diff --git a/2021_Algebra_I.tex b/2021_Algebra_I.tex index d6750cc..96d32a1 100644 --- a/2021_Algebra_I.tex +++ b/2021_Algebra_I.tex @@ -4,6 +4,8 @@ \lecturer{Prof.~Dr.~Jens Franke} \author{Josia Pietsch} +\title{title} + \usepackage{algebra} \begin{document} @@ -29,29 +31,29 @@ Fields which are not assumed to be algebraically closed have been renamed (usual \pagebreak -\section{Finiteness conditions} +\subseteqction{Finiteness conditions} \subsection{Finitely generated and Noetherian modules} \begin{definition}[Generated submodule] - Let $R$ be a ring, $M$ an $R$-module, $S \se M$. + Let $R$ be a ring, $M$ an $R$-module, $S \subseteq M$. Then the following sets coincide \begin{enumerate} - \item $\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \se S' \text{finite}, r_s \in R, \right\}$ - \item $\bigcap_{\substack{S \se N \se M\\N \text{submodule}}} N$ - \item The $\se$-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 $N \se M$ 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 $:\iff \E S \se M$ 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 $:\iff \E S \subseteq M$ 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: \begin{enumerate} - \item Every submodule $N \se M$ is finitely generated. + \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 $\fM \neq \emptyset$ of submodules of $M$ has a $\se$-largest element. + \item Every set $\fM \neq \emptyset$ of submodules of $M$ has a $\subseteq$-largest element. \end{enumerate} \end{definition} \begin{proposition}[Hilbert's Basissatz]\label{basissatz} @@ -88,7 +90,7 @@ Fields which are not assumed to be algebraically closed have been renamed (usual 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, $N \se M$ a submodule, then $N' \coloneqq f\inv(N)$ 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 $N' \coloneqq f\inv(N)$ 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} @@ -98,7 +100,7 @@ Fields which are not assumed to be algebraically closed have been renamed (usual 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 $\alpha(R) \se A' \se A$.\\ + 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} @@ -113,8 +115,8 @@ A morphism of $R$-algebras $A \xrightarrow{f} \tilde{A}$ is a ring homomorphism 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 $S \se A$ the subalgebra generated by $S$ is the intersection of all subalgebras containing $S$ \\ - $=$ the union of subalgebras generated by finite $S' \se S$\\ + 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} @@ -175,7 +177,7 @@ This generalizes some facts about matrices to matrices with elements from commut Let $A$ be an $R$-algebra, $a \in A$. Then the following are equivalent: \begin{enumerate}[A] \item $\E n \in \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = \sum_{i=0}^{n-1} r_i a^i$ - \item There exists a subalgebra $B \se A$ finite over $R$ and containing $a$. + \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 $a_i$. \end{proposition} @@ -187,7 +189,7 @@ This generalizes some facts about matrices to matrices with elements from commut \begin{proof} \hskip 10pt \begin{enumerate} - {\color{gray} \item[B $\implies$ A] Let $a \in A$ such that there is a subalgebra $B \se A$ containing $a$ and finite over $R$. + {\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 \\ @@ -197,8 +199,8 @@ This generalizes some facts about matrices to matrices with elements from commut Then for all $v \in R^n: q(\fA \cdot v) = a \cdot q(v)$. By induction it follows that $q(P(\fA) \cdot v) = P(a)q(v)$ for all $P \in R[T]$. Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \fA)$ 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 $M_n \se B$ 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 $B \se A$ be the sub-$R$-module generated by $a^\alpha = \prod_{i=1}^n a_i^{\alpha_i}$ with $0 \le \alpha_i < d_i$. + 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 @@ -220,8 +222,8 @@ This generalizes some facts about matrices to matrices with elements from commut \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} \se A$ finite over $R$, such that all $a_i \in \tilde{A}$. - $b$ is integral over $\tilde{A} \implies \E \tilde{B} \se 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[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 \E \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} @@ -231,7 +233,7 @@ This generalizes some facts about matrices to matrices with elements from commut 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 \se 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] @@ -249,11 +251,11 @@ This generalizes some facts about matrices to matrices with elements from commut Let $B$ be a field and $a \in A \sm \{0\} $. Then $a\inv \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\inv = \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 \sm \{0\}$. As $B$ is integral over $A$, there is a sub-$A$-algebra $\tilde{B} \se 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 $\E 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 \sm \{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 $\E x \in \tilde{B} : b \cdot x \cdot 1$. \end{proof} \subsection{Noether normalization theorem} \begin{lemma}\label{nntechlemma} - Let $S \se \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$, + 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} @@ -311,9 +313,9 @@ is injective. $n$ and the $a_i$ can be chosen such that $A$ is finite over the i This contradicts the minimality of $n$, as $B$ can be generated by $< n$ elements $b_i$. \end{proof} -\section{The Nullstellensatz and the Zariski topology} +\subseteqction{The Nullstellensatz and the Zariski topology} \subsection{The Nullstellensatz} %LECTURE 1 -Let $\mathfrak{k}$ be a field, $R \coloneqq \mathfrak{k}[X_1,\ldots,X_n], I \se R$ an ideal. +Let $\mathfrak{k}$ be a field, $R \coloneqq \mathfrak{k}[X_1,\ldots,X_n], I \subseteq R$ an ideal. \begin{definition}[zero] $x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$} if $\A x \in I: P(x) = 0$. Let $\Va(I)$ denote the set of zeros if $I$ in $\mathfrak{k}^n$. @@ -361,7 +363,7 @@ Equivalent\footnote{used in a vague sense here} formulation: \end{theorem} \begin{proof} (HNS2 (\ref{hns2}) $\implies$ HNS1b (\ref{hns1b})) - $I \se \mathfrak{m}$ for some maximal ideal. $R / \mathfrak{m}$ is a field, since $\mathfrak{m}$ is maximal. + $I \subseteq \mathfrak{m}$ for some maximal ideal. $R / \mathfrak{m}$ is a field, since $\mathfrak{m}$ is maximal. $R / \mathfrak{m}$ is of finite type, since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra. There are thus a field extension $\mathfrak{i} / \mathfrak{l}$ and an isomorphism $R / \mathfrak{m} \xrightarrow{\iota} \mathfrak{i}$ of $\mathfrak{l}$-algebras. By HNS2 (\ref{hns2}), $\mathfrak{i} / \mathfrak{l}$ is a finite field extension. @@ -371,7 +373,7 @@ Equivalent\footnote{used in a vague sense here} formulation: \] Both sides are morphisms $R \to \mathfrak{i}$ of $\mathfrak{l}$-algebras. For for $P = X_i$ the equality is trivial. It follows in general, since the $X_i$ generate $R$ as a $\mathfrak{l}$-algebra. - Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \mathfrak{m} = 0$ for $P \in I \se \mathfrak{m}$). + Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \mathfrak{m} = 0$ for $P \in I \subseteq \mathfrak{m}$). HNS1 (\ref{hns1}) can easily be derived from HNS1b. \end{proof} @@ -401,12 +403,12 @@ The following proof of the Nullstellensatz only works for uncountable fields, bu \end{theorem} \begin{proof} If $(x_i)_{i=1}^{n}$ generate $L$ as an $K$-algebra, then the countably many monomials $x^{\alpha} = \prod_{i = 1}^{n} x_i^{\alpha_i} $ in the $x_i$ with $\alpha \in \N^n$ generate $L$ as a $K$-vector space. - Thus $\dim_K L \le \aleph_0$ and the same holds for any intermediate field $K \se M \se L$ . If $l \in L$ is transcendent over $K$ and $M = K(l)$, then $M \cong K(T)$ has uncountable dimension by \ref{dimrfunc}. Thus $L / K$ is algebraic, hence integral, hence finite (\ref{ftaiimplf}). + Thus $\dim_K L \le \aleph_0$ and the same holds for any intermediate field $K \subseteq M \subseteq L$ . If $l \in L$ is transcendent over $K$ and $M = K(l)$, then $M \cong K(T)$ has uncountable dimension by \ref{dimrfunc}. Thus $L / K$ is algebraic, hence integral, hence finite (\ref{ftaiimplf}). \end{proof} \subsection{The Zariski topology} \subsubsection{Operations on ideals and \texorpdfstring{$\Va\left( I \right) $}{V(I)}} -Let $R$ be a ring and $I,J, I_\lambda \se R$ ideals, $\lambda \in \Lambda$. +Let $R$ be a ring and $I,J, I_\lambda \subseteq R$ ideals, $\lambda \in \Lambda$. \begin{definition}[Radical, product and sum of ideals] \[ \sqrt{I} \coloneqq \bigcap_{n=0} ^{\infty} \{ f \in R | f^n \in I\} @@ -416,7 +418,7 @@ Let $R$ be a ring and $I,J, I_\lambda \se R$ ideals, $\lambda \in \Lambda$. \] \[ - \sum_{\lambda \in \Lambda} I_\lambda \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda | \Lambda' \se \Lambda \text{ finite}\right\} + \sum_{\lambda \in \Lambda} I_\lambda \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda | \Lambda' \subseteq \Lambda \text{ finite}\right\} \] \end{definition} \begin{fact} @@ -432,7 +434,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an algebraically Let $I, J, (I_{\lambda})_{\lambda \in \Lambda}$ be ideals in $R$. $\Lambda$ may be infinite. \begin{enumerate}[A] \item $\Va(I) = \Va(\sqrt{I})$ - \item $\sqrt{J} \se \sqrt{I} \implies \Va(I) \se \Va(J)$ + \item $\sqrt{J} \subseteq \sqrt{I} \implies \Va(I) \subseteq \Va(J)$ \item $\Va(R) = \emptyset, \Va(\{0\} =\mathfrak{k}^n$ \item $\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$ \item $\Va(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ @@ -441,14 +443,14 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an algebraically \begin{proof} \begin{enumerate} \item[A-C] trivial - \item[D] $I \cdot J \se I \cap J \se I$. Thus $\Va(I) \se \Va(I \cap J) \se \Va(I \cdot J)$. By symmetry we have $\Va(I) \cup \Va(J) \se \Va(I \cap J) \se \Va(I \cdot J)$. + \item[D] $I \cdot J \subseteq I \cap J \subseteq I$. Thus $\Va(I) \subseteq \Va(I \cap J) \subseteq \Va(I \cdot J)$. By symmetry we have $\Va(I) \cup \Va(J) \subseteq \Va(I \cap J) \subseteq \Va(I \cdot J)$. Let $x \not\in \Va(I) \cup \Va(J)$. Then there are $f \in I, g \in J$ such that $f(x) \neq 0, g(x) \neq 0$ thus $(f \cdot g)(x) \neq 0 \implies x \not\in \Va(I\cdot J)$. Therefore \[ - \Va(I) \cup \Va(J) \se \Va(I \cap J) \se \Va(I \cdot J) \se \Va(I) \cup \Va(J) + \Va(I) \cup \Va(J) \subseteq \Va(I \cap J) \subseteq \Va(I \cdot J) \subseteq \Va(I) \cup \Va(J) \] - \item[E] $I_\lambda \se \sum_{\lambda \in \Lambda} I_\lambda \implies \Va(\sum_{\lambda \in \Lambda} I_\lambda) \se \Va(I_\lambda)$. - Thus $\Va(\sum_{\lambda \in \Lambda} I_\lambda) \se \bigcap_{\lambda \in \Lambda} \Va(I_\lambda)$. - On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$ we have $f = \sum_{\lambda \in \Lambda} f_\lambda$. Thus $f$ vanishes on $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ and we have $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda) \se \Va(\sum_{\lambda \in \Lambda} I_\lambda)$. + \item[E] $I_\lambda \subseteq \sum_{\lambda \in \Lambda} I_\lambda \implies \Va(\sum_{\lambda \in \Lambda} I_\lambda) \subseteq \Va(I_\lambda)$. + Thus $\Va(\sum_{\lambda \in \Lambda} I_\lambda) \subseteq \bigcap_{\lambda \in \Lambda} \Va(I_\lambda)$. + On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$ we have $f = \sum_{\lambda \in \Lambda} f_\lambda$. Thus $f$ vanishes on $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ and we have $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda) \subseteq \Va(\sum_{\lambda \in \Lambda} I_\lambda)$. \end{enumerate} \end{proof} \begin{remark} @@ -458,7 +460,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an algebraically \subsubsection{Definition of the Zariski topology} Let $\mathfrak{k}$ be algebraically closed, $R = \mathfrak{k}[X_1,\ldots,X_n]$. \begin{corollary} (of \ref{fvop}) - There is a topology on $\mathfrak{k}^n$ for which the set of closed sets coincides with the set $\fA$ of subsets of the form $\Va\left(I \right) $ for ideals $I \se R$. + There is a topology on $\mathfrak{k}^n$ for which the set of closed sets coincides with the set $\fA$ of subsets of the form $\Va\left(I \right) $ for ideals $I \subseteq R$. This topology is called the \vocab{Zariski-Topology} \end{corollary} @@ -472,9 +474,9 @@ As $\Va(0) = \mathfrak{k}$ and $\Va(P)$ finite for $P \neq 0$ and $\{x_1,\ldots, \begin{definition} Let $X$ be a topological space. $X$ satisfies the separation properties $T_{0-2}$ if for any $x \neq y \in X$ \begin{enumerate} - \item[$T_0$ ] $\E U \se X$ open such that $|U \cap \{x,y\}| = 1$ - \item[$T_1$ ] $\E U \se X$ open such that $x \in U, y \not\in U$. - \item[$T_2$ ] There are disjoined open sets $U, V \se X$ such that $x \in U, y \in V$. (Hausdorff) + \item[$T_0$ ] $\E U \subseteq X$ open such that $|U \cap \{x,y\}| = 1$ + \item[$T_1$ ] $\E U \subseteq X$ open such that $x \in U, y \not\in U$. + \item[$T_2$ ] There are disjoined open sets $U, V \subseteq X$ such that $x \in U, y \in V$. (Hausdorff) \end{enumerate} \end{definition} \begin{remark} @@ -502,15 +504,15 @@ Let $X$ be a topological space. \begin{enumerate}[A] \item Every open subset of $X$ is quasi-compact. \item Every descending sequence $A_0 \supseteq A_1 \supseteq \ldots$ of closed subsets of $X$ stabilizes. - \item Every non-empty set $\cM$ of closed subsets of $X$ has a $\se$-minimal element. + \item Every non-empty set $\cM$ of closed subsets of $X$ has a $\subseteq$-minimal element. \end{enumerate} \end{definition} \begin{proof}\, \begin{enumerate} \item[A $\implies$ B] Let $A_j$ be a descending chain of closed subsets. Define $A \coloneqq \bigcap_{j = 0}^{\infty} A_j$. If A holds, the covering $X \sm A = \bigcup_{j = 0}^{\infty} (X \sm A_j)$ has a finite subcovering. - \item[B $\implies$ C] Suppose $\cM$ does not have a $\se$-minimal element. Using DC, one can construct a counterexample $A_1 \subsetneq A_2 \supsetneq \ldots$ to B. - \item[C $\implies$ A] Let $\bigcup_{i \in I} V_i$ be an open covering of an open subset $U \se X$. - By C, the set $\cM \coloneqq \{X \sm \bigcup_{i \in F} V_i | F \se I \text{ finite} \}$ has a $\se$-minimal element. + \item[B $\implies$ C] Suppose $\cM$ does not have a $\subseteq$-minimal element. Using DC, one can construct a counterexample $A_1 \subsetneq A_2 \supsetneq \ldots$ to B. + \item[C $\implies$ A] Let $\bigcup_{i \in I} V_i$ be an open covering of an open subset $U \subseteq X$. + By C, the set $\cM \coloneqq \{X \sm \bigcup_{i \in F} V_i | F \subseteq I \text{ finite} \}$ has a $\subseteq$-minimal element. \end{enumerate} \end{proof} @@ -518,12 +520,12 @@ Let $X$ be a topological space. Let $\mathfrak{k}$ be algebraically closed, $R = \mathfrak{k}[X_1,\ldots,X_n]$. For $f \in R$ let $V(f) = V(fR)$. \begin{theorem}[Hilbert's Nullstellensatz (3)] \label{hns3} - Let $I \se R$ be an ideal. Then $V(I) \se V(f)$ iff $f \in \sqrt{I}$. + Let $I \subseteq R$ be an ideal. Then $V(I) \subseteq V(f)$ iff $f \in \sqrt{I}$. \end{theorem} \begin{proof} Suppose $f$ vanishes on all zeros of $I$. Let $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n,T]$, $g(X_1,\ldots,X_n,T) \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)$ - and $J \se R'$ the ideal generated by $g$ and the elements of $I$ (viewed as elements of $R'$ which are constant in the $T$-direction). + and $J \subseteq R'$ the ideal generated by $g$ and the elements of $I$ (viewed as elements of $R'$ which are constant in the $T$-direction). If $f$ vanishes on all zeros of $I$, then $J$ has no zeros in $\mathfrak{k}^{n+1}$. @@ -544,11 +546,11 @@ For $f \in R$ let $V(f) = V(fR)$. \begin{corollary}\label{antimonbij} \begin{align} - f: \{I \se R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \se \mathfrak{k}^n | A \text{ Zariski-closed}\} \\ + f: \{I \subseteq R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\} \\ I &\longmapsto V(I)\\ - \{f \in R | A \se V(f)\} &\longmapsfrom A + \{f \in R | A \subseteq V(f)\} &\longmapsfrom A \end{align} - is a $\se$-antimonotonic bijection. + is a $\subseteq$-antimonotonic bijection. \end{corollary} \begin{corollary} The topological space $\mathfrak{k}^n$ is Noetherian. @@ -568,9 +570,9 @@ Let $X$ be a topological space. \begin{definition} $X$ is called \vocab[Topological space!irreducible]{irreducible}, if $X \neq \emptyset$ and the following equivalent conditions hold: \begin{enumerate}[A] - \item Every open $\emptyset \neq U \se X$ is dense. - \item The intersection of non-empty, open subsets $U, V \se X$ is non-empty. - \item If $A, B \se X$ are closed, $X = A \cup B$ then $X = A$ or $X = B$. + \item Every open $\emptyset \neq U \subseteq X$ is dense. + \item The intersection of non-empty, open subsets $U, V \subseteq X$ is non-empty. + \item If $A, B \subseteq X$ are closed, $X = A \cup B$ then $X = A$ or $X = B$. \item Every open subset of $X$ is connected. \end{enumerate} \end{definition} @@ -608,7 +610,7 @@ Let $X$ be a topological space. \subsubsection{Irreducible components} \begin{fact} - If $D \se X$ is dense, then $X$ is irreducible iff $D$ is irreducible with its induced topology. + If $D \subseteq X$ is dense, then $X$ is irreducible iff $D$ is irreducible with its induced topology. \end{fact} \begin{proof} $X = \emptyset$ iff $D = \emptyset$. @@ -617,12 +619,12 @@ Let $X$ be a topological space. On the other hand, if $U$ and $V$ are disjoint non-empty open subsets of $X$, then $U \cap D$ and $V \cap D$ are disjoint non-empty open subsets of $D$. \end{proof} \begin{definition}[Irreducible subsets] - A subset $Z \se X$ is called \vocab[Topological space!irreducible]{irreducible} if it is irreducible with its induced topology. - $Z$ is called an \vocab{irreducible component} of $X$, if it is irreducible and if every irreducible subset $Z \se Y \se X$ coincides with $Z$. + A subset $Z \subseteq X$ is called \vocab[Topological space!irreducible]{irreducible} if it is irreducible with its induced topology. + $Z$ is called an \vocab{irreducible component} of $X$, if it is irreducible and if every irreducible subset $Z \subseteq Y \subseteq X$ coincides with $Z$. \end{definition} \begin{corollary} \begin{enumerate} - \item $Z \se X$ is irreducible iff $\overline{Z} \se X$ is irreducible. + \item $Z \subseteq X$ is irreducible iff $\overline{Z} \subseteq X$ is irreducible. \item Every irreducible component of $X$ is a closed subset of $X$. \end{enumerate} \end{corollary} @@ -639,24 +641,24 @@ Let $X$ be a topological space. \begin{proof} % i = ic Let $\fM$ be the set of closed subsets of $X$ which cannot be decomposed as a union of finitely many irreducible subsets. - Suppose $\fM \neq \emptyset$. Then there exists a $\se$-minimal $Y \in \fM$. $Y$ cannot be empty or irreducible. Hence $Y = A \cup B$ where $A,B$ are proper closed subsets of $ Y$. By the minimality of $Y$, $A$ and $B$ can be written as a union of proper closed subsets $\lightning$. + Suppose $\fM \neq \emptyset$. Then there exists a $\subseteq$-minimal $Y \in \fM$. $Y$ cannot be empty or irreducible. Hence $Y = A \cup B$ where $A,B$ are proper closed subsets of $ Y$. By the minimality of $Y$, $A$ and $B$ can be written as a union of proper closed subsets $\lightning$. Let $X = \bigcup_{i = 1}^n Z_i$, where there are no inclusions between the $Z_i$. If $Y$ is an irreducible subsets of $X$, $Y = \bigcup_{i = 1}^n (Y \cap Z_i)$ and there exists $1 \le i \le n$ such that $Y = Y \cap Z_i$. - Hence $Y \se Z_i$. Thus the $Z_i$ are irreducible components. Conversely, if $Y$ is an irreducible component of $X$, $Y \se Z_i$ for some $i$ and $Y = Z_i$ by the definition of irreducible component. + Hence $Y \subseteq Z_i$. Thus the $Z_i$ are irreducible components. Conversely, if $Y$ is an irreducible component of $X$, $Y \subseteq Z_i$ for some $i$ and $Y = Z_i$ by the definition of irreducible component. \end{proof} \begin{remark} - The proof of existence was an example of \vocab{Noetherian induction} : If $E$ is an assertion about closed subsets of a Noetherian topological space $X$ and $E$ holds for $A$ if it holds for all proper subsets of $A$, then $E(A)$ holds for every closed subset $A \se X$. + The proof of existence was an example of \vocab{Noetherian induction} : If $E$ is an assertion about closed subsets of a Noetherian topological space $X$ and $E$ holds for $A$ if it holds for all proper subsets of $A$, then $E(A)$ holds for every closed subset $A \subseteq X$. \end{remark} \begin{proposition}\label{bijiredprim} By \ref{antimonbij} there exists a bijection \begin{align} - f: \{I \se R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \se \mathfrak{k}^n | A \text{ Zariski-closed}\} \\ + f: \{I \subseteq R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\} \\ I &\longmapsto V(I)\\ - \{f \in R | A \se V(f)\} &\longmapsfrom A + \{f \in R | A \subseteq V(f)\} &\longmapsfrom A \end{align} - Under this correspondence $A \se \mathfrak{k}^n$ is irreducible iff $I \coloneqq f\inv(A)$ is a prime ideal. + Under this correspondence $A \subseteq \mathfrak{k}^n$ is irreducible iff $I \coloneqq f\inv(A)$ is a prime ideal. Moreover, $\#A = 1$ iff $I$ is a maximal ideal. \end{proposition} \begin{proof} @@ -665,16 +667,16 @@ Let $X$ be a topological space. On the other hand suppose that $fg \in I, f \notin I, g \not\in I$. By the Nullstellensatz (\ref{hns3}) and $I = \sqrt{I} $ neither $f$ nor $g$ vanishes on all of $A$. Thus $(A \cap V(f)) \cup (A \cap V(g))$ is a decomposition and $A$ fails to be irreducible. - The remaining assertion follows from the fact, that the bijection is $\se$-antimonotonic and thus maximal ideals correspond to minimal irreducible closed subsets, which are the one-point subsets as $\mathfrak{k}^n$ is T${}_1$. + The remaining assertion follows from the fact, that the bijection is $\subseteq$-antimonotonic and thus maximal ideals correspond to minimal irreducible closed subsets, which are the one-point subsets as $\mathfrak{k}^n$ is T${}_1$. \end{proof} \subsection{Krull dimension} \begin{definition} - Let $Z $ be an irreducible subset of the topological space $X$. Let $\codim(Z,X)$ be the maximum of the length $n$ of strictly increasing chains $Z \se Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n$ of irreducible closed subsets of $X$ containing $Z$ or $\infty$ if such chains can be found for arbitrary $n$. + Let $Z $ be an irreducible subset of the topological space $X$. Let $\codim(Z,X)$ be the maximum of the length $n$ of strictly increasing chains $Z \subseteq Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n$ of irreducible closed subsets of $X$ containing $Z$ or $\infty$ if such chains can be found for arbitrary $n$. Let \[ \dim X \coloneqq \begin{cases} - \infty &\text{if } X = \emptyset\\ - \sup_{\substack{Z \se X\\ Z \text{ irreducible}}} \codim(Z,X) & \text{otherwise} + \sup_{\substack{Z \subseteq X\\ Z \text{ irreducible}}} \codim(Z,X) & \text{otherwise} \end{cases} \] \end{definition} @@ -693,9 +695,9 @@ Let $X$ be a topological space. For every $x \in \mathfrak{k}$, $\codim( \{x\} ,\mathfrak{k}) = 1$. The only other irreducible closed subset of $\mathfrak{k}$ is $\mathfrak{k}$ itself, which has codimension zero. Thus $\dim \mathfrak{k} = 1$. \end{fact} \begin{fact} - Let $Y \se X$ be irreducible and $U \se X$ an open subset such that $U \cap Y \neq \emptyset$. Then we have a bijection + Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that $U \cap Y \neq \emptyset$. Then we have a bijection \begin{align} - f: \{A \se X | A \text{ irreducible, closed and } Y \se A\} &\longrightarrow \{B \se U | B \text{ irreducible, closed and } Y \cap U \se B\} \\ + f: \{A \subseteq X | A \text{ irreducible, closed and } Y \subseteq A\} &\longrightarrow \{B \subseteq U | B \text{ irreducible, closed and } Y \cap U \subseteq B\} \\ A&\longmapsto A \cap U\\ \overline{B}&\longmapsfrom B \end{align} @@ -705,11 +707,11 @@ Let $X$ be a topological space. If $A$ is given and $B = A \cap U$, then $B \neq \emptyset$ and B is open hence (irreducibility of $A$) dense in $A$, hence $A = \overline{B}$. The fact that $B = \overline{B} \cap U$ is a general property of the closure operator. \end{proof} \begin{corollary}[Locality of Krull codimension] \label{lockrullcodim} - Let $Y \se X$ be irreducible and $U \se X$ an open subset such that $U \cap Y \neq \emptyset$. + Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that $U \cap Y \neq \emptyset$. Then $\codim(Y,X) = \codim(Y \cap U, U)$. \end{corollary} \begin{fact} - Let $Z \se Y \se X$ be irreducible closed subsets of the topological space $X$. Then + Let $Z \subseteq Y \subseteq X$ be irreducible closed subsets of the topological space $X$. Then \[ \codim(Z,Y) + \codim(Y,X) \le \codim(Z,X) \tag{CD+}\label{eq:cdp} \] @@ -756,15 +758,15 @@ In general, these inequalities may be strict. \end{proof} \begin{proposition}[Irreducible subsets of codimension one]\label{irredcodimone} - Let $p \in R = \mathfrak{k}[X_1,\ldots, X_n]$ be a prime element. Then the irreducible subset $X = V(p) \se \mathfrak{k}^n$ has codimension one, and every codimension one subset of $\mathfrak{k}^n$ has this form. + Let $p \in R = \mathfrak{k}[X_1,\ldots, X_n]$ be a prime element. Then the irreducible subset $X = V(p) \subseteq \mathfrak{k}^n$ has codimension one, and every codimension one subset of $\mathfrak{k}^n$ has this form. \end{proposition} \begin{proof} Since $pR$ is a prime ideal, $X = V(p)$ is irreducible. Since $p \neq 0$, $X$ is a proper subset of $\mathfrak{k}^n$. - If $X \se Y \se \mathfrak{k}^n$ is irreducible and closed, then $Y = V(\fq)$ for some prime ideal $\fp \se pR$. - If $Y \neq \mathfrak{k}^n$, then $\fp \neq \{0\}$. By \ref{ufdprimeideal} there exists a prime element $q \in \fq$. As $\fq \se pR$ we have $p \divides q$. + If $X \subseteq Y \subseteq \mathfrak{k}^n$ is irreducible and closed, then $Y = V(\fq)$ for some prime ideal $\fp \subseteq pR$. + If $Y \neq \mathfrak{k}^n$, then $\fp \neq \{0\}$. By \ref{ufdprimeideal} there exists a prime element $q \in \fq$. As $\fq \subseteq pR$ we have $p \divides q$. By the irreducibility of $p$ and $q$ it follows that $p \sim q$. Hence $\fq = pR$ and $X = Y$. - Suppose $X = V(\fp) \se \mathfrak{k}^n$ is closed, irreducible and of codimension one. + Suppose $X = V(\fp) \subseteq \mathfrak{k}^n$ is closed, irreducible and of codimension one. Then $\fp \neq \{0\}$, hence $X \neq \mathfrak{k}^n$. By \ref{ufdprimeideal} there is a prime element $p \in \fp$. If $\fp \neq pR$, then $X \subsetneq V(p) \subsetneq \mathfrak{k}^n$ contradicts $\codim(X, \mathfrak{k}^n) = 1$. \end{proof} @@ -776,17 +778,17 @@ In general, these inequalities may be strict. Let $X$ be a set, $\cP(X)$ the power set of $X$. A \vocab{Hull operator} on $X$ is a map $\cP(X) \xrightarrow{\cH} \cP(X)$ such that \begin{enumerate} - \item[H1] $\A A \in \cP(X) ~ A \se \cH(A)$. - \item[H2] $A \se B \se X \implies \cH(A) \se \cH(B)$. + \item[H1] $\A A \in \cP(X) ~ A \subseteq \cH(A)$. + \item[H2] $A \subseteq B \subseteq X \implies \cH(A) \subseteq \cH(B)$. \item[H3] $\cH(\cH(X)) = \cH(X)$. \end{enumerate} We call $\cH$ \vocab{matroidal} if in addition the following conditions hold: \begin{enumerate} - \item[M] If $m,n \in X$ and $A \se X$ then $m \in \cH( \{n\} \cup A) \sm \cH(A) \iff n \in \cH(\{m\} \cup A) \sm \cH(A).$ - \item[F] $\cH(A) = \bigcup_{F \se A \text{ finite}} \cH(F)$. + \item[M] If $m,n \in X$ and $A \subseteq X$ then $m \in \cH( \{n\} \cup A) \sm \cH(A) \iff n \in \cH(\{m\} \cup A) \sm \cH(A).$ + \item[F] $\cH(A) = \bigcup_{F \subseteq A \text{ finite}} \cH(F)$. \end{enumerate} - In this case, $S \se X$ is called \vocab{Independent subset}, if $s \not\in \cH(S \sm \{s\})$ for all $s \in S$ and + In this case, $S \subseteq X$ is called \vocab{Independent subset}, if $s \not\in \cH(S \sm \{s\})$ for all $s \in S$ and \vocab[Generating subset]{generating} if $X = \cH(S)$. $S$ is called a \vocab{base}, if it is both generating and independent. \end{definition} @@ -797,7 +799,7 @@ In general, these inequalities may be strict. \begin{example} - Let $K$ be a field, $V$ a $K$-vector space and $\cL(T)$ the $K$-linear hull of $T$ for $T \se V$. + Let $K$ be a field, $V$ a $K$-vector space and $\cL(T)$ the $K$-linear hull of $T$ for $T \subseteq V$. Then $\cL$ is a matroidal hull operator on $V$. \end{example} @@ -807,10 +809,10 @@ In general, these inequalities may be strict. Then $\cH$ is a matroidal hull operator. \end{lemma} \begin{proof} - H1, H2 and F are trivial. For an algebraically closed subfield $K \se M \se L$ we have $\cH(M) = M$. Thus $\cH(\cH(T)) = \cH(T)$ (H3). + H1, H2 and F are trivial. For an algebraically closed subfield $K \subseteq M \subseteq L$ we have $\cH(M) = M$. Thus $\cH(\cH(T)) = \cH(T)$ (H3). - Let $x,y \in L$, $T \se L$ and $x \in \cH(T \cup \{y\}) \sm \cH(T)$. We have to show that $y \in \cH(T \cup \{x\}) \sm \cH(T)$. - If $y \in \cH(T)$ we have $\cH(T \cup \{y\}) \se \cH(\cH(T)) = \cH(T) \implies x \in \cH(T) \sm \cH(T) \lightning$. + Let $x,y \in L$, $T \subseteq L$ and $x \in \cH(T \cup \{y\}) \sm \cH(T)$. We have to show that $y \in \cH(T \cup \{x\}) \sm \cH(T)$. + If $y \in \cH(T)$ we have $\cH(T \cup \{y\}) \subseteq \cH(\cH(T)) = \cH(T) \implies x \in \cH(T) \sm \cH(T) \lightning$. Hence it is sufficient to show $y \in \cH(T \cup \{x\})$. \Wlog $T = \emptyset$ (replace $K$ be the subfield generated by $K \cup T$). Then $x$ is algebraic over the subfield $M$ of $L$ generated by $K \cup \{y\}$. Thus there exists $0 \neq P \in M[T]$ with $P(x) = 0$. The coefficients $p_i$ of $P$ belong to the field of quotients of the $K$-subalgebra of $L$ generated by $y$. There are thus polynomials $Q_i, R \in K[Y]$ such that $p_i = \frac{Q_i(y)}{R(y)}$, $R(y) \neq 0$. @@ -838,7 +840,7 @@ The following will lead to another proof of the Nullstellensatz, which uses the Let $A$ be a subring of the Noetherian ring $B$. If the ring extension $B / A$ is finite (i.e. $B$ finitely generated as an $A$-module) then $A$ is Noetherian. \end{theorem} \begin{dfact}\label{noethersubalg} - Let $R$ be Noetherian and let $B$ be a finite $R$-algebra. Then every $R$-subalgebra $A \se B$ is finite over $R$. + Let $R$ be Noetherian and let $B$ be a finite $R$-algebra. Then every $R$-subalgebra $A \subseteq B$ is finite over $R$. \end{dfact} \begin{proof} Since $B$ a finitely generated $R$-module and $R$ a Noetherian ring, $B$ is a Noetherian $R$-module (this is a stronger assertion than Noetherian algebra). @@ -852,7 +854,7 @@ The following will lead to another proof of the Nullstellensatz, which uses the \begin{figure}[H] \centering \begin{tikzcd} - A \arrow[hookrightarrow]{rr}{\se}& & B \\ + A \arrow[hookrightarrow]{rr}{\subseteq}& & B \\ &R \arrow{ul}{\alpha} \arrow{ur}{\alpha} \text{~(Noeth.)