrak{l}) + \] + and it remains to show the opposite inequality. -\begin{claim} - For any maximal ideal $\fp \in \MaxSpec A$ \[ - \hght(\mathfrak{m}) \ge \trdeg(K / \mathfrak{l}) - \] -\end{claim} -\begin{subproof} - By the Noether normalization theorem (\ref{noenort}), there are $(x_i)_{i=1}^d \in A^d$ which are algebraically independent over $\mathfrak{l}$ such that $A$ is finite over the subalgebra $S$ generated by the $x_i$. We have $d = \trdeg(K / \mathfrak{l})$ as the $x_i$ form a transcendence base of $K / \mathfrak{l}$. -\begin{claim} - We can choose $x_i \in \mathfrak{m}$ -\end{claim} -\begin{subproof} - By the Nullstellensatz (\ref{hns2}), $\mathfrak{k}(\mathfrak{m}) = A / \mathfrak{m}$ is a finite field extension of $\mathfrak{l}$. Hence there exists a normed polynomial $P_i \in \mathfrak{l}[T]$ with $P_i(x_i \mod \mathfrak{m}) = 0$ in $\mathfrak{k}(\mathfrak{m})$. - Let $\tilde x_i \coloneqq P_i(x_i) \in \mathfrak{m}$ and $\tilde S$ the subalgebra generated by the $\tilde x_i$. As $P_i(x_i) - \tilde x_i = 0$, $x_i$ is integral over $\tilde S$ and so is $S / \tilde S$. It follows that $A / \tilde S$ is integral, hence finite by \ref{ftaiimplf}. Replacing $x_i$ by $\tilde x_i$, we may thus assume that $x_i \in \mathfrak{m}$. + \begin{claim} + For any maximal ideal $\fp \in \MaxSpec A$ + \[ + \hght(\mathfrak{m}) \ge \trdeg(K + / \mathfrak{l}) + \] + \end{claim} + \begin{subproof} + By the Noether normalization + theorem ( + \ref{noenort}), there are + $(x_i)_{i=1}^d \in A^d$ which are + algebraically independent over $\mathfrak{l}$ such that $A$ is + finite over the + subalgebra $S$ generated by the $x_i$. + We have $d = \trdeg(K / \mathfrak{l})$ as the $x_i$ form a transcendence base + of $K / \mathfrak{l}$. + \begin{claim} + We can choose $x_i \in \mathfrak{m}$ + \end{claim} + \begin{subproof} + By the + Nullstellensatz ( + \ref{hns2}), $\mathfrak{k}(\mathfrak{m}) = A / + \mathfrak{m}$ + is a finite field extension of $\mathfrak{l}$. + Hence there exists a normed polynomial $P_i \in \mathfrak{l}[T]$ with + $P_i(x_i \mod \mathfrak{m}) = 0$ in $\mathfrak{k}(\mathfrak{m})$. + Let $\tilde x_i \coloneqq P_i(x_i) \in \mathfrak{m}$ and $\tilde + S$ the + subalgebra generated by the $\tilde x_i$. + As $P_i(x_i) - \tilde x_i = 0$, $x_i$ is integral over $\tilde S$ and so is $S + / \tilde S$. + It follows that $A / \tilde S$ is integral, hence finite by + \ref{ftaiimplf}. + Replacing $x_i$ by $\tilde x_i$, we may thus assume that $x_i \in + \mathfrak{m}$. - -\end{subproof} -% TODO: fix names A_1 = A_S, k_1 = R_S - The ring homomorphism $\ev_x : R = \mathfrak{l}[X_1,\ldots,X_d] \xrightarrow{P \mapsto P(x_1,\ldots,x_d)} A$ is injective. Because $R$ is a UFD, $R$ is normal (\ref{ufdnormal}). Thus the going-down theorem (\ref{gdkrull}) applies to the integral $R$-algebra $A$. - For $0 \le i \le d$, let $\fp_i \subseteq R$ be the ideal generated by $(X_j)_{j=i+1}^d$. We have $\mathfrak{m} \sqcap R = \fp_0$ as all $X_i \in \mathfrak{m}$, hence $X_i \in \mathfrak{m} \sqcap R$ and $\fp_0$ is a maximal ideal. - By applying going-down and induction on $i$, there is a chain $\mathfrak{m} = \fq_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_d$ of elements of $\Spec A$ such that $\fq_i \sqcap R = \fp_i$. - It follows that $\hght(\mathfrak{m}) \ge d$. -\end{subproof} -This finishes the proof in the case of $\fp \in \MaxSpec A$. -To reduce the general case to that special case, we proceed as in \ref{trdegresfield}: -By lemma \ref{ltrdegresfieldtrbase} there are $a_1,\ldots,a_n \in A$ whose images in $A / \fp$ form a transcendence base for $\mathfrak{k}(\fp) / \mathfrak{l}$. -As these images are $\mathfrak{l}$-algebraically independent, the same holds for the $a_i$ themselves. + \end{subproof} + % TODO: fix names A_1 = A_S, k_1 = R_S + The ring homomorphism $\ev_x : R = \mathfrak{l}[X_1,\ldots,X_d] + \xrightarrow{P + \mapsto P(x_1,\ldots,x_d)} A$ is injective. + Because $R$ is a UFD, $R$ is normal ( + \ref{ufdnormal}). + Thus the going-down theorem ( + \ref{gdkrull}) applies to the integral + $R$-algebra + $A$. + For $0 \le i \le d$, let $\fp_i \subseteq R$ be the ideal generated by + $(X_j)_{j=i+1}^d$. + We have $\mathfrak{m} \sqcap R = \fp_0$ as all $X_i \in + \mathfrak{m}$, hence + $X_i \in \mathfrak{m} \sqcap R$ and $\fp_0$ is a maximal ideal. + By applying going-down and induction on $i$, there is a chain + $\mathfrak{m} = + \fq_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_d$ of elements of + $\Spec A$ such that $\fq_i \sqcap R = \fp_i$. + It follows that $\hght(\mathfrak{m}) \ge d$. + \end{subproof} + This finishes the proof in the case of $\fp \in \MaxSpec A$. -By lemma \ref{extendtotrbase} we can extend $(a_{i})_{i=1}^n$ to a transcendence base $(a_i)_{i=1}^m \in A^m$ of $K / \mathfrak{l}$. -Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and let $S \coloneqq R \setminus \{0\}$. -Let $A_1 \coloneqq A_S$ and $\fp_S$ the prime ideal corresponding to $\fp$ under $\Spec(A_1) \cong \{\fr \in \Spec A | \fr \cap S = \emptyset\}$ (\ref{idealslocbij}). -As in \ref{locandquot}, $A_1$ is a domain with $Q(A_1) \cong K = Q(A)$ and by \ref{locandfactor} $A_1 / \fp_S \cong (A / \fp)_{\overline{S}}$, where $\overline{S}$ denotes the image of $S$ in $A / \fp$. -As in \ref{trdegresfield}, $\mathfrak{k}(\fp_S) \cong \mathfrak{k}(\fp)$ is integral over $A_1 / \fp_S$. -From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 / \fp_S$ is a field. Hence $\fp_S \in \MaxSpec(A_1)$ and the special case can be applied to $\fp_S$ and $A_1 / \mathfrak{l}_1$, showing that $\hght(\fp_S) \ge e = \trdeg(K / \mathfrak{l}_1)$. We have $\trdeg(K / \mathfrak{l}_1) = m - n$, as $(a_i)_{i = n+1}^m$ is a transcendence base for $K / \mathfrak{l}_1$. By the description of $\Spec A_S$ (\ref{idealslocbij}), a chain $\fp_S = \fq_0 \supsetneq \ldots \supsetneq \fp_e$ of prime ideals in $A_S$ defines a similar chain $\fp_i \coloneqq \fq_i \sqcap A$ in $A$ with $\fp_0 = \fp$. Thus $\hght(\fp) \ge e$. + To reduce the general case to that special case, we proceed as in + \ref{trdegresfield}: By lemma + \ref{ltrdegresfieldtrbase} there are + $a_1,\ldots,a_n \in A$ whose images in $A / \fp$ form a transcendence base for + $\mathfrak{k}(\fp) / \mathfrak{l}$. + As these images are $\mathfrak{l}$-algebraically independent, the + same holds + for the $a_i$ themselves. + + By lemma + \ref{extendtotrbase} we can extend + $(a_{i})_{i=1}^n$ to + a + transcendence base $(a_i)_{i=1}^m \in A^m$ + of $K / \mathfrak{l}$. + Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated + by + $a_1,\ldots,a_n$ and let $S \coloneqq R \setminus \{0\}$. + Let $A_1 \coloneqq A_S$ and $\fp_S$ the prime ideal corresponding to $\fp$ + under $\Spec(A_1) \cong \{\fr \in \Spec A | \fr \cap S = + \emptyset\}$ + ( + \ref{idealslocbij}). + As in + \ref{locandquot}, $A_1$ is a domain with $Q(A_1) \cong K = Q(A)$ + and by + \ref{locandfactor} $A_1 / \fp_S + \cong (A / \fp)_{\overline{S}}$, where + $\overline{S}$ denotes the image of $S$ in $A / \fp$. + As in + \ref{trdegresfield}, $\mathfrak{k}(\fp_S) \cong + \mathfrak{k}(\fp)$ is + integral over $A_1 / \fp_S$. + From the fact about integrality and fields ( + \ref{fintaf}), it + follows that $A_1 + / \fp_S$ is a field. + Hence $\fp_S \in \MaxSpec(A_1)$ and the special case can be applied + to $\fp_S$ + and $A_1 / \mathfrak{l}_1$, showing that $\hght(\fp_S) + \ge e = \trdeg(K / + \mathfrak{l}_1)$. + We have $\trdeg(K / \mathfrak{l}_1) = m - n$, as + $(a_i)_{i = n+1}^m$ is a + transcendence base for $K / \mathfrak{l}_1$. + By the description of $\Spec A_S$ ( + \ref{idealslocbij}), a chain $\fp_S = + \fq_0 + \supsetneq \ldots \supsetneq \fp_e$ of prime ideals in $A_S$ defines a similar + chain $\fp_i \coloneqq \fq_i \sqcap A$ in $A$ with $\fp_0 = \fp$. + Thus $\hght(\fp) \ge e$. \end{proof} \begin{remark} - As a consequence of his principal ideal theorem, Krull has shown the finiteness of $\hght(\fp)$ for $\fp \in \Spec A$ when $A$ is a Noetherian ring. But $\dim A = \sup_{\fp \in \Spec A} \hght(\fp) = \sup_{\mathfrak{m} \in \MaxSpec A} \hght(\mathfrak{m})$, the Krull dimension of the Noetherian topological space $\Spec A$ may nevertheless be infinite. + As a consequence of his principal ideal theorem, Krull has shown the finiteness + of $\hght(\fp)$ for $\fp \in \Spec A$ when $A$ is a Noetherian + ring. + But $\dim A = \sup_{\fp \in \Spec A} \hght(\fp) = + \sup_{\mathfrak{m} \in + \MaxSpec A} \hght(\mathfrak{m})$, the Krull dimension of the + Noetherian + topological space $\Spec A$ may nevertheless be infinite. \end{remark} -\begin{example}+[Noetherian ring with infinite dimension]\footnote{\url{https://math.stackexchange.com/questions/1109732/noetherian-ring-with-infinite-krull-dimension-nagatas-example}} - Let $A = \mathfrak{k}[X_i | i \in \N]$ and $m_1, m_2, \ldots \in \N$ an increasing sequence such that $m_{i+1}-m_i > m_i - m_{i-1}$. - Let $\fp_i \coloneqq (X_{m_{i}+1},\ldots,X_{m_{i+1}})$ and $S \coloneqq A \setminus \bigcup_{i \in \N} \fp_i$. +\begin{example} + +[Noetherian ring with infinite dimension]\footnote{\url{https://math.stackexchange.com/questions/1109732/noetherian-ring-with-infinite-krull-dimension-nagatas-example}} + Let $A = \mathfrak{k}[X_i | i \in \N]$ and $m_1, m_2, \ldots \in \N$ an + increasing sequence such that $m_{i+1}-m_i > m_i - + m_{i-1}$. + Let $\fp_i \coloneqq (X_{m_{i}+1},\ldots,X_{m_{i+1}})$ and $S \coloneqq A + \setminus \bigcup_{i \in \N} \fp_i$. $S$ is multiplicatively closed. - $A_S$ is Noetherian but $\hght((\fp_i)_S) = m_{i+1}- m_{i}$ hence $\dim(A_S) = \infty$. + $A_S$ is Noetherian but $\hght((\fp_i)_S) = m_{i+1}- + m_{i}$ hence $\dim(A_S) = \infty$. \end{example} % Lecture 10 @@ -1381,25 +2705,59 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 \subsection{Dimension of products} -\begin{proposition}\label{dimprod} - Let $X \subseteq \mathfrak{k}^n$ and $Y \subseteq \mathfrak{k}^n$ be irreducible and closed. Then $X \times Y$ is also an irreducible closed subset of $\mathfrak{k}^{m+n}$. - Moreover, $\dim(X \times Y) = \dim(X) + \dim(Y)$ and $\codim(X \times Y, \mathfrak{k}^{m+n}) = \codim(X, \mathfrak{k}^m) + \codim(Y, \mathfrak{k}^n)$. +\begin{proposition} + \label{dimprod} + Let $X \subseteq \mathfrak{k}^n$ and $Y \subseteq + \mathfrak{k}^n$ be + irreducible and closed. + Then $X \times Y$ is also an irreducible closed subset of + $\mathfrak{k}^{m+n}$. + Moreover, $\dim(X \times Y) = \dim(X) + + \dim(Y)$ and $\codim(X \times Y, + \mathfrak{k}^{m+n}) = \codim(X, \mathfrak{k}^m) + + \codim(Y, \mathfrak{k}^n)$. \end{proposition} \begin{proof} - Let $X = V(\fp)$ and $Y = V(\fq)$ where $\fp \in \Spec \mathfrak{k}[X_1,\ldots,X_m]$ and $\fq \in \Spec \mathfrak{k}[X_1,\ldots,X_n]$. - We denote points of $\mathfrak{k}^{m+n}$ as $x = (x',x'')$ with $x' \in \mathfrak{k}^m, x''\in\mathfrak{k}^n$. Then $X \times Y$ is the set of zeroes of the ideal in $\mathfrak{k}[X_1,\ldots,X_{m+n}]$ generated by the polynomials $f(x) = \phi(x')$, with $\phi$ running over $\fp$ and $g(x) = \gamma(x'')$ with $\gamma$ running over $\fq$. + Let $X = V(\fp)$ and $Y = V(\fq)$ where $\fp \in \Spec + \mathfrak{k}[X_1,\ldots,X_m]$ and $\fq \in \Spec \mathfrak{k}[X_1,\ldots,X_n]$. + We denote points of $\mathfrak{k}^{m+n}$ as $x = (x',x'')$ with $x' \in + \mathfrak{k}^m, x''\in\mathfrak{k}^n$. + Then $X \times Y$ is the set of zeroes of the ideal in + $\mathfrak{k}[X_1,\ldots,X_{m+n}]$ generated by the polynomials $f(x) = + \phi(x')$, with $\phi$ running over $\fp$ and $g(x) = + \gamma(x'')$ with + $\gamma$ running over $\fq$. Thus $X \times Y$ is closed in $\mathfrak{k}^{m+n}$. - We must also show irreducibility. $X \times Y \neq \emptyset$ is obvious. + We must also show irreducibility. + $X \times Y \neq \emptyset$ is obvious. - Assume that $X \times Y = A_1 \cup A_2$, where the $A_i \subseteq \mathfrak{k}^{m+n}$ are closed. - For $x' \in \mathfrak{k}^m, x' \times Y$ is homeomorphic to the irreducible $Y$. Thus $X = X_1 \cup X_2$ where $X_i = \{x \in X | \{x\} \times Y \subseteq A_i\} $. - Because $X_i = \bigcap_{y \in Y} \{x \in X | (x,y) \in A_i\}$, this is closed. As $X$ is irreducible, there is $i \in \{1;2\} $ which $X_i = X$. Then $X \times Y = A_i$ confirming the irreducibility of $X \times Y$. + Assume that $X \times Y = A_1 \cup A_2$, where the $A_i \subseteq + \mathfrak{k}^{m+n}$ are closed. + For $x' \in \mathfrak{k}^m, x' \times Y$ is homeomorphic to the + irreducible + $Y$. + Thus $X = X_1 \cup X_2$ where $X_i = \{x \in X | \{x\} \times Y \subseteq + A_i\} $. + Because $X_i = \bigcap_{y \in Y} \{x \in X | (x,y) \in A_i\}$, this is + closed. + As $X$ is irreducible, there is $i \in \{1;2\} $ which $X_i = X$. + Then $X \times Y = A_i$ confirming the irreducibility of $X \times Y$. - Let $a = \dim X$ and $b = \dim Y$ and $X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_a = X$,$Y_0 \subsetneq Y_1 \subsetneq \ldots \subsetneq Y_b = Y$ be chains of irreducible subsets. By the previous result, - $X_0 \times Y_0 \subsetneq X_1 \times Y_0 \subsetneq \ldots \subsetneq X_a \times Y_0 \subsetneq X_a \times Y_1 \subsetneq \ldots \subsetneq X_a \times Y_a = X \times Y$ is a chain of irreducible subsets. + Let $a = \dim X$ and $b = \dim Y$ and $X_0 \subsetneq X_1 \subsetneq \ldots + \subsetneq X_a = X$,$Y_0 \subsetneq Y_1 \subsetneq \ldots \subsetneq Y_b = Y$ + be chains of irreducible subsets. + By the previous result, $X_0 \times Y_0 \subsetneq X_1 \times Y_0 \subsetneq + \ldots \subsetneq X_a \times Y_0 \subsetneq X_a \times Y_1 \subsetneq \ldots + \subsetneq X_a \times Y_a = X \times Y$ is a chain of irreducible subsets. Thus $\dim(X \times Y) \ge a + b = \dim X + \dim Y$. - Similarly one derives $\codim(X \times Y, \mathfrak{k}^{m+n}) \ge \codim(X, \mathfrak{k}^m) + \codim(Y, \mathfrak{k}^n)$. - By \ref{trdegandkdim} we have $\dim(A) + \codim(A, \mathfrak{k}^l) = l$ for irreducible subsets of $\mathfrak{k}^l$. Thus equality must hold in the previous two inequalities. + Similarly one derives $\codim(X \times Y, \mathfrak{k}^{m+n}) \ge \codim(X, + \mathfrak{k}^m) + + \codim(Y, \mathfrak{k}^n)$. + By + \ref{trdegandkdim} we have $\dim(A) + + \codim(A, \mathfrak{k}^l) = l$ for + irreducible subsets of $\mathfrak{k}^l$. + Thus equality must hold in the previous two inequalities. \end{proof} \subsection{The nil radical} @@ -1408,38 +2766,80 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 \end{notation} \begin{proposition}[Nil radical] - For a ring $A$, $\bigcap_{\fp \in \Spec A} \fp = \sqrt{\{0\} } = \{a \in A | \exists k \in \N ~ a^k = 0\} \text{\reflectbox{$\coloneqq$}} \nil(A)$, the set of nilpotent elements of $A$. - This is called the \vocab{nil radical} of $A$. + For a ring $A$, $\bigcap_{\fp \in \Spec A} \fp = + \sqrt{\{0\} } = \{a \in A | \exists k \in \N ~ a^k = 0\} + \text{\reflectbox{$\coloneqq$}} \nil(A)$, the set of nilpotent elements + of $A$. + This is called the \vocab{nil radical} of $A$. \end{proposition} \begin{proof} - It is clear that elements of $\sqrt{\{0\} } $ must belong to all prime ideals. Conversely, let $a \in A \setminus \sqrt{\{0\} }$. Then $S = a^{\N}$ is a multiplicative subset of $A$ not containing $0$. - The localisation $A_S$ of $A$ is thus not the null ring. Hence $\Spec A_S \neq \emptyset$. If $\fq \in \Spec A_S$, then by the description of $\Spec A_S$ (\ref{idealslocbij}), $\fp \coloneqq \fq \sqcap A$ is a prime ideal of $A$ disjoint from $S$, hence $a \not\in \fp$. + It is clear that elements of $\sqrt{\{0\} } $ must belong to all + prime ideals. + Conversely, let $a \in A \setminus \sqrt{\{0\} }$. + Then $S = a^{\N}$ is a multiplicative subset of $A$ + not containing $0$. + The localisation $A_S$ of $A$ is thus not the null ring. + Hence $\Spec A_S \neq \emptyset$. + If $\fq \in \Spec A_S$, then by the description of $\Spec A_S$ + ( + \ref{idealslocbij}), $\fp \coloneqq \fq \sqcap A$ is a prime ideal of $A$ + disjoint from $S$, hence $a \not\in \fp$. \end{proof} -\begin{corollary}\label{sqandvspec} - For an ideal $I$ of $R$, $\sqrt{I} = \bigcap_{\fp \in \Vspec(I)} \fp$. +\begin{corollary} + \label{sqandvspec} + For an ideal $I$ of $R$, $\sqrt{I} = \bigcap_{\fp \in \Vspec(I)} + \fp$. \end{corollary} \begin{proof} - This is obtained by applying the proposition to $A = R / I $ and using the bijection $\Spec( R / I) \cong V(I)$ sending $\fp \in V(I)$ to $\fp \coloneqq \fp / I$ and $\fq \in \Spec(R / I)$ to its inverse image $\fp$ in $R$. + This is obtained by applying the proposition to $A = R / I $ and using the + bijection $\Spec( R / I) \cong V(I)$ sending $\fp \in V(I)$ to $\fp \coloneqq + \fp / I$ and $\fq \in \Spec(R / I)$ to its inverse image $\fp$ in $R$. \end{proof} \subsubsection{Closed subsets of \texorpdfstring{$\Spec R$}{Spec R}} -\begin{proposition}\label{bijspecideal} +\begin{proposition} + \label{bijspecideal} There is a bijection \begin{align} - f: \{A \subseteq \Spec R | A\text{ closed}\} &\longrightarrow \{I \subseteq R | I \text{ ideal and } I = \sqrt{I} \} \\ - A &\longmapsto \bigcap_{\fp \in A} \fp\\ - \Vspec(I) &\longmapsfrom I + f: \{A \subseteq \Spec R | A\text{ closed}\} + & \longrightarrow \{I \subseteq R | I \text{ ideal and } I = \sqrt{I} \} \\ A + & \longmapsto \bigcap_{\fp \in A} \fp \\ \Vspec(I) & \longmapsfrom I \end{align} - Under this bijection, the irreducible subsets correspond to the prime ideals and the closed points $\{\mathfrak{m}\}, \mathfrak{m} \in \Spec A$ to the maximal ideals. + Under this bijection, the irreducible subsets correspond to the prime ideals + and the closed points $\{\mathfrak{m}\}, \mathfrak{m} + \in \Spec A$ to the + maximal ideals. \end{proposition} \begin{proof} - If $A = \Vspec(I)$, then by \ref{sqandvspec} $\sqrt{I} = \bigcap_{\fp \in A} \fp$. Thus, an ideal with $\sqrt{I} = I$ can be recovered from $\Vspec( I)$. Since $\Vspec(J) = \Vspec(\sqrt{J})$, the map from ideals with $\sqrt{I} = I$ to closed subsets is surjective. + If $A = \Vspec(I)$, then by + \ref{sqandvspec} + $\sqrt{I} = \bigcap_{\fp \in A} + \fp$. + Thus, an ideal with $\sqrt{I} = I$ can be recovered from + $\Vspec( I)$. + Since $\Vspec(J) = \Vspec(\sqrt{J})$, the map from ideals + with $\sqrt{I} = I$ + to closed subsets is surjective. - Sine $R$ corresponds to $\emptyset$, the proper ideals correspond to non-empty subsets of $\Spec R$. Assume that $\Vspec(I) = \Vspec(J_1) \cup \Vspec(J_2)$, where the decomposition is proper and the ideals coincide with their radicals. - Let $g = f_1f_2$ with $f_k \in J_k \setminus I$. Since $\Vspec(g) \supseteq \Vspec(f_k) \supseteq \Vspec(I_k), \Vspec(I) \subseteq \Vspec(g)$. Hence $g \in \sqrt{I} = I$. + Sine $R$ corresponds to $\emptyset$, the proper ideals correspond to non-empty + subsets of $\Spec R$. + Assume that $\Vspec(I) = \Vspec(J_1) \cup + \Vspec(J_2)$, where the decomposition + is proper and the ideals coincide with their radicals. + Let $g = f_1f_2$ with $f_k \in J_k \setminus I$. + Since $\Vspec(g) \supseteq \Vspec(f_k) \supseteq + \Vspec(I_k), \Vspec(I) + \subseteq \Vspec(g)$. + Hence $g \in \sqrt{I} = I$. As $f_k \not\in I$, $I$ fails to be a prime ideal. - Conversely, assume that $f_1f_2 \in I$ while the factors are not in $I$. Since $I = \sqrt{I}, \Vspec(f_k) \not\supseteq \Vspec(I)$. But $\Vspec(f_1) \cup \Vspec(f_2) = \Vspec(f_1f_2) \supseteq \Vspec(I)$. - The proper decomposition $\Vspec(I) = \left( \Vspec(I) \cap \Vspec(f_1) \right) \cup \left( \Vspec(I) \cap \Vspec(f_2) \right) $ now shows that $\Vspec(I)$ fails to be irreducible. + Conversely, assume that $f_1f_2 \in I$ while the factors are not in $I$. + Since $I = \sqrt{I}, \Vspec(f_k) \not\supseteq + \Vspec(I)$. + But $\Vspec(f_1) \cup \Vspec(f_2) = + \Vspec(f_1f_2) \supseteq \Vspec(I)$. + The proper decomposition $\Vspec(I) = \left( \Vspec(I) \cap \Vspec(f_1) \right) + \cup \left( \Vspec(I) \cap \Vspec(f_2) \right) $ now shows that $\Vspec(I)$ + fails to be irreducible. The final assertion is trivial. \end{proof} @@ -1447,23 +2847,49 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 If $R$ is a Noetherian ring, then $\Spec R$ is a Noetherian topological space. \end{corollary} \begin{remark} - It is not particularly hard to come up with examples which show that the converse implication does not hold. + It is not particularly hard to come up with examples which show that the + converse implication does not hold. \end{remark} -\begin{example}+ - Let $A = \mathfrak{k}[X_n | n \in \N] / I$ where $I$ denotes the ideal generated by $\{X_i^2 | i \in \N\}$. - $A$ is not Noetherian, since the ideal $J$ generated by $\{X_i | i \in \N\} $ is not finitely generated. - $A / J \cong \mathfrak{k}$, hence $J$ is maximal. As every prime ideal must contain $\nil(A) \supseteq J$, $J$ is the only prime ideal. +\begin{example} + + + Let $A = \mathfrak{k}[X_n | n \in \N] / I$ where $I$ denotes the ideal generated by + $\{X_i^2 | i \in \N\}$. + $A$ is not Noetherian, since the ideal $J$ generated by $\{X_i | i \in \N\} $ + is not finitely generated. + $A / J \cong \mathfrak{k}$, hence $J$ is maximal. + As every prime ideal must contain $\nil(A) \supseteq J$, $J$ is + the only prime + ideal. Thus $\Spec A$ contains only one element and is hence Noetherian. \end{example} -\begin{corollary}[About the smallest prime ideals containing $I$ ]\label{smallestprimesvi} - If $R$ is Noetherian and $I \subseteq R$ an ideal, then the set $\Vspec(I) = \{\fp \in \Spec R | I \subseteq \fp\}$ has finitely many $\subseteq$-minimal elements $(\fp_i)_{i=1}^k$ and every element of $V(I)$ contains at least one $\fp_i$. - The $\Vspec(\fp_i)$ are precisely the irreducible components of $V(I)$. Moreover $\bigcap_{i=1}^k \fp_i = \sqrt{I}$ and $k > 0$ if $I$ is a proper ideal. +\begin{corollary}[About the smallest prime ideals containing $I$ ] + \label{smallestprimesvi} + If $R$ is Noetherian and $I \subseteq R$ an ideal, then the set + $\Vspec(I) = + \{\fp \in \Spec R | I \subseteq \fp\}$ has finitely many $\subseteq$-minimal + elements $(\fp_i)_{i=1}^k$ and every element + of $V(I)$ contains at least one + $\fp_i$. + The $\Vspec(\fp_i)$ are precisely the irreducible components of + $V(I)$. + Moreover $\bigcap_{i=1}^k \fp_i = \sqrt{I}$ and $k > + 0$ if $I$ is a proper + ideal. \end{corollary} \begin{proof} - If $\Vspec(I) = \bigcup_{i=1}^k \Vspec(\fp_i)$ is the decomposition into irreducible components then every $\fq \in \Vspec(I)$ must belong to at least one $\Vspec(\fp_i)$, hence $\fp_i \subseteq \fq$. Also $\fp_i \in \Vspec(\fp_i) \subseteq \Vspec(I)$. - It follows that the sets of $\subseteq$-minimal elements of $\Vspec(I)$ and of $\{\fp_1,\ldots,\fp_k\} $ coincide. - As there are no non-trivial inclusions between the $\Vspec(\fp_i)$, there are no non-trivial inclusions between the $\fp_i$ and the assertion follows. + If $\Vspec(I) = \bigcup_{i=1}^k + \Vspec(\fp_i)$ is the decomposition into + irreducible components then every $\fq \in \Vspec(I)$ must belong + to at least + one $\Vspec(\fp_i)$, hence $\fp_i \subseteq \fq$. + Also $\fp_i \in \Vspec(\fp_i) \subseteq \Vspec(I)$. + It follows that the sets of $\subseteq$-minimal elements of + $\Vspec(I)$ and of + $\{\fp_1,\ldots,\fp_k\} $ coincide. + As there are no non-trivial inclusions between the $\Vspec(\fp_i)$, + there are + no non-trivial inclusions between the $\fp_i$ and the assertion follows. The final remark is trivial. \end{proof} \begin{corollary} @@ -1473,121 +2899,229 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 \subsection{The principal ideal theorem} Krull was able to show: -\begin{theorem}[Principal ideal theorem / Hauptidealsatz]\label{pitheorem} - Let $A$ be a Noetherian ring, $a \in A$ and $\fp \in \Spec A$ a $\subseteq$-minimal element of $\Vspec(a)$. Then $\hght(\fp) \le 1$. +\begin{theorem}[Principal ideal theorem / + Hauptidealsatz] + \label{pitheorem} Let $A$ be a Noetherian ring, + $a \in A$ and + $\fp \in \Spec A$ a $\subseteq$-minimal element of $\Vspec(a)$. + Then $\hght(\fp) \le 1$. \end{theorem} \begin{proof} Probably not relevant for the exam. \end{proof} \begin{remark} - Intuitively, the theorem says that by imposing a single equation one ends up in codimension at most $1$. This would not be true in real analysis (or real algebraic geometry) as the equation $\sum_{i=1}^{n} X_i^2 = 0$ shows. By \ref{smallestprimesvi}, if $a$ is a non-unit then a $\fp \in \Spec A$ to which the theorem applies can always be found. + Intuitively, the theorem says that by imposing a single equation one ends up in + codimension at most $1$. + This would not be true in real analysis (or real algebraic geometry) as the + equation $\sum_{i=1}^{n} X_i^2 = 0$ shows. + By + \ref{smallestprimesvi}, if $a$ is a non-unit then a $\fp \in \Spec A$ to + which the theorem applies can always be found. Using induction on $k$, Krull was able to derive: \end{remark} -\begin{theorem}[Generalized principal ideal theorem] - Let $A$ be a Noetherian ring, $(a_i)_{i=1}^k \in A$ and $\fp \in \Spec A$ a $\subseteq$-minimal element of $\bigcap_{i=1}^k V(a_i)$, the set of prime ideals containing all $a_i$. +\begin{theorem}[Generalized principal ideal theorem] Let $A$ be a Noetherian + ring, $(a_i)_{i=1}^k \in A$ and $\fp \in + \Spec A$ a $\subseteq$-minimal element + of $\bigcap_{i=1}^k V(a_i)$, the set of prime ideals containing all + $a_i$. Then $\hght(\fp) \le k$. \end{theorem} -Modern approaches to the principal ideal theorem usually give a direct proof of this more general theorem. +Modern approaches to the principal ideal theorem usually give a direct proof of +this more general theorem. \begin{corollary} If $R$ is a Noetherian ring and $\fp \in \Spec R$, then $\hght(\fp) < \infty$. \end{corollary} \begin{proof} - If $\fp$ is generated by $(f_i)_{i=1}^k$, then $\hght(\fp) \le k$. + If $\fp$ is generated by $(f_i)_{i=1}^k$, + then $\hght(\fp) \le k$. \end{proof} \subsubsection{Application to the dimension of intersections} -\begin{remark}\label{smallestprimeandirredcomp} -Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ an ideal. +\begin{remark} + \label{smallestprimeandirredcomp} + Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ an ideal. - If $(\fp_i)_{i=1}^k$ are the smallest prime ideals of $R$ containing $I$, then $(\Va(\fp_i))_{i=1}^k$ are the irreducible components of $\Va(I)$. + If $(\fp_i)_{i=1}^k$ are the smallest prime + ideals of $R$ containing $I$, then + $(\Va(\fp_i))_{i=1}^k$ are the + irreducible components of $\Va(I)$. \end{remark} \begin{proof} - The $\Va(\fp_i)$ are irreducible, there are no non-trivial inclusions between them and $ \Va(I) = \Va(\sqrt{I}) = \Va(\bigcap_{i=1}^k \fp_i) = \bigcup_{i=1}^k \Va(\fp_i)$. + The $\Va(\fp_i)$ are irreducible, there are no non-trivial + inclusions between + them and $ \Va(I) = \Va(\sqrt{I}) = + \Va(\bigcap_{i=1}^k \fp_i) = + \bigcup_{i=1}^k \Va(\fp_i)$. \end{proof} \begin{corollary}[of the principal ideal theorem] \label{corpithm} - Let $X \subseteq \mathfrak{k}^n$ be irreducible, $(f_i)_{i=1}^k$ elements of $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $Y$ an irreducible component of $A = X \cap \bigcap_{i=1}^k V(f_i)$. + Let $X \subseteq \mathfrak{k}^n$ be irreducible, + $(f_i)_{i=1}^k$ elements of $R + = \mathfrak{k}[X_1,\ldots,X_n]$ and $Y$ an irreducible component of $A = X \cap + \bigcap_{i=1}^k V(f_i)$. Then $\codim(Y,X) \le k$. \end{corollary} \begin{remark} - This confirms the naive geometric intuition that by imposing $k$ equations one ends up in codimension at most $k$. + This confirms the naive geometric intuition that by imposing $k$ equations one + ends up in codimension at most $k$. \end{remark} \begin{proof} - If $X = v(\fp), X \cap \bigcap_{i=1}^k V(f_i) = V(I)$ where $I \subseteq R$ is the ideal generated by $\fp$ and the $f_i$. - By \ref{smallestprimeandirredcomp}, $Y = V(\fq)$ where $\fq$ is the smallest prime ideal containing $I$. - Then $\fq / \fp$ is a smallest prime ideal of $R / \fp$ containing all $(f_i \mod \fp)_{i=1}^k$. - By the principal ideal theorem (\ref{pitheorem}), $\hght(\fq / \fp) \le k$ and the assertion follows from example \ref{htandcodim}. + If $X = v(\fp), X \cap \bigcap_{i=1}^k V(f_i) = V(I)$ where $I + \subseteq R$ is + the ideal generated by $\fp$ and the $f_i$. + By + \ref{smallestprimeandirredcomp}, $Y = V(\fq)$ where $\fq$ is the smallest + prime ideal containing $I$. + Then $\fq / \fp$ is a smallest prime ideal of $R / \fp$ containing all $(f_i + \mod \fp)_{i=1}^k$. + By the principal ideal theorem ( + \ref{pitheorem}), + $\hght(\fq / \fp) \le k$ and + the assertion follows from example + \ref{htandcodim}. \end{proof} -\begin{remark}\label{affineproblem} - Note that the intersection $X \cap \bigcap_{i=1}^k V(f_i)$ can easily be empty, even when $k$ is much smaller than $\dim X$. +\begin{remark} + \label{affineproblem} + Note that the intersection $X \cap \bigcap_{i=1}^k V(f_i)$ can easily + be empty, + even when $k$ is much smaller than $\dim X$. \end{remark} -\begin{corollary}\label{codimintersection} - Let $A$ and $B$ be irreducible subsets of $\mathfrak{k}^n$. If $C$ is an irreducible component of $A \cap B$, then $\codim(C, \mathfrak{k}^n) \le \codim(A, \mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)$. +\begin{corollary} + \label{codimintersection} + Let $A$ and $B$ be irreducible subsets of $\mathfrak{k}^n$. + If $C$ is an irreducible component of $A \cap B$, then $\codim(C, + \mathfrak{k}^n) + \le \codim(A, \mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)$. \end{corollary} -\begin{remark}+ - Equivalently, $\dim(C) \ge \dim(A) + \dim(B)-n$. +\begin{remark} + + + Equivalently, $\dim(C) \ge \dim(A) + \dim(B)-n$. \end{remark} \begin{proof} - Let $X = A \times B \subseteq \mathfrak{k}^{2n}$, where we use $(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$ as coordinates of $\mathfrak{k}^{2n}$. - Let $\Delta \coloneqq \{(x_1,\ldots,x_n,x_1,\ldots,x_n) | x \in \mathfrak{k}^n\} $ be the diagonal in $\mathfrak{k}^n \times \mathfrak{k}^n$. - The projection $\mathfrak{k}^{2n}\to \mathfrak{k}^n$ to the $X$-coordinates defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$. - Thus, $C$ is homeomorphic to an irreducible component $C'$ of $(A \times B) \cap \Delta$ and + Let $X = A \times B \subseteq \mathfrak{k}^{2n}$, where we use + $(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$ as coordinates of $\mathfrak{k}^{2n}$. + Let $\Delta \coloneqq \{(x_1,\ldots,x_n,x_1,\ldots,x_n) | x \in + \mathfrak{k}^n\} $ be the diagonal in $\mathfrak{k}^n + \times \mathfrak{k}^n$. + The projection $\mathfrak{k}^{2n}\to \mathfrak{k}^n$ to the + $X$-coordinates + defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$. + Thus, $C$ is homeomorphic to an irreducible component $C'$ of $(A \times B) + \cap \Delta$ and \begin{align} - \codim(C, \mathfrak{k}^n) = n - \dim(C) = n - \dim(C') = n - \dim(A \times B) + \codim(C', A \times B)\\ - \overset{\text{\ref{corpithm}}}{\le }2n - \dim(A \times B) \overset{\text{\ref{dimprod}}}{=} 2n - \dim(A) - \dim(B) = \codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n) + \codim(C, \mathfrak{k}^n) = n - \dim(C) = n - + \dim(C') = n - \dim(A \times B) + \codim(C', A \times B) \\ + \overset{\text{ + \ref{corpithm}}}{\le }2n - \dim(A \times B) + \overset{\text{ + \ref{dimprod}}}{=} 2n - \dim(A) - \dim(B) = + \codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n) \end{align} - by the general properties of dimension and codimension, \ref{corpithm} applied to $(X_i - Y_i)_{i=1}^n$, - the result about the dimension of products (\ref{dimprod}) and again the general properties of dimension and codimension. + by the general + properties of dimension and codimension, + \ref{corpithm} applied to + $(X_i - + Y_i)_{i=1}^n$, the result about the dimension + of products ( + \ref{dimprod}) and + again the general properties of dimension and codimension. \end{proof} \begin{remark} - As in \ref{affineproblem}, $A \cap B$ can easily be empty, even when $A$ and $B$ have codimension $1$ and $n$ is very large. + As in + \ref{affineproblem}, $A \cap B$ can easily be empty, even when $A$ + and + $B$ have codimension $1$ and $n$ is very large. \end{remark} \subsubsection{Application to the property of being a UFD} \begin{proposition} - Let $R$ be a Noetherian domain. Then $R$ is a UFD iff every $\fp \in \Spec R$ with $\hght(\fp)= 1$\footnote{In other words, every $\subseteq$-minimal element of the set of non-zero prime ideals of $R$ } is a principal ideal. + Let $R$ be a Noetherian domain. + Then $R$ is a UFD iff every $\fp \in \Spec R$ with $\hght(\fp)= + 1$\footnote{In + other words, every $\subseteq$-minimal element of the set of non-zero prime + ideals of $R$ } is a principal ideal. \end{proposition} \begin{proof} - Every element of every Noetherian domain can be written as a product of irreducible elements.\footnote{Consider the set of principal ideals $rR$ where $r$ is not a product of irreducible elements.} + Every element of every Noetherian domain can be written as a product of + irreducible elements. + \footnote{Consider the set of principal ideals $rR$ where $r$ is not a product of + irreducible elements.} Thus, $R$ is a UFD iff every irreducible element of $R$ is prime. - Assume that this is the case. Let $\fp \in \Spec R, \hght(\fp) = 1$. - Let $p \in \fp \setminus \{0\}$. Replacing $p$ by a prime factor of $p$, we may assume $p$ to be prime. Thus $\{0\} \subsetneq pR \subseteq \fp$ is a chain of prime ideals and since $\hght(\fp) = 1$ it follows that $\fp = pR$. + Assume that this is the case. + Let $\fp \in \Spec R, \hght(\fp) = 1$. + Let $p \in \fp \setminus \{0\}$. + Replacing $p$ by a prime factor of $p$, we may assume $p$ to be prime. + Thus $\{0\} \subsetneq pR \subseteq \fp$ is a chain of prime ideals and since + $\hght(\fp) = 1$ it follows that $\fp = pR$. - Conversely, assume that every $\fp \in \Spec R$ with $\hght(\fp)=1$ is a principal ideal. Let $f \in R$ be irreducible. - Let $\fp \in \Spec R$ be a $\subseteq$-minimal element of $V(f)$. By the principal ideal theorem (\ref{pitheorem}), $\hght(\fp)=1$. - Thus $\fp = pR$ for some prime element $p$. We have $p | f$ since $f \in \fp$. As $f$ is irreducible, $p$ and $f$ are multiplicatively equivalent. Thus $f$ is a prime element. + Conversely, assume that every $\fp \in \Spec R$ with + $\hght(\fp)=1$ is a + principal ideal. + Let $f \in R$ be irreducible. + Let $\fp \in \Spec R$ be a $\subseteq$-minimal element of $V(f)$. + By the principal ideal theorem ( + \ref{pitheorem}), + $\hght(\fp)=1$. + Thus $\fp = pR$ for some prime element $p$. + We have $p | f$ since $f \in \fp$. + As $f$ is irreducible, $p$ and $f$ are multiplicatively equivalent. + Thus $f$ is a prime element. \end{proof} \subsection{The Jacobson radical} \begin{proposition} - For a ring $A, \bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m} = \{a \in A | \forall x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$, the \vocab{Jacobson radical} of $A$. + For a ring $A, \bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m} = \{a \in A | + \forall x \in A ~ 1 - ax \in A^{\times }\} + \text{\reflectbox{$\coloneqq$}} + \rad(A)$, the \vocab{Jacobson radical} of $A$. \end{proposition} \begin{proof} - Suppose $\mathfrak{m} \in \MaxSpec A$ and $a \in A \setminus \mathfrak{m}$. Then $a \mod \mathfrak{m} \neq 0$ and $A / \mathfrak{m}$ is a field. Hence $a \mod \mathfrak{m}$ has an inverse $x \mod \mathfrak{m}$. - $1 - ax \in \mathfrak{m}$, hence $1 - ax \not\in A^{\times}$ and $a $ is not al element of the RHS. + Suppose $\mathfrak{m} \in \MaxSpec A$ and $a \in A \setminus + \mathfrak{m}$. + Then $a \mod \mathfrak{m} \neq 0$ and $A / \mathfrak{m}$ + is a field. + Hence $a \mod \mathfrak{m}$ has an inverse $x \mod + \mathfrak{m}$. + $1 - ax \in \mathfrak{m}$, hence $1 - ax \not\in + A^{\times}$ and $a $ is not al element of the RHS. - Conversely, let $a \in A$ belong to all $\mathfrak{m} \in \MaxSpec A$. If there exists $x \in A$ such that $1 - ax \not\in A^{\times }$ then $(1-ax) A$ was a proper ideal in $A$, hence contained in a maximal ideal $\mathfrak{m}$. As $a \in \mathfrak{m}, 1 = (1-ax) + ax \in \mathfrak{m}$, a contradiction. - Hence every element of $\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m}$ belongs to the right hand side. + Conversely, let $a \in A$ belong to all $\mathfrak{m} \in \MaxSpec + A$. + If there exists $x \in A$ such that $1 - ax \not\in + A^{\times }$ then $(1-ax) + A$ was a proper ideal in $A$, hence contained in a maximal ideal + $\mathfrak{m}$. + As $a \in \mathfrak{m}, 1 = (1-ax) + ax \in + \mathfrak{m}$, a contradiction. + Hence every element of $\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m}$ + belongs to the right hand side. \end{proof} \begin{example} - If $A$ is a local ring, then $\rad(A) = \mathfrak{m}_A$. + If $A$ is a local ring, then $\rad(A) = + \mathfrak{m}_A$. \end{example} \begin{example} - If $A$ is a PID with infinitely many multiplicative equivalence classes of prime elements (e.g. $\Z$ of $\mathfrak{k}[X]$), then $\rad(A) = \{0\}$: - Prime ideals of a PID are maximal. Thus if $x \in \rad(A)$, every prime element divides $x$. If $x \neq 0$, it follows that $x$ has infinitely many prime divisors. + If $A$ is a PID with infinitely many multiplicative equivalence classes of + prime elements (e.g. $\Z$ of $\mathfrak{k}[X]$), then + $\rad(A) = \{0\}$: Prime + ideals of a PID are maximal. + Thus if $x \in \rad(A)$, every prime element divides $x$. + If $x \neq 0$, it follows that $x$ has infinitely many prime divisors. However every PID is a UFD. \end{example} \begin{example} -If $A$ is a PID for which $p_1,\ldots,p_n$ is a list of representatives of the multiplicative equivalence classes of prime elements, then -$\rad(A) = f A$ where $f = \prod_{i=1}^{n} p_i$. + If $A$ is a PID for which $p_1,\ldots,p_n$ is a list of representatives of the + multiplicative equivalence classes of prime elements, then + $\rad(A) = f A$ + where $f = \prod_{i=1}^{n} p_i$. \end{example} % proof of the pitheorem probably won't be relevant in the exam diff --git a/inputs/preamble.tex b/inputs/preamble.tex index cfb99cf..3904d25 100644 --- a/inputs/preamble.tex +++ b/inputs/preamble.tex @@ -1,13 +1,21 @@ \begin{warning} This is not an official script! - This document was written in preparation for the oral exam. It mostly follows the way \textsc{Prof. Franke} presented the material in his lecture rather closely. + This document was written in preparation for the oral exam. + It mostly follows the way \textsc{Prof. + Franke} presented the material in his + lecture rather closely. There are probably errors. \end{warning} -\noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the MIT-License and can be obtained from \url{https://github.com/kesslermaximilian/LatexPackages}. % TODO +\noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the +MIT-License and can be obtained from \url{https://github.com/kesslermaximilian/LatexPackages}. +% TODO \newline -\noindent $\mathfrak{k}$ is {\color{red} always} an algebraically closed field and $\mathfrak{k}^n$ is equipped with the Zariski-topology. -Fields which are not assumed to be algebraically closed have been renamed (usually to $\mathfrak{l}$). +\noindent $\mathfrak{k}$ is {\color{red} always} an +algebraically closed field and $\mathfrak{k}^n$ is equipped with the +Zariski-topology. +Fields which are not assumed to be algebraically closed have been renamed +(usually to $\mathfrak{l}$). diff --git a/inputs/projective_spaces.tex b/inputs/projective_spaces.tex index 12250f6..b63accd 100644 --- a/inputs/projective_spaces.tex +++ b/inputs/projective_spaces.tex @@ -1,25 +1,65 @@ Let $\mathfrak{l}$ be any field. \begin{definition} - For a $\mathfrak{l}$-vector space $V$, let $\mathbb{P}(V)$ be the set of one-dimensional subspaces of $V$. - Let $\mathbb{P}^n(\mathfrak{l}) \coloneqq \mathbb{P}(\mathfrak{l}^{n+1})$, the \vocab[Projective space]{$n$-dimensional projective space over $\mathfrak{l}$}. + For a $\mathfrak{l}$-vector space $V$, let $\mathbb{P}(V)$ be + the set of + one-dimensional subspaces of $V$. + Let $\mathbb{P}^n(\mathfrak{l}) \coloneqq + \mathbb{P}(\mathfrak{l}^{n+1})$, the + \vocab[Projective space]{$n$-dimensional projective space over $\mathfrak{l}$}. - If $\mathfrak{l}$ is kept fixed, we will often write $\mathbb{P}^n$ for $\mathbb{P}^n(\mathfrak{l})$. - - When dealing with $\mathbb{P}^n$, the usual convention is to use $0$ as the index of the first coordinate. - - We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\}$ by $[x_0,\ldots,x_n] \in \mathbb{P}^n$. - If $x = [x_0,\ldots,x_n] \in \mathbb{P}^n$, the $(x_{i})_{i=0}^n$ are called \vocab{homogeneous coordinates} of $x$. + If $\mathfrak{l}$ is kept fixed, we will often write + $\mathbb{P}^n$ for + $\mathbb{P}^n(\mathfrak{l})$. + + When dealing with $\mathbb{P}^n$, the usual convention is to use + $0$ as the + index of the first coordinate. + + We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in + \mathfrak{k}^{n+1} \setminus \{0\}$ by + $[x_0,\ldots,x_n] \in \mathbb{P}^n$. + If $x = [x_0,\ldots,x_n] \in \mathbb{P}^n$, the + $(x_{i})_{i=0}^n$ are + called + \vocab{homogeneous coordinates} of $x$. At least one of the $x_{i}$ must be $\neq 0$. \end{definition} \begin{remark} - There are points $[1,0], [0,1] \in \mathbb{P}^1$ but there is no point $[0,0] \in \mathbb{P}^1$. + There are points $[1,0], + [0,1] \in \mathbb{P}^1$ but there + is no point $[0,0] + \in \mathbb{P}^1$. \end{remark} \begin{definition}[Infinite hyperplane] - For $0 \le i \le n$ let $U_i \subseteq \mathbb{P}^n$ denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$. - This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the