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\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}

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\input{inputs/varieties}
\iffalse
\section{Übersicht}
\input{inputs/uebersicht}
\section{Übersicht}
\input{inputs/uebersicht}
\fi
\cleardoublepage

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inputs/finiteness_conditions.tex
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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}

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inputs/nullstellensatz_and_zariski_topology.tex
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@ -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 $\mathfrak{l}$).

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inputs/projective_spaces.tex
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@ -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}

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@ -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.
$b_i \coloneqq a_{i+1} - a_1^{k_{i+1}}, 1 \le i <n$. Claim: $A$ is integral ober subalgebra $B$ generated by the $b_i$ ($\lightning$ minimality)
$Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n})$. $Q(a_1) = P(\vec a) = 0$.
For suitable $\beta_{\alpha, l} in B$:
\[
T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_{i+1}})^{\alpha_{i+1}} = T^{w_k(\alpha)} + \sum_{l=0}^{w_k(\alpha) - 1} \beta_{\alpha,l} T^l
\]
Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where $\alpha \in S$ such that $w_k(\alpha)$ is maximal. Thus, $Q$ is normed.
% TODO Artin-Tate
%
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 $S \coloneqq R \setminus \{0\}$.
%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
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\section{Übersicht}
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@ -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.
$b_i \coloneqq a_{i+1} - a_1^{k_{i+1}}, 1 \le i <n$. Claim: $A$ is integral ober subalgebra $B$ generated by the $b_i$ ($\lightning$ minimality)
$Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n})$. $Q(a_1) = P(\vec a) = 0$.
$b_i \coloneqq a_{i+1} -
a_1^{k_{i+1}}, 1 \le i <n$.
Claim: $A$ is integral ober subalgebra $B$ generated by the $b_i$ ($\lightning$
minimality) $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots,
b_{n-1} + T^{k_n})$.
$Q(a_1) = P(\vec a) = 0$.
For suitable $\beta_{\alpha, l} in B$:
\[
T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_{i+1}})^{\alpha_{i+1}} = T^{w_k(\alpha)} + \sum_{l=0}^{w_k(\alpha) - 1} \beta_{\alpha,l} T^l
T^{\alpha_1} \prod_{i=1}^{n-1} (b_i +
T^{k_{i+1}})^{\alpha_{i+1}} =
T^{w_k(\alpha)} + \sum_{l=0}^{w_k(\alpha) - 1}
\beta_{\alpha,l} T^l
\]
Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where $\alpha \in S$ such that $w_k(\alpha)$ is maximal. Thus, $Q$ is normed.
Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where
$\alpha \in S$ such that $w_k(\alpha)$ is maximal.
Thus, $Q$ is normed.
% TODO Artin-Tate
@ -55,14 +86,24 @@ Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where $\alpha \in S$ such that $
%
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 $\implies \fp = \fq \lightning$.
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
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\section{Übersicht}
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