} \end{tikzcd} \end{figure} @@ -861,12 +863,12 @@ The following will lead to another proof of the Nullstellensatz, which uses the \begin{proof} Let $(b_i)_{i=1}^{m}$ generate $B$ as an $A$-module and $(\beta_j)_{j=1}^m$ as an $R$-algebra. There are $a_{ijk} \in A$ such that $b_i b_j = \sum_{k=1}^{m} a_{ijk}b_k$. And $\alpha_{ij} \in A$ such that $\beta_i = \sum_{j=1}^{m} \alpha_{ij}b_j$. Let $\tilde{A}$ be the sub- $R$-algebra of $A$ generated by the $a_{ijk}$ and $\alpha_{ij}$. $\tilde{A}$ is of finite type over $ R$, hence Noetherian. The $\tilde{A}$-submodule generated by $1$ and the $b_i$ is a sub-$R$-algebra containing the $\beta_i$ and thus coincides with $B$. - Hence $B / \tilde{A}$ is finite. Since $A \se B, A / \tilde{A}$ is finite (\ref{noethersubalg}). + Hence $B / \tilde{A}$ is finite. Since $A \subseteq B, A / \tilde{A}$ is finite (\ref{noethersubalg}). Hence $A / \tilde{A}$ is of finite type. By the transitivity of ``of finite type'', it follows that $A / R$ is of finite type. \begin{figure}[H] \centering \begin{tikzcd} - \tilde A \arrow[hookrightarrow]{r}{\se}& A \arrow[hookrightarrow]{r}{\se} & B \\ + \tilde A \arrow[hookrightarrow]{r}{\subseteq}& A \arrow[hookrightarrow]{r}{\subseteq} & B \\ &R \arrow[bend left, swap]{ul}{\alpha} \arrow{u}{\alpha} \arrow[bend right]{ur}{\alpha} \end{tikzcd} \end{figure} @@ -904,7 +906,7 @@ The Nullstellensatz (\ref{hns2}) can be reduced to the case of \ref{rfuncnft}: \begin{proof}(Artin-Tate proof of HNS) Let $(l_i)_{i=1}^n$ be a transcendence base of $L / K$. If $n = 0$ then $L / K$ is algebraic, hence an integral ring extension, hence a finite ring extension (\ref{ftaiimplf}). - Suppose $n > 0$. Let $\tilde R \se L$ be the $K$-subalgebra generated by the $l_i$. $\tilde R \cong R \coloneqq K[X_1,\ldots,X_n]$, as the $l_i$ are algebraically independent. + Suppose $n > 0$. Let $\tilde R \subseteq L$ be the $K$-subalgebra generated by the $l_i$. $\tilde R \cong R \coloneqq K[X_1,\ldots,X_n]$, as the $l_i$ are algebraically independent. As they are a transcendence base, $L$ is algebraic over the field of quotients $Q(\tilde R)$, hence integral over $Q(\tilde R)$. As $L / K$ is of finite type, so is $L / Q(\tilde R)$ and it follows that $L / Q(\tilde R)$ is a finite ring extension. @@ -916,15 +918,15 @@ The Nullstellensatz (\ref{hns2}) can be reduced to the case of \ref{rfuncnft}: Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. %i = ic \begin{notation} - Let $X \se \mathfrak{k}^n$ be an irreducible closed subset. Then $X = V(\fp)$ for a unique prime ideal $\fp \se R$. + Let $X \subseteq \mathfrak{k}^n$ be an irreducible closed subset. Then $X = V(\fp)$ for a unique prime ideal $\fp \subseteq R$. Let $\fK(X) \coloneqq Q(R / \fp)$ denote the field of quotients of $R / \fp$. \end{notation} \begin{remark} As the elements of $\fp$ vanish on $X$, $R / \fp$ may be viewed as the ring of polynomials and $\fK(X)$ as the field of rational functions on $X$. \end{remark} \begin{theorem}\label{trdegandkdim} - If $X \se \mathfrak{k}^n$ is irreducible, then $\dim X = \trdeg (\mathfrak{k}(X) / \mathfrak{k})$ and $\codim(X, \mathfrak{k}^n) = n - \trdeg(\fK(X) / \mathfrak{k})$. - More generally if $Y \se \mathfrak{k}^n$ is irreducible and $X \se Y$, then $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$. + If $X \subseteq \mathfrak{k}^n$ is irreducible, then $\dim X = \trdeg (\mathfrak{k}(X) / \mathfrak{k})$ and $\codim(X, \mathfrak{k}^n) = n - \trdeg(\fK(X) / \mathfrak{k})$. + More generally if $Y \subseteq \mathfrak{k}^n$ is irreducible and $X \subseteq Y$, then $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$. \end{theorem} \begin{proof} % DIMT @@ -950,8 +952,8 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. \begin{definition}[Spectrum] Let $R$ be a commutative ring. \begin{itemize} - \item Let $\Spec R$ denote the set of prime ideals and $\mSpec R \se \Spec R$ the set of maximal ideals of $R$. - \item For an ideal $I \se R$ let $V(I) \coloneqq \{\fp \in \Spec R | I \se \fp\}$ + \item Let $\Spec R$ denote the set of prime ideals and $\mSpec R \subseteq \Spec R$ the set of maximal ideals of $R$. + \item For an ideal $I \subseteq R$ let $V(I) \coloneqq \{\fp \in \Spec R | I \subseteq \fp\}$ \item We equip $\Spec R$ with the \vocab{Zariski-Topology} for which the closed subsets are the subsets of the form $V(I)$, where $I$ runs over the set of ideals in $R$. \end{itemize} \end{definition} @@ -970,11 +972,11 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. \subsection{Localization of rings} \begin{definition}[Multiplicative subset] - A \vocab{multiplicative subset} of a ring $R$ is a subset $S \se R$ such that $\prod_{i=1}^{n} f_i \in S $ when $n \in \N$ and all $f_i \in S$. + A \vocab{multiplicative subset} of a ring $R$ is a subset $S \subseteq R$ such that $\prod_{i=1}^{n} f_i \in S $ when $n \in \N$ and all $f_i \in S$. \end{definition} \begin{proposition} - Let $S \se R$ be a multiplicative subset. Then there is a ring homomorphism $R \xrightarrow{i} R_S$ such that $i(S) \se R_S^{\times }$ and $i$ has the \vocab{universal property} for such ring homomorphisms: - If $R \xrightarrow{j} T$ is a ring homomorphism with $j(S) \se T^{\times }$, then there is a unique ring homomorphism $R_S \xrightarrow{\iota} T$ with $j = \iota i$. + Let $S \subseteq R$ be a multiplicative subset. Then there is a ring homomorphism $R \xrightarrow{i} R_S$ such that $i(S) \subseteq R_S^{\times }$ and $i$ has the \vocab{universal property} for such ring homomorphisms: + If $R \xrightarrow{j} T$ is a ring homomorphism with $j(S) \subseteq T^{\times }$, then there is a unique ring homomorphism $R_S \xrightarrow{\iota} T$ with $j = \iota i$. \begin{figure}[H] \centering @@ -1000,21 +1002,21 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. In particular $(r = 1)$, $R_S$ is the null ring iff $0 \in S$. \end{remark} \begin{notation} - Let $S \se R$ be a multiplicative subset of $R$. We write $\frac{r}{s}$ for $[r,s]$. + Let $S \subseteq R$ be a multiplicative subset of $R$. We write $\frac{r}{s}$ for $[r,s]$. The ring homomorphism $R \xrightarrow{i} R_S$ i given by $i(r) = \frac{r}{1}$. - For $X \se R_S$ let $X \sqcap R$ denote $i\inv(X)$. + For $X \subseteq R_S$ let $X \sqcap R$ denote $i\inv(X)$. \end{notation} \begin{definition}[$S$-saturated ideal] - An ideal $I \se R$ is called \vocab[Ideal!S-saturated]{$S$-saturated} if for all $s \in S, r \in R$ + An ideal $I \subseteq R$ is called \vocab[Ideal!S-saturated]{$S$-saturated} if for all $s \in S, r \in R$ $rs \in I \implies r \in I$. \end{definition} \begin{fact}\label{primeidealssat} - A prime ideal $\fp \se \Spec R$ is $S$-saturated iff $\fp \cap S = \emptyset$. + A prime ideal $\fp \subseteq \Spec R$ is $S$-saturated iff $\fp \cap S = \emptyset$. \end{fact} Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$. \begin{fact}\label{ssatiis} - Let $I \se R$ be an $S$-saturated ideal and let $I_S$ denote the ideal $\{\frac{r}{s} | r \in R, s \in S\} \se R_S$. + Let $I \subseteq R$ be an $S$-saturated ideal and let $I_S$ denote the ideal $\{\frac{r}{s} | r \in R, s \in S\} \subseteq R_S$. Then for all $r \in R, s \in S$ we have $\frac{r}{s} \in I_S \iff r \in I$. \end{fact} @@ -1028,7 +1030,7 @@ Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-satura \begin{proposition}\label{idealslocbij} \begin{align} - f: \{I \se R | I \text{ $S$-saturated ideal}\} &\longrightarrow \left\{J \se R_S | J \text{ ideal}\right\} \\ + f: \{I \subseteq R | I \text{ $S$-saturated ideal}\} &\longrightarrow \left\{J \subseteq R_S | J \text{ ideal}\right\} \\ I &\longmapsto I_S \coloneqq \left\{\frac{i}{s} | i \in I, s \in S\right\}\\ J \sqcap R &\longmapsfrom J\\ \end{align} @@ -1054,7 +1056,7 @@ Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-satura \begin{remark}\label{locandquot} Let $R$ be a domain. If $S = R \sm \{0\}$, then $R_S$ is the field of quotients $Q(R)$. - If $S \se R \sm \{0\} $, then + If $S \subseteq R \sm \{0\} $, then \[ R_S \cong \left\{ \frac{a}{s} \in K | a \in R, s \in S\right\} \] @@ -1062,7 +1064,7 @@ Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-satura \end{remark} \begin{definition}[$S$-saturation]\label{ssaturation} - Let $R$ be any ring, $I \se R$ an ideal. Even if $I$ is not $S$-saturated, $J = I_S \coloneqq \{\frac{i}{s} | i \in I, s \in S\}$ is an ideal in $R_S$, and $I_S \sqcap R = \{r \in R | s\cdot r \in I, s \in S\}$ is called the \vocab[Ideal!$S$-saturation]{$S$-saturation of $I$ } which is the smallest $S$-saturated ideal containing $I$. + Let $R$ be any ring, $I \subseteq R$ an ideal. Even if $I$ is not $S$-saturated, $J = I_S \coloneqq \{\frac{i}{s} | i \in I, s \in S\}$ is an ideal in $R_S$, and $I_S \sqcap R = \{r \in R | s\cdot r \in I, s \in S\}$ is called the \vocab[Ideal!$S$-saturation]{$S$-saturation of $I$ } which is the smallest $S$-saturated ideal containing $I$. \end{definition} @@ -1070,9 +1072,9 @@ Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-satura In the situation of \ref{ssaturation}, if $\overline{S}$ denotes the image of $S$ in $R / I$, there is a canonical isomorphism $R_S / I_S \cong (R / I)_{\overline{S}}$. \end{lemma} \begin{proof} - We show that both rings have the universal property for ring homomorphisms $R \xrightarrow{\tau} T$ with $\tau(I) = \{0\} $ and $\tau(S) \se T^{\times }$. + We show that both rings have the universal property for ring homomorphisms $R \xrightarrow{\tau} T$ with $\tau(I) = \{0\} $ and $\tau(S) \subseteq T^{\times }$. For such $\tau$, by the fundamental theorem on homomorphisms (Homomorphiesatz) there is a unique $R/I \xrightarrow{\tau_1} T$ such that $\tau = \tau_1 \pi_{R,I}$. - We have $\tau_1(\overline{S}) = \tau(S) \se T^{\times }$, hence there is a unique $(R / I)_{\overline{S}} \xrightarrow{\tau_2} T$ such that the composition $R / I \to (R / I)_{\overline{S}} \xrightarrow{\tau_2} T $ equals $\tau_1$. It is easy to see that this is the only one for which $R \to R / I \to (R / I)_{\overline{S}} \xrightarrow{\tau_2} T$ equals $\tau$. + We have $\tau_1(\overline{S}) = \tau(S) \subseteq T^{\times }$, hence there is a unique $(R / I)_{\overline{S}} \xrightarrow{\tau_2} T$ such that the composition $R / I \to (R / I)_{\overline{S}} \xrightarrow{\tau_2} T $ equals $\tau_1$. It is easy to see that this is the only one for which $R \to R / I \to (R / I)_{\overline{S}} \xrightarrow{\tau_2} T$ equals $\tau$. Similarly, by the universal property of $R_S$ there is a unique $R_S \xrightarrow{\tau_3} T$ whose composition with $R \to R_S$ equals $\tau$. @@ -1126,7 +1128,7 @@ Then \[ Let $\fq$ be any prime ideal. Take $a_1,\ldots,a_n \in A$ as in the lemma. As the $a_i \mod \fq$ are $\mathfrak{l}$-algebraically independent, the same holds for the $a_i$ themselves. Thus $\trdeg(Q(A) / \mathfrak{l}) \ge n$ and the inequality is strict, if it can be shown that the $a_i$ fail to be a transcendence base of $Q(A) / \mathfrak{l}$. - Let $R \se A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and $S \coloneqq R \sm \{0\} $. + Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and $S \coloneqq R \sm \{0\} $. We must show, that $Q(A)$ fails to be algebraic over $\mathfrak{l}_1 \coloneqq R_S = Q(R)$. Let $A_1 \coloneqq A_S$ and $\fq_S$ the prime ideal corresponding to $\fq$ as in \ref{idealslocbij}. We have $\fq_S \neq \{0\} $ as $\{0_{A}\}_S = \{0_{A_S}\}$. @@ -1137,7 +1139,7 @@ Then \[ \end{proof} \begin{corollary}\label{upperboundcodim} - Let $X, Y \se \mathfrak{k}^n$ be irreducible and closed. Then $\codim(X,Y) \le \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$. + Let $X, Y \subseteq \mathfrak{k}^n$ be irreducible and closed. Then $\codim(X,Y) \le \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$. \end{corollary} \begin{proof} Let $X = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_c = Y$ be a chain of irreducible closed subsets between $X$ and $Y$. @@ -1150,7 +1152,7 @@ Then \[ $$\codim(X,Y) \le \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$$ \end{proof} \begin{corollary}\label{upperbounddim} - Let $Z \se \mathfrak{k}^n$ be irreducible and closed. + Let $Z \subseteq \mathfrak{k}^n$ be irreducible and closed. Then \[\dim Z \le \trdeg(\fK(Z) / \mathfrak{k})\] and \[\codim(Z, \mathfrak{k}^n) \le n - \trdeg(\fK(Z) / \mathfrak{k}\] \end{corollary} \begin{proof} @@ -1171,7 +1173,7 @@ Then \[ Suppose $\mSpec R = \{\mathfrak{m}\}$. If $x \in \mathfrak{m}$, then $x \not\in R^{\times }$ as otherwise $xR = R \implies \mathfrak{m} = R$. If $x \not\in R^{\times }$ then $xR$ is a proper ideal, hence contained in some maximal ideal. Thus $x \in \mathfrak{m}$. - Assume that $\mathfrak{m} = R \sm R^{\times }$ is an ideal in $R$. As $1 \in R^{\times }$ this is a proper ideal. If $I$ is any proper ideal and $x \in I$, then $x \in \mathfrak{m}$. Hence $R = xR \se I \se \mathfrak{m}$. It follows that $\mathfrak{m}$ is the only maximal ideal of $R$. + Assume that $\mathfrak{m} = R \sm R^{\times }$ is an ideal in $R$. As $1 \in R^{\times }$ this is a proper ideal. If $I$ is any proper ideal and $x \in I$, then $x \in \mathfrak{m}$. Hence $R = xR \subseteq I \subseteq \mathfrak{m}$. It follows that $\mathfrak{m}$ is the only maximal ideal of $R$. \end{proof} \begin{remark} \begin{itemize} @@ -1187,7 +1189,7 @@ Many questons of commutative algebra are easier in the case of local rings. Loca We have a bijection \begin{align} - f: \Spec A_S &\longrightarrow \{\fq \in \Spec A | \fq \se \fp\} \\ + f: \Spec A_S &\longrightarrow \{\fq \in \Spec A | \fq \subseteq \fp\} \\ \fr &\longmapsto \fr \sqcap A\\ \fq_S \coloneqq \left\{\frac{q}{s} | q \in \fq, s \in S\right\} &\longmapsfrom \fq \end{align} @@ -1196,7 +1198,7 @@ Many questons of commutative algebra are easier in the case of local rings. Loca It is clear that $S$ is a multiplicative subset and that $\fp_S$ is an ideal. By \ref{ssatiis} $\frac{a}{s} \in \fp_S \iff a \in \fp \iff a \in A \sm S$ for all $a \in A, s \in S$. Thus, if $\frac{a}{s} \not\in \fp_S$ then it is a unit in $A_S$ with inverse $\frac{s}{a}$. Hence $A_S$ is a local ring with maximal ideal $\fp_S$. - The claim about $\Spec A_S$ follows from \ref{idealslocbij} using the fact (\ref{primeidealssat}) that a prime ideal $\fr \in \Spec A$ is $S$-saturated iff it is disjoint from $S = A \sm \fp$ iff $\fr \se \fp$. + The claim about $\Spec A_S$ follows from \ref{idealslocbij} using the fact (\ref{primeidealssat}) that a prime ideal $\fr \in \Spec A$ is $S$-saturated iff it is disjoint from $S = A \sm \fp$ iff $\fr \subseteq \fp$. \end{proof} \begin{definition} The ring $A_S$ as in \ref{locatprime} is called the \vocab[Localization]{localization of $A$ at the prime ideal $\fp$} and denoted $A_\fp$. @@ -1214,11 +1216,11 @@ Many questons of commutative algebra are easier in the case of local rings. Loca %Hence the name localization. \end{remark} \begin{remark} - Let $Y = V(\fp) \se \mathfrak{k}^n$ be an irreducible subset of $\mathfrak{k}^n$. Elements of $B_\fp$ are the fractions $\frac{b}{s}, s \not\in \fp$, i.e. $s$ does not vanish identically on $Y$. + Let $Y = V(\fp) \subseteq \mathfrak{k}^n$ be an irreducible subset of $\mathfrak{k}^n$. Elements of $B_\fp$ are the fractions $\frac{b}{s}, s \not\in \fp$, i.e. $s$ does not vanish identically on $Y$. Thus, $B_\fp$ is the ring of rational functions on $\mathfrak{k}^n$ which are well defined on some open subset $U$ intersecting $Y$. As $Y$ is irreducible, the intersection of two such subsets still intersects $Y$. \end{remark} \begin{remark} - For arbitrary $A$, we have a bijection $\Spec A_\fp \cong N = \{\fq \in \Spec A | \fp \se \fp\} $. One can show that $N$ is the intersection of all neighbourhoods of $\fp$ in $\Spec A$, confirming the intuition that ``the localization sees things which go on in arbitrarily small neighbourhoods of $\fp$''. + For arbitrary $A$, we have a bijection $\Spec A_\fp \cong N = \{\fq \in \Spec A | \fp \subseteq \fp\} $. One can show that $N$ is the intersection of all neighbourhoods of $\fp$ in $\Spec A$, confirming the intuition that ``the localization sees things which go on in arbitrarily small neighbourhoods of $\fp$''. \end{remark} \begin{remark} @@ -1230,27 +1232,27 @@ Many questons of commutative algebra are easier in the case of local rings. Loca \begin{definition}[Going-up and going-down]\label{goupgodown} Let $R$ be a ring and $A$ an $R$-algebra. - \vocab{Going-up} holds for $A / R$ if for arbitrary $\fq \in \Spec A$ and arbitrary $\tilde \fp \in\Spec R$ with $\tilde \fp \supseteq \fq \sqcap R$ there exists $\tilde \fq \in \Spec A$ with $\fq \se \tilde \fq$ and $\tilde \fp = \tilde \fq \sqcap R$. + \vocab{Going-up} holds for $A / R$ if for arbitrary $\fq \in \Spec A$ and arbitrary $\tilde \fp \in\Spec R$ with $\tilde \fp \supseteq \fq \sqcap R$ there exists $\tilde \fq \in \Spec A$ with $\fq \subseteq \tilde \fq$ and $\tilde \fp = \tilde \fq \sqcap R$. - (We are given $\fp \se \tilde \fp$ and $\fq$ such that $\fp = \fq \sqcap R$ and must make $\fq$ larger). + (We are given $\fp \subseteq \tilde \fp$ and $\fq$ such that $\fp = \fq \sqcap R$ and must make $\fq$ larger). \begin{figure}[H] \centering \begin{tikzcd} - \fq \arrow[mapsto]{d}{\cdot \sqcap R}& \se &{\color{blue}\tilde\fq}\arrow[mapsto]{d}{\cdot \sqcap R} & \in \Spec A\\ - \fq \sqcap R = \fp & \se & \tilde \fp & \in \Spec R + \fq \arrow[mapsto]{d}{\cdot \sqcap R}& \subseteq &{\color{blue}\tilde\fq}\arrow[mapsto]{d}{\cdot \sqcap R} & \in \Spec A\\ + \fq \sqcap R = \fp & \subseteq & \tilde \fp & \in \Spec R \end{tikzcd} \end{figure} -\vocab{Going-down} holds for $A / R$ if for arbitrary $\tilde \fq \in \Spec A$ and arbitrary $\fp \in \Spec R$ with $\fp \se \tilde \fq \sqcap R$, there exists $\fq \in \Spec A$ with $\fq \se \tilde \fq$ and $\fp = \fq \sqcap R$. +\vocab{Going-down} holds for $A / R$ if for arbitrary $\tilde \fq \in \Spec A$ and arbitrary $\fp \in \Spec R$ with $\fp \subseteq \tilde \fq \sqcap R$, there exists $\fq \in \Spec A$ with $\fq \subseteq \tilde \fq$ and $\fp = \fq \sqcap R$. -(We are given $\fp \se \tilde \fp$ and $\tilde \fq$ such that $\tilde \fp = \tilde \fq \sqcap R$ and must make $\tilde \fq$ smaller). +(We are given $\fp \subseteq \tilde \fp$ and $\tilde \fq$ such that $\tilde \fp = \tilde \fq \sqcap R$ and must make $\tilde \fq$ smaller). \begin{figure}[H] \centering \begin{tikzcd} - {\color{blue}\fq} \arrow[mapsto]{d}{\cdot \sqcap R}& \se &\tilde\fq\arrow[mapsto]{d}{\cdot \sqcap R} & \in \Spec A\\ - \fp & \se & \tilde \fp = \tilde \fq \sqcap R & \in \Spec R + {\color{blue}\fq} \arrow[mapsto]{d}{\cdot \sqcap R}& \subseteq &\tilde\fq\arrow[mapsto]{d}{\cdot \sqcap R} & \in \Spec A\\ + \fp & \subseteq & \tilde \fp = \tilde \fq \sqcap R & \in \Spec R \end{tikzcd} \end{figure} \end{definition} @@ -1261,7 +1263,7 @@ Many questons of commutative algebra are easier in the case of local rings. Loca \subsubsection{Going-up for integral ring extensions} \begin{theorem}[Krull, Cohen-Seidenberg] \label{cohenseidenberg} - Let $A$ be a ring and $R \se A$ a subring such that $A$ is integral over $R$. + Let $A$ be a ring and $R \subseteq A$ a subring such that $A$ is integral over $R$. \begin{enumerate}[A] \item The map $\Spec A \xrightarrow{\fq \mapsto \fq \cap R} \Spec R$ is surjective. \item For $\fp \in \Spec R$, there are no inclusions between the prime ideals $\fp \in \Spec A$ lying over $\fp$. @@ -1280,7 +1282,7 @@ Many questons of commutative algebra are easier in the case of local rings. Loca \begin{figure}[H] \centering \begin{tikzcd} - R \arrow{r}{\rho}\arrow[hookrightarrow]{d}{\se}& R_\fp \arrow[hookrightarrow, dotted]{d}{\Eone i}\\ + R \arrow{r}{\rho}\arrow[hookrightarrow]{d}{\subseteq}& R_\fp \arrow[hookrightarrow, dotted]{d}{\Eone i}\\ A \arrow{r}{\alpha} & A_\fp \end{tikzcd} \end{figure} @@ -1296,16 +1298,16 @@ Many questons of commutative algebra are easier in the case of local rings. Loca \[ \fq \cap R = \alpha\inv(\mathfrak{m}) \cap R = \rho\inv(i\inv (\mathfrak{m})) = \rho\inv(\fp_\fp) = \fp \] - \item[B] The map $\Spec A_\fp \xrightarrow{\alpha\inv} \Spec A$ is injective with image equal to $\{\fq \in \Spec A | \fq \cap R \se \fp\}$. In particular, it contains the set of all $\fq$ lying over $\fp$. + \item[B] The map $\Spec A_\fp \xrightarrow{\alpha\inv} \Spec A$ is injective with image equal to $\{\fq \in \Spec A | \fq \cap R \subseteq \fp\}$. In particular, it contains the set of all $\fq$ lying over $\fp$. If $\fq = \alpha\inv(\fr)$ lies over $\fp$, then \[\rho\inv(i\inv(\fr)) = (\alpha\inv(\fr)) \cap R = \fq \cap R = \fp = \rho\inv(\fp_\fp)\] hence $i\inv(\fr) = \fp_\fp$ by the injectivity of $\Spec R_\fp \xrightarrow{\rho\inv} \Spec R$. Because D applies to the integral ring extension $A_\fp / R_\fp$ and $\fp_\fp \in \mSpec R_\fp$, $\fr$ is a maximal ideal. - There are thus no inclusions between different such $\fr$. Because $\Spec A_\fp \xrightarrow{\alpha\inv} \Spec A$ is $\se$-monotonic and injective, there are no inclusions between different $\fp \in \Spec A$ lying over $\fp$. + There are thus no inclusions between different such $\fr$. Because $\Spec A_\fp \xrightarrow{\alpha\inv} \Spec A$ is $\subseteq$-monotonic and injective, there are no inclusions between different $\fp \in \Spec A$ lying over $\fp$. - \item[C] Let $\fp \se \tilde \fp$ be prime ideals of $R$ and $\fq \in \Spec A$ such that $\fq \cap R = \fp$. + \item[C] Let $\fp \subseteq \tilde \fp$ be prime ideals of $R$ and $\fq \in \Spec A$ such that $\fq \cap R = \fp$. By applying A to the ring extension $A / \fq$ of $R / \fp$, there is $\fr \in \Spec A /\fq$ such that $\fr \sqcap R / \fp = \tilde \fp / \fp$. - The preimage $\tilde \fq$ of $\fr$ under $A \to A / \fq$ satisfies $\fq \se \tilde \fq$ and $\tilde \fq \cap R = \tilde \fp$. + The preimage $\tilde \fq$ of $\fr$ under $A \to A / \fq$ satisfies $\fq \subseteq \tilde \fq$ and $\tilde \fq \cap R = \tilde \fp$. \end{enumerate} \end{proof} \begin{remark} @@ -1317,7 +1319,7 @@ This is part of the proof of \ref{trdegandkdim}. %It uses going-up. %TODO: relate to \ref{htandcodim} \begin{proof} - Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and let $X \se Y \se \mathfrak{k}^n$ be irreducible closed subsets of $\mathfrak{k}^n$. + Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and let $X \subseteq Y \subseteq \mathfrak{k}^n$ be irreducible closed subsets of $\mathfrak{k}^n$. We have to show $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) \sm \mathfrak{k})$. The inequality \[ @@ -1329,7 +1331,7 @@ This is part of the proof of \ref{trdegandkdim}. We have $Y = V(\fp)$ for a unique $\fp \in \Spec B$. Let $A = B / \fp$ be the ring of polynomials on $Y$. Apply the Noether normaization theorem to $A$. This yields $(f_i)_{i=1}^d \in A^d$ which are algebraically independent over $\mathfrak{k}$ and such that $A$ is finite over the subalgebra generated by the $f_i$. - Let $L$ be the algebraic closure in $\fK(Y)$ of the subfield of $\fK(Y)$ generated by $\mathfrak{k}$ and the $f_i$. We have $A \se L$ and since $\fK(Y) = Q(B / \fp) = Q(A)$\footnote{by definition} it follows that $\fK(Y) = L$. Hence $(f_i)_{i=1}^d$ is a transcendence base for $\fK(y) / \mathfrak{k}$ and $d = \trdeg \fK(Y) / \mathfrak{k}$. + Let $L$ be the algebraic closure in $\fK(Y)$ of the subfield of $\fK(Y)$ generated by $\mathfrak{k}$ and the $f_i$. We have $A \subseteq L$ and since $\fK(Y) = Q(B / \fp) = Q(A)$\footnote{by definition} it follows that $\fK(Y) = L$. Hence $(f_i)_{i=1}^d$ is a transcendence base for $\fK(y) / \mathfrak{k}$ and $d = \trdeg \fK(Y) / \mathfrak{k}$. \begin{align} @@ -1337,7 +1339,7 @@ This is part of the proof of \ref{trdegandkdim}. P &\longmapsto P(f_1,\ldots,f_d) \end{align} is an isomorphism and in $\mathfrak{k}[X_1,\ldots,X_d]$ there is a strictly ascending chain of prime ideals corresponding to $\mathfrak{k}^d \supsetneq \{0\} \times \mathfrak{k}^{d-1} \supsetneq \ldots \supsetneq \{0\}$. Thus there is a strictly ascending chain $\{0\} = \fp_0 \subsetneq \fp_1 \subsetneq \ldots \subsetneq \fp_d$ of elements of $\Spec R$. - Let $\fq_0 = \{0\} \in \Spec A$. If $0 < i \le d$ and a chain $\fq_0 \subsetneq \ldots \subsetneq \fq_{i-1}$ in $\Spec A$ with $\fq_j \cap R = \fp_j$ for $0 \le j < i$ has been selected, we may apply going-up (\ref{cohenseidenberg}) to $A / R$ to extend this chain by a $\fq_i \in \Spec A$ with $\fq_{i-1} \se \fq_i$ and $\fq_i \cap R = \fp_i$ (thus $\fq_{i-1} \subsetneq \fq_i$ as $\fp-i \neq \fp_{i-1})$. + Let $\fq_0 = \{0\} \in \Spec A$. If $0 < i \le d$ and a chain $\fq_0 \subsetneq \ldots \subsetneq \fq_{i-1}$ in $\Spec A$ with $\fq_j \cap R = \fp_j$ for $0 \le j < i$ has been selected, we may apply going-up (\ref{cohenseidenberg}) to $A / R$ to extend this chain by a $\fq_i \in \Spec A$ with $\fq_{i-1} \subseteq \fq_i$ and $\fq_i \cap R = \fp_i$ (thus $\fq_{i-1} \subsetneq \fq_i$ as $\fp-i \neq \fp_{i-1})$. Thus, we have a chain $\fq_0 = \{0\} \subsetneq \ldots \subsetneq \fq_d$ in $\Spec A$. Let $\tilde \fq_i \coloneqq \pi_{B,\fp}\inv(\fq_i), Y_i \coloneqq V(\tilde \fq_i)$. This is a chain $Y = Y_0 \supsetneq Y_1 \supsetneq \ldots \supsetneq Y_d$ of irreducible subsets of $\mathfrak{k}^n$. @@ -1346,7 +1348,7 @@ This is part of the proof of \ref{trdegandkdim}. The general case of $\codim(X,Y) \ge \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) \sm \mathfrak{k})$ is shown in \ref{proofcodimletrdeg}. % TODO: reorder - % TODO: Motivation: "Morphism" (AlGeo) and Lift of {0} x k \se \ldots + % TODO: Motivation: "Morphism" (AlGeo) and Lift of {0} x k \subseteq \ldots @@ -1359,7 +1361,7 @@ This is part of the proof of \ref{trdegandkdim}. \subsubsection{Prime avoidance} \begin{proposition}[Prime avoidance]\label{primeavoidance} - Let $A$ be a ring and $I \se A$ a subset which is closed under arbitrary finite sums and non-empty products, for instance, an ideal in $A$. Let $(\fp_i)_{i=1}^n$ be a finite list of ideals in $A$ of which at most two fail to be prime ideals and such that there is no $i$ with $I \se \fp_i$. Then $I \not\se \bigcup_{i=1}^n \fp_i$. + Let $A$ be a ring and $I \subseteq A$ a subset which is closed under arbitrary finite sums and non-empty products, for instance, an ideal in $A$. Let $(\fp_i)_{i=1}^n$ be a finite list of ideals in $A$ of which at most two fail to be prime ideals and such that there is no $i$ with $I \subseteq \fp_i$. Then $I \not\subseteq \bigcup_{i=1}^n \fp_i$. \end{proposition} \begin{proof} Induction on $n$. The case of $n < 2$ is trivial. @@ -1402,7 +1404,7 @@ Recall the definition of a normal field extension in the case of finite field ex \begin{definition} Suppose $L / K$ is an arbitrary field extension. Let $\Aut( L / K)$ be the set of automorphisms of $L$ leaving all elements of (the image in $L$ of) $K$ fixed. - Let $G \se \Aut(L / K)$ be a subgroup. Then the \vocab{fixed field } is definied as + Let $G \subseteq \Aut(L / K)$ be a subgroup. Then the \vocab{fixed field } is definied as \[ L^G \coloneqq \{l \in L | \A g \in G : g(l) = l\} \] @@ -1415,7 +1417,7 @@ Recall the definition of a normal field extension in the case of finite field ex \begin{proof} In both cases $L^G \supseteq$ is easy to see. - If $K \se M \se L$ is an intermediate field, then $L$ is normal over $M$. If $\sigma \in \Aut(M /K)$, an application of Zorn's lemma to the set of all $(N, \vartheta)$ where $N$ is an intermediate field $M \se N \se L$ and $N \xrightarrow{\vartheta} L$ a ring homomorphism such that $\vartheta\defon{M} = \sigma$ shows that $\sigma$ has an extension to an element of $\Aut(L / K)$. % TODO make this rigorous + If $K \subseteq M \subseteq L$ is an intermediate field, then $L$ is normal over $M$. If $\sigma \in \Aut(M /K)$, an application of Zorn's lemma to the set of all $(N, \vartheta)$ where $N$ is an intermediate field $M \subseteq N \subseteq L$ and $N \xrightarrow{\vartheta} L$ a ring homomorphism such that $\vartheta\defon{M} = \sigma$ shows that $\sigma$ has an extension to an element of $\Aut(L / K)$. % TODO make this rigorous If $M$ is normal over $K$, it is easily seen to be $\Aut(L / K)$ invariant. Thus $L^G$ is the union of $M^{\Aut(M / K)}$ over all intermediate fields which are finite and normal over $K$, and it is sufficient to show the proposition for finite normal extensions $L / K$. @@ -1427,7 +1429,7 @@ Recall the definition of a normal field extension in the case of finite field ex If $\deg(l / K) = 1$, we have $l \in K$. Otherwise, assume that the assertion has been shown for elements of $L^G$ whose degree over $K$ is smaller than $\deg( l / K)$. Let $\overline{L}$ be an algebraic closure of $L$ and $\lambda$ a zero of $P$ in $\overline{L}$. - If $M = K(l) \se L$, then there is a ring homomorphism $M - \overline{L}$ sending $l$ to $\lambda$. This can be extended to a ring homomorphism $L \xrightarrow{\sigma} \overline{L}$. We have $\sigma \in G$ because $L / K$ is normal. Hence $\lambda = \sigma(l) = l$, as $l \in L^G$. Thus $l$ is the only zero of $P$ in $\overline{L}$ and because $\deg P >1$ it is a multiple zero. + If $M = K(l) \subseteq L$, then there is a ring homomorphism $M - \overline{L}$ sending $l$ to $\lambda$. This can be extended to a ring homomorphism $L \xrightarrow{\sigma} \overline{L}$. We have $\sigma \in G$ because $L / K$ is normal. Hence $\lambda = \sigma(l) = l$, as $l \in L^G$. Thus $l$ is the only zero of $P$ in $\overline{L}$ and because $\deg P >1$ it is a multiple zero. It is shown in the Galois theory lecture % TODO: link to EinfAlg that this is possible only when $P(T) = Q(T^p)$ for some $Q \in K[T]$. Then $Q(l^p) = 0$ and the induction assumption can be applied to $x = l^p$ showing $x^{p^m} \in K$ hence $l^{p^{m+1}} \in K$ for some $m \in \N$. \end{itemize} @@ -1477,7 +1479,7 @@ Recall the definition of a normal field extension in the case of finite field ex \begin{proof} Let $\fq, \fr$ be prime ideals of $B$ above the given $\fp \in \Spec A$. We must show that there exists $\sigma \in G$ such that $\fq = \sigma(\fr)$. - This is equivalent to $\fq \se \sigma(\fr)$, since the Krull going-up theorem (\ref{cohenseidenberg}) applies to the integral ring extension $B / A$, showing that there are no inclusions between different elements of $\Spec B$ lying above $\fp \in \Spec A$. + This is equivalent to $\fq \subseteq \sigma(\fr)$, since the Krull going-up theorem (\ref{cohenseidenberg}) applies to the integral ring extension $B / A$, showing that there are no inclusions between different elements of $\Spec B$ lying above $\fp \in \Spec A$. If $L / K$ is finite and there is no such $\sigma$, then by prime avoidance (\ref{primeavoidance}) there is $ x \in \fq \sm \bigcup_{\sigma \in G} \sigma(\fr)$. As $\fr$ is a prime ideal, $y = \prod_{\sigma \in G} \sigma(x) \in \fq \sm \fr$.