same point $x \in \mathbb{P}^n$ differ by scaling with a $\lambda \in \mathfrak{l}^{\times}$, $x_i = \lambda \xi_i$. Since not all $x_i$ may be $0$, $\mathbb{P}^n = \bigcup_{i=0}^n U_i$. We identify $\mathbb{A}^n = \mathbb{A}^n(\mathfrak{l}) = \mathfrak{l}^n$ with $U_0$ by identifying $(x_1,\ldots,x_n) \in \mathbb{A}^n$ with $[1,x_1,\ldots,x_n] \in \mathbb{P}^n$. - Then $\mathbb{P}^1 = \mathbb{A}^1 \cup \{\infty\} $ where $\infty=[0,1]$. More generally, when $n > 0$ $\mathbb{P}^n \setminus \mathbb{A}^n$ can be identified with $\mathbb{P}^{n-1}$ identifying $[0,x_1,\ldots,x_n] \in \mathbb{P}^n \setminus \mathbb{A}^n$ with $[x_1,\ldots,x_n] \in \mathbb{P}^{n-1}$. + For $0 \le i \le n$ let $U_i \subseteq + \mathbb{P}^n$ denote the set of $[x_0,\ldots,x_{n}]$ with + $x_{i}\neq 0$. + This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and + $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the + same point $x \in + \mathbb{P}^n$ differ by scaling with a $\lambda \in + \mathfrak{l}^{\times}$, + $x_i = \lambda \xi_i$. + Since not all $x_i$ may be $0$, $\mathbb{P}^n = + \bigcup_{i=0}^n U_i$. + We identify $\mathbb{A}^n = + \mathbb{A}^n(\mathfrak{l}) = \mathfrak{l}^n$ + with + $U_0$ by identifying $(x_1,\ldots,x_n) \in \mathbb{A}^n$ with + $[1,x_1,\ldots,x_n] \in \mathbb{P}^n$. + Then $\mathbb{P}^1 = \mathbb{A}^1 \cup \{\infty\} $ + where $\infty=[0,1]$. + More generally, when $n > 0$ $\mathbb{P}^n \setminus + \mathbb{A}^n$ can be + identified with $\mathbb{P}^{n-1}$ identifying + $[0,x_1,\ldots,x_n] \in + \mathbb{P}^n \setminus \mathbb{A}^n$ with + $[x_1,\ldots,x_n] \in + \mathbb{P}^{n-1}$. - Thus $\mathbb{P}^n$ is $\mathbb{A}^n \cong \mathfrak{l}^n$ with a copy of $\mathbb{P}^{n-1}$ added as an \vocab{infinite hyperplane} . + Thus $\mathbb{P}^n$ is $\mathbb{A}^n \cong + \mathfrak{l}^n$ with a copy of + $\mathbb{P}^{n-1}$ added as an \vocab{infinite hyperplane} . \end{definition} \subsubsection{Graded rings and homogeneous ideals} @@ -27,18 +67,40 @@ Let $\mathfrak{l}$ be any field. Let $\mathbb{I} = \N$ or $\mathbb{I} = \Z$. \end{notation} \begin{definition} - By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$ we understand a ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of subgroups of the additive group $(A, +)$ such that $A_a \cdot A_b \subseteq A_{a + b}$ for $a,b \in \mathbb{I}$ and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in the sense that every $r \in A$ has a unique decomposition $r = \sum_{d \in \mathbb{I}} r_d$ with $r_d \in A_d$ and but finitely many $r_d \neq 0$. + By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$ we understand a + ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of + subgroups of the + additive group $(A, +)$ such that $A_a \cdot A_b \subseteq + A_{a + b}$ for $a,b + \in \mathbb{I}$ and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in + the + sense that every $r \in A$ has a unique decomposition $r = + \sum_{d \in + \mathbb{I}} r_d$ with $r_d \in A_d$ and but finitely many $r_d + \neq 0$. We call the $r_d$ the \vocab{homogeneous components} of $r$. - An ideal $I \subseteq A$ is called \vocab{homogeneous} if $r \in I \implies \forall d \in \mathbb{I} ~ r_d \in I_d$ where $I_d \coloneqq I \cap A_d$. + An ideal $I \subseteq A$ is called \vocab{homogeneous} if $r \in I + \implies + \forall d \in \mathbb{I} ~ r_d \in I_d$ where $I_d \coloneqq I + \cap A_d$. - By a \vocab{graded ring} we understand an $\N$-graded ring. Tin this case, $A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\} $ is called the \vocab{augmentation ideal} of $A$. + By a \vocab{graded ring} we understand an $\N$-graded ring. + Tin this case, $A_{+} \coloneqq \bigoplus_{d=1}^{\infty} + A_d = \{r \in A | r_0 + = 0\} $ is called the \vocab{augmentation ideal} of $A$. \end{definition} \begin{remark}[Decomposition of $1$] - If $1 = \sum_{d \in \mathbb{I}} \varepsilon_d$ is the decomposition into homogeneous components, then $\varepsilon_a = 1 \cdot \varepsilon_a = \sum_{b \in \mathbb{I}} \varepsilon_a\varepsilon_b$ with $\varepsilon_a\varepsilon_b \in A_{a+b}$. - By the uniqueness of the decomposition into homogeneous components, $\varepsilon_a \varepsilon_0 = \varepsilon_a$ and $b \neq 0 \implies \varepsilon_a \varepsilon_b = 0$. - Applying the last equation with $a = 0$ gives $b\neq 0 \implies \varepsilon_b = \varepsilon_0 \varepsilon _b = 0$. + If $1 = \sum_{d \in \mathbb{I}} \varepsilon_d$ is the + decomposition into homogeneous components, then $\varepsilon_a = 1 \cdot + \varepsilon_a = \sum_{b \in \mathbb{I}} \varepsilon_a\varepsilon_b$ with + $\varepsilon_a\varepsilon_b \in A_{a+b}$. + By the uniqueness of the decomposition into homogeneous components, + $\varepsilon_a \varepsilon_0 = \varepsilon_a$ and $b \neq 0 \implies + \varepsilon_a \varepsilon_b = 0$. + Applying the last equation with $a = 0$ gives $b\neq 0 \implies \varepsilon_b = + \varepsilon_0 \varepsilon _b = 0$. Thus $1 = \varepsilon_0 \in A_0$. \end{remark} \begin{remark} @@ -47,112 +109,194 @@ Let $\mathfrak{l}$ be any field. % Graded rings and homogeneous ideals (2) -\begin{proposition}\footnote{This holds for both $\Z$-graded and $\N$-graded rings.} +\begin{proposition} + \footnote{This holds for both $\Z$-graded and $\N$-graded rings.} \begin{itemize} - \item A principal ideal generated by a homogeneous element is homogeneous. - \item The operations $\sum, \bigcap, \sqrt{}$ preserve homogeneity. - \item An ideal is homogeneous iff it can be generated by a family of homogeneous elements. + \item + A principal ideal generated by a homogeneous element is homogeneous. + \item + The operations $\sum, \bigcap, \sqrt{}$ preserve homogeneity. + \item + An ideal is homogeneous iff it can be generated by a family of homogeneous + elements. \end{itemize} \end{proposition} \begin{proof} - Most assertions are trivial. We only show that $J$ homogeneous $\implies \sqrt{J} $ homogeneous. - Let $A$ be $\mathbb{I}$-graded, $f \in \sqrt{J} $ and $f = \sum_{d \in \mathbb{I}} f_d$ the decomposition. - To show that all $f_d \in \sqrt{J} $, we use induction on $N_f \coloneqq \# \{d \in \mathbb{I} | f_d \neq 0\}$. - $N_f = 0$ is trivial. Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq 0$. - For $l \in \N$, the $le$-th homogeneous component of $f^l$ is $f_e^l$. Choosing $l$ large enough such that $f^l \in J$ and using the homogeneity of $J$, we find $f_e \in \sqrt{J}$. - As $\sqrt{J} $ is an ideal, $\tilde f \coloneqq f - f_e \in \sqrt{J} $. As $N_{\tilde f} = N_f -1$, the induction assumption may be applied to $\tilde f$ and shows $f_d \in \sqrt{J} $ for $d \neq e$. + Most assertions are trivial. + We only show that $J$ homogeneous $\implies \sqrt{J} $ + homogeneous. + Let $A$ be $\mathbb{I}$-graded, $f \in \sqrt{J} $ and + $f = \sum_{d \in + \mathbb{I}} f_d$ the decomposition. + To show that all $f_d \in \sqrt{J} $, we use induction on $N_f + \coloneqq \# \{d + \in \mathbb{I} | f_d \neq 0\}$. + $N_f = 0$ is trivial. + Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq + 0$. + For $l \in \N$, the $le$-th homogeneous component of $f^l$ is $f_e^l$. + Choosing $l$ large enough such that $f^l \in J$ and using the homogeneity of + $J$, we find $f_e \in \sqrt{J}$. + As $\sqrt{J} $ is an ideal, $\tilde f \coloneqq f - f_e \in + \sqrt{J} $. + As $N_{\tilde f} = N_f -1$, the induction assumption may be + applied to $\tilde + f$ and shows $f_d \in \sqrt{J} $ for $d \neq e$. \end{proof} \begin{fact} - A homogeneous ideal is finitely generated iff it can be generated by finitely many of its homogeneous elements. + A homogeneous ideal is finitely generated iff it can be generated by finitely + many of its homogeneous elements. In particular, this is always the case when $A$ is a Noetherian ring. \end{fact} \subsubsection{The Zariski topology on $\mathbb{P}^n$} \begin{notation} - Recall that for $\alpha \in \N^{n+1}$ $|\alpha| = \sum_{i=0}^{n} \alpha_i$ and $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$. + Recall that for $\alpha \in \N^{n+1}$ $|\alpha| = + \sum_{i=0}^{n} \alpha_i$ and + $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$. \end{notation} \begin{definition}[Homogeneous polynomials] - Let $R$ be any ring and $f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$. - We say that $f$ is \vocab{homogeneous of degree $d$} if $|\alpha| \neq d \implies f_\alpha = 0$ . - We denote the subset of homogeneous polynomials of degree $d$ by $R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n]$. + Let $R$ be any ring and $f = \sum_{\alpha \in \N^{n+1}} + f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$. + We say that $f$ is \vocab{homogeneous of degree $d$} if $|\alpha| \neq d + \implies f_\alpha = 0$ . + We denote the subset of homogeneous polynomials of degree $d$ by + $R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n]$. \end{definition} \begin{remark} This definition gives $R$ the structure of a graded ring. \end{remark} -\begin{definition}[Zariski topology on $\mathbb{P}^n(\mathfrak{k})$]\label{ztoppn} - Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.\footnote{As always, $\mathfrak{k}$ is algebraically closed} - For $f \in A_d = \mathfrak{k}[X_0,\ldots,X_n]_d$, the validity of the equation $f(x_0,\ldots,x_{n}) = 0$ does not depend on the choice of homogeneous coordinates, as - \[ - f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d f(x_0,\ldots,x_n) - \] - Let $\Vp(f) \coloneqq \{x \in \mathbb{P}^n | f(x) = 0\}$. +\begin{definition}[Zariski topology on $\mathbb{P}^n(\mathfrak{k})$] + \label{ztoppn} + Let $A = \mathfrak{k}[X_0,\ldots,X_n]$. + \footnote{As always, $\mathfrak{k}$ is algebraically closed} + For $f \in A_d = \mathfrak{k}[X_0,\ldots,X_n]_d$, the validity of the equation + $f(x_0,\ldots,x_{n}) = 0$ does not depend on the choice of homogeneous + coordinates, as + \[ + f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d + f(x_0,\ldots,x_n) + \] + Let $\Vp(f) \coloneqq \{x \in \mathbb{P}^n | f(x) + = 0\}$. - We call a subset $X \subseteq \mathbb{P}^n$ Zariski-closed if it can be represented as + We call a subset $X \subseteq \mathbb{P}^n$ Zariski-closed if it + can be + represented as \[ X = \bigcap_{i=1}^k \Vp(f_i) - \] - where the $f_i \in A_{d_i}$ are homogeneous polynomials. + \] + where the $f_i \in A_{d_i}$ + are homogeneous polynomials. \end{definition} \pagebreak \begin{fact} - If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq \mathbb{P}^n$ is closed, then $Y = X \cap \mathbb{A}^n$ can be identified with the closed subset + If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq + \mathbb{P}^n$ is closed, then $Y + = X \cap \mathbb{A}^n$ can be identified with the closed subset \[ - \{(x_1,\ldots,x_n) \in \mathfrak{k}^n | f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\} \subseteq \mathfrak{k}^n + \{(x_1,\ldots,x_n) \in \mathfrak{k}^n | f_i(1,x_1,\ldots,x_n) = 0, 1 + \le i \le + k\} \subseteq \mathfrak{k}^n \] - Conversely, if $Y \subseteq \mathfrak{k}^n$ is closed it has the form + Conversely, if $Y \subseteq \mathfrak{k}^n$ is + closed it has the form \[ - \{(x_1,\ldots,x_n) \in \mathfrak{k}^n | g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\} - \] - and can thus be identified with $X \cap \mathbb{A}^n$ where $X \coloneqq \bigcap_{i=1}^k \Vp(f_i)$ is given by \[f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i \ge \deg(g_i)\] - Thus, the Zariski topology on $\mathfrak{k}^n$ can be identified with the topology induced by the Zariski topology on $\mathbb{A}^n = U_0$, and the same holds for $U_i$ with $0 \le i \le n$. + \{(x_1,\ldots,x_n) \in \mathfrak{k}^n | + g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\} + \] + and can thus be identified with $X + \cap \mathbb{A}^n$ where $X \coloneqq \bigcap_{i=1}^k + \Vp(f_i)$ is given by + \[ + f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i + \ge \deg(g_i) + \] + Thus, the Zariski topology on $\mathfrak{k}^n$ can be + identified with the topology induced by the Zariski topology on + $\mathbb{A}^n = + U_0$, and the same holds for $U_i$ with $0 \le i \le n$. - In this sense, the Zariski topology on $\mathbb{P}^n$ can be thought of as gluing the Zariski topologies on the $U_i \cong \mathfrak{k}^n$. + In this sense, the Zariski topology on $\mathbb{P}^n$ can be + thought of as + gluing the Zariski topologies on the $U_i \cong \mathfrak{k}^n$. \end{fact} % The Zariski topology on P^n (2) \begin{definition} Let $I \subseteq A = \mathfrak{k}[X_0,\ldots,X_n]$ be a homogeneous ideal. - Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \mathbb{P}^n | \forall f \in I ~ f(x_0,\ldots,x_n) = 0\}$ - As $I$ is homogeneous, it is sufficient to impose this condition for the homogeneous elements $f \in I$. - Because $A$ is Noetherian, $I$ can finitely generated by homogeneous elements $(f_i)_{i=1}^k$ and $\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as in \ref{ztoppn}. - Conversely, if the homogeneous $f_i$ are given, then $I = \langle f_1,\ldots,f_k \rangle_A$ is homogeneous. + Let $\Vp(I) \coloneqq + \{[x_0,\ldots,_n] \in \mathbb{P}^n | + \forall f \in I ~ + f(x_0,\ldots,x_n) = 0\}$ As $I$ is homogeneous, it is sufficient to impose this + condition for the homogeneous elements $f \in I$. + Because $A$ is Noetherian, $I$ can finitely generated by homogeneous elements + $(f_i)_{i=1}^k$ and + $\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as + in + \ref{ztoppn}. + Conversely, if the homogeneous $f_i$ are given, then $I = \langle + f_1,\ldots,f_k \rangle_A$ is homogeneous. \end{definition} \begin{remark} Note that $V(A) = V(A_+) = \emptyset$. \end{remark} \begin{fact} For homogeneous ideals in $A$ and $m \in \N$, we have: - \begin{itemize} - \item $\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Vp(I_\lambda)$ - \item $\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) = \bigcup_{k=1}^m \Vp(I_k)$ - \item $\Vp(\sqrt{I}) = \Vp(I)$ + \begin{itemize} + \item + $\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} + \Vp(I_\lambda)$ + \item + $\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) = + \bigcup_{k=1}^m \Vp(I_k)$ + \item + $\Vp(\sqrt{I}) = \Vp(I)$ \end{itemize} \end{fact} \begin{fact} - If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an open covering of a topological space then $X$ is Noetherian iff there is a finite subcovering and all $U_\lambda$ are Noetherian. + If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an + open covering of a topological space then $X$ is Noetherian iff there is a + finite subcovering and all $U_\lambda$ are Noetherian. \end{fact} \begin{proof} - By definition, a topological space is Noetherian $\iff$ all open subsets are quasi-compact. + By definition, a topological space is Noetherian $\iff$ all open subsets are + quasi-compact. \end{proof} \begin{corollary} The Zariski topology on $\mathbb{P}^n$ is indeed a topology. - The induced topology on the open set $\mathbb{A}^n = \mathbb{P}^n \setminus \Vp(X_0) \cong \mathfrak{k}^n$ is the Zariski topology on $\mathfrak{k}^n$. - The same holds for all $U_i = \mathbb{P}^n \setminus \Vp(X_i) \cong \mathfrak{k}^n$. + The induced topology on the open set $\mathbb{A}^n = + \mathbb{P}^n \setminus + \Vp(X_0) \cong \mathfrak{k}^n$ is the Zariski + topology on $\mathfrak{k}^n$. + The same holds for all $U_i = \mathbb{P}^n \setminus + \Vp(X_i) \cong + \mathfrak{k}^n$. Moreover, the topological space $\mathbb{P}^n$ is Noetherian. \end{corollary} \subsection{Noetherianness of graded rings} \begin{proposition} - For a graded ring $R_{\bullet}$, the following conditions are equivalent: + For a graded ring $R_{\bullet}$, the following conditions + are equivalent: \begin{enumerate}[A] - \item $R$ is Noetherian. - \item Every homogeneous ideal of $R_{\bullet}$ is finitely generated. - \item Every chain $I_0\subseteq I_1 \subseteq \ldots$ of homogeneous ideals terminates. - \item Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals has a $\subseteq$-maximal element. - \item $R_0$ is Noetherian and the ideal $R_+$ is finitely generated. - \item $R_0$ is Noetherian and $R / R_0$ is of finite type. + \item + $R$ is Noetherian. + \item + Every homogeneous ideal of $R_{\bullet}$ is finitely + generated. + \item + Every chain $I_0\subseteq I_1 \subseteq \ldots$ of homogeneous ideals + terminates. + \item + Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals has a + $\subseteq$-maximal element. + \item + $R_0$ is Noetherian and the ideal $R_+$ is finitely generated. + \item + $R_0$ is Noetherian and $R / R_0$ is of finite type. \end{enumerate} \end{proposition} \begin{proof} @@ -160,122 +304,234 @@ Let $\mathfrak{l}$ be any field. \noindent\textbf{B $\iff$ C $\iff$ D} similar to the proof about Noetherianness. - \noindent\textbf{B $\land$ C $\implies $E} B implies that $R_+$ is finitely generated. Since $I \oplus R_+$ is homogeneous for any homogeneous ideal $I \subseteq R_0$, C implies the Noetherianness of $R_0$. + \noindent\textbf{B $\land$ C $\implies $E} + B implies that $R_+$ is finitely generated. + Since $I \oplus R_+$ is homogeneous for any homogeneous ideal $I \subseteq + R_0$, C implies the Noetherianness of $R_0$. -\noindent\textbf{E $\implies$ F} Let $R_+$ be generated by $f_i \in R_{d_i}, d_i > 0$ as an ideal. - \begin{claim} - The $R_0$-subalgebra $\tilde R$ of $R$ generated by the $f_i$ equals $R$. - \end{claim} - \begin{subproof} - It is sufficient to show that every homogeneous $f \in R_d$ belongs to $\tilde R$. We use induction on $d$. The case of $d = 0$ is trivial. - Let $d > 0$ and $R_e \subseteq \tilde R$ for all $e < d$. - as $f \in R_+$, $f = \sum_{i=1}^{k} g_if_i$. Let $f_a = \sum_{i=1}^{k} g_{i, a-d_i} f_i$, where $g_i = \sum_{b=0}^{\infty} g_{i,b}$ is the decomposition into homogeneous components. - Then $f = \sum_{a=0}^{\infty} f_a$ is the decomposition of $f$ into homogeneous components, hence $a \neq d \implies f_a = 0 $. Thus we may assume $g_i \in R_{d-d_i}$. - As $d_i > 0$, the induction assumption may now be applied to $g_i$, hence $g_i \in \tilde R$, hence $f \in \tilde R$. - \end{subproof} + \noindent\textbf{E $\implies$ F} + Let $R_+$ be generated by $f_i \in R_{d_i}, d_i > 0$ as + an ideal. + \begin{claim} + The $R_0$-subalgebra $\tilde R$ of $R$ generated by the $f_i$ equals $R$. + \end{claim} + \begin{subproof} + It is sufficient to show that every homogeneous $f \in R_d$ belongs to $\tilde + R$. + We use induction on $d$. + The case of $d = 0$ is trivial. + Let $d > 0$ and $R_e \subseteq \tilde R$ for all $e < d$. + as $f \in R_+$, $f = \sum_{i=1}^{k} g_if_i$. + Let $f_a = \sum_{i=1}^{k} g_{i, a-d_i} f_i$, where + $g_i = \sum_{b=0}^{\infty} + g_{i,b}$ is the decomposition into homogeneous + components. + Then $f = \sum_{a=0}^{\infty} f_a$ is the decomposition of $f$ into + homogeneous + components, hence $a \neq d \implies f_a = 0 $. + Thus we may assume $g_i \in R_{d-d_i}$. + As $d_i > 0$, the induction assumption may now be applied to $g_i$, hence $g_i + \in \tilde R$, hence $f \in \tilde R$. + \end{subproof} - \noindent\textbf{F $\implies$ A} Hilbert's Basissatz (\ref{basissatz}) + \noindent\textbf{F $\implies$ A} + Hilbert's Basissatz ( + \ref{basissatz}) \end{proof} + +\subsection{The projective form of the Nullstellensatz and the closed subsets + of $\mathbb{P}^n$} Let $A = \mathfrak{k}[X_0,\ldots,X_n]$. % Lecture 12 - -\subsection{The projective form of the Nullstellensatz and the closed subsets of $\mathbb{P}^n$} -Let $A = \mathfrak{k}[X_0,\ldots,X_n]$. -\begin{proposition}[Projective form of the Nullstellensatz]\label{hnsp} - If $I \subseteq A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, then $\Vp(I) \subseteq \Vp(f) \iff f \in \sqrt{I}$. +\begin{proposition}[Projective form of the Nullstellensatz] + \label{hnsp} + If $I \subseteq A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, then + $\Vp(I) \subseteq \Vp(f) \iff f \in + \sqrt{I}$. \end{proposition} \begin{proof} - $\impliedby$ is clear. Let $\Vp(I) \subseteq \Vp(f)$. If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in which case $f(x) = 0$ since $d > 0$ - or the point $[x_0,\ldots,x_n] \in \mathbb{P}^n$ is well-defined and belongs to $\Vp(I) \subseteq \Vp(f)$, hence $f(x) = 0$. - Thus $\Va(I) \subseteq \Va(f)$ and $f \in \sqrt{I}$ be the Nullstellensatz (\ref{hns3}). + $\impliedby$ is clear. + Let $\Vp(I) \subseteq \Vp(f)$. + If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in + which case $f(x) = + 0$ since $d > 0$ or the point $[x_0,\ldots,x_n] \in + \mathbb{P}^n$ is + well-defined and belongs to $\Vp(I) \subseteq + \Vp(f)$, hence $f(x) = 0$. + Thus $\Va(I) \subseteq \Va(f)$ and $f \in + \sqrt{I}$ be the Nullstellensatz + ( + \ref{hns3}). \end{proof} -\begin{definition}\footnote{This definition is not too important, the characterization in the following remark suffices.