\footnote{$\prod_{\sigma \in G} \sigma(x) = \prod_{\sigma \in G} \sigma\inv(x)$} @@ -1500,7 +1502,7 @@ Recall the definition of a normal field extension in the case of finite field ex \begin{proof} It follows from the assumptions that the field of quotients $Q(B)$ is an algebraic field extension of $Q(A)$. There is an algebraic extension $L$ of $Q(B)$ such that $L / Q(A)$ is normal (for instance an algebraic closure of $Q(B)$). - Let $C$ be the integral closure of $A$ in $L$. Then $B \se C$ and $C / B$ is integral. + Let $C$ be the integral closure of $A$ in $L$. Then $B \subseteq C$ and $C / B$ is integral. \begin{figure}[H] \centering \begin{tikzcd} @@ -1513,13 +1515,13 @@ Recall the definition of a normal field extension in the case of finite field ex Going-down holds for $C / A$. \end{claim} \begin{subproof} - Let $\fp \se \tilde \fp$ be an inclusion of prime ideals of $A$ and $\tilde \fr \in \Spec C$ with $\tilde \fr \cap A = \tilde \fp$. - By going-up for integral ring extensions (\ref{cohenseidenberg}), $\Spec C \xrightarrow{\cdot \cap A} \Spec A$ is surjectiv. Thus there is $\fr' \in \Spec C$ such that $\fr' \cap A = \fp$. By going up for $C / A$ there is $\tilde \fr' \in \Spec C$ with $\tilde \fr' \cap A = \tilde \fp, \fr' \se \tilde \fr'$. - By the theorem about the action of the automorphism group on prime ideals of a normal ring extension (\ref{autonprime}) there exists a $\sigma \in \Aut(L / Q(A))$ with $\sigma(\tilde \fr') = \tilde \fr$. Then $\fr \coloneqq \sigma(\fr')$ satisfies $\fr \se \tilde \fr$ and $\fr \cap A = \fp$. + Let $\fp \subseteq \tilde \fp$ be an inclusion of prime ideals of $A$ and $\tilde \fr \in \Spec C$ with $\tilde \fr \cap A = \tilde \fp$. + By going-up for integral ring extensions (\ref{cohenseidenberg}), $\Spec C \xrightarrow{\cdot \cap A} \Spec A$ is surjectiv. Thus there is $\fr' \in \Spec C$ such that $\fr' \cap A = \fp$. By going up for $C / A$ there is $\tilde \fr' \in \Spec C$ with $\tilde \fr' \cap A = \tilde \fp, \fr' \subseteq \tilde \fr'$. + By the theorem about the action of the automorphism group on prime ideals of a normal ring extension (\ref{autonprime}) there exists a $\sigma \in \Aut(L / Q(A))$ with $\sigma(\tilde \fr') = \tilde \fr$. Then $\fr \coloneqq \sigma(\fr')$ satisfies $\fr \subseteq \tilde \fr$ and $\fr \cap A = \fp$. \end{subproof} - If $\fp \se \tilde \fp$ is an inclusion of elements of $\Spec A$ and $\tilde \fq \in \Spec B$ with $\tilde \fp \cap A = \tilde \fp$, by the surjectivity of $\Spec C \xrightarrow{\cdot \cap B} \Spec B$ (\ref{cohenseidenberg}) there is $\tilde \fr \in \Spec C$ with $\tilde \fr \cap B = \fq$. - By going-down for $C / A$, there is $\fr \in \Spec C$ with $\fr \se \tilde \fr$ and $\fr \cap A = \fp$. - Then $\fq \coloneqq \fr \cap B \in \Spec B, \fq \se \tilde \fq$ and $\fq \cap A = \fp$. Thus going-down holds for $B / A$. + If $\fp \subseteq \tilde \fp$ is an inclusion of elements of $\Spec A$ and $\tilde \fq \in \Spec B$ with $\tilde \fp \cap A = \tilde \fp$, by the surjectivity of $\Spec C \xrightarrow{\cdot \cap B} \Spec B$ (\ref{cohenseidenberg}) there is $\tilde \fr \in \Spec C$ with $\tilde \fr \cap B = \fq$. + By going-down for $C / A$, there is $\fr \in \Spec C$ with $\fr \subseteq \tilde \fr$ and $\fr \cap A = \fp$. + Then $\fq \coloneqq \fr \cap B \in \Spec B, \fq \subseteq \tilde \fq$ and $\fq \cap A = \fp$. Thus going-down holds for $B / A$. \end{proof} \begin{remark}[Universally Japanese rings] @@ -1532,8 +1534,8 @@ Recall the definition of a normal field extension in the case of finite field ex \begin{dexample}[Counterexample to going down] Let $R = \mathfrak{k}[X,Y]$ and $A = \mathfrak{k}[X,Y, \frac{X}{Y}]$. Then going down does not hold for $A / R$: - For any ideal $Y \in \fq \se A$ we have $X = \frac{X}{Y} \cdot Y \in \fq$. - Consider $(Y)_R \subsetneq (X,Y)_R \se \fq \cap R$. As $(X,Y)_R$ is maximal and the preimage of a prime ideal is prime and thus proper, we have $(X,Y)_R = \fq \cap R$. + For any ideal $Y \in \fq \subseteq A$ we have $X = \frac{X}{Y} \cdot Y \in \fq$. + Consider $(Y)_R \subsetneq (X,Y)_R \subseteq \fq \cap R$. As $(X,Y)_R$ is maximal and the preimage of a prime ideal is prime and thus proper, we have $(X,Y)_R = \fq \cap R$. The prime ideal $(\frac{X}{Y},Y)_A = (\frac{X}{Y}, X,Y)_A$ is lying over $(X,Y)_R$, so going down is violated. \end{dexample} @@ -1544,7 +1546,7 @@ This is part of the proof of \ref{trdegandkdim}. %TODO: reorder \begin{proof} % DIMT - Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and $X \se Y = V(\fp) \se \mathfrak{k}^n$ irreducible closed subsets of $\mathfrak{k}^n$. + Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and $X \subseteq Y = V(\fp) \subseteq \mathfrak{k}^n$ irreducible closed subsets of $\mathfrak{k}^n$. We want to show that $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$. $\le $ was shown in \ref{upperboundcodim}. $\dim Y \ge \trdeg(\fK(Y) / \mathfrak{k})$ was shown in \ref{lowerbounddimy} by @@ -1585,32 +1587,32 @@ To formulate a result which still applies in this context, we need the following \begin{example} Let $A = \mathfrak{k}[X_1,\ldots,X_n]$, $X = V(\fp)$ for a prime ideal $\fp$. By the correspondence between irreducible subsets of $\mathfrak{k}^n$ and prime ideals in $A$ (\ref{bijiredprim}), - the $\fp_i$ correspond to irreducible subsets $X_i \se \mathfrak{k}^n$ containing $X$. Thus $\hght(\fp) = \codim(X, \mathfrak{k}^n)$. + the $\fp_i$ correspond to irreducible subsets $X_i \subseteq \mathfrak{k}^n$ containing $X$. Thus $\hght(\fp) = \codim(X, \mathfrak{k}^n)$. \end{example} \begin{example}\label{htandcodim} Let $B = \mathfrak{k}[X_1,\ldots,X_n], \fq \in \Spec B$ and let $A \coloneqq B / \fp$. - Let $Y \coloneqq V(\fq) \se \mathfrak{k}^n$, $\tilde \fp \coloneqq \pi_{B, \fq}\inv(\fp)$, where $B \xrightarrow{\pi_{B, \fp}} A $ is the projection to the ring of residue classes, and let $X = V(\tilde \fp)$. - By \ref{idealslocbij} we have a bijection between the prime ideals $\fr \se \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime ideals $\tilde \fr \in \Spec B$ with $\fq \se \tilde \fr \se \tilde \fp$: + Let $Y \coloneqq V(\fq) \subseteq \mathfrak{k}^n$, $\tilde \fp \coloneqq \pi_{B, \fq}\inv(\fp)$, where $B \xrightarrow{\pi_{B, \fp}} A $ is the projection to the ring of residue classes, and let $X = V(\tilde \fp)$. + By \ref{idealslocbij} we have a bijection between the prime ideals $\fr \subseteq \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime ideals $\tilde \fr \in \Spec B$ with $\fq \subseteq \tilde \fr \subseteq \tilde \fp$: \begin{align} - f: \{\fr \in \Spec A | \fr \se \fp \} &\longrightarrow \{\tilde \fr \in \Spec B | \fq \se \tilde \fr \se \tilde \fp\} \\ + f: \{\fr \in \Spec A | \fr \subseteq \fp \} &\longrightarrow \{\tilde \fr \in \Spec B | \fq \subseteq \tilde \fr \subseteq \tilde \fp\} \\ \fr &\longmapsto \pi_{B, \fq}\inv(\fr)\\ \tilde \fr / \fq &\longmapsfrom \tilde \fr \end{align} By \ref{bijiredprim}, the $\tilde \fr$ are in canonical bijection with the irreducible subsets $Z$ of $Y$ containing $X$. - Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in canonical bijection with the chains $X = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \se Y$ of irreducible subsets and + Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in canonical bijection with the chains $X = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \subseteq Y$ of irreducible subsets and $\hght(\fp) = \codim(X,Y)$. \end{example} \begin{remark} - Let $A$ be an arbitrary ring. One can show that there is a bijection between $\Spec A$ and the set of irreducible subsets $Y \se \Spec A$: + Let $A$ be an arbitrary ring. One can show that there is a bijection between $\Spec A$ and the set of irreducible subsets $Y \subseteq \Spec A$: \begin{align} - f: \Spec A &\longrightarrow \{Y \se \Spec A | Y\text{irreducible}\} \\ + f: \Spec A &\longrightarrow \{Y \subseteq \Spec A | Y\text{irreducible}\} \\ \fp &\longmapsto \Vs(\fp)\\ \bigcup_{\fp \in Y} \fp &\longmapsfrom Y \end{align} - Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in canonical bijection with the chains $V(\fp) = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \se \Spec A$ of irreducible subsets, and $\hght(\fp) = \codim(V(\fp), \Spec A)$. + Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in canonical bijection with the chains $V(\fp) = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \subseteq \Spec A$ of irreducible subsets, and $\hght(\fp) = \codim(V(\fp), \Spec A)$. \end{remark} @@ -1643,7 +1645,7 @@ Thus \[ k \le \trdeg(\mathfrak{k}(\fp_k) / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) \le \trdeg(K / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) \] -where the last inequality is another application of \ref{trdegresfield} (using $K = Q(A) = Q(A / \{0\}) = \mathfrak{k}(\{0\})$ and the fact that $\{0\} \se \fp_k$ is a prime ideal). +where the last inequality is another application of \ref{trdegresfield} (using $K = Q(A) = Q(A / \{0\}) = \mathfrak{k}(\{0\})$ and the fact that $\{0\} \subseteq \fp_k$ is a prime ideal). Hence \[ \hght(\fp) \le \trdeg( K / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) \] @@ -1667,7 +1669,7 @@ and it remains to show the opposite inequality. \end{subproof} % TODO: fix names A_1 = A_S, k_1 = R_S The ring homomorphism $\ev_x : R = \mathfrak{l}[X_1,\ldots,X_d] \xrightarrow{P \mapsto P(x_1,\ldots,x_d)} A$ is injective. Because $R$ is a UFD, $R$ is normal (\ref{ufdnormal}). Thus the going-down theorem (\ref{gdkrull}) applies to the integral $R$-algebra $A$. - For $0 \le i \le d$, let $\fp_i \se R$ be the ideal generated by $(X_j)_{j=i+1}^d$. We have $\mathfrak{m} \sqcap R = \fp_0$ as all $X_i \in \mathfrak{m}$, hence $X_i \in \mathfrak{m} \sqcap R$ and $\fp_0$ is a maximal ideal. + For $0 \le i \le d$, let $\fp_i \subseteq R$ be the ideal generated by $(X_j)_{j=i+1}^d$. We have $\mathfrak{m} \sqcap R = \fp_0$ as all $X_i \in \mathfrak{m}$, hence $X_i \in \mathfrak{m} \sqcap R$ and $\fp_0$ is a maximal ideal. By applying going-down and induction on $i$, there is a chain $\mathfrak{m} = \fq_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_d$ of elements of $\Spec A$ such that $\fq_i \sqcap R = \fp_i$. It follows that $\hght(\mathfrak{m}) \ge d$. \end{subproof} @@ -1678,7 +1680,7 @@ By lemma \ref{ltrdegresfieldtrbase} there are $a_1,\ldots,a_n \in A$ whose image As these images are $\mathfrak{l}$-algebraically independent, the same holds for the $a_i$ themselves. By lemma \ref{extendtotrbase} we can extend $(a_{i})_{i=1}^n$ to a transcendence base $(a_i)_{i=1}^m \in A^m$ of $K / \mathfrak{l}$. -Let $R \se A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and let $S \coloneqq R \sm \{0\}$. +Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and let $S \coloneqq R \sm \{0\}$. Let $A_1 \coloneqq A_S$ and $\fp_S$ the prime ideal corresponding to $\fp$ under $\Spec(A_1) \cong \{\fr \in \Spec A | \fr \cap S = \emptyset\}$ (\ref{idealslocbij}). As in \ref{locandquot}, $A_1$ is a domain with $Q(A_1) \cong K = Q(A)$ and by \ref{locandfactor} $A_1 / \fp_S \cong (A / \fp)_{\overline{S}}$, where $\overline{S}$ denotes the image of $S$ in $A / \fp$. As in \ref{trdegresfield}, $\mathfrak{k}(\fp_S) \cong \mathfrak{k}(\fp)$ is integral over $A_1 / \fp_S$. @@ -1702,7 +1704,7 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 \begin{proposition}\label{dimprod} - Let $X \se \mathfrak{k}^n$ and $Y \se \mathfrak{k}^n$ be irreducible and closed. Then $X \times Y$ is also an irreducible closed subset of $\mathfrak{k}^{m+n}$. + Let $X \subseteq \mathfrak{k}^n$ and $Y \subseteq \mathfrak{k}^n$ be irreducible and closed. Then $X \times Y$ is also an irreducible closed subset of $\mathfrak{k}^{m+n}$. Moreover, $\dim(X \times Y) = \dim(X) + \dim(Y)$ and $\codim(X \times Y, \mathfrak{k}^{m+n}) = \codim(X, \mathfrak{k}^m) + \codim(Y, \mathfrak{k}^n)$. \end{proposition} \begin{proof} @@ -1711,8 +1713,8 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 Thus $X \times Y$ is closed in $\mathfrak{k}^{m+n}$. We must also show irreducibility. $X \times Y \neq \emptyset$ is obvious. - Assume that $X \times Y = A_1 \cup A_2$, where the $A_i \se \mathfrak{k}^{m+n}$ are closed. - For $x' \in \mathfrak{k}^m, x' \times Y$ is homeomorphic to the irreducible $Y$. Thus $X = X_1 \cup X_2$ where $X_i = \{x \in X | \{x\} \times Y \se A_i\} $. + Assume that $X \times Y = A_1 \cup A_2$, where the $A_i \subseteq \mathfrak{k}^{m+n}$ are closed. + For $x' \in \mathfrak{k}^m, x' \times Y$ is homeomorphic to the irreducible $Y$. Thus $X = X_1 \cup X_2$ where $X_i = \{x \in X | \{x\} \times Y \subseteq A_i\} $. Because $X_i = \bigcap_{y \in Y} \{x \in X | (x,y) \in A_i\}$, this is closed. As $X$ is irreducible, there is $i \in \{1;2\} $ which $X_i = X$. Then $X \times Y = A_i$ confirming the irreducibility of $X \times Y$. Let $a = \dim X$ and $b = \dim Y$ and $X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_a = X$,$Y_0 \subsetneq Y_1 \subsetneq \ldots \subsetneq Y_b = Y$ be chains of irreducible subsets. By the previous result, @@ -1746,7 +1748,7 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 \begin{proposition}\label{bijspecideal} There is a bijection \begin{align} - f: \{A \se \Spec R | A\text{ closed}\} &\longrightarrow \{I \se R | I \text{ ideal and } I = \sqrt{I} \} \\ + f: \{A \subseteq \Spec R | A\text{ closed}\} &\longrightarrow \{I \subseteq R | I \text{ ideal and } I = \sqrt{I} \} \\ A &\longmapsto \bigcap_{\fp \in A} \fp\\ \Vspec(I) &\longmapsfrom I \end{align} @@ -1756,7 +1758,7 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 If $A = \Vspec(I)$, then by \ref{sqandvspec} $\sqrt{I} = \bigcap_{\fp \in A} \fp$. Thus, an ideal with $\sqrt{I} = I$ can be recovered from $\Vspec( I)$. Since $\Vspec(J) = \Vspec(\sqrt{J})$, the map from ideals with $\sqrt{I} = I$ to closed subsets is surjective. Sine $R$ corresponds to $\emptyset$, the proper ideals correspond to non-empty subsets of $\Spec R$. Assume that $\Vspec(I) = \Vspec(J_1) \cup \Vspec(J_2)$, where the decomposition is proper and the ideals coincide with their radicals. - Let $g = f_1f_2$ with $f_k \in J_k \sm I$. Since $\Vspec(g) \supseteq \Vspec(f_k) \supseteq \Vspec(I_k), \Vspec(I) \se \Vspec(g)$. Hence $g \in \sqrt{I} = I$. + Let $g = f_1f_2$ with $f_k \in J_k \sm I$. Since $\Vspec(g) \supseteq \Vspec(f_k) \supseteq \Vspec(I_k), \Vspec(I) \subseteq \Vspec(g)$. Hence $g \in \sqrt{I} = I$. As $f_k \not\in I$, $I$ fails to be a prime ideal. Conversely, assume that $f_1f_2 \in I$ while the factors are not in $I$. Since $I = \sqrt{I}, \Vspec(f_k) \not\supseteq \Vspec(I)$. But $\Vspec(f_1) \cup \Vspec(f_2) = \Vspec(f_1f_2) \supseteq \Vspec(I)$. The proper decomposition $\Vspec(I) = \left( \Vspec(I) \cap \Vspec(f_1) \right) \cup \left( \Vspec(I) \cap \Vspec(f_2) \right) $ now shows that $\Vspec(I)$ fails to be irreducible. @@ -1777,12 +1779,12 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 \end{dexample} \begin{corollary}[About the smallest prime ideals containing $I$ ]\label{smallestprimesvi} - If $R$ is Noetherian and $I \se R$ an ideal, then the set $\Vspec(I) = \{\fp \in \Spec R | I \se \fp\}$ has finitely many $\se$-minimal elements $(\fp_i)_{i=1}^k$ and every element of $V(I)$ contains at least one $\fp_i$. + If $R$ is Noetherian and $I \subseteq R$ an ideal, then the set $\Vspec(I) = \{\fp \in \Spec R | I \subseteq \fp\}$ has finitely many $\subseteq$-minimal elements $(\fp_i)_{i=1}^k$ and every element of $V(I)$ contains at least one $\fp_i$. The $\Vspec(\fp_i)$ are precisely the irreducible components of $V(I)$. Moreover $\bigcap_{i=1}^k \fp_i = \sqrt{I}$ and $k > 0$ if $I$ is a proper ideal. \end{corollary} \begin{proof} - If $\Vspec(I) = \bigcup_{i=1}^k \Vspec(\fp_i)$ is the decomposition into irreducible components then every $\fq \in \Vspec(I)$ must belong to at least one $\Vspec(\fp_i)$, hence $\fp_i \se \fq$. Also $\fp_i \in \Vspec(\fp_i) \se \Vspec(I)$. - It follows that the sets of $\se$-minimal elements of $\Vspec(I)$ and of $\{\fp_1,\ldots,\fp_k\} $ coincide. + If $\Vspec(I) = \bigcup_{i=1}^k \Vspec(\fp_i)$ is the decomposition into irreducible components then every $\fq \in \Vspec(I)$ must belong to at least one $\Vspec(\fp_i)$, hence $\fp_i \subseteq \fq$. Also $\fp_i \in \Vspec(\fp_i) \subseteq \Vspec(I)$. + It follows that the sets of $\subseteq$-minimal elements of $\Vspec(I)$ and of $\{\fp_1,\ldots,\fp_k\} $ coincide. As there are no non-trivial inclusions between the $\Vspec(\fp_i)$, there are no non-trivial inclusions between the $\fp_i$ and the assertion follows. The final remark is trivial. \end{proof} @@ -1794,7 +1796,7 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 \subsection{The principal ideal theorem} Krull was able to show: \begin{theorem}[Principal ideal theorem / Hauptidealsatz]\label{pitheorem} - Let $A$ be a Noetherian ring, $a \in A$ and $\fp \in \Spec A$ a $\se$-minimal element of $\Vspec(a)$. Then $\hght(\fp) \le 1$. + Let $A$ be a Noetherian ring, $a \in A$ and $\fp \in \Spec A$ a $\subseteq$-minimal element of $\Vspec(a)$. Then $\hght(\fp) \le 1$. \end{theorem} \begin{proof} Probably not relevant for the exam. @@ -1805,7 +1807,7 @@ Krull was able to show: \end{remark} \begin{theorem}[Generalized principal ideal theorem] - Let $A$ be a Noetherian ring, $(a_i)_{i=1}^k \in A$ and $\fp \in \Spec A$ a $\se$-minimal element of $\bigcap_{i=1}^k V(a_i)$, the set of prime ideals containing all $a_i$. + Let $A$ be a Noetherian ring, $(a_i)_{i=1}^k \in A$ and $\fp \in \Spec A$ a $\subseteq$-minimal element of $\bigcap_{i=1}^k V(a_i)$, the set of prime ideals containing all $a_i$. Then $\hght(\fp) \le k$. \end{theorem} Modern approaches to the principal ideal theorem usually give a direct proof of this more general theorem. @@ -1819,7 +1821,7 @@ Modern approaches to the principal ideal theorem usually give a direct proof of \subsubsection{Application to the dimension of intersections} \begin{remark}\label{smallestprimeandirredcomp} -Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \se R$ an ideal. +Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ an ideal. If $(\fp_i)_{i=1}^k$ are the smallest prime ideals of $R$ containing $I$, then $(\Va(\fp_i))_{i=1}^k$ are the irreducible components of $\Va(I)$. \end{remark} @@ -1829,14 +1831,14 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \se R$ an ideal. \begin{corollary}[of the principal ideal theorem] \label{corpithm} - Let $X \se \mathfrak{k}^n$ be irreducible, $(f_i)_{i=1}^k$ elements of $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $Y$ an irreducible component of $A = X \cap \bigcap_{i=1}^k V(f_i)$. + Let $X \subseteq \mathfrak{k}^n$ be irreducible, $(f_i)_{i=1}^k$ elements of $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $Y$ an irreducible component of $A = X \cap \bigcap_{i=1}^k V(f_i)$. Then $\codim(Y,X) \le k$. \end{corollary} \begin{remark} This confirms the naive geometric intuition that by imposing $k$ equations one ends up in codimension at most $k$. \end{remark} \begin{proof} - If $X = v(\fp), X \cap \bigcap_{i=1}^k V(f_i) = V(I)$ where $I \se R$ is the ideal generated by $\fp$ and the $f_i$. + If $X = v(\fp), X \cap \bigcap_{i=1}^k V(f_i) = V(I)$ where $I \subseteq R$ is the ideal generated by $\fp$ and the $f_i$. By \ref{smallestprimeandirredcomp}, $Y = V(\fq)$ where $\fq$ is the smallest prime ideal containing $I$. Then $\fq / \fp$ is a smallest prime ideal of $R / \fp$ containing all $(f_i \mod \fp)_{i=1}^k$. By the principal ideal theorem (\ref{pitheorem}), $\hght(\fq / \fp) \le k$ and the assertion follows from example \ref{htandcodim}. @@ -1852,7 +1854,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \se R$ an ideal. Equivalently, $\dim(C) \ge \dim(A) + \dim(B)-n$. \end{dremark} \begin{proof} - Let $X = A \times B \se \mathfrak{k}^{2n}$, where we use $(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$ as coordinates of $\mathfrak{k}^{2n}$. + Let $X = A \times B \subseteq \mathfrak{k}^{2n}$, where we use $(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$ as coordinates of $\mathfrak{k}^{2n}$. Let $\Delta \coloneqq \{(x_1,\ldots,x_n,x_1,\ldots,x_n) | x \in \mathfrak{k}^n\} $ be the diagonal in $\mathfrak{k}^n \times \mathfrak{k}^n$. The projection $\mathfrak{k}^{2n}\to \mathfrak{k}^n$ to the $X$-coordinates defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$. Thus, $C$ is homeomorphic to an irreducible component $C'$ of $(A \times B) \cap \Delta$ and @@ -1870,7 +1872,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \se R$ an ideal. \subsubsection{Application to the property of being a UFD} \begin{proposition}\limrel - Let $R$ be a Noetherian domain. Then $R$ is a UFD iff every $\fp \in \Spec R$ with $\hght(\fp)= 1$\footnote{In other words, every $\se$-minimal element of the set of non-zero prime ideals of $R$ } is a principal ideal. + Let $R$ be a Noetherian domain. Then $R$ is a UFD iff every $\fp \in \Spec R$ with $\hght(\fp)= 1$\footnote{In other words, every $\subseteq$-minimal element of the set of non-zero prime ideals of $R$ } is a principal ideal. \end{proposition} \begin{proof} Every element of every Noetherian domain can be written as a product of irreducible elements.\footnote{Consider the set of principal ideals $rR$ where $r$ is not a product of irreducible elements.} @@ -1878,10 +1880,10 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \se R$ an ideal. Assume that this is the case. Let $\fp \in \Spec R, \hght(\fp) = 1$. - Let $p \in \fp \sm \{0\}$. Replacing $p$ by a prime factor of $p$, we may assume $p$ to be prime. Thus $\{0\} \subsetneq pR \se \fp$ is a chain of prime ideals and since $\hght(\fp) = 1$ it follows that $\fp = pR$. + Let $p \in \fp \sm \{0\}$. Replacing $p$ by a prime factor of $p$, we may assume $p$ to be prime. Thus $\{0\} \subsetneq pR \subseteq \fp$ is a chain of prime ideals and since $\hght(\fp) = 1$ it follows that $\fp = pR$. Conversely, assume that every $\fp \in \Spec R$ with $\hght(\fp)=1$ is a principal ideal. Let $f \in R$ be irreducible. - Let $\fp \in \Spec R$ be a $\se$-minimal element of $V(f)$. By the principal ideal theorem (\ref{pitheorem}), $\hght(\fp)=1$. + Let $\fp \in \Spec R$ be a $\subseteq$-minimal element of $V(f)$. By the principal ideal theorem (\ref{pitheorem}), $\hght(\fp)=1$. Thus $\fp = pR$ for some prime element $p$. We have $p | f$ since $f \in \fp$. As $f$ is irreducible, $p$ and $f$ are multiplicatively equivalent. Thus $f$ is a prime element. \end{proof} @@ -1917,7 +1919,7 @@ $\rad(A) = f A$ where $f = \prod_{i=1}^{n} p_i$. % Lecture 11 -\section{Projective spaces} +\subseteqction{Projective spaces} Let $\mathfrak{l}$ be any field. \begin{definition} For a $\mathfrak{l}$-vector space $V$, let $\bP(V)$ be the set of one-dimensional subspaces of $V$. @@ -1935,7 +1937,7 @@ Let $\mathfrak{l}$ be any field. There are points $[1,0], [0,1] \in \bP^1$ but there is no point $[0,0] \in \bP^1$. \end{remark} \begin{definition}[Infinite hyperplane] - For $0 \le i \le n$ let $U_i \se \bP^n$ denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$. + For $0 \le i \le n$ let $U_i \subseteq \bP^n$ 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 $x \in \bP^n$ differ by scaling with a $\lambda \in \mathfrak{l}^{\times}$, $x_i = \lambda \xi_i$. Since not all $x_i$ may be $0$, $\bP^n = \bigcup_{i=0}^n U_i$. We identify $\bA^n = \bA^n(\mathfrak{l}) = \mathfrak{l}^n$ with $U_0$ by identifying $(x_1,\ldots,x_n) \in \bA^n$ with $[1,x_1,\ldots,x_n] \in \bP^n$. Then $\bP^1 = \bA^1 \cup \{\infty\} $ where $\infty=[0,1]$. More generally, when $n > 0$ $\bP^n \sm \bA^n$ can be identified with $\bP^{n-1}$ identifying $[0,x_1,\ldots,x_n] \in \bP^n \sm \bA^n$ with $[x_1,\ldots,x_n] \in \bP^{n-1}$. @@ -1947,11 +1949,11 @@ Let $\mathfrak{l}$ be any field. Let $\bI = \N$ or $\bI = \Z$. \end{notation} \begin{definition} - By an \vocab[Graded ring]{$\bI$-graded ring} $A_\bullet$ we understand a ring $A$ with a collection $(A_d)_{d \in \bI}$ of subgroups of the additive group $(A, +)$ such that $A_a \cdot A_b \se A_{a + b}$ for $a,b \in \bI$ and such that $A = \bigoplus_{d \in \bI} A_d$ in the sense that every $r \in A$ has a unique decomposition $r = \sum_{d \in \bI} r_d$ with $r_d \in A_d$ and but finitely many $r_d \neq 0$. + By an \vocab[Graded ring]{$\bI$-graded ring} $A_\bullet$ we understand a ring $A$ with a collection $(A_d)_{d \in \bI}$ of subgroups of the additive group $(A, +)$ such that $A_a \cdot A_b \subseteq A_{a + b}$ for $a,b \in \bI$ and such that $A = \bigoplus_{d \in \bI} A_d$ in the sense that every $r \in A$ has a unique decomposition $r = \sum_{d \in \bI} r_d$ 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 $I \se A$ is called \vocab{homogeneous} if $r \in I \implies \A d \in \bI ~ r_d \in I_d$ where $I_d \coloneqq I \cap A_d$. + An ideal $I \subseteq A$ is called \vocab{homogeneous} if $r \in I \implies \A d \in \bI ~ r_d \in I_d$ 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 $A$. \end{definition} @@ -1995,7 +1997,7 @@ Let $\mathfrak{l}$ be any field. \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 $R[X_0,\ldots,X_n]_d \se R[X_0,\ldots,X_n]$. + 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. @@ -2008,7 +2010,7 @@ Let $\mathfrak{l}$ be any field. \] Let $\Vp(f) \coloneqq \{x \in \bP^n | f(x) = 0\}$. - We call a subset $X \se \bP^n$ Zariski-closed if it can be represented as + We call a subset $X \subseteq \bP^n$ Zariski-closed if it can be represented as \[ X = \bigcap_{i=1}^k \Vp(f_i) \] @@ -2016,11 +2018,11 @@ Let $\mathfrak{l}$ be any field. \end{definition} \pagebreak \begin{fact} - If $X = \bigcap_{i = 1}^k \Vp(f_i) \se \bP^n$ is closed, then $Y = X \cap \bA^n$ can be identified with the closed subset + If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq \bP^n$ is closed, then $Y = X \cap \bA^n$ can be identified with the closed subset \[ - \{(x_1,\ldots,x_n) \in \mathfrak{k}^n | f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\} \se \mathfrak{k}^n + \{(x_1,\ldots,x_n) \in \mathfrak{k}^n | f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\} \subseteq \mathfrak{k}^n \] - Conversely, if $Y \se \mathfrak{k}^n$ is closed it has the form + Conversely, if $Y \subseteq \mathfrak{k}^n$ is closed it has the form \[ \{(x_1,\ldots,x_n) \in \mathfrak{k}^n | g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\} \] @@ -2033,7 +2035,7 @@ Let $\mathfrak{l}$ be any field. % The Zariski topology on P^n (2) \begin{definition} - Let $I \se A = \mathfrak{k}[X_0,\ldots,X_n]$ be a homogeneous ideal. + Let $I \subseteq A = \mathfrak{k}[X_0,\ldots,X_n]$ be a homogeneous ideal. Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \bP^n | \A f \in I ~ f(x_0,\ldots,x_n) = 0\}$ 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}. @@ -2069,8 +2071,8 @@ Let $\mathfrak{l}$ be any field. \begin{enumerate}[A] \item $R$ is Noetherian. \item Every homogeneous ideal of $R_{\bullet}$ is finitely generated. - \item Every chain $I_0\se I_1 \se \ldots$ of homogeneous ideals terminates. - \item Every set $\fM \neq \emptyset$ of homogeneous ideals has a $\se$-maximal element. + \item Every chain $I_0\subseteq I_1 \subseteq \ldots$ of homogeneous ideals terminates. + \item Every set $\fM \neq \emptyset$ 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} @@ -2080,7 +2082,7 @@ 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 $I \se R_0$, 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} @@ -2088,7 +2090,7 @@ Let $\mathfrak{l}$ be any field. \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 $R_e \se \tilde R$ for all $e < d$. + 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 $f \in \tilde R$. @@ -2105,48 +2107,48 @@ Let $\mathfrak{l}$ be any field. \subsection{The projective form of the Nullstellensatz and the closed subsets of $\bP^n$} Let $A = \mathfrak{k}[X_0,\ldots,X_n]$. \begin{proposition}[Projective form of the Nullstellensatz]\label{hnsp} - If $I \se A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, then $\Vp(I) \se \Vp(f) \iff f \in \sqrt{I}$. + 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 $\Vp(I) \se \Vp(f)$. 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 $[x_0,\ldots,x_n] \in \bP^n$ is well-defined and belongs to $\Vp(I) \se \Vp(f)$, hence $f(x) = 0$. - Thus $\Va(I) \se \Va(f)$ 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 $[x_0,\ldots,x_n] \in \bP^n$ 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 $\fp \not\supseteq R_+$. \end{definition} \begin{remark}\label{proja} - As the elements of $A_0 \sm \{0\}$ are units in $A$ it follows that for every homogeneous ideal $I$ we have $I \se A_+$ or $I = A$. + As the elements of $A_0 \sm \{0\}$ 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 \sm A_+ | \fp \text{ is homogeneous}\} $. \end{remark} \begin{proposition}\label{bijproj} There is a bijection \begin{align} - f: \{I \se A_+ | I \text{ homogeneous ideal}, I = \sqrt{I}\} &\longrightarrow \{X \se \bP^n | X \text{ closed}\} \\ + f: \{I \subseteq A_+ | I \text{ homogeneous ideal}, I = \sqrt{I}\} &\longrightarrow \{X \subseteq \bP^n | X \text{ closed}\} \\ I &\longmapsto \Vp(I)\\ - \langle \{f \in A_d | d > 0, X \se \Vp(f)\} \rangle & \longmapsfrom X + \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 $\Proj(A_\bullet)$. \end{proposition} \begin{proof} From the projective form of the Nullstellensatz it follows that $f$ is injective and that $f\inv(\Vp\left( I \right)) = \sqrt{I} = I$. - If $X \se \bP^n$ is closed, then $X = \Vp(J)$ for some homogeneous ideal $J \se A$. \Wlog $J = \sqrt{J}$. If $J \not\se A_+$, then $J = A$ (\ref{proja}), hence $X = \Vp(J) = \emptyset = \Vp(A_+)$. - Thus we may assume $J \se A_+$, and $f$ is surjective. + If $X \subseteq \bP^n$ is closed, then $X = \Vp(J)$ for some homogeneous ideal $J \subseteq A$. \Wlog $J = \sqrt{J}$. 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 $I_k \se A_+, I_k = \sqrt{I_k}$. Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \sm \fp$. + 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 $f_k \in I_k \sm \fp$. 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 $f_i \not\in A_{d_i} \sm \fp$. - Then $X \not \se \Vp(f_i)$ by the projective Nullstellensatz when $d_i > 0$ and because $\Vp(1) = \emptyset$ when $d_i = 0$. + 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 $I \se A_{\color{red} +}$, 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} $\bP^n$ is irreducible. @@ -2158,7 +2160,7 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$. \subsection{Some remarks on homogeneous prime ideals} \begin{lemma}\label{homprime} Let $R_\bullet$ be an $\bI$ graded ring ($\bI = \N$ or $\bI = \Z$). - A homogeneous ideal $I \se R$ is a prime ideal iff $1 \not\in I$ and for homogeneous elements $f, g \in R , fg \in I \implies f \in I \lor g \in I$. + 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. @@ -2182,15 +2184,15 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$. \begin{itemize} \item $\bP^n$ is catenary. \item $\dim(\bP^n) = n$. Moreover, $\codim(\{x\} ,\bP^n) = n$ for every $x \in \bP^n$. - \item If $X \se \bP^n$ is irreducible and $x \in X$, then $\codim(\{x\}, X) = \dim(X) = n - \codim(X, \bP^n)$. - \item If $X \se Y \se \bP^n$ are irreducible subsets, then $\codim(X,Y) = \dim(Y) - \dim(X)$. + \item If $X \subseteq \bP^n$ is irreducible and $x \in X$, then $\codim(\{x\}, X) = \dim(X) = n - \codim(X, \bP^n)$. + \item If $X \subseteq Y \subseteq \bP^n$ are irreducible subsets, then $\codim(X,Y) = \dim(Y) - \dim(X)$. \end{itemize} \end{proposition} \begin{proof} - Let $X \se \bP^n$ be irreducible. If $x \in X$, there is an integer $0 \le i \le n$ and $X \in U_i = \bP^n \sm \Vp(X_i)$. + Let $X \subseteq \bP^n$ be irreducible. If $x \in X$, there is an integer $0 \le i \le n$ and $X \in U_i = \bP^n \sm \Vp(X_i)$. \Wlog $i = 0$. Then $\codim(X, \bP^n) = \codim(X \cap \bA^n, \bA^n)$ 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 $X \se Y \se Z$, then $\codim(X,Y) = \codim(X \cap \bA^n, Y \cap \bA^n)$, $\codim(X,Z) = \codim(X \cap \bA^n, Z \cap \bA^n)$ and $\codim(Y,Z) = \codim(Y \cap \bA^n, Z \cap \bA^n)$. + If $Y$ and $Z$ are also irreducible with $X \subseteq Y \subseteq Z$, then $\codim(X,Y) = \codim(X \cap \bA^n, Y \cap \bA^n)$, $\codim(X,Z) = \codim(X \cap \bA^n, Z \cap \bA^n)$ and $\codim(Y,Z) = \codim(Y \cap \bA^n, Z \cap \bA^n)$. Thus \begin{align} \codim(X,Y) + \codim(Y,Z) &= \codim(X \cap \bA^n, Y \cap \bA^n) + \codim(Y \cap \bA^n, Z \cap \bA^n)\\ @@ -2203,11 +2205,11 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$. \subsection{The cone $C(X)$} \begin{definition} - If $X \se \bP^n$ is closed, we define the \vocab{affine cone over $X$} + If $X \subseteq \bP^n$ is closed, we define the \vocab{affine cone over $X$} \[ C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \sm \{0\} | [x_0,\ldots,x_n] \in X\} \] - If $X = \Vp(I)$ where $I \se A_+ = \mathfrak{k}[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) = \Va(I)$. + 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{itemize} @@ -2245,17 +2247,17 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$. 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 $\bP^n$. 