}. - For a graded ring $R_\bullet$, let $\Proj(R_\bullet)$ be the set of $\fp \in \Spec R$ such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$. +\begin{definition} + \footnote{This definition is not too important, the characterization in the following remark suffices.}. + For a graded ring $R_\bullet$, let $\Proj(R_\bullet)$ be the set of + $\fp \in + \Spec R$ such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$. \end{definition} -\begin{remark}\label{proja} - As the elements of $A_0 \setminus \{0\}$ are units in $A$ it follows that for every homogeneous ideal $I$ we have $I \subseteq A_+$ or $I = A$. - In particular, $\Proj(A_\bullet) = \{\fp \in \Spec A \setminus A_+ | \fp \text{ is homogeneous}\} $. +\begin{remark} + \label{proja} + As the elements of $A_0 \setminus \{0\}$ are units in $A$ it follows that for + every homogeneous ideal $I$ we have $I \subseteq A_+$ or $I = A$. + In particular, $\Proj(A_\bullet) = \{\fp \in \Spec A \setminus A_+ | + \fp + \text{ is homogeneous}\} $. \end{remark} -\begin{proposition}\label{bijproj} - There is a bijection +\begin{proposition} + \label{bijproj} + There is a bijection \begin{align} - f: \{I \subseteq A_+ | I \text{ homogeneous ideal}, I = \sqrt{I}\} &\longrightarrow \{X \subseteq \mathbb{P}^n | X \text{ closed}\} \\ - I &\longmapsto \Vp(I)\\ - \langle \{f \in A_d | d > 0, X \subseteq \Vp(f)\} \rangle & \longmapsfrom X + f: \{I \subseteq A_+ | I \text{ homogeneous + ideal}, I = \sqrt{I}\} & \longrightarrow \{X \subseteq \mathbb{P}^n | X \text{ + closed}\} \\ I & \longmapsto \Vp(I)\\ \langle \{f \in A_d | d > 0, X + \subseteq \Vp(f)\} \rangle & \longmapsfrom X \end{align} - Under this bijection, the irreducible subsets correspond to the elements of $\Proj(A_\bullet)$. + Under this bijection, + the irreducible subsets correspond to the elements of + $\Proj(A_\bullet)$. \end{proposition} \begin{proof} - From the projective form of the Nullstellensatz it follows that $f$ is injective and that $f^{-1}(\Vp\left( I \right)) = \sqrt{I} = I$. - If $X \subseteq \mathbb{P}^n$ is closed, then $X = \Vp(J)$ for some homogeneous ideal $J \subseteq A$. Without loss of generality loss of generality $J = \sqrt{J}$. If $J \not\subseteq A_+$, then $J = A$ (\ref{proja}), hence $X = \Vp(J) = \emptyset = \Vp(A_+)$. + From the projective form of the Nullstellensatz it follows that $f$ is + injective and that $f^{-1}(\Vp\left( I \right)) + = \sqrt{I} = I$. + If $X \subseteq \mathbb{P}^n$ is closed, then $X = + \Vp(J)$ for some homogeneous + ideal $J \subseteq A$. + Without loss of generality loss of generality $J = \sqrt{J}$. + If $J \not\subseteq A_+$, then $J = A$ ( + \ref{proja}), hence $X = + \Vp(J) = + \emptyset = \Vp(A_+)$. Thus we may assume $J \subseteq A_+$, and $f$ is surjective. - Suppose $\fp \in \Proj(A_\bullet)$. Then $\fp \neq A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the proven part of the proposition. - Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets, where $X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k = \sqrt{I_k}$. Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \setminus \fp$. - We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq \Vp(I_k)$ hence $\Vp(f_1f_2) \supseteq \Vp(I_1) \cup \Vp(I_2) = X = \Vp(\fp)$ and it follows that $f_1f_2\in \sqrt{\fp} = \fp \lightning$. + Suppose $\fp \in \Proj(A_\bullet)$. + Then $\fp \neq A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the + proven part of + the proposition. + Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets, where + $X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k = + \sqrt{I_k}$. + Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \setminus \fp$. + We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq + \Vp(I_k)$ hence $\Vp(f_1f_2) + \supseteq \Vp(I_1) \cup \Vp(I_2) = X = + \Vp(\fp)$ and it follows that $f_1f_2\in + \sqrt{\fp} = \fp \lightning$. - Assume $X = \Vp(\fp)$ is irreducible, where $\fp = \sqrt{\fp} \in A_+$ is homogeneous. The $\fp \neq A_+$ as $X = \emptyset$ otherwise. Assume that $f_1f_2 \in \fp$ but $f_i \not\in A_{d_i} \setminus \fp$. - Then $X \not \subseteq \Vp(f_i)$ by the projective Nullstellensatz when $d_i > 0$ and because $\Vp(1) = \emptyset$ when $d_i = 0$. - Thus $X = (X \cap \Vp\left( f_1 \right)) \cup (X \cap \Vp(f_2))$ is a proper decomposition $\lightning$. - By lemma \ref{homprime}, $\fp$ is a prime ideal. + Assume $X = \Vp(\fp)$ is irreducible, where $\fp = + \sqrt{\fp} \in A_+$ is + homogeneous. + The $\fp \neq A_+$ as $X = \emptyset$ otherwise. + Assume that $f_1f_2 \in \fp$ but $f_i \not\in A_{d_i} + \setminus \fp$. + Then $X \not \subseteq \Vp(f_i)$ by the projective + Nullstellensatz when $d_i > + 0$ and because $\Vp(1) = \emptyset$ when $d_i = 0$. + Thus $X = (X \cap \Vp\left( f_1 \right)) \cup (X \cap \Vp(f_2))$ is a + proper + decomposition $\lightning$. + By lemma + \ref{homprime}, $\fp$ is a prime ideal. \end{proof} \begin{remark} - It is important that $I \subseteq A_{\color{red} +}$, since $\Vp(A) = \Vp(A_+) = \emptyset$ would be a counterexample. + It is important that $I \subseteq A_{\color{red} +}$, since + $\Vp(A) = \Vp(A_+) + = \emptyset$ would be a counterexample. \end{remark} \begin{corollary} $\mathbb{P}^n$ is irreducible. \end{corollary} \begin{proof} - Apply \ref{bijproj} to $\{0\} \in \Proj(A_\bullet)$. + Apply + \ref{bijproj} to $\{0\} \in \Proj(A_\bullet)$. \end{proof} \subsection{Some remarks on homogeneous prime ideals} -\begin{lemma}\label{homprime} - Let $R_\bullet$ be an $\mathbb{I}$ graded ring ($\mathbb{I} = \N$ or $\mathbb{I} = \Z$). - A homogeneous ideal $I \subseteq R$ is a prime ideal iff $1 \not\in I$ and for homogeneous elements $f, g \in R , fg \in I \implies f \in I \lor g \in I$. +\begin{lemma} + \label{homprime} + Let $R_\bullet$ be an $\mathbb{I}$ graded ring + ($\mathbb{I} = \N$ or + $\mathbb{I} = \Z$). + A homogeneous ideal $I \subseteq R$ is a prime ideal iff $1 \not\in I$ and for + homogeneous elements $f, g \in R , fg \in I \implies f \in I \lor g \in I$. \end{lemma} \begin{proof} $\implies$ is trivial. - It suffices to show that for arbitrary $f,g \in R fg \in I \implies f \in I \lor g \in I$. - Let $f = \sum_{d \in \mathbb{I}} f_d, g = \sum_{d \in \mathbb{I}} g_d $ be the decompositions into homogeneous components. - If $f \not\in I$ and $g \not\in I$ there are $d,e \in I$ with $f_d \in I, g_e \in I$, and they may assumed to be maximal with this property. - As $I$ is homogeneous and $fg \in I$, we have $(fg)_{d+e} \in I$ but + It suffices to show that for arbitrary $f,g \in R fg \in I \implies f \in I + \lor g \in I$. + Let $f = \sum_{d \in \mathbb{I}} f_d, g = \sum_{d \in \mathbb{I}} g_d $ be the + decompositions into homogeneous components. + If $f \not\in I$ and $g \not\in I$ there are $d,e \in I$ with $f_d \in I, g_e + \in I$, and they may assumed to be maximal with this property. + As $I$ is homogeneous and $fg \in I$, we have + $(fg)_{d+e} \in I$ but \[ - (fg)_{d+e} = f_dg_e + \sum_{\delta = 1}^{\infty} (f_{d + \delta} g_{e - \delta} + f_{d - \delta} g_{e + \delta}) - \] - where $f_dg_e \not\in I$ by our assumption on $I$ and all other summands on the right hand side are $\in I$ (as $f_{d+ \delta} \in I$ and $g_{e + \delta} \in I$ by the maximality of $d$ and $e$), a contradiction. + (fg)_{d+e} = f_dg_e + \sum_{\delta = 1}^{\infty} (f_{d + \delta} g_{e - \delta} + + f_{d - \delta} g_{e + \delta}) + \] + where $f_dg_e \not\in I$ by our assumption + on $I$ and all other summands on the right hand side are $\in I$ (as + $f_{d+ + \delta} \in I$ and $g_{e + \delta} \in + I$ by the maximality of $d$ and $e$), a + contradiction. \end{proof} \begin{remark} - If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, then $\fp \oplus R_+ = \{r \in R | r_0 \in \fp\} $ is a homogeneous prime ideal of $R$. - \[\{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } R_\bullet\} = \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec R_0\}\] + If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, then $\fp \oplus R_+ = + \{r \in R | r_0 \in \fp\} $ is a homogeneous prime ideal of $R$. + \[ + \{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } + R_\bullet\} = \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec + R_0\} + \] \end{remark} \subsection{Dimension of $\mathbb{P}^n$} \begin{proposition} \begin{itemize} - \item $\mathbb{P}^n$ is catenary. - \item $\dim(\mathbb{P}^n) = n$. Moreover, $\codim(\{x\} ,\mathbb{P}^n) = n$ for every $x \in \mathbb{P}^n$. - \item If $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$, then $\codim(\{x\}, X) = \dim(X) = n - \codim(X, \mathbb{P}^n)$. - \item If $X \subseteq Y \subseteq \mathbb{P}^n$ are irreducible subsets, then $\codim(X,Y) = \dim(Y) - \dim(X)$. + \item + $\mathbb{P}^n$ is catenary. + \item + $\dim(\mathbb{P}^n) = n$. + Moreover, $\codim(\{x\} ,\mathbb{P}^n) = n$ for every $x \in + \mathbb{P}^n$. + \item + If $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$, then + $\codim(\{x\}, X) = \dim(X) = n - + \codim(X, \mathbb{P}^n)$. + \item + If $X \subseteq Y \subseteq \mathbb{P}^n$ are irreducible subsets, + then $\codim(X,Y) = \dim(Y) - + \dim(X)$. \end{itemize} \end{proposition} \begin{proof} - Let $X \subseteq \mathbb{P}^n$ be irreducible. If $x \in X$, there is an integer $0 \le i \le n$ and $X \in U_i = \mathbb{P}^n \setminus \Vp(X_i)$. - Without loss of generality loss of generality $i = 0$. Then $\codim(X, \mathbb{P}^n) = \codim(X \cap \mathbb{A}^n, \mathbb{A}^n)$ by the locality of Krull codimension (\ref{lockrullcodim}). - Applying this with $X = \{x\}$ and our results about the affine case gives the second assertion. - If $Y$ and $Z$ are also irreducible with $X \subseteq Y \subseteq Z$, then $\codim(X,Y) = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)$, $\codim(X,Z) = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$ and $\codim(Y,Z) = \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$. + Let $X \subseteq \mathbb{P}^n$ be irreducible. + If $x \in X$, there is an integer $0 \le i \le n$ and $X \in U_i = + \mathbb{P}^n \setminus \Vp(X_i)$. + Without loss of generality loss of generality $i = 0$. + Then $\codim(X, \mathbb{P}^n) = \codim(X \cap \mathbb{A}^n, \mathbb{A}^n)$ by + the locality of Krull codimension ( + \ref{lockrullcodim}). + Applying this with $X = \{x\}$ and our results about the affine case gives the + second assertion. + If $Y$ and $Z$ are also irreducible with $X \subseteq Y \subseteq Z$, then + $\codim(X,Y) = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)$, $\codim(X,Z) + = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$ and $\codim(Y,Z) = + \codim(Y + \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$. Thus \begin{align} - \codim(X,Y) + \codim(Y,Z) &= \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n) + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)\\ - &= \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)\\ - &= \codim(X, Z) + \codim(X,Y) + \codim(Y,Z) & = \codim(X \cap \mathbb{A}^n, Y + \cap \mathbb{A}^n) + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\ & = + \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\ & = \codim(X, Z) \end{align} because $\mathfrak{k}^n$ is catenary and the first point follows. The remaining assertions can easily be derived from the first two. @@ -283,74 +539,142 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$. \subsection{The cone $C(X)$} \begin{definition} - If $X \subseteq \mathbb{P}^n$ is closed, we define the \vocab{affine cone over $X$} + If $X \subseteq \mathbb{P}^n$ is closed, we define the + \vocab{affine cone over + $X$} \[ - C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\} | [x_0,\ldots,x_n] \in X\} - \] - If $X = \Vp(I)$ where $I \subseteq A_+ = \mathfrak{k}[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) = \Va(I)$. + C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus + \{0\} | [x_0,\ldots,x_n] \in X\} + \] + If $X = \Vp(I)$ where $I \subseteq A_+ = + \mathfrak{k}[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) = + \Va(I)$. \end{definition} -\begin{proposition}\label{conedim} +\begin{proposition} + \label{conedim} \begin{itemize} - \item $C(X)$ is irreducible iff $X$ is irreducible or $X = \emptyset$. - \item If $X$ is irreducible, then + \item + $C(X)$ is irreducible iff $X$ is irreducible or $X = \emptyset$. + \item + If $X$ is irreducible, then - $\dim(C(X)) = \dim(X) + 1$ and - - $\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \mathbb{P}^n)$ + $\dim(C(X)) = \dim(X) + 1$ and + + $\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \mathbb{P}^n)$ \end{itemize} \end{proposition} \begin{proof} - The first assertion follows from \ref{bijproj} and \ref{bijiredprim} (bijection of irreducible subsets and prime ideals in the projective and affine case). + The first assertion follows from + \ref{bijproj} and + \ref{bijiredprim} (bijection + of irreducible subsets and prime ideals in the projective and affine case). Let $d = \dim(X)$ and \[ - X_0 \subsetneq \ldots \subsetneq X_d = X \subsetneq X_{d+1} \subsetneq \ldots \subsetneq X_n = \mathbb{P}^n - \] - be a chain of irreducible subsets of $\mathbb{P}^n$. Then + X_0 \subsetneq \ldots \subsetneq X_d = X \subsetneq + X_{d+1} \subsetneq \ldots \subsetneq X_n = + \mathbb{P}^n + \] + be a chain of + irreducible subsets of $\mathbb{P}^n$. + Then \[ - \{0\} \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X) \subsetneq \ldots \subsetneq C(X_n) = \mathfrak{k}^{n+1} - \] - is a chain of irreducible subsets of $\mathfrak{k}^{n+1}$. Hence $\dim(C(X)) \ge 1 + d$ and $\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$. Since $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) = n+1$, the two inequalities must be equalities. + \{0\} \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X) + \subsetneq \ldots \subsetneq C(X_n) = \mathfrak{k}^{n+1} + \] + is a chain of + irreducible subsets of $\mathfrak{k}^{n+1}$. + Hence $\dim(C(X)) \ge 1 + d$ and $\codim(C(X), \mathfrak{k}^{n+1}) \ge + n-d$. + Since $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = + \dim(\mathfrak{k}^{n+1}) + = n+1$, the two inequalities must be equalities. \end{proof} \subsubsection{Application to hypersurfaces in $\mathbb{P}^n$} \begin{definition}[Hypersurface] Let $n > 0$. - By a \vocab{hypersurface} in $\mathbb{P}^n$ or $\mathbb{A}^n$ we understand an irreducible closed subset of codimension $1$. + By a \vocab{hypersurface} in $\mathbb{P}^n$ or + $\mathbb{A}^n$ we understand an + irreducible closed subset of codimension $1$. \end{definition} \begin{corollary} - If $P \in A_d$ is a prime element, then $H = \Vp(P)$ is a hypersurface in $\mathbb{P}^n$ and every hypersurface $H$ in $\mathbb{P}^n$ can be obtained in this way. + If $P \in A_d$ is a prime element, then $H = \Vp(P)$ is a + hypersurface in + $\mathbb{P}^n$ and every hypersurface $H$ in + $\mathbb{P}^n$ can be obtained in + this way. \end{corollary} \begin{proof} - If $H = \Vp(P)$ then $C(H) = \Va(P)$ is a hypersurface in $\mathfrak{k}^{n+1}$ by \ref{irredcodimone}. By \ref{conedim}, $H$ is irreducible and of codimension $1$. + If $H = \Vp(P)$ then $C(H) = \Va(P)$ is a + hypersurface in $\mathfrak{k}^{n+1}$ + by + \ref{irredcodimone}. + By + \ref{conedim}, $H$ is irreducible and of codimension $1$. - Conversely, let $H$ be a hypersurface in $\mathbb{P}^n$. By \ref{conedim}, $C(H)$ is a hypersurface in $\mathfrak{k}^{n+1}$, hence $C(H) = \Vp(P)$ for some prime element $P \in A$ (again by \ref{irredcodimone}). - We have $H = \Vp(\fp)$ for some $\fp \in \Proj(A)$ and $C(H) = \Va(\fp)$. By the bijection between closed subsets of $\mathfrak{k}^{n+1}$ and ideals $I = \sqrt{I} \subseteq A$ (\ref{antimonbij}), $\fp = P \cdot A$. -Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition into homogeneous components. -If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ contradicting the homogeneity of $\fp = P \cdot A$. Thus, $P$ is homogeneous of degree $d$. + Conversely, let $H$ be a hypersurface in $\mathbb{P}^n$. + By + \ref{conedim}, $C(H)$ is a hypersurface in + $\mathfrak{k}^{n+1}$, hence $C(H) + = \Vp(P)$ for some prime element $P \in A$ (again by + \ref{irredcodimone}). + We have $H = \Vp(\fp)$ for some $\fp \in + \Proj(A)$ and $C(H) = \Va(\fp)$. + By the bijection between closed subsets of $\mathfrak{k}^{n+1}$ and ideals + $I = + \sqrt{I} \subseteq A$ ( + \ref{antimonbij}), $\fp = P \cdot + A$. + Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition + into + homogeneous components. + If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ + contradicting the homogeneity of $\fp = P \cdot A$. + Thus, $P$ is homogeneous of degree $d$. \end{proof} \begin{definition} - A hypersurface $H \subseteq \mathbb{P}^n$ has \vocab{degree $d$} if $H = \Vp(P)$ where $P \in A_d$ is an irreducible polynomial. + A hypersurface $H \subseteq \mathbb{P}^n$ has + \vocab{degree $d$} if $H = + \Vp(P)$ where $P \in A_d$ is an irreducible polynomial. \end{definition} -\subsubsection{Application to intersections in $\mathbb{P}^n$ and Bezout's theorem} +\subsubsection{Application to intersections in $\mathbb{P}^n$ and Bezout's + theorem} \begin{corollary} - Let $A \subseteq \mathbb{P}^n$ and $B \subseteq \mathbb{P}^n$ be irreducible subsets of dimensions $a$ and $b$. If $a+ b \ge n$, then $A \cap B \neq \emptyset$ and every irreducible component of $A \cap B$ as dimension $\ge a + b - n$. + Let $A \subseteq \mathbb{P}^n$ and $B \subseteq + \mathbb{P}^n$ be irreducible + subsets of dimensions $a$ and $b$. + If $a+ b \ge n$, then $A \cap B \neq \emptyset$ and every irreducible component + of $A \cap B$ as dimension $\ge a + b - n$. \end{corollary} \begin{remark} - This shows that $\mathbb{P}^n$ indeed fulfilled the goal of allowing for nicer results of algebraic geometry because ``solutions at infinity'' to systems of algebraic equations are present in $\mathbb{P}^n$ - (see \ref{affineproblem}). + This shows that $\mathbb{P}^n$ indeed fulfilled the goal of + allowing for nicer + results of algebraic geometry because ``solutions at infinity'' to systems of + algebraic equations are present in $\mathbb{P}^n$ (see + \ref{affineproblem}). \end{remark} \begin{proof} - The lower bound on the dimension of irreducible components of $A \cap B$ is easily derived from the similar affine result (corollary of the principal ideal theorem, \ref{codimintersection}). - From the definition of the affine cone it follows that $C(A \cap B) = C(A) \cap C(B)$. - We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by \ref{conedim}. - If $A \cap B = \emptyset$, then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an irreducible component, contradicting the lower bound $a + b + 1 - n > 0$ for the dimension of irreducible components of $C(A) \cap C(B)$ (again \ref{codimintersection}). + The lower bound on the dimension of irreducible components of $A \cap B$ is + easily derived from the similar affine result (corollary of the principal ideal + theorem, + \ref{codimintersection}). + From the definition of the affine cone it follows that $C(A \cap B) = C(A) \cap + C(B)$. + We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by + \ref{conedim}. + If $A \cap B = \emptyset$, then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an + irreducible component, contradicting the lower bound $a + b + 1 - n > 0$ for + the dimension of irreducible components of $C(A) \cap C(B)$ (again + \ref{codimintersection}). \end{proof} \begin{remark}[Bezout's theorem] - If $A \neq B$ are hypersurfaces of degree $a$ and $b$ in $\mathbb{P}^2$, then $A \cap B$ has $ab$ points counted by (suitably defined) multiplicity. + If $A \neq B$ are hypersurfaces of degree $a$ and $b$ + in $\mathbb{P}^2$, then $A \cap B$ has $ab$ points counted by + (suitably defined) multiplicity. \end{remark} diff --git a/inputs/uebersicht.bak0 b/inputs/uebersicht.bak0 new file mode 100644 index 0000000..7128c6b --- /dev/null +++ b/inputs/uebersicht.bak0 @@ -0,0 +1,125 @@ +% TODO REMARK ABOUT ZORNS LEMMA (LECTURE 1) + +% TODO REMARK ABOUT FIN PRESENTED MODULES (LECTURE 2) + + +% TODO: LECTURE 9 LEMMA + + + + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +% ÜBERSICHT % +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + + +% List of forms of HNS + + +\begin{itemize} + \item[HNS2 $\implies$ HNS1b] Let $I \subseteq \mathfrak{l}[X_1,\ldots,X_n]$. $I \subseteq \mathfrak{m}$ maximal. $R / \mathfrak{m}$ is isomorphic to a field extension of $\mathfrak{l}$. Finite by HNS2. + \item[NNT $\implies$ HNS2] Apply NNT to $L / K$ $\leadsto$ alg. independent $a_i$ such that $L$ is finite over the image of $K[X_1,\ldots,X_n] \xrightarrow{\ev_a} L$ ($\ev_a : P \to P(a_1,\ldots,a_n)$). + $\ev_a(K[X_1,\ldots,X_n])$ is a field by fact about integrality and fields \ref{fintaf}. Hence $n = 0$ and $L / K$ is finite. + \item[UNCHNS2] $K$ uncountable, $L / K$ fin. type. Then $\dim_K L$ is countable. Suppose $l \in L$ is not integral. Then $K(l) \cong K(T)$ and $\dim_K L \ge \dim_K K(T) \ge \aleph_1$. + Thus $L / K$ algebraic $\implies$ integral $\implies$ finite. + \item[HNS3] ($V(I) \subseteq V(f) \iff f \in \sqrt{I} $). Suppose $V(I) \subseteq V(f)$. $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n, T], J \subseteq R'$ the ideal generated by $I$ and $g(X_1,\ldots,X_n,T) \coloneqq 1 - Tf(X_1,\ldots,X_n)$. +\end{itemize} + + +% Proofs +Def of integrality (<=>) + + +Fact about integrality and field: + % TODO + + +Technical lemma for Noether normalization: For $S \subseteq \N^n$ finite, there exists $k \in \N^n$ such that $k_1 = 1$ and $s_1 \neq s_2 \in S \implies \langle k, s_1 \rangle \neq \langle k, s_2 \rangle$: +For $s_1 \neq s_2$, % TODO + +Noether normalization: +$a_i \in A$ minimal such that $A$ is integral over the subalgebra genereted by the $a_i$. +Suppose $\exists P \in K[X_1,\ldots,X_n] \setminus \{ 0\} ~ P(a_1,\ldots,a_n) = 0$. $P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in \N^n | p_\alpha \neq 0\}$. +Choose $k$ as in the lemma. +$b_i \coloneqq a_{i+1} - a_1^{k_{i+1}}, 1 \le i \trdeg(\mathfrak{k}(\fq) / \mathfrak{l})$: +Without loss of generality loss of generality $\fp = \{0\}$ and $A$ a domain ($A' \coloneqq A / \fp$). +For $\fq \in \MaxSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence finite (HNS) $\implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$. +$\trdeg(Q(A) / \mathfrak{l}) = 0 \implies A$ integral over $\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$. + +If $\fq \not\in \MaxSpec A$, let $a_1,\ldots,a_n \in A$ alg. independent such that the $\overline{a_i}$ are a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{k}$ +Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. Localize with respect to $S \coloneqq R \setminus \{0\}$. +%TODO +% TODO: LERNEN + + +% Dim k^n +$\dim(\mathfrak{k}^n)$ +$ \ge n$ build chian +$\le n$ a first result in dim T ($\fp \subsetneq \fq \implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})$. Thus $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{l}) - \trdeg(\mathfrak{K}(X) / \mathfrak{l})$. + +TODO +% List of proofs of HNS + + +% Going up + + +% TODO proof of dim Y = trdeg(K(Y) / k) +$\dim Y \ge \trdeg(\mathfrak{k}(Y) / \mathfrak{k})$: Noether normalization. Subalgebra $\cong \mathfrak{k}[X_1,\ldots,X_d]$. Lift chain of prime ideals using going up. + +% TODO prime avoidance + + +Action of $\Aut(L/K)$ on prime ideals of a normal ring extension. $A$ normal domain, $L / Q(A)$ normal field extension, $B$ int closure of $A$ in $L$, $\fp \in \Spec A$. +Then $\Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$ : + +\begin{itemize} + \item $\fq, \fr \in \Spec B$ lying over $\fp$. + \item only need to show $\fq \subseteq \sigma(\fr)$ for some $\sigma \in G$ (Krull going-up, no inclusions) + \item Suppose not. Then $x \in \fq \setminus \bigcup_{\sigma \in G} \sigma(\fr)$ (prime aviodance) + \item $y = \prod_{\sigma \in G} \sigma(x) \in \fq \setminus \fr$ ($\fr$ prime ideal) + \item $\exists k \in \N$ s.t. $y^k \in K$ ($y \in L^G$) + \item $y^k \in K \cap B = A $ ($A$ normal). Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp$. + \item $L / K$ infinite: Apply Zorn to pairs $(M, \sigma)$ where $K \subseteq M \subseteq L$ and $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap M$. +\end{itemize} + + +Going down Krull %TODO + +The ht p and trdeg +================== +% TODO % TODO % TODO % + +% Definitions +Zariski-Topology, Spec, $\mathfrak{k}^n$ +Residue field $\mathfrak{k}(\fp) \coloneqq Q(A / \fp), \mathfrak{K}(V(\fp)) \coloneqq \mathfrak{k}(\fp)$. TODO? +% Counterexamples + no going-up +% list of definitions of codim, dim, trdeg, ht +Original (Noether normalization) +Artin-Tate +Uncountable fields +\begin{landscape} +\section{Übersicht} +{\rowcolors{2}{gray!10}{white} + \begin{longtable}{lll} + \end{longtable} +} + +\end{landscape} + diff --git a/inputs/uebersicht.tex b/inputs/uebersicht.tex index 7128c6b..3058692 100644 --- a/inputs/uebersicht.tex +++ b/inputs/uebersicht.tex @@ -17,37 +17,68 @@ \begin{itemize} - \item[HNS2 $\implies$ HNS1b] Let $I \subseteq \mathfrak{l}[X_1,\ldots,X_n]$. $I \subseteq \mathfrak{m}$ maximal. $R / \mathfrak{m}$ is isomorphic to a field extension of $\mathfrak{l}$. Finite by HNS2. - \item[NNT $\implies$ HNS2] Apply NNT to $L / K$ $\leadsto$ alg. independent $a_i$ such that $L$ is finite over the image of $K[X_1,\ldots,X_n] \xrightarrow{\ev_a} L$ ($\ev_a : P \to P(a_1,\ldots,a_n)$). - $\ev_a(K[X_1,\ldots,X_n])$ is a field by fact about integrality and fields \ref{fintaf}. Hence $n = 0$ and $L / K$ is finite. - \item[UNCHNS2] $K$ uncountable, $L / K$ fin. type. Then $\dim_K L$ is countable. Suppose $l \in L$ is not integral. Then $K(l) \cong K(T)$ and $\dim_K L \ge \dim_K K(T) \ge \aleph_1$. + \item[HNS2 $\implies$ HNS1b] + Let $I \subseteq \mathfrak{l}[X_1,\ldots,X_n]$. $I \subseteq \mathfrak{m}$ + maximal. $R / \mathfrak{m}$ is isomorphic to a field extension of + $\mathfrak{l}$. + Finite by HNS2. + \item[NNT $\implies$ HNS2] + Apply NNT to $L / K$ $\leadsto$ alg. independent $a_i$ such that $L$ is finite + over the image of $K[X_1,\ldots,X_n] \xrightarrow{\ev_a} + L$ ($\ev_a : P \to P(a_1,\ldots,a_n)$). + $\ev_a(K[X_1,\ldots,X_n])$ is a field by fact about integrality and fields + \ref{fintaf}. + Hence $n = 0$ and $L / K$ is finite. + \item[UNCHNS2] + $K$ uncountable, $L / K$ fin. type. + Then $\dim_K L$ is countable. + Suppose $l \in L$ is not integral. + Then $K(l) \cong K(T)$ and $\dim_K L \ge \dim_K K(T) \ge \aleph_1$. Thus $L / K$ algebraic $\implies$ integral $\implies$ finite. - \item[HNS3] ($V(I) \subseteq V(f) \iff f \in \sqrt{I} $). Suppose $V(I) \subseteq V(f)$. $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n, T], J \subseteq R'$ the ideal generated by $I$ and $g(X_1,\ldots,X_n,T) \coloneqq 1 - Tf(X_1,\ldots,X_n)$. + \item[HNS3] + ($V(I) \subseteq V(f) \iff f \in \sqrt{I} $). + Suppose $V(I) \subseteq V(f)$. + $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n, T], J \subseteq R'$ the ideal generated by $I$ + and $g(X_1,\ldots,X_n,T) \coloneqq 1 - Tf(X_1,\ldots,X_n)$. \end{itemize} % Proofs -Def of integrality (<=>) +Def of integrality (<=>) -Fact about integrality and field: - % TODO +Fact about integrality and field: -Technical lemma for Noether normalization: For $S \subseteq \N^n$ finite, there exists $k \in \N^n$ such that $k_1 = 1$ and $s_1 \neq s_2 \in S \implies \langle k, s_1 \rangle \neq \langle k, s_2 \rangle$: -For $s_1 \neq s_2$, % TODO +Technical lemma for Noether normalization: For $S \subseteq \N^n$ finite, there +exists $k \in \N^n$ such that $k_1 = 1$ and $s_1 \neq s_2 \in S \implies +\langle k, s_1 \rangle \neq \langle k, s_2 \rangle$: For $s_1 \neq s_2$, -Noether normalization: -$a_i \in A$ minimal such that $A$ is integral over the subalgebra genereted by the $a_i$. -Suppose $\exists P \in K[X_1,\ldots,X_n] \setminus \{ 0\} ~ P(a_1,\ldots,a_n) = 0$. $P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in \N^n | p_\alpha \neq 0\}$. +Noether normalization: $a_i \in A$ minimal such that $A$ is integral over the +subalgebra genereted by the $a_i$. +% TODO% TODO +Suppose $\exists P \in K[X_1,\ldots,X_n] \setminus \{ 0\} ~ +P(a_1,\ldots,a_n) = +0$. +$P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in \N^n +| p_\alpha \neq 0\}$. Choose $k$ as in the lemma. -$b_i \coloneqq a_{i+1} - a_1^{k_{i+1}}, 1 \le i \trdeg(\mathfrak{k}(\fq) / \mathfrak{l})$: -Without loss of generality loss of generality $\fp = \{0\}$ and $A$ a domain ($A' \coloneqq A / \fp$). -For $\fq \in \MaxSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence finite (HNS) $\implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$. -$\trdeg(Q(A) / \mathfrak{l}) = 0 \implies A$ integral over $\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$. +A first result of dimension theory: $A \mathfrak{l}$-algebra of +finite type, +$\fp, \fq \in \Spec A, \fp \subsetneq \fq$. +Then $\trdeg(\mathfrak{k}(\fp) /\mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) / +\mathfrak{l})$: Without loss of +generality loss of generality $\fp = \{0\}$ +and $A$ a domain ($A' \coloneqq A / \fp$). +For $\fq \in \MaxSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence +finite (HNS) $\implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$. +$\trdeg(Q(A) / \mathfrak{l}) = 0 \implies A$ integral over +$\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq +\lightning$. -If $\fq \not\in \MaxSpec A$, let $a_1,\ldots,a_n \in A$ alg. independent such that the $\overline{a_i}$ are a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{k}$ -Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. Localize with respect to $S \coloneqq R \setminus \{0\}$. +If $\fq \not\in \MaxSpec A$, let $a_1,\ldots,a_n \in A$ alg. +independent such that the $\overline{a_i}$ are a transcendence base for +$\mathfrak{k}(\fq) / \mathfrak{k}$ +Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. +Localize with respect to $S \coloneqq R \setminus \{0\}$. %TODO % TODO: LERNEN @@ -70,56 +111,77 @@ Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. Localize with res % Dim k^n $\dim(\mathfrak{k}^n)$ $ \ge n$ build chian -$\le n$ a first result in dim T ($\fp \subsetneq \fq \implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})$. Thus $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{l}) - \trdeg(\mathfrak{K}(X) / \mathfrak{l})$. +$\le n$ a first result in dim T ($\fp \subsetneq \fq \implies +\trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})$. +Thus $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{l}) - +\trdeg(\mathfrak{K}(X) / \mathfrak{l})$. -TODO -% List of proofs of HNS +TODO -% Going up -% TODO proof of dim Y = trdeg(K(Y) / k) -$\dim Y \ge \trdeg(\mathfrak{k}(Y) / \mathfrak{k})$: Noether normalization. Subalgebra $\cong \mathfrak{k}[X_1,\ldots,X_d]$. Lift chain of prime ideals using going up. +$\dim Y \ge \trdeg(\mathfrak{k}(Y) / \mathfrak{k})$: Noether normalization. +% List of proofs of HNS% Going up% TODO proof of dim Y = trdeg(K(Y) / k) +Subalgebra $\cong \mathfrak{k}[X_1,\ldots,X_d]$. +Lift chain of prime ideals using going up. % TODO prime avoidance -Action of $\Aut(L/K)$ on prime ideals of a normal ring extension. $A$ normal domain, $L / Q(A)$ normal field extension, $B$ int closure of $A$ in $L$, $\fp \in \Spec A$. -Then $\Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$ : +Action of $\Aut(L/K)$ on prime ideals of a normal ring extension. +$A$ normal domain, $L / Q(A)$ normal field extension, $B$ int closure of $A$ in +$L$, $\fp \in \Spec A$. +Then $\Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq +\cap A = +\fp\}$ : \begin{itemize} - \item $\fq, \fr \in \Spec B$ lying over $\fp$. - \item only need to show $\fq \subseteq \sigma(\fr)$ for some $\sigma \in G$ (Krull going-up, no inclusions) - \item Suppose not. Then $x \in \fq \setminus \bigcup_{\sigma \in G} \sigma(\fr)$ (prime aviodance) - \item $y = \prod_{\sigma \in G} \sigma(x) \in \fq \setminus \fr$ ($\fr$ prime ideal) - \item $\exists k \in \N$ s.t. $y^k \in K$ ($y \in L^G$) - \item $y^k \in K \cap B = A $ ($A$ normal). Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp$. - \item $L / K$ infinite: Apply Zorn to pairs $(M, \sigma)$ where $K \subseteq M \subseteq L$ and $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap M$. + \item + $\fq, \fr \in \Spec B$ lying over $\fp$. + \item + only need to show $\fq \subseteq \sigma(\fr)$ for some $\sigma \in G$ (Krull + going-up, no inclusions) + \item + Suppose not. + Then $x \in \fq \setminus \bigcup_{\sigma \in G} \sigma(\fr)$ + (prime + aviodance) + \item + $y = \prod_{\sigma \in G} \sigma(x) \in \fq \setminus \fr$ + ($\fr$ prime ideal) + \item + $\exists k \in \N$ s.t. + $y^k \in K$ ($y \in L^G$) + \item + $y^k \in K \cap B = A $ ($A$ normal). + Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp$. + \item + $L / K$ infinite: Apply Zorn to pairs $(M, \sigma)$ where $K \subseteq M + \subseteq L$ and $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap + M$. \end{itemize} -Going down Krull %TODO +Going down Krull -The ht p and trdeg -================== -% TODO % TODO % TODO % +The ht p and trdeg ================== -% Definitions -Zariski-Topology, Spec, $\mathfrak{k}^n$ -Residue field $\mathfrak{k}(\fp) \coloneqq Q(A / \fp), \mathfrak{K}(V(\fp)) \coloneqq \mathfrak{k}(\fp)$. TODO? +Zariski-Topology, Spec, $\mathfrak{k}^n$ Residue field +$\mathfrak{k}(\fp) +\coloneqq Q(A / \fp), \mathfrak{K}(V(\fp)) \coloneqq \mathfrak{k}(\fp)$. +%TODO% TODO % TODO % TODO %% Definitions +TODO? % Counterexamples - no going-up +no going-up % list of definitions of codim, dim, trdeg, ht -Original (Noether normalization) -Artin-Tate -Uncountable fields +Original (Noether normalization) Artin-Tate Uncountable fields \begin{landscape} -\section{Übersicht} -{\rowcolors{2}{gray!10}{white} - \begin{longtable}{lll} - \end{longtable} -} + \section{Übersicht} {\rowcolors{2}{gray! + 10}{white} + \begin{longtable}{lll} + \end{longtable} + } \end{landscape} diff --git a/inputs/varieties.tex b/inputs/varieties.tex index 535d310..1d1d05a 100644 --- a/inputs/varieties.tex +++ b/inputs/varieties.tex @@ -3,43 +3,88 @@ \begin{definition}[Sheaf] Let $X$ be any topological space. - A \vocab{presheaf} $\mathcal{G}$ of sets (or rings, (abelian) groups) on $X$ associates a set (or rings, or (abelian) group) $\mathcal{G}(U)$ to every open subset $U$ of $X$, and a map (or ring or group homomorphism) $\mathcal{G}(U) \xrightarrow{r_{U,V}} \mathcal{G}(V)$ to every inclusion $V \subseteq U$ of open subsets of $X$ such that $r_{U,W} = r_{V,W} r_{U,V}$ for inclusions $U \subseteq V \subseteq W$ of open subsets. + A \vocab{presheaf} $\mathcal{G}$ of sets (or rings, + (abelian) groups) on $X$ + associates a set (or rings, or (abelian) group) $\mathcal{G}(U)$ to + every open + subset $U$ of $X$, and a map (or ring or group homomorphism) + $\mathcal{G}(U) + \xrightarrow{r_{U,V}} \mathcal{G}(V)$ to every inclusion $V + \subseteq U$ of + open subsets of $X$ such that $r_{U,W} = + r_{V,W} r_{U,V}$ for inclusions + $U + \subseteq V \subseteq W$ of open subsets. - Elements of $\mathcal{G}(U)$ are often called \vocab{sections} of $\mathcal{G}$ on $U$ or \vocab{global sections} when $U = X$. + Elements of $\mathcal{G}(U)$ are often called \vocab{sections} + of + $\mathcal{G}$ on $U$ or \vocab{global sections} when $U = X$. - Let $U \subseteq X$ be open and $U = \bigcup_{i \in I} U_i$ an open covering. - A family $(f_i)_{i \in I} \in \prod_{i \in I} \mathcal{G}(U_i)$ is called \vocab[Sections!compatible]{compatible} if $r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j)$ for all $i,j \in I$. + Let $U \subseteq X$ be open and $U = \bigcup_{i \in I} U_i$ an open + covering. + A family $(f_i)_{i \in I} \in + \prod_{i \in I} \mathcal{G}(U_i)$ is called + \vocab[Sections! + compatible]{compatible} if $r_{U_i, U_i \cap U_j}(f_i) = + r_{U_j, U_i \cap U_j}(f_j)$ for all $i,j \in I$. - Consider the map + Consider the map \begin{align} - \phi_{U, (U_i)_{i \in I}}: \mathcal{G}(U) &\longrightarrow \{(f_i)_{i \in I} \in \prod_{i \in I} \mathcal{G}(U_i) | r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j) \text{ for } i,j \in I \} \\ - f &\longmapsto (r_{U, U_i}( f))_{i \in I} + \phi_{U, (U_i)_{i \in I}}: \mathcal{G}(U) + & \longrightarrow \{(f_i)_{i \in I} \in \prod_{i \in I} \mathcal{G}(U_i) | + r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j) \text{ for } i,j \in I + \} \\ f & \longmapsto (r_{U, U_i}( f))_{i \in I} \end{align} - A presheaf is called \vocab[Presheaf!separated]{separated} if $\phi_{U, (U_i)_{i \in I}}$ is injective for all such $U$ and $(U_i)_{i \in I}$.\footnote{This also called ``locality''.} - It satisfies \vocab{gluing} if $\phi_{U, (U_i)_{i \in I}}$ is surjective. + A presheaf is called \vocab[Presheaf! + separated]{separated} if $\phi_{U, (U_i)_{i \in I}}$ is + injective for all such $U$ and + $(U_i)_{i \in I}$.\footnote{This also called ``locality''.} + It satisfies \vocab{gluing} if $\phi_{U, (U_i)_{i \in I}}$ is + surjective. - A presheaf is called a \vocab{sheaf} if it is separated and satisfies gluing. + A presheaf is called a \vocab{sheaf} if it is separated and + satisfies gluing. - The bijectivity of the $\phi_{U, (U_i)_{i \in I}}$ is called the \vocab{sheaf axiom}. + The bijectivity of the $\phi_{U, (U_i)_{i \in I}}$ is called the + \vocab{sheaf + axiom}. \end{definition} -\begin{trivial}+ - A presheaf is a contravariant functor $\mathcal{G} : \mathcal{O}(X) \to C$ where $\mathcal{O}(X)$ denotes the category of open subsets of $X$ with inclusions as morphisms and $C$ is the category of sets, rings or (abelian) groups. +\begin{trivial} + + + A presheaf is a contravariant functor $\mathcal{G} : + \mathcal{O}(X) \to C$ where $\mathcal{O}(X)$ denotes the category + of open subsets of $X$ with inclusions as morphisms and $C$ is the category of + sets, rings or (abelian) groups. \end{trivial} \begin{definition} - A subsheaf $\mathcal{G}'$ is defined by subsets (resp. subrings or subgroups) $\mathcal{G}'(U) \subseteq \mathcal{G}(U)$ for all open $U \subseteq X$ such that the sheaf axioms still hold. + A subsheaf $\mathcal{G}'$ is defined by subsets (resp. + subrings or subgroups) $\mathcal{G}'(U) \subseteq + \mathcal{G}(U)$ for all open $U \subseteq X$ such that the sheaf axioms + still hold. \end{definition} \begin{remark} - If $\mathcal{G}$ is a sheaf on $X$ and $\Omega \subseteq X$ open, then $\mathcal{G}\defon{\Omega}(U) \coloneqq \mathcal{G}(U)$ for open $U \subseteq \Omega$ and $r_{U,V}^{(\mathcal{G}\defon{\Omega})}(f) \coloneqq r_{U,V}^{(\mathcal{G})}(f)$ is a sheaf of the same kind as $\mathcal{G}$ on $\Omega$. + If $\mathcal{G}$ is a sheaf on $X$ and $\Omega \subseteq X$ open, + then + $\mathcal{G}\defon{\Omega}(U) \coloneqq \mathcal{G}(U)$ + for open $U \subseteq + \Omega$ and $r_{U,V}^{(\mathcal{G}\defon{\Omega})}(f) \coloneqq + r_{U,V}^{(\mathcal{G})}(f)$ is a sheaf of the same kind as + $\mathcal{G}$ on + $\Omega$. \end{remark} \begin{remark} - The notion of restriction of a sheaf to a closed subset, or of preimages under general continuous maps, can be defined but this is a bit harder. + The notion of restriction of a sheaf to a closed subset, or of preimages under + general continuous maps, can be defined but this is a bit harder. \end{remark} \begin{notation} - It is often convenient to write $f \defon{V}$ instead of $r_{U,V}(f)$. + It is often convenient to write $f \defon{V}$ instead of + $r_{U,V}(f)$. \end{notation} \begin{remark} - Applying the \vocab{sheaf axiom} to the empty covering of $U = \emptyset$, one finds that $\mathcal{G}(\emptyset) = \{0\} $. + Applying the \vocab{sheaf axiom} to the empty covering of $U = \emptyset$, + one + finds that $\mathcal{G}(\emptyset) = \{0\} $. \end{remark} @@ -47,72 +92,137 @@ \subsubsection{Examples of sheaves} \begin{example} - Let $G$ be a set and let $\mathfrak{G}(U)$ be the set of arbitrary maps $U \xrightarrow{f} G$. We put $r_{U,V}(f) = f\defon{V}$. + Let $G$ be a set and let $\mathfrak{G}(U)$ be the set of arbitrary maps + $U + \xrightarrow{f} G$. + We put $r_{U,V}(f) = f\defon{V}$. It is easy to see that this defines a sheaf. - If $\cdot $ is a group operation on $G$, then $(f\cdot g)(x) \coloneqq f(x)\cdot g(x)$ defines the structure of a sheaf of group on $\mathfrak{G}$. - Similarly, a ring structure on $G$ can be used to define the structure of a sheaf of rings on $\mathfrak{G}$. + If $\cdot $ is a group operation on $G$, then $(f\cdot g)(x) \coloneqq + f(x)\cdot g(x)$ defines the structure of a sheaf of group on + $\mathfrak{G}$. + Similarly, a ring structure on $G$ can be used to define the structure of a + sheaf of rings on $\mathfrak{G}$. \end{example} \begin{example} - If in the previous example $G$ carries a topology and $\mathcal{G}(U) \subseteq \mathfrak{G}(U)$ is the subset (subring, subgroup) of continuous functions $U \xrightarrow{f} G$, then $\mathcal{G}$ is a subsheaf of $\mathfrak{G}$, called the sheaf of continuous $G$-valued functions on (open subsets of) $X$. + If in the previous example $G$ carries a topology and $\mathcal{G}(U) + \subseteq + \mathfrak{G}(U)$ is the subset (subring, subgroup) of continuous + functions $U + \xrightarrow{f} G$, then $\mathcal{G}$ is a subsheaf of + $\mathfrak{G}$, called + the sheaf of continuous $G$-valued functions on (open subsets of) $X$. \end{example} \begin{example} - If $X = \R^n$, $\mathbb{K} \in \{\R, \C\}$ and $\mathcal{O}(U)$ is the sheaf of $\mathbb{K}$-valued $C^{\infty}$-functions on $U$, then $\mathcal{O}$ is a subsheaf of the sheaf (of rings) of $\mathbb{K}$-valued continuous functions on $X$. + If $X = \R^n$, $\mathbb{K} \in \{\R, \C\}$ and + $\mathcal{O}(U)$ is the sheaf + of $\mathbb{K}$-valued $C^{\infty}$-functions on + $U$, then $\mathcal{O}$ is a + subsheaf of the sheaf (of rings) of $\mathbb{K}$-valued continuous + functions on + $X$. \end{example} \begin{example} - If $X = \C^n$ and $\mathcal{O}(U)$ the set of holomorphic functions on $X$, then $\mathcal{O}$ is a subsheaf of the sheaf of $\C$-valued $C^{\infty}$-functions on $X$. + If $X = \C^n$ and $\mathcal{O}(U)$ the set of holomorphic functions + on $X$, + then $\mathcal{O}$ is a subsheaf of the sheaf of $\C$-valued + $C^{\infty}$-functions on $X$. \end{example} \subsubsection{The structure sheaf on a closed subset of $\mathfrak{k}^n$} -Let $X \subseteq \mathfrak{k}^n$ be open. Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. -\begin{definition}\label{structuresheafkn} - For open subsets $U \subseteq X$, let $\mathcal{O}_X(U)$ be the set of functions $U \xrightarrow{\phi} \mathfrak{k}$ such that every $x \in U$ has a neighbourhood $V$ such that there are $f,g \in R$ such that for $y \in V$ we have $g(y) \neq 0$ and $\phi(y) = \frac{f(y)}{g(y)}$. +Let $X \subseteq \mathfrak{k}^n$ be open. +Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. +\begin{definition} + \label{structuresheafkn} + For open subsets $U \subseteq X$, let $\mathcal{O}_X(U)$ be the set + of + functions $U \xrightarrow{\phi} \mathfrak{k}$ such that every $x + \in U$ has a + neighbourhood $V$ such that there are $f,g \in R$ such that for $y \in V$ we + have $g(y) \neq 0$ and $\phi(y) = \frac{f(y)}{g(y)}$. \end{definition} -\begin{remark}\label{structuresheafcontinuous} - $\mathcal{O}_X$ is a subsheaf (of rings) of the sheaf of $\mathfrak{k}$-valued functions on $X$. - The elements of $\mathcal{O}_X(U)$ are continuous: - Let $M \subseteq \mathfrak{k}$ be closed. We must show the closedness of $N \coloneqq \phi^{-1}(M)$ in $U$. For $M = \mathfrak{k}$ this is trivial. Otherwise $M$ is finite and we may assume $M = \{t\} $ for some $t \in \mathfrak{k}$. For $x \in U$, there are open $V_x \subseteq U$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $V_x$. - Then $N \cap V_x = V(f_x - t\cdot g_x) \cap V_x)$ is closed in $V_x$. As the $V_x$ cover $U$ and $U$ is quasi-compact, $N$ is closed in $U$. +\begin{remark} + \label{structuresheafcontinuous} + $\mathcal{O}_X$ is a subsheaf (of rings) of the sheaf of + $\mathfrak{k}$-valued functions on $X$. + The elements of $\mathcal{O}_X(U)$ are continuous: Let $M \subseteq + \mathfrak{k}$ be closed. + We must show the closedness of $N \coloneqq \phi^{-1}(M)$ in $U$. + For $M = \mathfrak{k}$ this is trivial. + Otherwise $M$ is finite and we may assume $M = \{t\} $ for some $t \in + \mathfrak{k}$. + For $x \in U$, there are open $V_x \subseteq U$ and $f_x, g_x \in R$ such that + $\phi = \frac{f_x}{g_x}$ on $V_x$. + Then $N \cap V_x = V(f_x - t\cdot g_x) \cap V_x)$ is closed in $V_x$. + As the $V_x$ cover $U$ and $U$ is quasi-compact, $N$ is closed in $U$. \end{remark} -\begin{proposition}\label{structuresheafri} - Let $X = V(I)$ where $I = \sqrt{I} \subseteq R$ is an ideal. Let $A = R / I$. Then +\begin{proposition} + \label{structuresheafri} + Let $X = V(I)$ where $I = \sqrt{I} \subseteq R$ is an ideal. + Let $A = R / I$. + Then \begin{align} - \phi: A &\longrightarrow \mathcal{O}_X(X) \\ - f \mod I &\longmapsto f\defon{X} + \phi: A & \longrightarrow \mathcal{O}_X(X) \\ f \mod I + & \longmapsto f\defon{X} \end{align} -is an isomorphism. + is an isomorphism. \end{proposition} \begin{proof} - It is easy to see that the map $A \to \mathcal{O}_X(X)$ is well-defined and a ring homomorphism. - Its injectivity follows from the Nullstellensatz and $I = \sqrt{I}$ (\ref{hns3}). + It is easy to see that the map $A \to \mathcal{O}_X(X)$ is + well-defined and a + ring homomorphism. + Its injectivity follows from the Nullstellensatz and $I = + \sqrt{I}$ + ( + \ref{hns3}). - Let $\phi \in \mathcal{O}_X(X)$. for $x \in X$, there are an open subset $U_x \subseteq X$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $U_x$. + Let $\phi \in \mathcal{O}_X(X)$. + for $x \in X$, there are an open subset $U_x \subseteq X$ and $f_x, g_x \in R$ + such that $\phi = \frac{f_x}{g_x}$ on $U_x$. \begin{claim} - Without loss of generality loss of generality we can assume $U_x = X \setminus V(g_x)$. + Without loss of generality loss of generality we can assume $U_x = X \setminus + V(g_x)$. \end{claim} \begin{subproof} - The closed subsets $(X \setminus U_x) \subseteq \mathfrak{k}^n$ has the form $X\setminus U_x = V(J_x)$ for some ideal $J_x \subseteq R$. + The closed subsets $(X \setminus U_x) \subseteq \mathfrak{k}^n$ has + the form + $X\setminus U_x = V(J_x)$ for some ideal $J_x \subseteq R$. As $x \not\in X \setminus V_x$ there is $h_x \in J_x$ with $h_x(x) \neq 0$. - Replacing $U_x$ by $X \setminus V(h_x)$, $f_x$ by $f_xh_x$ and $g_x$ by $g_xh_x$, we may assume $U_x = X \setminus V(g_x)$. + Replacing $U_x$ by $X \setminus V(h_x)$, $f_x$ by $f_xh_x$ and $g_x$ by + $g_xh_x$, we may assume $U_x = X \setminus V(g_x)$. \end{subproof} \begin{claim} - Without loss of generality loss of generality we can assume $V(g_x) \subseteq V(f_x)$. + Without loss of generality loss of generality we can assume $V(g_x) \subseteq + V(f_x)$. \end{claim} \begin{subproof} Replace $f_x$ by $f_xg_x$ and $g_x$ by $g_x^2$. \end{subproof} - As $X$ is quasi-compact, there are finitely many points $(x_i)_{i=1}^m$ such that the $U_{x_i}$ cover $X$. - Let $U_i \coloneqq U_{x_i}, f_i \coloneqq f_{x_i}, g_i \coloneqq g_{x_i}$. + As $X$ is quasi-compact, there are finitely many points + $(x_i)_{i=1}^m$ such + that the $U_{x_i}$ cover $X$. + Let $U_i \coloneqq U_{x_i}, f_i \coloneqq + f_{x_i}, g_i \coloneqq g_{x_i}$. + + As the $U_i = X \setminus V(g_i)$ cover $X$, $V(I) \cap + \bigcap_{i=1}^m V(g_i) + = X \cap \bigcap_{i=1}^m V(g_i) = \emptyset$. + By the Nullstellensatz ( + \ref{hns1}) the ideal of $R$ generated by + $I$ and the + $a_i$ equals $R$. + There are thus $n \ge m \in \N$ and elements + $(g_i)_{i = m+1}^n$ of $I$ and + $(a_i)_{i=1}^n \in R^n$ such that $1 = + \sum_{i=1}^{n} a_ig_i$. + Let for $i > m$ $f_i \coloneqq 0$, $F = \sum_{i=1}^{n} a_if_i = + \sum_{i=1}^{m} + a_if_i \in R$. - As the $U_i = X \setminus V(g_i)$ cover $X$, $V(I) \cap \bigcap_{i=1}^m V(g_i) = X \cap \bigcap_{i=1}^m V(g_i) = \emptyset$. - By the Nullstellensatz (\ref{hns1}) the ideal of $R$ generated by $I$ and the $a_i$ equals $R$. - There are thus $n \ge m \in \N$ and elements $(g_i)_{i = m+1}^n$ of $I$ and $(a_i)_{i=1}^n \in R^n$ such that $1 = \sum_{i=1}^{n} a_ig_i$. - Let for $i > m$ $f_i \coloneqq 0$, $F = \sum_{i=1}^{n} a_if_i = \sum_{i=1}^{m} a_if_i \in R$. - \begin{claim} For all $x \in X $ ~ $f_i(x) = \phi(x) g_i(x)$. \end{claim} @@ -122,32 +232,66 @@ is an isomorphism. \end{subproof} It follows that \[ - \phi(x) = \phi(x) \cdot 1 = \phi(x) \cdot \sum_{i=1}^{n} a_i(x) g_i(x) = \sum_{i=1}^{n} a_i(x) f_i(x) = F(x) - \] - Hence $\phi = F\defon{X}$. + \phi(x) = \phi(x) \cdot 1 = + \phi(x) \cdot \sum_{i=1}^{n} + a_i(x) g_i(x) = \sum_{i=1}^{n} a_i(x) f_i(x) = F(x) + \] + Hence $\phi = + F\defon{X}$. \end{proof} \subsubsection{The structure sheaf on closed subsets of $\mathbb{P}^n$} -Let $X \subseteq \mathbb{P}^n$ be closed and $R_\bullet = \mathfrak{k}[X_0,\ldots,X_n]$ with its usual grading. +Let $X \subseteq \mathbb{P}^n$ be closed and $R_\bullet = +\mathfrak{k}[X_0,\ldots,X_n]$ with its usual grading. -\begin{definition}\label{structuresheafpn} - For open $U \subseteq X$, let $\mathcal{O}_X(U)$ be the set of functions $U \xrightarrow{\phi} \mathfrak{k}$ such that for every $x \in U$, there are an open subset $W \subseteq U$, a natural number $d$ and $f,g \in R_d$ such that $W \cap \Vp(g) = \emptyset$ and $\phi(y) = \frac{f(y_0,\ldots,y_n)}{g(y_0,\ldots,y_n)}$ for $y = [y_0,\ldots,y_n] \in W$. +\begin{definition} + \label{structuresheafpn} + For open $U \subseteq X$, let $\mathcal{O}_X(U)$ be the set of + functions $U + \xrightarrow{\phi} \mathfrak{k}$ such that for every $x \in U$, + there are an + open subset $W \subseteq U$, a natural number $d$ and $f,g \in R_d$ such that + $W \cap \Vp(g) = \emptyset$ and $\phi(y) = + \frac{f(y_0,\ldots,y_n)}{g(y_0,\ldots,y_n)}$ for $y = [y_0,\ldots,y_n] \in W$. \end{definition} \begin{remark} - This is a subsheaf of rings of the sheaf of $\mathfrak{k}$-valued functions on $X$. -Under the identification $\mathbb{A}^n =\mathfrak{k}^n$ with $\mathbb{P}^n \setminus \Vp(X_0)$, one has $\mathcal{O}_X \defon{X \setminus \Vp(X_0)} = \mathcal{O}_{X \cap \mathbb{A}^n}$ as subsheaves of the sheaf of $\mathfrak{k}$-valued functions, where the second sheaf is a sheaf on a closed subset of $\mathfrak{k}^n$: + This is a subsheaf of rings of the sheaf of $\mathfrak{k}$-valued + functions on + $X$. + Under the identification $\mathbb{A}^n =\mathfrak{k}^n$ + with $\mathbb{P}^n + \setminus \Vp(X_0)$, one has $\mathcal{O}_X + \defon{X \setminus \Vp(X_0)} = + \mathcal{O}_{X \cap \mathbb{A}^n}$ as subsheaves of the + sheaf of + $\mathfrak{k}$-valued functions, where the second sheaf is a sheaf on + a closed + subset of $\mathfrak{k}^n$: -Indeed, if $W$ is as in the definition then $\phi([1,y_1,\ldots,y_n]) = \frac{f(1,y_1,\ldots,y_n)}{g(1,y_1,\ldots,y_n)}$ for $[1,y_1,\ldots,y_n] \in W$. -Conversely if $\phi([1,y_1,\ldots,y_n]) = \frac{f(y_1,\ldots,y_n)}{g(y_1,\ldots,y_n)}$ on an open subset $W $ of $X \cap \mathbb{A}^n$ then -$\phi([y_0,\ldots,y_n]) = \frac{F(y_0,\ldots,y_n)}{G(y_0,\ldots,y_n)}$ on $W$ where $F(X_0,\ldots,X_n) \coloneqq X_0^d f(\frac{X_1}{X_0}, \ldots, \frac{X_n}{X_0})$ and $G(X_0,\ldots,X_n) = X_0^d g(\frac{X_1}{X_0},\ldots, \frac{X_n}{X_0})$ with a sufficiently large $d \in \N$. + Indeed, if $W$ is as in the definition then $\phi([1,y_1,\ldots,y_n]) = + \frac{f(1,y_1,\ldots,y_n)}{g(1,y_1,\ldots,y_n)}$ for $[1,y_1,\ldots,y_n] \in + W$. + Conversely if $\phi([1,y_1,\ldots,y_n]) = + \frac{f(y_1,\ldots,y_n)}{g(y_1,\ldots,y_n)}$ on an open subset $W $ of $X \cap + \mathbb{A}^n$ then $\phi([y_0,\ldots,y_n]) = + \frac{F(y_0,\ldots,y_n)}{G(y_0,\ldots,y_n)}$ on $W$ where $F(X_0,\ldots,X_n) + \coloneqq X_0^d f(\frac{X_1}{X_0}, \ldots, \frac{X_n}{X_0})$ and + $G(X_0,\ldots,X_n) = X_0^d g(\frac{X_1}{X_0},\ldots, + \frac{X_n}{X_0})$ with a + sufficiently large $d \in \N$. \end{remark} \begin{remark} - It follows from the previous remark and the similar result in the affine case that the elements of $\mathcal{O}_X(U)$ are continuous on $U \setminus V(X_0)$. - Since the situation is symmetric in the homogeneous coordinates, they are continuous on all of $U$. + It follows from the previous remark and the similar result in the affine case + that the elements of $\mathcal{O}_X(U)$ are continuous on $U + \setminus V(X_0)$. + Since the situation is symmetric in the homogeneous coordinates, they are + continuous on all of $U$. \end{remark} The following is somewhat harder than in the affine case: \begin{proposition} - If $X$ is connected (e.g. irreducible), then the elements of $\mathcal{O}_X\left( X \right) $ are constant functions on $X$. + If $X$ is connected (e.g. irreducible), then the elements of + $\mathcal{O}_X\left( X \right) $ are constant functions on + $X$. \end{proposition} @@ -158,70 +302,163 @@ The following is somewhat harder than in the affine case: \begin{definition} A \vocab{category} $\mathcal{A}$ consists of: \begin{itemize} - \item A class $\Ob \mathcal{A}$ of \vocab[Objects]{objects of $\mathcal{A}$}. - \item For two arbitrary objects $A, B \in \Ob \mathcal{A}$, a \textbf{set} $\Hom_\mathcal{A}(A,B)$ of \vocab[Morphism]{morphisms for $A$ to $B$ in $\mathcal{A}$}. - \item A map $\Hom_\mathcal{A}(B,C) \times \Hom_\mathcal{A}(A,B) \xrightarrow{\circ} \Hom_\mathcal{A}(A,C)$, the composition of morphisms, for arbitrary triples $(A,B,C)$ of objects of $\mathcal{A}$. + \item + A class + $\Ob \mathcal{A}$ of \vocab[Objects]{objects of $\mathcal{A}$}. + \item + For two arbitrary objects $A, B \in \Ob \mathcal{A}$, a + \textbf{set} $\Hom_\mathcal{A}(A,B)$ of + \vocab[Morphism]{morphisms for $A$ to $B$ in $\mathcal{A}$}. + \item + A map $\Hom_\mathcal{A}(B,C) \times \Hom_\mathcal{A}(A,B) + \xrightarrow{\circ} \Hom_\mathcal{A}(A,C)$, the composition of + morphisms, for arbitrary triples $(A,B,C)$ of objects of + $\mathcal{A}$. \end{itemize} The following conditions must be satisfied: \begin{enumerate}[A] - \item For morphisms $A \xrightarrow{f} B\xrightarrow{g} C \xrightarrow{h} D$, we have $h \circ (g \circ f) = (h \circ g) \circ f$. - \item For every $A \in \Ob(\mathcal{A})$, there is an $\Id_A \in \Hom_{\mathcal{A}}(A,A)$ such that $\Id_A \circ f = f$ (reps. $g \circ \Id_A = g$) for arbitrary morphisms $B \xrightarrow{f} A$ (reps. $A \xrightarrow{g} C).$ + \item + For + morphisms $A \xrightarrow{f} B\xrightarrow{g} C + \xrightarrow{h} D$, we have $h + \circ (g \circ f) = (h \circ g) \circ f$. + \item + For every $A \in \Ob(\mathcal{A})$, there is an $\Id_A \in + \Hom_{\mathcal{A}}(A,A)$ such that $\Id_A \circ f = f$ (reps. $g \circ \Id_A = + g$) for arbitrary morphisms $B \xrightarrow{f} + A$ (reps. + $A \xrightarrow{g} + C). + $ \end{enumerate} - A morphism $X \xrightarrow{f} Y$ is called an \vocab[Isomorphism]{isomorphism (in $\mathcal{A} $)} if there is a morphism $Y \xrightarrow{g} X$ (called the \vocab[Inverse morphism]{inverse $f^{-1}$ of $f$)} such that $g \circ f = \Id_X$ and $f \circ g = \Id_Y$. + A morphism $X \xrightarrow{f} Y$ is called an \vocab[Isomorphism]{isomorphism + (in $\mathcal{A} $)} + if there is a morphism $Y \xrightarrow{g} X$ (called the + \vocab[Inverse morphism]{inverse $f^{-1}$ of $f$)} such that $g \circ f = + \Id_X$ and $f \circ g = \Id_Y$. \end{definition} \begin{remark} \begin{itemize} - \item The distinction between classes and sets is important here. - \item We will usually omit the composition sign $\circ$. - \item It is easy to see that $\Id_A$ is uniquely determined by the above condition $B$, and that the inverse $f^{-1}$ of an isomorphism $f$ is uniquely determined. + \item + The distinction between classes and sets is important here. + \item + We will usually omit the composition sign $\circ$. + \item + It is easy to see that $\Id_A$ is uniquely determined by the above condition + $B$, and that the inverse $f^{-1}$ of an isomorphism + $f$ is uniquely determined. \end{itemize} \end{remark} \subsubsection{Examples of categories} \begin{example} \begin{itemize} - \item The category of sets. - \item The category of groups. - \item The category of rings. - \item If $R$ is a ring, the category of $R$-modules and the category $\Alg_R$ of $R$-algebras - \item The category of topological spaces - \item The category $\Var_\mathfrak{k}$ of varieties over $\mathfrak{k}$ (see \ref{defvariety}) - \item If $\mathcal{A}$ is a category, then the \vocab{opposite category} or \vocab{dual category} is defined by $\Ob(\mathcal{A}\op) = \Ob(\mathcal{A})$ and $\Hom_{\mathcal{A}\op}(X,Y) = \Hom_\mathcal{A}(Y,X)$. + \item + The category of sets. + \item + The category of groups. + \item + The category of rings. + \item + If $R$ is a ring, the category of $R$-modules and the category $\Alg_R$ of + $R$-algebras + \item + The category of topological spaces + \item + The category $\Var_\mathfrak{k}$ of varieties over + $\mathfrak{k}$ (see + \ref{defvariety}) + \item + If $\mathcal{A}$ is a category, then the \vocab{opposite category} + or \vocab{dual category} is defined by $\Ob(\mathcal{A}\op) = + \Ob(\mathcal{A})$ and $\Hom_{\mathcal{A}\op}(X,Y) = + \Hom_\mathcal{A}(Y,X)$. \end{itemize} - In most of these cases, isomorphisms in the category were just called `isomorphism'. The isomorphisms in the category of topological spaces are the homeomophisms. + In most of these cases, isomorphisms in the category were just called + `isomorphism'. + The isomorphisms in the category of topological spaces are the homeomophisms. \end{example} \subsubsection{Subcategories} \begin{definition}[Subcategories] - A \vocab{subcategory} of $\mathcal{A}$ is a category $\mathcal{B}$ such that $\Ob(\mathcal{B}) \subseteq \Ob(\mathcal{A})$, such that $\Hom_\mathcal{B}(X,Y) \subseteq \Hom_\mathcal{A}(X,Y)$ for objects $X$ and $Y$ of $\mathcal{B}$, such that for every object $X \in \Ob(\mathcal{B})$, the identity $\Id_X$ of $X$ is the same in $\mathcal{B}$ as in $\mathcal{A}$, and such that for composable morphisms in $\mathcal{B}$, their compositions in $\mathcal{A}$ and $\mathcal{B}$ coincide. - We call $\mathcal{B}$ a \vocab{full subcategory} of $\mathcal{A}$ if in addition $\Hom_\mathcal{B}(X,Y) = \Hom_\mathcal{A}(X,Y)$ for arbitrary $X,Y \in \Ob(\mathcal{B})$. + A \vocab{subcategory} of + $\mathcal{A}$ is a category $\mathcal{B}$ such that + $\Ob(\mathcal{B}) \subseteq \Ob(\mathcal{A})$, such that + $\Hom_\mathcal{B}(X,Y) \subseteq \Hom_\mathcal{A}(X,Y)$ for + objects $X$ and $Y$ of $\mathcal{B}$, such that for every object $X + \in \Ob(\mathcal{B})$, the identity $\Id_X$ of $X$ is the same in + $\mathcal{B}$ as in $\mathcal{A}$, and such that for + composable morphisms in $\mathcal{B}$, their compositions in + $\mathcal{A}$ and $\mathcal{B}$ coincide. + We call $\mathcal{B}$ a \vocab{full subcategory} of + $\mathcal{A}$ if in + addition $\Hom_\mathcal{B}(X,Y) = \Hom_\mathcal{A}(X,Y)$ for + arbitrary $X,Y \in + \Ob(\mathcal{B})$. \end{definition} \begin{example} \begin{itemize} - \item The category of abelian groups is a full subcategory of the category of groups. - It can be identified with the category of $\Z$-modules. - \item The category of finitely generated $R$-modules as a full subcategory of the category of $R$-modules. - \item The category of $R$-algebras of finite type as a full subcategory of $\Alg_R$. - \item The category of affine varieties over $\mathfrak{k}$ as a full subcategory of the category of varieties over $\mathfrak{k}$. + \item + The category of abelian groups is a full subcategory of the category of groups. + It can be identified with the category of $\Z$-modules. + \item + The category of finitely generated $R$-modules as a full subcategory of the + category of $R$-modules. + \item + The category of $R$-algebras of finite type as a full subcategory of $\Alg_R$. + \item + The category of affine varieties over $\mathfrak{k}$ as a full + subcategory of the category of varieties over $\mathfrak{k}$. \end{itemize} \end{example} \subsubsection{Functors and equivalences of categories} \begin{definition} - A \vocab[Functor!covariant]{(covariant) functor} (resp. \vocab[Functor!contravariant]{contravariant functor}) between categories $\mathcal{A} \xrightarrow{F} \mathcal{B}$ is a map $\Ob(\mathcal{A}) \xrightarrow{F} \Ob(\mathcal{B})$ with a family of maps $\Hom_\mathcal{A}(X,Y) \xrightarrow{F} \Hom_\mathcal{B}(F(X),F(Y))$ (resp. $\Hom_\mathcal{A}(X,Y) \xrightarrow{F} \Hom_\mathcal{B}(F(Y),F(X))$ in the case of contravariant functors), where $X$ and $Y$ are arbitrary objects of $\mathcal{A}$, such that the following conditions hold: - \begin{itemize} - \item $F(\Id_X) = \Id_{F(X)}$ - \item For morphisms $X \xrightarrow{f} Y \xrightarrow{g} Z$ in $\mathcal{A}$, we have $F(gf) = F(g)F(f)$ ( resp. $F(gf) = F(f)F(g)$) + A \vocab[Functor! + covariant]{(covariant) functor} (resp. \vocab[Functor!contravariant]{contravariant functor}) between categories $\mathcal{A} + \xrightarrow{F} \mathcal{B}$ is a map + $\Ob(\mathcal{A}) \xrightarrow{F} \Ob(\mathcal{B})$ with + a family of maps $\Hom_\mathcal{A}(X,Y) \xrightarrow{F} + \Hom_\mathcal{B}(F(X),F(Y))$ (resp. $\Hom_\mathcal{A}(X,Y) + \xrightarrow{F} \Hom_\mathcal{B}(F(Y),F(X))$ in the case of + contravariant functors), where $X$ and $Y$ are arbitrary objects of + $\mathcal{A}$, such that the following conditions hold: + \begin{itemize} + \item + $F(\Id_X) = \Id_{F(X)}$ + \item + For morphisms $X \xrightarrow{f} + Y \xrightarrow{g} Z$ in $\mathcal{A}$, we have $F(gf) = + F(g)F(f)$ ( resp. + $F(gf) = F(f) + F(g)$) \end{itemize} - A functor is called \vocab[Functor!essentially surjective]{essentially surjective} if every object of $\mathcal{B}$ is isomorphic to an element of the image of $\Ob(\mathcal{A}) \xrightarrow{F} \Ob(\mathcal{B})$. - A functor is called \vocab[Functor!full]{full} (resp. \vocab[Functor!faithful]{faithful}) if it induces surjective (resp. injective) maps between sets of morphisms. - It is called an \vocab{equivalence of categories} if it is full, faithful and essentially surjective. + A functor is called \vocab[Functor! + essentially surjective]{essentially surjective} if every object of + $\mathcal{B}$ is isomorphic to an element of the image of + $\Ob(\mathcal{A}) \xrightarrow{F} \Ob(\mathcal{B})$. + A functor is called \vocab[Functor! + full]{full} (resp. \vocab[Functor!faithful]{faithful}) if it induces surjective (resp. + injective) maps between sets of morphisms. + It is called an \vocab{equivalence of categories} if it is full, faithful and + essentially surjective. \end{definition} \begin{example} \begin{itemize} - \item There are \vocab[Functor!forgetful]{forgetful functors} from rings to abelian groups or from abelian groups to sets which drop the multiplicative structure of a ring or the group structure of a group. - \item If $\mathfrak{k}$ is any vector space there is a contravariant functor from $\mathfrak{k}$-vector spaces to itself sending $V$ to its dual vector space $V\subseteq$ and $V \xrightarrow{f} W$ to the dual linear map $W^{\ast} \xrightarrow{f^{\ast}} V^{\ast}$. - When restricted to the full subcategory of finite-dimensional vector spaces it becomes a contravariant self-equivalence of that category. - \item The embedding of a subcategory is a faithful functor. In the case of a full subcategory it is also full. + \item + There are \vocab[Functor!forgetful]{forgetful functors} from rings to abelian groups or from abelian + groups to sets which drop the multiplicative structure of a ring or the group + structure of a group. + \item + If $\mathfrak{k}$ is any vector space there is a contravariant + functor from $\mathfrak{k}$-vector spaces to itself sending $V$ to + its dual vector space $V\subseteq$ and $V \xrightarrow{f} + W$ to the dual linear map $W^{\ast} + \xrightarrow{f^{\ast}} V^{\ast}$. + When restricted to the full subcategory of finite-dimensional vector spaces it + becomes a contravariant self-equivalence of that category. + \item + The embedding of a subcategory is a faithful functor. + In the case of a full subcategory it is also full. \end{itemize} \end{example} @@ -229,79 +466,184 @@ The following is somewhat harder than in the affine case: \subsection{The category of varieties} -\begin{definition}[Algebraic variety]\label{defvariety} - An \vocab{algebraic variety} or \vocab{prevariety} over $\mathfrak{k}$ is a pair $(X, \mathcal{O}_X)$, where $X$ is a topological space and $\mathcal{O}_X$ a subsheaf of the sheaf of $\mathfrak{k}$-valued functions on $X$ such that for every $x \in X$, there are a neighbourhood $U_x$ of $x$ in $X$, an open subset $V_x$ of a closed subset $Y_x$ of $\mathfrak{k}^{n_x}$\footnote{By the result of \ref{affopensubtopbase} it can be assumed that $V_x = Y_x$ without altering the definition.