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 $I = \sqrt{I} \se A$ (\ref{antimonbij}), $\fp = P \cdot A$. + 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 $H \se \bP^n$ has \vocab{degree $d$} if $H = \Vp(P)$ where $P \in A_d$ is an irreducible polynomial. + A hypersurface $H \subseteq \bP^n$ has \vocab{degree $d$} if $H = \Vp(P)$ where $P \in A_d$ is an irreducible polynomial. \end{definition} \subsubsection{Application to intersections in $\bP^n$ and Bezout's theorem} \begin{corollary} - Let $A \se \bP^n$ and $B \se \bP^n$ 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 $A \subseteq \bP^n$ and $B \subseteq \bP^n$ 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} @@ -2282,18 +2284,18 @@ If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ contradic % Lecture 13 -\section{Varieties} +\subseteqction{Varieties} \subsection{Sheaves} \begin{definition}[Sheaf] Let $X$ be any topological space. - A \vocab{presheaf} $\cG$ of sets (or rings, (abelian) groups) on $X$ associates a set (or rings, or (abelian) group) $\cG(U)$ to every open subset $U$ of $X$, and a map (or ring or group homomorphism) $\cG(U) \xrightarrow{r_{U,V}} \cG(V)$ to every inclusion $V \se U$ of open subsets of $X$ such that $r_{U,W} = r_{V,W} r_{U,V}$ for inclusions $U \se V \se W$ of open subsets. + A \vocab{presheaf} $\cG$ of sets (or rings, (abelian) groups) on $X$ associates a set (or rings, or (abelian) group) $\cG(U)$ to every open subset $U$ of $X$, and a map (or ring or group homomorphism) $\cG(U) \xrightarrow{r_{U,V}} \cG(V)$ to every inclusion $V \subseteq U$ of open subsets of $X$ such that $r_{U,W} = r_{V,W} r_{U,V}$ for inclusions $U \subseteq V \subseteq W$ of open subsets. Elements of $\cG(U)$ are often called \vocab{sections} of $\cG$ on $U$ or \vocab{global sections} when $U = X$. - Let $U \se X$ be open and $U = \bigcup_{i \in I} U_i$ an open covering. + Let $U \subseteq X$ be open and $U = \bigcup_{i \in I} U_i$ an open covering. A family $(f_i)_{i \in I} \in \prod_{i \in I} \cG(U_i)$ is called \vocab[Sections!compatible]{compatible} if $r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j)$ for all $i,j \in I$. Consider the map @@ -2313,10 +2315,10 @@ If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ contradic A presheaf is a contravariant functor $\cG : \cO(X) \to C$ where $\cO(X)$ denotes the category of open subsets of $X$ with inclusions as morphisms and $C$ is the category of sets, rings or (abelian) groups. \end{dtrivial} \begin{definition} - A subsheaf $\cG'$ is defined by subsets (resp. subrings or subgroups) $\cG'(U) \se \cG(U)$ for all open $U \se X$ such that the sheaf axioms still hold. + A subsheaf $\cG'$ is defined by subsets (resp. subrings or subgroups) $\cG'(U) \subseteq \cG(U)$ for all open $U \subseteq X$ such that the sheaf axioms still hold. \end{definition} \begin{remark} - If $\cG$ is a sheaf on $X$ and $\Omega \se X$ open, then $\cG\defon{\Omega}(U) \coloneqq \cG(U)$ for open $U \se \Omega$ and $r_{U,V}^{(\cG\defon{\Omega})}(f) \coloneqq r_{U,V}^{(\cG)}(f)$ is a sheaf of the same kind as $\cG$ on $\Omega$. + If $\cG$ is a sheaf on $X$ and $\Omega \subseteq X$ open, then $\cG\defon{\Omega}(U) \coloneqq \cG(U)$ for open $U \subseteq \Omega$ and $r_{U,V}^{(\cG\defon{\Omega})}(f) \coloneqq r_{U,V}^{(\cG)}(f)$ is a sheaf of the same kind as $\cG$ on $\Omega$. \end{remark} \begin{remark} The notion of restriction of a sheaf to a closed subset, or of preimages under general continuous maps, can be defined but this is a bit harder. @@ -2339,7 +2341,7 @@ If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ contradic Similarly, a ring structure on $G$ can be used to define the structure of a sheaf of rings on $\fG$. \end{example} \begin{example} - If in the previous example $G$ carries a topology and $\cG(U) \se \fG(U)$ is the subset (subring, subgroup) of continuous functions $U \xrightarrow{f} G$, then $\cG$ is a subsheaf of $\fG$, called the sheaf of continuous $G$-valued functions on (open subsets of) $X$. + If in the previous example $G$ carries a topology and $\cG(U) \subseteq \fG(U)$ is the subset (subring, subgroup) of continuous functions $U \xrightarrow{f} G$, then $\cG$ is a subsheaf of $\fG$, called the sheaf of continuous $G$-valued functions on (open subsets of) $X$. \end{example} \begin{example} @@ -2350,20 +2352,20 @@ If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ contradic \end{example} \subsubsection{The structure sheaf on a closed subset of $\mathfrak{k}^n$} -Let $X \se \mathfrak{k}^n$ be open. Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. +Let $X \subseteq \mathfrak{k}^n$ be open. Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. \begin{definition}\label{structuresheafkn} - For open subsets $U \se X$, let $\cO_X(U)$ be the set of functions $U \xrightarrow{\phi} \mathfrak{k}$ such that every $x \in U$ has a neighbourhood $V$ such that there are $f,g \in R$ such that for $y \in V$ we have $g(y) \neq 0$ and $\phi(y) = \frac{f(y)}{g(y)}$. + For open subsets $U \subseteq X$, let $\cO_X(U)$ be the set of functions $U \xrightarrow{\phi} \mathfrak{k}$ such that every $x \in U$ has a neighbourhood $V$ such that there are $f,g \in R$ such that for $y \in V$ we have $g(y) \neq 0$ and $\phi(y) = \frac{f(y)}{g(y)}$. \end{definition} \begin{remark}\label{structuresheafcontinuous} $\cO_X$ is a subsheaf (of rings) of the sheaf of $\mathfrak{k}$-valued functions on $X$. The elements of $\cO_X(U)$ are continuous: - Let $M \se \mathfrak{k}$ be closed. We must show the closedness of $N \coloneqq \phi\inv(M)$ in $U$. For $M = \mathfrak{k}$ this is trivial. Otherwise $M$ is finite and we may assume $M = \{t\} $ for some $t \in \mathfrak{k}$. For $x \in U$, there are open $V_x \se U$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $V_x$. + Let $M \subseteq \mathfrak{k}$ be closed. We must show the closedness of $N \coloneqq \phi\inv(M)$ in $U$. For $M = \mathfrak{k}$ this is trivial. Otherwise $M$ is finite and we may assume $M = \{t\} $ for some $t \in \mathfrak{k}$. For $x \in U$, there are open $V_x \subseteq U$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $V_x$. Then $N \cap V_x = V(f_x - t\cdot g_x) \cap V_x)$ is closed in $V_x$. As the $V_x$ cover $U$ and $U$ is quasi-compact, $N$ is closed in $U$. \end{remark} \begin{proposition}\label{structuresheafri} - Let $X = V(I)$ where $I = \sqrt{I} \se R$ is an ideal. Let $A = R / I$. Then + Let $X = V(I)$ where $I = \sqrt{I} \subseteq R$ is an ideal. Let $A = R / I$. Then \begin{align} \phi: A &\longrightarrow \cO_X(X) \\ f \mod I &\longmapsto f\defon{X} @@ -2376,17 +2378,17 @@ is an isomorphism. Its injectivity follows from the Nullstellensatz and $I = \sqrt{I}$ (\ref{hns3}). - Let $\phi \in \cO_X(X)$. for $x \in X$, there are an open subset $U_x \se X$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $U_x$. + Let $\phi \in \cO_X(X)$. for $x \in X$, there are an open subset $U_x \subseteq X$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $U_x$. \begin{claim} \Wlog we can assume $U_x = X \sm V(g_x)$. \end{claim} \begin{subproof} - The closed subsets $(X \sm U_x) \se \mathfrak{k}^n$ has the form $X\sm U_x = V(J_x)$ for some ideal $J_x \se R$. + The closed subsets $(X \sm U_x) \subseteq \mathfrak{k}^n$ has the form $X\sm U_x = V(J_x)$ for some ideal $J_x \subseteq R$. As $x \not\in X \sm V_x$ there is $h_x \in J_x$ with $h_x(x) \neq 0$. Replacing $U_x$ by $X \sm V(h_x)$, $f_x$ by $f_xh_x$ and $g_x$ by $g_xh_x$, we may assume $U_x = X \sm V(g_x)$. \end{subproof} \begin{claim} - \Wlog we can assume $V(g_x) \se V(f_x)$. + \Wlog we can assume $V(g_x) \subseteq V(f_x)$. \end{claim} \begin{subproof} Replace $f_x$ by $f_xg_x$ and $g_x$ by $g_x^2$. @@ -2413,10 +2415,10 @@ is an isomorphism. Hence $\phi = F\defon{X}$. \end{proof} \subsubsection{The structure sheaf on closed subsets of $\bP^n$} -Let $X \se \bP^n$ be closed and $R_\bullet = \mathfrak{k}[X_0,\ldots,X_n]$ with its usual grading. +Let $X \subseteq \bP^n$ be closed and $R_\bullet = \mathfrak{k}[X_0,\ldots,X_n]$ with its usual grading. \begin{definition}\label{structuresheafpn} - For open $U \se X$, let $\cO_X(U)$ be the set of functions $U \xrightarrow{\phi} \mathfrak{k}$ such that for every $x \in U$, there are an open subset $W \se U$, a natural number $d$ and $f,g \in R_d$ such that $W \cap \Vp(g) = \emptyset$ and $\phi(y) = \frac{f(y_0,\ldots,y_n)}{g(y_0,\ldots,y_n)}$ for $y = [y_0,\ldots,y_n] \in W$. + For open $U \subseteq X$, let $\cO_X(U)$ be the set of functions $U \xrightarrow{\phi} \mathfrak{k}$ such that for every $x \in U$, there are an open subset $W \subseteq U$, a natural number $d$ and $f,g \in R_d$ such that $W \cap \Vp(g) = \emptyset$ and $\phi(y) = \frac{f(y_0,\ldots,y_n)}{g(y_0,\ldots,y_n)}$ for $y = [y_0,\ldots,y_n] \in W$. \end{definition} \begin{remark} @@ -2478,7 +2480,7 @@ The following is somewhat harder than in the affine case: \end{example} \subsubsection{Subcategories} \begin{definition}[Subcategories] - A \vocab{subcategory} of $\cA$ is a category $\cB$ such that $\Ob(\cB) \se \Ob(\cA)$, such that $\Hom_\cB(X,Y) \se \Hom_\cA(X,Y)$ for objects $X$ and $Y$ of $\cB$, such that for every object $X \in \Ob(\cB)$, the identity $\Id_X$ of $X$ is the same in $\cB$ as in $\cA$, and such that for composable morphisms in $\cB$, their compositions in $\cA$ and $\cB$ coincide. + A \vocab{subcategory} of $\cA$ is a category $\cB$ such that $\Ob(\cB) \subseteq \Ob(\cA)$, such that $\Hom_\cB(X,Y) \subseteq \Hom_\cA(X,Y)$ for objects $X$ and $Y$ of $\cB$, such that for every object $X \in \Ob(\cB)$, the identity $\Id_X$ of $X$ is the same in $\cB$ as in $\cA$, and such that for composable morphisms in $\cB$, their compositions in $\cA$ and $\cB$ coincide. We call $\cB$ a \vocab{full subcategory} of $\cA$ if in addition $\Hom_\cB(X,Y) = \Hom_\cA(X,Y)$ for arbitrary $X,Y \in \Ob(\cB)$. \end{definition} \begin{example} @@ -2505,7 +2507,7 @@ The following is somewhat harder than in the affine case: \begin{example} \begin{itemize} \item There are \vocab[Functor!forgetful]{forgetful functors} from rings to abelian groups or from abelian groups to sets which drop the multiplicative structure of a ring or the group structure of a group. - \item If $\mathfrak{k}$ is any vector space there is a contravariant functor from $\mathfrak{k}$-vector spaces to itself sending $V$ to its dual vector space $V\se$ and $V \xrightarrow{f} W$ to the dual linear map $W\st \xrightarrow{f\st} V\st$. + \item If $\mathfrak{k}$ is any vector space there is a contravariant functor from $\mathfrak{k}$-vector spaces to itself sending $V$ to its dual vector space $V\subseteq$ and $V \xrightarrow{f} W$ to the dual linear map $W\st \xrightarrow{f\st} V\st$. When restricted to the full subcategory of finite-dimensional vector spaces it becomes a contravariant self-equivalence of that category. \item The embedding of a subcategory is a faithful functor. In the case of a full subcategory it is also full. \end{itemize} @@ -2516,21 +2518,21 @@ The following is somewhat harder than in the affine case: \subsection{The category of varieties} \begin{definition}[Algebraic variety]\label{defvariety} - An \vocab{algebraic variety} or \vocab{prevariety} over $\mathfrak{k}$ is a pair $(X, \cO_X)$, where $X$ is a topological space and $\cO_X$ a subsheaf of the sheaf of $\mathfrak{k}$-valued functions on $X$ such that for every $x \in X$, there are a neighbourhood $U_x$ of $x$ in $X$, an open subset $V_x$ of a closed subset $Y_x$ of $\mathfrak{k}^{n_x}$\footnote{By the result of \ref{affopensubtopbase} it can be assumed that $V_x = Y_x$ without altering the definition.} and a homeomorphism $V_x \xrightarrow{\iota_x} U_x$ such that for every open subset $V \se U_x$ and every function $V\xrightarrow{f} \mathfrak{k}$, we have $f \in \cO_X(V) \iff \iota\st_x(f) \in \cO_{Y_x}(\iota_x\inv(V))$, + An \vocab{algebraic variety} or \vocab{prevariety} over $\mathfrak{k}$ is a pair $(X, \cO_X)$, where $X$ is a topological space and $\cO_X$ a subsheaf of the sheaf of $\mathfrak{k}$-valued functions on $X$ such that for every $x \in X$, there are a neighbourhood $U_x$ of $x$ in $X$, an open subset $V_x$ of a closed subset $Y_x$ of $\mathfrak{k}^{n_x}$\footnote{By the result of \ref{affopensubtopbase} it can be assumed that $V_x = Y_x$ without altering the definition.} and a homeomorphism $V_x \xrightarrow{\iota_x} U_x$ such that for every open subset $V \subseteq U_x$ and every function $V\xrightarrow{f} \mathfrak{k}$, we have $f \in \cO_X(V) \iff \iota\st_x(f) \in \cO_{Y_x}(\iota_x\inv(V))$, In this, the \vocab{pull-back} $\iota_x\st(f)$ of $f$ is defined by $(\iota_x\st(f))(\xi) \coloneqq f(\iota_x(\xi))$. - A morphism $(X, \cO_X) \to (Y, \cO_Y)$ of varieties is a continuous map $X \xrightarrow{\phi} Y$ such that for all open $U \se Y$ and $f \in \cO_Y(U)$, $\phi\st(f) \in \cO_X(\phi\inv(U))$. + A morphism $(X, \cO_X) \to (Y, \cO_Y)$ of varieties is a continuous map $X \xrightarrow{\phi} Y$ such that for all open $U \subseteq Y$ and $f \in \cO_Y(U)$, $\phi\st(f) \in \cO_X(\phi\inv(U))$. An isomorphism is a morphism such that $\phi$ is bijective and $\phi\inv$ also is a morphism of varieties. \end{definition} \begin{example} \begin{itemize} - \item If $(X, \cO_X)$ is a variety and $U \se X$ open, then $(U, \cO_X\defon{U})$ is a variety (called an \vocab{open subvariety} of $X$), and the embedding $U \to X$ is a morphism of varieties. + \item If $(X, \cO_X)$ is a variety and $U \subseteq X$ open, then $(U, \cO_X\defon{U})$ is a variety (called an \vocab{open subvariety} of $X$), and the embedding $U \to X$ is a morphism of varieties. \item If $X$ is a closed subset of $\mathfrak{k}^n$ or $\bP^n$, then $(X, \cO_X)$ is a variety, where $\cO_X$ is the structure sheaf on $X$ (\ref{structuresheafkn}, reps. \ref{structuresheafpn}). A variety is called \vocab[Variety!affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of this form, with $X $ closed in $\mathfrak{k}^n$ (resp. $\bP^n$). A variety which is isomorphic to and open subvariety of $X$ is called \vocab[Variety!quasi-affine]{quasi-affine} (resp. \vocab[Variety!quasi-projective]{quasi-projective}). - \item If $X = V(X^2 - Y^3) \se \mathfrak{k}^2$ then $\mathfrak{k} \xrightarrow{t \mapsto (t^3,t^2)} X$ is a morphism which is a homeomorphism of topological spaces but not an isomorphism of varieties. + \item If $X = V(X^2 - Y^3) \subseteq \mathfrak{k}^2$ then $\mathfrak{k} \xrightarrow{t \mapsto (t^3,t^2)} X$ is a morphism which is a homeomorphism of topological spaces but not an isomorphism of varieties. % TODO \item The composition of two morphisms $X \to Y \to Z$ of varieties is a morphism of varieties. @@ -2540,10 +2542,10 @@ The following is somewhat harder than in the affine case: \subsubsection{The category of affine varieties} \begin{lemma}\label{localinverse} - Let $X$ be any $\mathfrak{k}$-variety and $U \se X$ open. + Let $X$ be any $\mathfrak{k}$-variety and $U \subseteq X$ open. \begin{enumerate}[i)] \item All elements of $\cO_X(U)$ are continuous. - \item If $U \se X$ is open, $U \xrightarrow{\lambda} \mathfrak{k}$ any function and every $x \in U$ has a neighbourhood $V_x \se U$ such that $\lambda \defon{V_x} \in \cO_X(V_x)$, then $\lambda \in \cO_X(U)$. + \item If $U \subseteq X$ is open, $U \xrightarrow{\lambda} \mathfrak{k}$ any function and every $x \in U$ has a neighbourhood $V_x \subseteq U$ such that $\lambda \defon{V_x} \in \cO_X(V_x)$, then $\lambda \in \cO_X(U)$. \item If $\vartheta \in \cO_X(U)$ and $\vartheta(x) \neq 0$ for all $x \in U$, then $\vartheta \in \cO_X(U)^{\times }$. \end{enumerate} \end{lemma} @@ -2553,8 +2555,8 @@ The following is somewhat harder than in the affine case: \item For the second part, let $\lambda_x \coloneqq \lambda \defon{V_x} $. We have $\lambda_x\defon{V_x \cap V_y} = \lambda \defon{V_x \cap V_y} = \lambda_y \defon{V_x \cap V_y} $. The $V_x$ cover $U$. By the sheaf axiom for $\cO_X$ there is $\ell \in \cO_X(U)$ with $\ell\defon{V_x} =\lambda_x$. It follows that $\ell=\lambda$. - \item By the definition of variety, every $x \in U$ has a quasi-affine neighbourhood $V \se U$. We can assume $U$ to be quasi-affine and $X = V(I) \se \mathfrak{k}^n$, as the general assertion follows by an application of ii). - If $x \in U$ there are a neighbourhood $x \in W \se U$ and $a,b \in R = \mathfrak{k}[X_1,\ldots,X_n]$ such that $\vartheta(y) = \frac{a(y)}{b(y)}$ for $y \in W$, with $b(y) \neq 0$. + \item By the definition of variety, every $x \in U$ has a quasi-affine neighbourhood $V \subseteq U$. We can assume $U$ to be quasi-affine and $X = V(I) \subseteq \mathfrak{k}^n$, as the general assertion follows by an application of ii). + If $x \in U$ there are a neighbourhood $x \in W \subseteq U$ and $a,b \in R = \mathfrak{k}[X_1,\ldots,X_n]$ such that $\vartheta(y) = \frac{a(y)}{b(y)}$ for $y \in W$, with $b(y) \neq 0$. Then $a(x) \neq 0$ as $\vartheta(x) \neq 0$. Replacing $W$ by $W \sm V(a)$, we may assume that $a$ has no zeroes on $W$. Then $\lambda(y) = \frac{b(y)}{a(y)}$ for $y \in W$ has a non-vanishing denominator and $\lambda \in \cO_X(U)$. We have $\lambda \cdot \vartheta = 1$, thus $\vartheta \in \cO_X(U)^{\times}$. @@ -2586,7 +2588,7 @@ The following is somewhat harder than in the affine case: It is clear that $\nil(\cO_X(X)) = \{0\}$ for arbitrary varieties. For general varieties it is however not true that $\cO_X(X)$ is a $\mathfrak{k}$-algebra of finite type. There are counterexamples even for quasi-affine $X$. %TODO - If, however, $X$ is affine, we may assume w.l.o.g. that $X = V(I)$ where $I = \sqrt{I} \se R$ is an ideal with $R = \mathfrak{k}[X_1,\ldots,X_n]$. + If, however, $X$ is affine, we may assume w.l.o.g. that $X = V(I)$ where $I = \sqrt{I} \subseteq R$ is an ideal with $R = \mathfrak{k}[X_1,\ldots,X_n]$. Then $\cO_X(X) \cong R / I$ (see \ref{structuresheafri}) is a $\mathfrak{k}$-algebra of finite type. \end{remark} @@ -2595,10 +2597,10 @@ The following is somewhat harder than in the affine case: - It suffices to investigate $\phi$ when $Y$ is an open subset of $V(I) \se \mathfrak{k}^n$, where $I = \sqrt{I} \se R$ is an ideal and $Y = V(I)$ when $Y$ is affine. - Let $(f_1,\ldots,f_n)$ be the components of $X \xrightarrow{f} Y \se \mathfrak{k}^n$. Let $Y \xrightarrow{\xi_i} \mathfrak{k}$ be the $i$-th coordinate. + It suffices to investigate $\phi$ when $Y$ is an open subset of $V(I) \subseteq \mathfrak{k}^n$, where $I = \sqrt{I} \subseteq R$ is an ideal and $Y = V(I)$ when $Y$ is affine. + Let $(f_1,\ldots,f_n)$ be the components of $X \xrightarrow{f} Y \subseteq \mathfrak{k}^n$. Let $Y \xrightarrow{\xi_i} \mathfrak{k}$ be the $i$-th coordinate. By definition $f_i = f\st(\xi_i) $. Thus $f$ is uniquely determined by $\cO_Y(Y) \xrightarrow{f\st} \cO_X(X)$. - Conversely, let $Y = V(I)$ and $\cO_Y(Y) \xrightarrow{\phi} \cO_X(X)$ be a morphism of $\mathfrak{k}$-algebras. Define $f_i \coloneqq \phi(\xi_i)$ and consider $X \xrightarrow{f = (f_1,\ldots,f_n)} Y\se \mathfrak{k}^n$. + Conversely, let $Y = V(I)$ and $\cO_Y(Y) \xrightarrow{\phi} \cO_X(X)$ be a morphism of $\mathfrak{k}$-algebras. Define $f_i \coloneqq \phi(\xi_i)$ and consider $X \xrightarrow{f = (f_1,\ldots,f_n)} Y\subseteq \mathfrak{k}^n$. \begin{claim} $f$ has image contained in $Y$. \end{claim} @@ -2610,7 +2612,7 @@ The following is somewhat harder than in the affine case: $f$ is a morphism in $\Var_\mathfrak{k}$ \end{claim} \begin{subproof} - For open $\Omega \se Y, U = f\inv(\Omega) = \{x \in X | \A \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y \sm \Omega = V(J)$. + For open $\Omega \subseteq Y, U = f\inv(\Omega) = \{x \in X | \A \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y \sm \Omega = V(J)$. If $\lambda \in \cO_Y(\Omega)$ and $x \in U$, then $f(x)$ has a neighbourhood $V$ such that there are $a,b \in R$ with $\lambda(v) = \frac{a(v)}{b(v)}$ and $b(v) \neq 0$ for all $v \in V$. Let $W \coloneqq f\inv(V)$. Then $\alpha \coloneqq \phi(a)\defon{W} \in \cO_X(W)$, $\beta \coloneqq \phi(b)\defon{W} \in \cO_X(W)$. By the second part of \ref{localinverse} $\beta \in \cO_X(W)^{\times}$ and $f\st(\lambda)\defon{W} = \frac{\alpha}{\beta} \in \cO_X(W)$. @@ -2642,7 +2644,7 @@ The following is somewhat harder than in the affine case: A set $\cB$ of open subsets of a topological space $X$ is called a \vocab{topology base} for $X$ if every open subset of $X$ can be written as a (possibly empty) union of elements of $\cB$. \end{definition} \begin{fact} -If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X = \bigcup_{U \in \cB} U$ and for arbitrary $U, V \in \cB, U \cap V$ is a union of elements of $\cB$. +If $X$ is a set, then $\cB \subseteq \cP(X)$ is a base for some topology on $X$ iff $X = \bigcup_{U \in \cB} U$ and for arbitrary $U, V \in \cB, U \cap V$ is a union of elements of $\cB$. \end{fact} \begin{definition} Let $X$ be a variety. @@ -2668,9 +2670,9 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X X \arrow[hookrightarrow]{r}{}& U \arrow[swap]{u}{\sigma} & \cO_X(U) \end{tikzcd} \end{figure} - For the rest of the proof, we may assume $X = V(I) \se \mathfrak{k}^n$ where $I = \sqrt{I} \se R \coloneqq\mathfrak{k}[X_1,\ldots,X_n]$ is an ideal. + For the rest of the proof, we may assume $X = V(I) \subseteq \mathfrak{k}^n$ where $I = \sqrt{I} \subseteq R \coloneqq\mathfrak{k}[X_1,\ldots,X_n]$ is an ideal. Then $A \coloneqq \cO_X(X) \cong R / I$ and there is $\ell \in R$ such that $\ell\defon{X} = \lambda$. - Let $Y = V(J) \se \mathfrak{k}^{n+1}$ where $J \se \mathfrak{k}[Z,X_1,\ldots,X_n]$ is generated by the elements of $I$ and $1 - Z\ell(X_1,\ldots,X_n)$. + Let $Y = V(J) \subseteq \mathfrak{k}^{n+1}$ where $J \subseteq \mathfrak{k}[Z,X_1,\ldots,X_n]$ is generated by the elements of $I$ and $1 - Z\ell(X_1,\ldots,X_n)$. Then $\cO_Y(Y) \cong \mathfrak{k}[Z,X_1,\ldots,X_n] / J \cong A[Z] / (1 -\lambda Z) \cong A_\lambda$. By the proposition about affine varieties (\ref{propaffvar}), the morphism $\mathfrak{s}: \cO_Y(Y) \cong A_\lambda \to \cO_X(U)$ corresponds to a morphism $U \xrightarrow{\sigma} Y$. @@ -2687,11 +2689,11 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X The affine open subsets of a variety $X$ are a topology base on $X$. \end{corollary} \begin{proof} - Let $X = V(I) \se \mathfrak{k}^n$ with $I = \sqrt{I}$. If $U \se X$ is open then $X \sm U = V(J)$ with $J \supseteq I$ and $U = \bigcup_{f \in J} (X \sm V(f))$. + Let $X = V(I) \subseteq \mathfrak{k}^n$ with $I = \sqrt{I}$. If $U \subseteq X$ is open then $X \sm U = V(J)$ with $J \supseteq I$ and $U = \bigcup_{f \in J} (X \sm V(f))$. Thus $U$ is a union of affine open subsets. The same then holds for arbitrary quasi-affine varieties. - Let $X$ be any variety, $U \se X$ open and $x \in U$. - By the definition of variety, $x$ has a neighbourhood $V_x$ which is quasi-affine, and replacing $V_x$ by $U \cap V_x$ which is also quasi-affine we may assume $V_x \se U$. + Let $X$ be any variety, $U \subseteq X$ open and $x \in U$. + By the definition of variety, $x$ has a neighbourhood $V_x$ which is quasi-affine, and replacing $V_x$ by $U \cap V_x$ which is also quasi-affine we may assume $V_x \subseteq U$. $V_x$ is a union of its affine open subsets. Because $U$ is the union of the $V_x$, $U$ as well is a union of affine open subsets. \end{proof} @@ -2708,7 +2710,7 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X Let $\cG$ be a presheaf of sets on the topological space $X$, and let $x \in X$. The \vocab{stalk} (\vocab[Stalk]{Halm}) of $\cG$ at $x$ is the set of equivalence classes of pairs $(U, \gamma)$, where $U$ is an open neighbourhood of $x$ and $\gamma \in \cG(U)$ and the equivalence relation $\sim $ is defined as follows: - $( U , \gamma) \sim (V, \delta)$ iff there exists an open neighbourhood $W \se U \cap V$ of $x$ such that $\gamma \defon{W} = \delta \defon{W}$. + $( U , \gamma) \sim (V, \delta)$ iff there exists an open neighbourhood $W \subseteq U \cap V$ of $x$ such that $\gamma \defon{W} = \delta \defon{W}$. If $\cG$ is a presheaf of groups, one can define a groups structure on $\cG_x$ by @@ -2731,10 +2733,10 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X In the case of a sheaf, the image of the injective map $\cG(U) \xrightarrow{\gamma \mapsto (\gamma_x)_{x \in U}} \prod_{x \in U} \cG_x$ is the set of all $(g_x)_{x \in U} \in \prod_{x \in U} \cG_x $ satisfying the following \vocab{coherence condition}: - For every $x \in U$, there are an open neighbourhood $W_x \se U$ of $x$ and $g^{(x)} \in \cG(W_x)$ with $g_y^{(x)} = g_y$ for all $y \in W_x$. + For every $x \in U$, there are an open neighbourhood $W_x \subseteq U$ of $x$ and $g^{(x)} \in \cG(W_x)$ with $g_y^{(x)} = g_y$ for all $y \in W_x$. \end{fact} \begin{proof} - Because of $\gamma_x = \delta_x$, there is $x \in W_x \se U$ open such that $\gamma\defon{W_x} = \delta\defon{W_x}$. As the $W_x$ cover $U$, $\gamma = \delta$ by the sheaf axiom. + Because of $\gamma_x = \delta_x$, there is $x \in W_x \subseteq U$ open such that $\gamma\defon{W_x} = \delta\defon{W_x}$. As the $W_x$ cover $U$, $\gamma = \delta$ by the sheaf axiom. \end{proof} \begin{definition} Let $\cG$ be a sheaf of functions. @@ -2759,8 +2761,8 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X \end{proof} \begin{proposition}\label{proplocalring} - Let $X = \Va(I) \se \mathfrak{k}^n$ be equipped with its usual structure sheaf, where $I = \sqrt{I} \se R = \mathfrak{k}[X_1,\ldots,X_n]$ . Let $x \in X$ and $A = \cO_X(X) \cong R / I$. - $\{P \in R | P(x) = 0\} \text{\reflectbox{$\coloneqq$}} \fn_x \se R$ is maximal, $I \se \fn_x$ and $\mathfrak{m}_x \coloneqq \fn_x / I$ is the maximal ideal of elements of $A$ vanishing at $x$. + Let $X = \Va(I) \subseteq \mathfrak{k}^n$ be equipped with its usual structure sheaf, where $I = \sqrt{I} \subseteq R = \mathfrak{k}[X_1,\ldots,X_n]$ . Let $x \in X$ and $A = \cO_X(X) \cong R / I$. + $\{P \in R | P(x) = 0\} \text{\reflectbox{$\coloneqq$}} \fn_x \subseteq R$ is maximal, $I \subseteq \fn_x$ and $\mathfrak{m}_x \coloneqq \fn_x / I$ is the maximal ideal of elements of $A$ vanishing at $x$. If $\lambda \in A \sm \mathfrak{m}_x$, we have $\lambda_x \in \cO_{X,x}^{\times}$, where $\lambda_x$ denotes the image under $A \cong \cO_X(X) \to \cO_{X,x}$. By the universal property of the localization, there exists a unique ring homomorphism $A_{\mathfrak{m}_x} \xrightarrow{\iota} \cO_{X,x}$ such that @@ -2778,7 +2780,7 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X \end{proposition} \begin{proof} To show surjectivity, let $\ell = (U, \lambda) / \sim \in \cO_{X,x}$, where $U$ is an open neighbourhood of $x$ in $X$. - We have $X \sm U = V(J)$ where $J \se A$ is an ideal. As $x \in U$ there is $f \in J$ with $f(x) \neq 0$. Replacing $U $ by $X \sm V(f)$ we may assume $U = X \sm V(f)$. + We have $X \sm U = V(J)$ where $J \subseteq A$ is an ideal. As $x \in U$ there is $f \in J$ with $f(x) \neq 0$. Replacing $U $ by $X \sm V(f)$ we may assume $U = X \sm V(f)$. By \ref{oxulocaf}, $\cO_X(U) \cong A_f$, and $\lambda = f^{-n}\vartheta$ for some $n \in \N$ and $\vartheta \in A$. Then $\ell = \iota(f^{-n} \vartheta)$ where the last fraction is taken in $A_{\mathfrak{m}_x}$. @@ -2786,7 +2788,7 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X Let $\lambda = \frac{\vartheta}{g} \in A_{\mathfrak{m}_x}$ with $\iota(\lambda) = 0$. It is easy to see that $\iota(\lambda) = (X \sm V(g), \frac{\vartheta}{g}) / \sim $. Thus there is an open neighbourhood $U$ of $x$ in $X \sm V(g)$ such that $\vartheta$ vanishes on $U$. - Similar as before there is $h \in A$ with $h(x) \neq 0$ and $W = X \sm V(h) \se U$. + Similar as before there is $h \in A$ with $h(x) \neq 0$ and $W = X \sm V(h) \subseteq U$. By the isomorphism $\cO_X(W) \cong A_h$, there is $n \in \N$ with $h^{n}\vartheta = 0$ in $A$. Since $h \not\in \mathfrak{m}_x$, $h$ is a unit and the image of $\vartheta$ in $A_{\mathfrak{m}_x}$ vanishes, implying $\lambda = 0$. \end{proof} \subsubsection{Intersection multiplicities and Bezout's theorem} @@ -2794,7 +2796,7 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X Let $R = \mathfrak{k}[X_0,X_1,X_2]$ equipped with its usual grading and let $x \in \bP^{2}$. Let $G \in R_g, H \in R_h$ be homogeneous polynomials with $x \in V(G) \cap V(h)$. Let $\ell\in R_1$ such that $\ell(x) \neq 0$. Then $x \in U = \bP^2 \sm V(\ell)$ and the rational functions $\gamma = \ell^{-g}G, \eta = \ell^{-h}H$ are elements of $\cO_{\bP^2}(U)$. -Let $I_x(G,H) \se \cO_{\bP^2,x}$ denote the ideal generated by $\gamma_x$ and $\eta_x$. +Let $I_x(G,H) \subseteq \cO_{\bP^2,x}$ denote the ideal generated by $\gamma_x$ and $\eta_x$. \noindent The dimension $\dim_{\mathfrak{k}}(\cO_{X,x} / I_x(G,H)) \text{\reflectbox{$\coloneqq$}} i_x(G,H)$ is called the \vocab{intersection multiplicity} of $G$ and $H$ at $x$. @@ -2804,7 +2806,7 @@ Let $I_x(G,H) \se \cO_{\bP^2,x}$ denote the ideal generated by $\gamma_x$ and $\ Thus $I_x(G,H)$ does not depend on the choice of $\ell \in R_1$ with $\ell(x) \neq 0$. \end{remark} \begin{theorem}[Bezout's theorem] - In the above situation, assume that $V(H)$ and $V(G)$ intersect properly in the sense that $V(G) \cap V(H) \se \bP^2$ has no irreducible component of dimension $\ge 1$. + In the above situation, assume that $V(H)$ and $V(G)$ intersect properly in the sense that $V(G) \cap V(H) \subseteq \bP^2$ has no irreducible component of dimension $\ge 1$. Then \[ \sum_{x \in V(G) \cap V(H)} i_x(G,H) = gh @@ -2834,12 +2836,12 @@ Let $I_x(G,H) \se \cO_{\bP^2,x}$ denote the ideal generated by $\gamma_x$ and $\ \begin{itemize} - \item[HNS2 $\implies$ HNS1b] Let $I \se \mathfrak{l}[X_1,\ldots,X_n]$. $I \se \mathfrak{m}$ maximal. $R / \mathfrak{m}$ is isomorphic to a field extension of $\mathfrak{l}$. Finite by HNS2. + \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) \se V(f) \iff f \in \sqrt{I} $). Suppose $V(I) \se V(f)$. $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n, T], J \se R'$ 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} @@ -2851,7 +2853,7 @@ Fact about integrality and field: % TODO -Technical lemma for Noether normalization: For $S \se \N^n$ finite, there exists $k \in \N^n$ such that $k_1 = 1$ and $s_1 \neq s_2 \in S \implies \langle k, s_1 \rangle \neq \langle k, s_2 \rangle$: +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: @@ -2907,12 +2909,12 @@ 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 $\fq \se \sigma(\fr)$ for some $\sigma \in G$ (Krull going-up, no inclusions) + \item only need to show $\fq \subseteq \sigma(\fr)$ for some $\sigma \in G$ (Krull going-up, no inclusions) \item Suppose not. Then $x \in \fq \sm \bigcup_{\sigma \in G} \sigma(\fr)$ (prime aviodance) \item $y = \prod_{\sigma \in G} \sigma(x) \in \fq \sm \fr$ ($\fr$ prime ideal) \item $\E k \in \N$ s.t. $y^k \in K$ ($y \in L^G$) \item $y^k \in K \cap B = A $ ($A$ normal). Thus $y^k \in (A \cap \fq) \sm (A \cap \fr) = \fp \sm \fp$. - \item $L / K$ infinite: Apply Zorn to pairs $(M, \sigma)$ where $K \se M \se L$ and $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap M$. + \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} @@ -2932,7 +2934,7 @@ Original (Noether normalization) Artin-Tate Uncountable fields \begin{landscape} -\section{Übersicht} +\subseteqction{Übersicht} {\rowcolors{2}{gray!10}{white} \begin{longtable}{lll} \end{longtable} diff --git a/algebra.sty b/algebra.sty index 0eb48b1..b8e2ca5 100644 --- a/algebra.sty +++ b/algebra.sty @@ -7,7 +7,6 @@ \RequirePackage{hyperref} \RequirePackage[english, index]{mkessler-vocab} \RequirePackage{mkessler-hypersetup} -\input{/home/jrpie/templates/latex/math.tex} \RequirePackage[utf8x]{inputenc} \RequirePackage{babel} @@ -42,16 +41,12 @@ \newcommand{\Vp}{\ensuremath V_{\mathbb{P}}}%\Spec}} \newcommand{\Pn}{\bP^n}%\Spec}} -\newcommand{\Span}[1]{\langle#1\rangle} \newcommand{\npr}{\footnote{Not relevant for the exam.}} \newcommand{\limrel}{\footnote{Limited relevance for the exam.}} % may appear in 3x questions -\DeclareMathOperator{\Mat}{Mat} -\DeclareMathOperator{\ev}{ev} +%\DeclareMathOperator{\ev}{ev} \DeclareMathOperator{\Ker}{Ker} -\DeclareMathOperator{\Aut}{Aut} -\DeclareMathOperator{\Gal}{Gal} \DeclareMathOperator{\nil}{\mathfrak{nil}} -\DeclareMathOperator{\rad}{\mathfrak{rad}} +%\DeclareMathOperator{\rad}{\mathfrak{rad}} \RequirePackage{stackengine} \stackMath \usetikzlibrary{arrows.meta,