} and a homeomorphism $V_x \xrightarrow{\iota_x} U_x$ such that for every open subset $V \subseteq U_x$ and every function $V\xrightarrow{f} \mathfrak{k}$, we have $f \in \mathcal{O}_X(V) \iff \iota^{\ast}_x(f) \in \mathcal{O}_{Y_x}(\iota_x^{-1}(V))$, +\begin{definition}[Algebraic variety] + \label{defvariety} + An \vocab{algebraic variety} or \vocab{prevariety} over + $\mathfrak{k}$ is a + pair $(X, \mathcal{O}_X)$, where $X$ is a topological space and + $\mathcal{O}_X$ + a subsheaf of the sheaf of $\mathfrak{k}$-valued functions on $X$ + such that + for every $x \in X$, there are a neighbourhood $U_x$ of $x$ in $X$, an open + subset $V_x$ of a closed subset $Y_x$ of + $\mathfrak{k}^{n_x}$\footnote{By the + result of + \ref{affopensubtopbase} it can be assumed that $V_x = Y_x$ without + altering the definition. + } and a homeomorphism $V_x + \xrightarrow{\iota_x} + U_x$ such that for every open subset $V \subseteq U_x$ and every function + $V\xrightarrow{f} \mathfrak{k}$, we have $f \in + \mathcal{O}_X(V) \iff + \iota^{\ast}_x(f) \in + \mathcal{O}_{Y_x}(\iota_x^{-1}(V))$, - In this, the \vocab{pull-back} $\iota_x^{\ast}(f)$ of $f$ is defined by $(\iota_x^{\ast}(f))(\xi) \coloneqq f(\iota_x(\xi))$. + In this, the \vocab{pull-back} $\iota_x^{\ast}(f)$ of $f$ is + defined by + $(\iota_x^{\ast}(f))(\xi) \coloneqq f(\iota_x(\xi))$. - A morphism $(X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ of varieties is a continuous map $X \xrightarrow{\phi} Y$ such that for all open $U \subseteq Y$ and $f \in \mathcal{O}_Y(U)$, $\phi^{\ast}(f) \in \mathcal{O}_X(\phi^{-1}(U))$. - An isomorphism is a morphism such that $\phi$ is bijective and $\phi^{-1}$ also is a morphism of varieties. + A morphism $(X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ of + varieties is a + continuous map $X \xrightarrow{\phi} Y$ such that for all open $U + \subseteq Y$ + and $f \in \mathcal{O}_Y(U)$, $\phi^{\ast}(f) \in + \mathcal{O}_X(\phi^{-1}(U))$. + An isomorphism is a morphism such that $\phi$ is bijective and + $\phi^{-1}$ also + is a morphism of varieties. \end{definition} \begin{example} \begin{itemize} - \item If $(X, \mathcal{O}_X)$ is a variety and $U \subseteq X$ open, then $(U, \mathcal{O}_X\defon{U})$ is a variety (called an \vocab{open subvariety} of $X$), and the embedding $U \to X$ is a morphism of varieties. - \item If $X$ is a closed subset of $\mathfrak{k}^n$ or $\mathbb{P}^n$, then $(X, \mathcal{O}_X)$ is a variety, where $\mathcal{O}_X$ is the structure sheaf on $X$ (\ref{structuresheafkn}, reps. \ref{structuresheafpn}). - A variety is called \vocab[Variety!affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of this form, with $X $ closed in $\mathfrak{k}^n$ (resp. $\mathbb{P}^n$). - A variety which is isomorphic to and open subvariety of $X$ is called \vocab[Variety!quasi-affine]{quasi-affine} (resp. \vocab[Variety!quasi-projective]{quasi-projective}). - \item If $X = V(X^2 - Y^3) \subseteq \mathfrak{k}^2$ then $\mathfrak{k} \xrightarrow{t \mapsto (t^3,t^2)} X$ is a morphism which is a homeomorphism of topological spaces but not an isomorphism of varieties. - % TODO + \item + If $(X, \mathcal{O}_X)$ is a variety and $U \subseteq X$ open, then + $(U, \mathcal{O}_X\defon{U})$ is a variety (called an + \vocab{open subvariety} of $X$), and the embedding $U \to X$ is a morphism of + varieties. + \item + If $X$ is a closed subset of $\mathfrak{k}^n$ or + $\mathbb{P}^n$, then $(X, \mathcal{O}_X)$ is a + variety, where $\mathcal{O}_X$ is the structure sheaf on $X$ + ( + \ref{structuresheafkn}, reps. + \ref{structuresheafpn}). + A variety is called \vocab[Variety! + affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of + this form, with $X $ closed in $\mathfrak{k}^n$ (resp. + $\mathbb{P}^n$). + A variety which is isomorphic to and open subvariety of $X$ is called + \vocab[Variety! + quasi-affine]{quasi-affine} (resp. \vocab[Variety!quasi-projective]{quasi-projective}). + \item + If $X = V(X^2 - Y^3) \subseteq \mathfrak{k}^2$ then + $\mathfrak{k} \xrightarrow{t \mapsto (t^3,t^2)} + X$ is a morphism which is a homeomorphism of topological spaces but not an + isomorphism of varieties. + % TODO - \item The composition of two morphisms $X \to Y \to Z$ of varieties is a morphism of varieties. - \item $X\xrightarrow{\Id_X} X$ is a morphism of varieties. + \item + The composition of two morphisms $X \to Y \to Z$ of varieties is a morphism of + varieties. + \item + $X\xrightarrow{\Id_X} + X$ is a morphism of varieties. \end{itemize} \end{example} \subsubsection{The category of affine varieties} -\begin{lemma}\label{localinverse} +\begin{lemma} + \label{localinverse} Let $X$ be any $\mathfrak{k}$-variety and $U \subseteq X$ open. \begin{enumerate}[i)] - \item All elements of $\mathcal{O}_X(U)$ are continuous. - \item If $U \subseteq X$ is open, $U \xrightarrow{\lambda} \mathfrak{k}$ any function and every $x \in U$ has a neighbourhood $V_x \subseteq U$ such that $\lambda \defon{V_x} \in \mathcal{O}_X(V_x)$, then $\lambda \in \mathcal{O}_X(U)$. - \item If $\vartheta \in \mathcal{O}_X(U)$ and $\vartheta(x) \neq 0$ for all $x \in U$, then $\vartheta \in \mathcal{O}_X(U)^{\times }$. + \item + All elements of $\mathcal{O}_X(U)$ are continuous. + \item + If $U \subseteq X$ is open, $U \xrightarrow{\lambda} \mathfrak{k}$ + any function and every $x \in U$ has a neighbourhood $V_x \subseteq U$ such + that $\lambda \defon{V_x} \in \mathcal{O}_X(V_x)$, + then $\lambda \in \mathcal{O}_X(U)$. + \item + If $\vartheta \in \mathcal{O}_X(U)$ and $\vartheta(x) + \neq 0$ for all $x \in U$, then $\vartheta \in + \mathcal{O}_X(U)^{\times }$. \end{enumerate} \end{lemma} \begin{proof} \begin{enumerate}[i)] - \item The property is local on $U$, hence it is sufficient to show it in the quasi-affine case. This was done in \ref{structuresheafcontinuous}. - \item For the second part, let $\lambda_x \coloneqq \lambda \defon{V_x} $. - We have $\lambda_x\defon{V_x \cap V_y} = \lambda \defon{V_x \cap V_y} = \lambda_y \defon{V_x \cap V_y} $. - The $V_x$ cover $U$. By the sheaf axiom for $\mathcal{O}_X$ there is $\ell \in \mathcal{O}_X(U)$ with $\ell\defon{V_x} =\lambda_x$. It follows that $\ell=\lambda$. - \item By the definition of variety, every $x \in U$ has a quasi-affine neighbourhood $V \subseteq U$. We can assume $U$ to be quasi-affine and $X = V(I) \subseteq \mathfrak{k}^n$, as the general assertion follows by an application of ii). - If $x \in U$ there are a neighbourhood $x \in W \subseteq U$ and $a,b \in R = \mathfrak{k}[X_1,\ldots,X_n]$ such that $\vartheta(y) = \frac{a(y)}{b(y)}$ for $y \in W$, with $b(y) \neq 0$. - Then $a(x) \neq 0$ as $\vartheta(x) \neq 0$. Replacing $W$ by $W \setminus V(a)$, we may assume that $a$ has no zeroes on $W$. - Then $\lambda(y) = \frac{b(y)}{a(y)}$ for $y \in W$ has a non-vanishing denominator and $\lambda \in \mathcal{O}_X(U)$. - We have $\lambda \cdot \vartheta = 1$, thus $\vartheta \in \mathcal{O}_X(U)^{\times}$. + \item + The property is local on $U$, hence it is sufficient to show it in the + quasi-affine case. + This was done in + \ref{structuresheafcontinuous}. + \item + For the second part, let $\lambda_x \coloneqq \lambda \defon{V_x} + $. + We have $\lambda_x\defon{V_x \cap V_y} = \lambda \defon{V_x \cap V_y} = + \lambda_y \defon{V_x \cap V_y} $. + The $V_x$ cover $U$. + By the sheaf axiom for $\mathcal{O}_X$ there is $\ell \in + \mathcal{O}_X(U)$ + with $\ell\defon{V_x} =\lambda_x$. + It follows that $\ell=\lambda$. + \item + By the definition of variety, every $x \in U$ has a quasi-affine neighbourhood + $V \subseteq U$. + We can assume $U$ to be quasi-affine and $X = V(I) \subseteq + \mathfrak{k}^n$, + as the general assertion follows by an application of ii). + If $x \in U$ there are a neighbourhood $x \in W \subseteq U$ and $a,b \in R = + \mathfrak{k}[X_1,\ldots,X_n]$ such that $\vartheta(y) = + \frac{a(y)}{b(y)}$ for + $y \in W$, with $b(y) \neq 0$. + Then $a(x) \neq 0$ as $\vartheta(x) \neq 0$. + Replacing $W$ by $W \setminus V(a)$, we may assume that $a$ has no zeroes on + $W$. + Then $\lambda(y) = \frac{b(y)}{a(y)}$ for $y \in W$ has a + non-vanishing + denominator and $\lambda \in \mathcal{O}_X(U)$. + We have $\lambda \cdot \vartheta = 1$, thus $\vartheta \in + \mathcal{O}_X(U)^{\times}$. \end{enumerate} - + \end{proof} \begin{proposition}[About affine varieties] \label{propaffvar} \begin{itemize} - \item Let $X,Y$ be varieties over $\mathfrak{k}$. Then the map - \begin{align} - \phi: \Hom_{\Var_\mathfrak{k}}(X,Y) &\longrightarrow \Hom_{\Alg_\mathfrak{k}}(\mathcal{O}_Y(Y), \mathcal{O}_X(X)) \\ - (X \xrightarrow{f} Y) &\longmapsto (\mathcal{O}_Y(Y) \xrightarrow{f^{\ast}} \mathcal{O}_X(X)) - \end{align} - is injective when $Y$ is quasi-affine and bijective when $Y$ is affine. - \item The contravariant functor - \begin{align} - F: \Var_\mathfrak{k} &\longrightarrow \Alg_\mathfrak{k} \\ - X &\longmapsto \mathcal{O}_X(X)\\ - (X\xrightarrow{f} Y) &\longmapsto (\mathcal{O}_X(X) \xrightarrow{f^{\ast}} \mathcal{O}_Y(Y)) - \end{align} - restricts to an equivalence of categories between the category of affine varieties over $\mathfrak{k}$ and the full subcategory $\mathcal{A}$ of $\Alg_\mathfrak{k}$, - having the $\mathfrak{k}$-algebras $A$ of finite type with $\nil A = \{0\} $ as objects. + \item + Let $X,Y$ be varieties over $\mathfrak{k}$. + Then the map + \begin{align} + \phi: \Hom_{\Var_\mathfrak{k}}(X,Y) & \longrightarrow + \Hom_{\Alg_\mathfrak{k}}(\mathcal{O}_Y(Y), \mathcal{O}_X(X)) \\ (X + \xrightarrow{f} Y) & \longmapsto (\mathcal{O}_Y(Y) \xrightarrow{f^{\ast}} + \mathcal{O}_X(X)) + \end{align} + is injective when $Y$ is quasi-affine and + bijective when $Y$ is affine. + \item + The contravariant functor + \begin{align} + F: \Var_\mathfrak{k} & \longrightarrow \Alg_\mathfrak{k} \\ X & \longmapsto + \mathcal{O}_X(X) \\ (X\xrightarrow{f} Y) & \longmapsto (\mathcal{O}_X(X) + \xrightarrow{f^{\ast}} \mathcal{O}_Y(Y)) + \end{align} + restricts to an + equivalence of categories between the category of affine varieties over + $\mathfrak{k}$ and the full subcategory $\mathcal{A}$ of + $\Alg_\mathfrak{k}$, + having the $\mathfrak{k}$-algebras $A$ of finite type with $\nil A = + \{0\} $ as + objects. \end{itemize} \end{proposition} \begin{remark} - It is clear that $\nil(\mathcal{O}_X(X)) = \{0\}$ for arbitrary varieties. For general varieties it is however not true that $\mathcal{O}_X(X)$ is a $\mathfrak{k}$-algebra of finite type. - There are counterexamples even for quasi-affine $X$. %TODO + It is clear that $\nil(\mathcal{O}_X(X)) = \{0\}$ for arbitrary + varieties. + For general varieties it is however not true that + $\mathcal{O}_X(X)$ is a + $\mathfrak{k}$-algebra of finite type. + There are counterexamples even for quasi-affine $X$. + %TODO - If, however, $X$ is affine, we may assume w.l.o.g. that $X = V(I)$ where $I = \sqrt{I} \subseteq R$ is an ideal with $R = \mathfrak{k}[X_1,\ldots,X_n]$. - Then $\mathcal{O}_X(X) \cong R / I$ (see \ref{structuresheafri}) is a $\mathfrak{k}$-algebra of finite type. + If, however, $X$ is affine, we may assume w.l.o.g. + that $X = V(I)$ where $I = \sqrt{I} \subseteq R$ is an ideal + with $R = \mathfrak{k}[X_1,\ldots,X_n]$. + Then $\mathcal{O}_X(X) \cong R / I$ (see + \ref{structuresheafri}) + is a + $\mathfrak{k}$-algebra of finite type. \end{remark} \begin{proof} @@ -309,104 +651,212 @@ The following is somewhat harder than in the affine case: - It suffices to investigate $\phi$ when $Y$ is an open subset of $V(I) \subseteq \mathfrak{k}^n$, where $I = \sqrt{I} \subseteq R$ is an ideal and $Y = V(I)$ when $Y$ is affine. - Let $(f_1,\ldots,f_n)$ be the components of $X \xrightarrow{f} Y \subseteq \mathfrak{k}^n$. Let $Y \xrightarrow{\xi_i} \mathfrak{k}$ be the $i$-th coordinate. - By definition $f_i = f^{\ast}(\xi_i) $. Thus $f$ is uniquely determined by $\mathcal{O}_Y(Y) \xrightarrow{f^{\ast}} \mathcal{O}_X(X)$. - Conversely, let $Y = V(I)$ and $\mathcal{O}_Y(Y) \xrightarrow{\phi} \mathcal{O}_X(X)$ be a morphism of $\mathfrak{k}$-algebras. Define $f_i \coloneqq \phi(\xi_i)$ and consider $X \xrightarrow{f = (f_1,\ldots,f_n)} Y\subseteq \mathfrak{k}^n$. + It suffices to investigate $\phi$ when $Y$ is an open subset of $V(I) \subseteq + \mathfrak{k}^n$, where $I = \sqrt{I} \subseteq R$ is + an ideal and $Y = V(I)$ + when $Y$ is affine. + Let $(f_1,\ldots,f_n)$ be the components of $X \xrightarrow{f} Y + \subseteq + \mathfrak{k}^n$. + Let $Y \xrightarrow{\xi_i} \mathfrak{k}$ be the $i$-th + coordinate. + By definition $f_i = f^{\ast}(\xi_i) $. + Thus $f$ is uniquely determined by $\mathcal{O}_Y(Y) + \xrightarrow{f^{\ast}} + \mathcal{O}_X(X)$. + Conversely, let $Y = V(I)$ and $\mathcal{O}_Y(Y) + \xrightarrow{\phi} + \mathcal{O}_X(X)$ be a morphism of + $\mathfrak{k}$-algebras. + Define $f_i \coloneqq \phi(\xi_i)$ and consider $X + \xrightarrow{f = + (f_1,\ldots,f_n)} Y\subseteq \mathfrak{k}^n$. \begin{claim} $f$ has image contained in $Y$. \end{claim} \begin{subproof} - For $x \in X, \lambda \in I$ we have $\lambda(f(x)) = (\phi(\lambda \mod I))(x) = 0$ as $\phi$ is a morphism of $\mathfrak{k}$-algebras. + For $x \in X, \lambda \in I$ we have $\lambda(f(x)) = + (\phi(\lambda \mod I))(x) + = 0$ as $\phi$ is a morphism of $\mathfrak{k}$-algebras. Thus $f(x) \in V(I) = Y$. \end{subproof} \begin{claim} $f$ is a morphism in $\Var_\mathfrak{k}$ \end{claim} \begin{subproof} - For open $\Omega \subseteq Y, U = f^{-1}(\Omega) = \{x \in X | \forall \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y \setminus \Omega = V(J)$. - If $\lambda \in \mathcal{O}_Y(\Omega)$ and $x \in U$, then $f(x)$ has a neighbourhood $V$ such that there are $a,b \in R$ with $\lambda(v) = \frac{a(v)}{b(v)}$ and $b(v) \neq 0$ for all $v \in V$. - Let $W \coloneqq f^{-1}(V)$. Then $\alpha \coloneqq \phi(a)\defon{W} \in \mathcal{O}_X(W)$, $\beta \coloneqq \phi(b)\defon{W} \in \mathcal{O}_X(W)$. - By the second part of \ref{localinverse} $\beta \in \mathcal{O}_X(W)^{\times}$ and $f^{\ast}(\lambda)\defon{W} = \frac{\alpha}{\beta} \in \mathcal{O}_X(W)$. - The first part of \ref{localinverse} shows that $f^{\ast}(\lambda) \in \mathcal{O}_X(U)$. + For open $\Omega \subseteq Y, U = f^{-1}(\Omega) = \{x + \in X | \forall \lambda + \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y + \setminus \Omega + = V(J)$. + If $\lambda \in \mathcal{O}_Y(\Omega)$ and $x \in U$, then $f(x)$ + has a + neighbourhood $V$ such that there are $a,b \in R$ with $\lambda(v) + = + \frac{a(v)}{b(v)}$ and $b(v) \neq 0$ for all $v \in V$. + Let $W \coloneqq f^{-1}(V)$. + Then $\alpha \coloneqq \phi(a)\defon{W} \in + \mathcal{O}_X(W)$, $\beta + \coloneqq \phi(b)\defon{W} \in + \mathcal{O}_X(W)$. + By the second part of + \ref{localinverse} $\beta \in + \mathcal{O}_X(W)^{\times}$ + and $f^{\ast}(\lambda)\defon{W} = + \frac{\alpha}{\beta} \in \mathcal{O}_X(W)$. + The first part of + \ref{localinverse} shows that + $f^{\ast}(\lambda) \in + \mathcal{O}_X(U)$. \end{subproof} - By definition of $f$, we have $f^{\ast} = \phi$. This finished the proof of the first point. + By definition of $f$, we have $f^{\ast} = \phi$. + This finished the proof of the first point. \begin{claim} - The functor in the second part maps affine varieties to objects of $\mathcal{A}$ and is essentially surjective. + The functor in the second part maps affine varieties to objects of + $\mathcal{A}$ and is essentially surjective. \end{claim} \begin{subproof} - It follows from the remark that the functor maps affine varieties to objects of $\mathcal{A}$. + It follows from the remark that the functor maps affine varieties to objects of + $\mathcal{A}$. - If $A \in \Ob(\mathcal{A})$ then $ A /\mathfrak{k}$ is of finite type, thus $A \cong R / I$ for some $n$. - Since $\nil(A) = \{0\}$ we have $I = \sqrt{I}$, as for $x \in \sqrt{I}$, $x \mod I \in \nil(R / I) \cong \nil(A) = \{0\}$. + If $A \in \Ob(\mathcal{A})$ then $ A /\mathfrak{k}$ is of + finite type, thus $A + \cong R / I$ for some $n$. + Since $\nil(A) = \{0\}$ we have $I = \sqrt{I}$, + as for $x \in \sqrt{I}$, $x + \mod I \in \nil(R / I) \cong \nil(A) = \{0\}$. Thus $A \cong\mathcal{O}_X(X)$ where $X = V(I)$. \end{subproof} Fullness and faithfulness of the functor follow from the first point. \end{proof} \begin{remark} - Note that giving a contravariant functor $\mathcal{C} \to \mathcal{D}$ is equivalent to giving a functor $\mathcal{C} \to \mathcal{D}\op$. We have thus shown that the category of affine varieties is equivalent to $\mathcal{A}\op$, where $\mathcal{A} \subsetneq \Alg_\mathfrak{k}$ is the full subcategory of $\mathfrak{k}$-algebras $A$ of finite type with $\nil(A) = \{0\}$. + Note that giving a contravariant functor $\mathcal{C} \to + \mathcal{D}$ is + equivalent to giving a functor $\mathcal{C} \to + \mathcal{D}\op$. + We have thus shown that the category of affine varieties is equivalent to + $\mathcal{A}\op$, where $\mathcal{A} \subsetneq + \Alg_\mathfrak{k}$ is the full + subcategory of $\mathfrak{k}$-algebras $A$ of finite type with + $\nil(A) = + \{0\}$. \end{remark} \subsubsection{Affine open subsets are a topology base} \begin{definition} - A set $\mathcal{B}$ of open subsets of a topological space $X$ is called a \vocab{topology base} for $X$ if every open subset of $X$ can be written as a (possibly empty) union of elements of $\mathcal{B}$. -\end{definition} + A set $\mathcal{B}$ of open subsets of a topological space $X$ is + called a + \vocab{topology base} for $X$ if every open subset of $X$ can be written as a + (possibly empty) union of elements of $\mathcal{B}$. +\end{definition} \begin{fact} -If $X$ is a set, then $\mathcal{B} \subseteq \mathcal{P}(X)$ is a base for some topology on $X$ iff $X = \bigcup_{U \in \mathcal{B}} U$ and for arbitrary $U, V \in \mathcal{B}, U \cap V$ is a union of elements of $\mathcal{B}$. + If $X$ is a set, then $\mathcal{B} \subseteq + \mathcal{P}(X)$ is a base for some + topology on $X$ iff $X = \bigcup_{U \in \mathcal{B}} U$ and for arbitrary $U, V + \in \mathcal{B}, U \cap V$ is a union of elements of + $\mathcal{B}$. \end{fact} \begin{definition} Let $X$ be a variety. - An \vocab{affine open subset} of $X$ is a subset which is an affine variety. - + An \vocab{affine open subset} of $X$ is a subset which is an affine variety. + \end{definition} -\begin{proposition}\label{oxulocaf} - Let $X$ be an affine variety over $\mathfrak{k}$, $\lambda \in \mathcal{O}_X(X)$ and $U = X \setminus V(\lambda)$. - Then $U$ is an affine variety and the morphism $\phi: \mathcal{O}_X(X)_\lambda \to \mathcal{O}_X(U)$ defined by the restriction $\mathcal{O}_X(X) \xrightarrow{\cdot |_U } \mathcal{O}_X(U)$ and the universal property of the localization is an isomorphism. +\begin{proposition} + \label{oxulocaf} + Let $X$ be an affine variety over $\mathfrak{k}$, $\lambda \in + \mathcal{O}_X(X)$ and $U = X \setminus V(\lambda)$. + Then $U$ is an affine variety and the morphism $\phi: + \mathcal{O}_X(X)_\lambda + \to \mathcal{O}_X(U)$ defined by the restriction + $\mathcal{O}_X(X) + \xrightarrow{\cdot |_U } \mathcal{O}_X(U)$ and the universal + property of the + localization is an isomorphism. \end{proposition} \begin{proof} - Let $X$ be an affine variety over $\mathfrak{k}, \lambda \in \mathcal{O}_X(X)$ and $U = X \setminus V(\lambda)$. The fact that $\lambda\defon{U} \in \mathcal{O}_x(U)^{\times}$ follows from \ref{localinverse}. - Thus the universal property of the localization $\mathcal{O}_X(X)_\lambda$ can be applied to $\mathcal{O}_X(X) \xrightarrow{\cdot |_U} \mathcal{O}_X(U)$. + Let $X$ be an affine variety over $\mathfrak{k}, \lambda \in + \mathcal{O}_X(X)$ + and $U = X \setminus V(\lambda)$. + The fact that $\lambda\defon{U} \in + \mathcal{O}_x(U)^{\times}$ follows + from + \ref{localinverse}. + Thus the universal property of the localization + $\mathcal{O}_X(X)_\lambda$ can + be applied to $\mathcal{O}_X(X) \xrightarrow{\cdot |_U} + \mathcal{O}_X(U)$. \[ \begin{tikzcd} \mathcal{O}_X(X) \arrow{d}{\cdot |_U}\arrow{r}{x \mapsto \frac{x}{1}} & \mathcal{O}_X(X)_\lambda \arrow[dotted, bend left]{dl}{\existsone \phi} \\ \mathcal{O}_X(U) & \end{tikzcd} - \] - \[ + \] + \[ \begin{tikzcd} &Y \arrow[bend right, swap]{ld}{\pi_0} \arrow[bend right, swap]{d}{\pi}&\mathcal{O}_Y(Y) \cong A_\lambda \arrow{d}{\mathfrak{s}}& \\ X \arrow[hookrightarrow]{r}{}& U \arrow[swap]{u}{\sigma} & \mathcal{O}_X(U) \end{tikzcd} - \] - For the rest of the proof, we may assume $X = V(I) \subseteq \mathfrak{k}^n$ where $I = \sqrt{I} \subseteq R \coloneqq\mathfrak{k}[X_1,\ldots,X_n]$ is an ideal. - Then $A \coloneqq \mathcal{O}_X(X) \cong R / I$ and there is $\ell \in R$ such that $\ell\defon{X} = \lambda$. - Let $Y = V(J) \subseteq \mathfrak{k}^{n+1}$ where $J \subseteq \mathfrak{k}[Z,X_1,\ldots,X_n]$ is generated by the elements of $I$ and $1 - Z\ell(X_1,\ldots,X_n)$. + \] + For the rest of the proof, we may assume $X = V(I) \subseteq + \mathfrak{k}^n$ where $I = \sqrt{I} \subseteq R + \coloneqq\mathfrak{k}[X_1,\ldots,X_n]$ is an ideal. + Then $A \coloneqq \mathcal{O}_X(X) \cong R / I$ and there is $\ell + \in R$ such + that $\ell\defon{X} = \lambda$. + Let $Y = V(J) \subseteq \mathfrak{k}^{n+1}$ where $J \subseteq + \mathfrak{k}[Z,X_1,\ldots,X_n]$ is generated by the elements of $I$ and $1 - + Z\ell(X_1,\ldots,X_n)$. - Then $\mathcal{O}_Y(Y) \cong \mathfrak{k}[Z,X_1,\ldots,X_n] / J \cong A[Z] / (1 -\lambda Z) \cong A_\lambda$. - By the proposition about affine varieties (\ref{propaffvar}), the morphism $\mathfrak{s}: \mathcal{O}_Y(Y) \cong A_\lambda \to \mathcal{O}_X(U)$ corresponds to a morphism $U \xrightarrow{\sigma} Y$. - We have $\mathfrak{s}(Z \mod J) = \lambda^{-1}$ and $\mathfrak{s}(X_i \mod J) = X_i \mod I$. + Then $\mathcal{O}_Y(Y) \cong \mathfrak{k}[Z,X_1,\ldots,X_n] / J \cong + A[Z] / (1 + -\lambda Z) \cong A_\lambda$. + By the proposition about affine varieties ( + \ref{propaffvar}), the + morphism + $\mathfrak{s}: \mathcal{O}_Y(Y) \cong A_\lambda \to + \mathcal{O}_X(U)$ + corresponds to a morphism $U \xrightarrow{\sigma} Y$. + We have $\mathfrak{s}(Z \mod J) = \lambda^{-1}$ and + $\mathfrak{s}(X_i \mod J) = + X_i \mod I$. Thus $\sigma(x) = (\lambda(x)^{-1}, x)$ for $x \in U$. - Moreover, the projection $Y \xrightarrow{\pi_0} X$ dropping the $Z$-coordinate has image contained in $U$, as for $(z,x) \in Y$ the equation + Moreover, the projection $Y \xrightarrow{\pi_0} X$ dropping the + $Z$-coordinate + has image contained in $U$, as for $(z,x) \in Y$ the equation \[ - 1 = z\lambda(x) - \] - implies $\lambda(x) \neq 0$. It thus defines a morphism $Y \xrightarrow{\pi} U$ and by the description of $\sigma$ it follows that $\sigma \pi = \Id_U$. - Similarly it follows that $\sigma \pi = \Id_Y$. Thus, $\sigma$ and $\pi$ are inverse to each other. + 1 = + z\lambda(x) + \] + implies $\lambda(x) \neq 0$. + It thus defines a morphism $Y \xrightarrow{\pi} U$ and by the description + of + $\sigma$ it follows that $\sigma \pi = \Id_U$. + Similarly it follows that $\sigma \pi = \Id_Y$. + Thus, $\sigma$ and $\pi$ are inverse to each other. \end{proof} -\begin{corollary}\label{affopensubtopbase} +\begin{corollary} + \label{affopensubtopbase} The affine open subsets of a variety $X$ are a topology base on $X$. \end{corollary} \begin{proof} - Let $X = V(I) \subseteq \mathfrak{k}^n$ with $I = \sqrt{I}$. If $U \subseteq X$ is open then $X \setminus U = V(J)$ with $J \supseteq I$ and $U = \bigcup_{f \in J} (X \setminus V(f))$. - Thus $U$ is a union of affine open subsets. The same then holds for arbitrary quasi-affine varieties. - + Let $X = V(I) \subseteq \mathfrak{k}^n$ with $I = + \sqrt{I}$. + If $U \subseteq X$ is open then $X \setminus U = V(J)$ with $J \supseteq I$ and + $U = \bigcup_{f \in J} (X \setminus V(f))$. + Thus $U$ is a union of affine open subsets. + The same then holds for arbitrary quasi-affine varieties. + Let $X$ be any variety, $U \subseteq X$ open and $x \in U$. - By the definition of variety, $x$ has a neighbourhood $V_x$ which is quasi-affine, and replacing $V_x$ by $U \cap V_x$ which is also quasi-affine we may assume $V_x \subseteq U$. - $V_x$ is a union of its affine open subsets. Because $U$ is the union of the $V_x$, $U$ as well is a union of affine open subsets. + By the definition of variety, $x$ has a neighbourhood $V_x$ which is + quasi-affine, and replacing $V_x$ by $U \cap V_x$ which is also quasi-affine we + may assume $V_x \subseteq U$. + $V_x$ is a union of its affine open subsets. + Because $U$ is the union of the $V_x$, $U$ as well is a union of affine open + subsets. \end{proof} @@ -419,110 +869,214 @@ If $X$ is a set, then $\mathcal{B} \subseteq \mathcal{P}(X)$ is a base for some \subsection{Stalks of sheaves} \begin{definition}[Stalk] - Let $\mathcal{G}$ be a presheaf of sets on the topological space $X$, and let $x \in X$. - The \vocab{stalk} (\vocab[Stalk]{Halm}) of $\mathcal{G}$ at $x$ is the set of equivalence classes of pairs $(U, \gamma)$, where $U$ is an open neighbourhood of $x$ and $\gamma \in \mathcal{G}(U)$ - and the equivalence relation $\sim $ is defined as follows: - $( U , \gamma) \sim (V, \delta)$ iff there exists an open neighbourhood $W \subseteq U \cap V$ of $x$ such that $\gamma \defon{W} = \delta \defon{W}$. + Let $\mathcal{G}$ be a presheaf of sets on + the topological space $X$, and let $x \in X$. + The \vocab{stalk} (\vocab[Stalk]{Halm}) of $\mathcal{G}$ at $x$ is the set of + equivalence classes of pairs $(U, \gamma)$, where $U$ is an open neighbourhood + of $x$ and $\gamma \in \mathcal{G}(U)$ and the equivalence relation + $\sim $ is + defined as follows: $( U , \gamma) \sim (V, \delta)$ iff there exists + an open + neighbourhood $W \subseteq U \cap V$ of $x$ such that $\gamma + \defon{W} = + \delta \defon{W}$. - If $\mathcal{G}$ is a presheaf of groups, one can define a groups structure on $\mathcal{G}_x$ by + If $\mathcal{G}$ is a presheaf of groups, one can define a groups + structure on + $\mathcal{G}_x$ by \[ - ((U, \gamma) / \sim ) \cdot \left( (V,\delta) / \sim \right) = (U \cap V, \gamma \defon{U \cap V} \cdot \delta\defon{U \cap V}) / \sim - \] + ((U, \gamma) / \sim ) \cdot \left( (V,\delta) / \sim + \right) = (U \cap V, \gamma \defon{U \cap V} \cdot + \delta\defon{U \cap V}) / + \sim + \] - If $\mathcal{G}$ is a presheaf of rings, one can similarly define a ring structure on $\mathcal{G}_x$. - + If $\mathcal{G}$ is a presheaf of rings, one can similarly define a + ring + structure on $\mathcal{G}_x$. - If $U$ is an open neighbourhood of $x \in X$, then we have a map (resp. homomorphism) + + If $U$ is an open neighbourhood of $x \in X$, then we have a map (resp. + homomorphism) \begin{align} - \cdot_x : \mathcal{G}(U) &\longrightarrow \mathcal{G}_x \\ - \gamma &\longmapsto \gamma_x \coloneqq (U, \gamma) / \sim + \cdot_x : \mathcal{G}(U) & \longrightarrow \mathcal{G}_x \\ + \gamma & \longmapsto \gamma_x \coloneqq (U, \gamma) / \sim \end{align} \end{definition} \begin{fact} - Let $\gamma,\delta \in \mathcal{G}(U)$. If $\mathcal{G}$ is a sheaf\footnote{or, more generally, a separated presheaf} and if for all $x \in U$, we have $\gamma_x = \delta_x$, then $\gamma = \delta$. + Let $\gamma,\delta \in \mathcal{G}(U)$. + If $\mathcal{G}$ is a sheaf\footnote{or, more generally, a separated presheaf} + and if for all $x \in U$, we have $\gamma_x = \delta_x$, then $\gamma = + \delta$. - In the case of a sheaf, the image of the injective map $\mathcal{G}(U) \xrightarrow{\gamma \mapsto (\gamma_x)_{x \in U}} \prod_{x \in U} \mathcal{G}_x$ - is the set of all $(g_x)_{x \in U} \in \prod_{x \in U} \mathcal{G}_x $ satisfying the following \vocab{coherence condition}: - For every $x \in U$, there are an open neighbourhood $W_x \subseteq U$ of $x$ and $g^{(x)} \in \mathcal{G}(W_x)$ with $g_y^{(x)} = g_y$ for all $y \in W_x$. + In the case of a sheaf, the image of the injective map $\mathcal{G}(U) + \xrightarrow{\gamma \mapsto (\gamma_x)_{x \in U}} \prod_{x \in U} + \mathcal{G}_x$ is the set of all + $(g_x)_{x \in U} \in \prod_{x \in U} + \mathcal{G}_x $ satisfying the following \vocab{coherence condition}: + For + every $x \in U$, there are an open neighbourhood $W_x \subseteq U$ of $x$ and + $g^{(x)} \in \mathcal{G}(W_x)$ with $g_y^{(x)} = + g_y$ for all $y \in W_x$. \end{fact} \begin{proof} - Because of $\gamma_x = \delta_x$, there is $x \in W_x \subseteq U$ open such that $\gamma\defon{W_x} = \delta\defon{W_x}$. As the $W_x$ cover $U$, $\gamma = \delta$ by the sheaf axiom. + Because of $\gamma_x = \delta_x$, there is $x \in W_x \subseteq U$ open such + that $\gamma\defon{W_x} = \delta\defon{W_x}$. + As the $W_x$ cover $U$, $\gamma = \delta$ by the sheaf axiom. \end{proof} \begin{definition} Let $\mathcal{G}$ be a sheaf of functions. - Then $\gamma_x$ is called the \vocab{germ} of the function $\gamma$ at $x$. - The \vocab[Germ!value at $x$]{value at $x$ } of $g = (U, \gamma) / \sim \in \mathcal{G}_x$ defined as $g(x) \coloneqq \gamma(x)$, which is independent of the choice of the representative $\gamma$. + Then $\gamma_x$ is called the \vocab{germ} of the function $\gamma$ + at $x$. + The \vocab[Germ! + value at $x$]{value at $x$ } of $g = (U, \gamma) / \sim \in + \mathcal{G}_x$ defined as $g(x) \coloneqq \gamma(x)$, + which is independent of the choice of the representative $\gamma$. \end{definition} \begin{remark} - If $\mathcal{G}$ is a sheaf of $C^{\infty}$-functions (resp. holomorphic functions), then $\mathcal{G}_x$ is called the ring of germs of $C^\infty$-functions (resp. of holomorphic functions) at $x$. + If $\mathcal{G}$ is a sheaf of + $C^{\infty}$-functions (resp. + holomorphic functions), then $\mathcal{G}_x$ is called the ring of + germs of $C^\infty$-functions (resp. of holomorphic functions) at $x$. \end{remark} \subsubsection{The local ring of an affine variety} \begin{definition} - If $X$ is a variety, the stalk $\mathcal{O}_{X,x}$ of the structure sheaf at $x$ is called the \vocab{local ring} of $X$ at $x$. - This is indeed a local ring, with maximal ideal $\mathfrak{m}_x = \{f \in \mathcal{O}_{X,x} | f(x) = 0\}$. + If $X$ is a variety, the stalk $\mathcal{O}_{X,x}$ of the structure sheaf + at + $x$ is called the \vocab{local ring} of $X$ at $x$. + This is indeed a local ring, with maximal ideal $\mathfrak{m}_x = + \{f \in + \mathcal{O}_{X,x} | f(x) = 0\}$. \end{definition} \begin{proof} - By \ref{localring} it suffices to show that $\mathfrak{m}_x$ is a proper ideal, which is trivial, and that the elements of $\mathcal{O}_{X,x} \setminus \mathfrak{m}_x$ are units in $\mathcal{O}_{X,x}$. + By + \ref{localring} it suffices to show that $\mathfrak{m}_x$ + is a proper ideal, + which is trivial, and that the elements of $\mathcal{O}_{X,x} \setminus + \mathfrak{m}_x$ are units in $\mathcal{O}_{X,x}$. Let $g = (U, \gamma)/\sim \in \mathcal{O}_{X,x}$ and $g(x) \neq 0$. - $\gamma$ is Zariski continuous (first point of \ref{localinverse}). Thus $V(\gamma)$ is closed. By replacing $U$ by $U \setminus V(\gamma)$ we may assume that $\gamma$ vanishes nowhere on $U$. - By the third point of \ref{localinverse} we have $\gamma \in \mathcal{O}_X(U)^{\times}$. + $\gamma$ is Zariski continuous (first point of + \ref{localinverse}). + Thus $V(\gamma)$ is closed. + By replacing $U$ by $U \setminus V(\gamma)$ we may assume that $\gamma$ + vanishes nowhere on $U$. + By the third point of + \ref{localinverse} we have $\gamma \in + \mathcal{O}_X(U)^{\times}$. $(\gamma^{-1})_x$ is an inverse to $g$. \end{proof} -\begin{proposition}\label{proplocalring} - Let $X = \Va(I) \subseteq \mathfrak{k}^n$ be equipped with its usual structure sheaf, where $I = \sqrt{I} \subseteq R = \mathfrak{k}[X_1,\ldots,X_n]$ . Let $x \in X$ and $A = \mathcal{O}_X(X) \cong R / I$. - $\{P \in R | P(x) = 0\} \text{\reflectbox{$\coloneqq$}} \fn_x \subseteq R$ is maximal, $I \subseteq \fn_x$ and $\mathfrak{m}_x \coloneqq \fn_x / I$ is the maximal ideal of elements of $A$ vanishing at $x$. - If $\lambda \in A \setminus \mathfrak{m}_x$, we have $\lambda_x \in \mathcal{O}_{X,x}^{\times}$, where $\lambda_x$ denotes the image under $A \cong \mathcal{O}_X(X) \to \mathcal{O}_{X,x}$. - By the universal property of the localization, there exists a unique ring homomorphism $A_{\mathfrak{m}_x} \xrightarrow{\iota} \mathcal{O}_{X,x}$ - such that +\begin{proposition} + \label{proplocalring} + Let $X = \Va(I) \subseteq \mathfrak{k}^n$ be + equipped with its usual structure + sheaf, where $I = \sqrt{I} \subseteq R = + \mathfrak{k}[X_1,\ldots,X_n]$ . + Let $x \in X$ and $A = \mathcal{O}_X(X) \cong R / I$. + $\{P \in R | P(x) = 0\} \text{\reflectbox{$\coloneqq$}} \fn_x \subseteq R$ is maximal, + $I \subseteq \fn_x$ and $\mathfrak{m}_x \coloneqq \fn_x / I$ is the + maximal ideal of elements of $A$ vanishing at $x$. + If $\lambda \in A \setminus \mathfrak{m}_x$, we have $\lambda_x \in + \mathcal{O}_{X,x}^{\times}$, where $\lambda_x$ denotes the image under $A \cong + \mathcal{O}_X(X) \to \mathcal{O}_{X,x}$. + By the universal property of the localization, there exists a unique ring + homomorphism $A_{\mathfrak{m}_x} \xrightarrow{\iota} + \mathcal{O}_{X,x}$ such + that \[ \begin{tikzcd} - A \arrow{r}{} \arrow{d}{\lambda \mapsto \lambda_x} & A_{\mathfrak{m}_x} \arrow[dotted, bend left]{ld}{\existsone \iota} \\ + A \arrow{r}{} \arrow{d}{\lambda \mapsto \lambda_x} & + A_{\mathfrak{m}_x} \arrow[dotted, bend left]{ld}{\existsone \iota} \\ \mathcal{O}_{X,x} \end{tikzcd} - \] + \] commutes. - - The morphism $A_{\mathfrak{m}_x}\xrightarrow{\iota} \mathcal{O}_{X,x}$ is an isomorphism. + + The morphism $A_{\mathfrak{m}_x}\xrightarrow{\iota} + \mathcal{O}_{X,x}$ is an + isomorphism. \end{proposition} \begin{proof} - To show surjectivity, let $\ell = (U, \lambda) / \sim \in \mathcal{O}_{X,x}$, where $U$ is an open neighbourhood of $x$ in $X$. - We have $X \setminus U = V(J)$ where $J \subseteq A$ is an ideal. As $x \in U$ there is $f \in J$ with $f(x) \neq 0$. Replacing $U $ by $X \setminus V(f)$ we may assume $U = X \setminus V(f)$. - By \ref{oxulocaf}, $\mathcal{O}_X(U) \cong A_f$, and $\lambda = f^{-n}\vartheta$ for some $n \in \N$ and $\vartheta \in A$. - Then $\ell = \iota(f^{-n} \vartheta)$ where the last fraction is taken in $A_{\mathfrak{m}_x}$. + To show surjectivity, let $\ell = (U, \lambda) / \sim \in + \mathcal{O}_{X,x}$, + where $U$ is an open neighbourhood of $x$ in $X$. + We have $X \setminus U = V(J)$ where $J \subseteq A$ is an ideal. + As $x \in U$ there is $f \in J$ with $f(x) \neq 0$. + Replacing $U $ by $X \setminus V(f)$ we may assume $U = X \setminus V(f)$. + By + \ref{oxulocaf}, $\mathcal{O}_X(U) \cong A_f$, and + $\lambda = + f^{-n}\vartheta$ for some $n \in \N$ and $\vartheta \in + A$. + Then $\ell = \iota(f^{-n} \vartheta)$ where the last fraction is taken in + $A_{\mathfrak{m}_x}$. - Let $\lambda = \frac{\vartheta}{g} \in A_{\mathfrak{m}_x}$ with $\iota(\lambda) = 0$. - It is easy to see that $\iota(\lambda) = (X \setminus V(g), \frac{\vartheta}{g}) / \sim $. - Thus there is an open neighbourhood $U$ of $x$ in $X \setminus V(g)$ such that $\vartheta$ vanishes on $U$. - Similar as before there is $h \in A$ with $h(x) \neq 0$ and $W = X \setminus V(h) \subseteq U$. - By the isomorphism $\mathcal{O}_X(W) \cong A_h$, there is $n \in \N$ with $h^{n}\vartheta = 0$ in $A$. Since $h \not\in \mathfrak{m}_x$, $h$ is a unit and the image of $\vartheta$ in $A_{\mathfrak{m}_x}$ vanishes, implying $\lambda = 0$. + Let $\lambda = \frac{\vartheta}{g} \in A_{\mathfrak{m}_x}$ + with + $\iota(\lambda) = 0$. + It is easy to see that $\iota(\lambda) = (X \setminus V(g), + \frac{\vartheta}{g}) / \sim $. + Thus there is an open neighbourhood $U$ of $x$ in $X \setminus V(g)$ such that + $\vartheta$ vanishes on $U$. + Similar as before there is $h \in A$ with $h(x) \neq 0$ and $W = X \setminus + V(h) \subseteq U$. + By the isomorphism $\mathcal{O}_X(W) \cong A_h$, there is $n \in + \N$ with + $h^{n}\vartheta = 0$ in $A$. + Since $h \not\in \mathfrak{m}_x$, $h$ is a unit and the image of + $\vartheta$ in + $A_{\mathfrak{m}_x}$ vanishes, implying $\lambda = 0$. \end{proof} \subsubsection{Intersection multiplicities and Bezout's theorem} \begin{definition} - Let $R = \mathfrak{k}[X_0,X_1,X_2]$ equipped with its usual grading and let $x \in \mathbb{P}^{2}$. -Let $G \in R_g, H \in R_h$ be homogeneous polynomials with $x \in V(G) \cap V(h)$. -Let $\ell\in R_1$ such that $\ell(x) \neq 0$. Then $x \in U = \mathbb{P}^2 \setminus V(\ell)$ and the rational functions $\gamma = \ell^{-g}G, \eta = \ell^{-h}H$ are elements of $\mathcal{O}_{\mathbb{P}^2}(U)$. -Let $I_x(G,H) \subseteq \mathcal{O}_{\mathbb{P}^2,x}$ denote the ideal generated by $\gamma_x$ and $\eta_x$. + Let $R = \mathfrak{k}[X_0,X_1,X_2]$ equipped with its usual grading and let $x + \in \mathbb{P}^{2}$. + Let $G \in R_g, H \in R_h$ be homogeneous polynomials with $x \in V(G) \cap + V(h)$. + Let $\ell\in R_1$ such that $\ell(x) \neq 0$. + Then $x \in U = \mathbb{P}^2 \setminus V(\ell)$ and the rational + functions + $\gamma = \ell^{-g}G, \eta = \ell^{-h}H$ are + elements of + $\mathcal{O}_{\mathbb{P}^2}(U)$. + Let $I_x(G,H) \subseteq \mathcal{O}_{\mathbb{P}^2,x}$ + denote the ideal + generated by $\gamma_x$ and $\eta_x$. -\noindent The dimension $\dim_{\mathfrak{k}}(\mathcal{O}_{X,x} / I_x(G,H)) \text{\reflectbox{$\coloneqq$}} i_x(G,H)$ is called the \vocab{intersection multiplicity} of $G$ and $H$ at $x$. + \noindent The dimension $\dim_{\mathfrak{k}}(\mathcal{O}_{X,x} + / I_x(G,H)) \text{\reflectbox{$\coloneqq$}} i_x(G,H)$ is called the + \vocab{intersection multiplicity} of $G$ and $H$ at $x$. \end{definition} \begin{remark} - If $\tilde \ell \in R_1$ also satisfies $\tilde \ell(x) \neq 0$, then the image of $\tilde \ell / \ell$ under $\mathcal{O}_{\mathbb{P}^2}(U) \to \mathcal{O}_{\mathbb{P}^2,x}$ is a unit, showing that the image of $\tilde \gamma = \tilde \ell^{-g} G$ in $\mathcal{O}_{\mathbb{P}^2,x}$ is multiplicatively equivalent to $\gamma_x$, and similarly for $\eta_x$. - Thus $I_x(G,H)$ does not depend on the choice of $\ell \in R_1$ with $\ell(x) \neq 0$. + If $\tilde \ell \in R_1$ also satisfies $\tilde \ell(x) \neq + 0$, then the image + of $\tilde \ell / \ell$ under + $\mathcal{O}_{\mathbb{P}^2}(U) \to + \mathcal{O}_{\mathbb{P}^2,x}$ is a unit, showing that + the image of $\tilde + \gamma = \tilde \ell^{-g} G$ in + $\mathcal{O}_{\mathbb{P}^2,x}$ is + multiplicatively equivalent to $\gamma_x$, and similarly for $\eta_x$. + Thus $I_x(G,H)$ does not depend on the choice of $\ell \in R_1$ with + $\ell(x) + \neq 0$. \end{remark} \begin{theorem}[Bezout's theorem] - In the above situation, assume that $V(H)$ and $V(G)$ intersect properly in the sense that $V(G) \cap V(H) \subseteq \mathbb{P}^2$ has no irreducible component of dimension $\ge 1$. - Then + In the above situation, assume that $V(H)$ and $V(G)$ + intersect properly in the sense that $V(G) \cap V(H) \subseteq + \mathbb{P}^2$ has no irreducible component of dimension $\ge 1$. + Then \[ \sum_{x \in V(G) \cap V(H)} i_x(G,H) = gh - \] - Thus, $V(G) \cap V(H)$ has $gh$ elements counted by multiplicity. + \] + Thus, $V(G) \cap V(H)$ has + $gh$ elements counted by multiplicity. \end{theorem} \printvocabindex \end{document}