From 0c91b657f2e4a2f2d5330f3d6326bb246ea286d4 Mon Sep 17 00:00:00 2001 From: zeramorphic Date: Thu, 23 Nov 2023 17:55:27 +0000 Subject: [PATCH] Lectures 22A Signed-off-by: zeramorphic --- ...tegrality_finiteness_finite_generation.tex | 6 +- iii/commalg/06_direct_and_inverse_limits.tex | 76 ++++++++++++++++++- iii/commalg/07_dimension_theory.tex | 45 +++++++++++ iii/commalg/main.tex | 2 + 4 files changed, 126 insertions(+), 3 deletions(-) create mode 100644 iii/commalg/07_dimension_theory.tex diff --git a/iii/commalg/04_integrality_finiteness_finite_generation.tex b/iii/commalg/04_integrality_finiteness_finite_generation.tex index 4265ae0..8a71bd8 100644 --- a/iii/commalg/04_integrality_finiteness_finite_generation.tex +++ b/iii/commalg/04_integrality_finiteness_finite_generation.tex @@ -227,6 +227,8 @@ \subsection{Integral and finite extensions} By the previous result, both extensions are finite. Then, by part (i), \( A \subseteq A'[c] \) is finite, so \( c \) is integral over \( A \). \end{proof} + +\subsection{Integral closure} \begin{definition} Let \( A \subseteq B \) be rings. The \emph{integral closure} of \( A \) in \( B \) is the set \( \overline A \) of elements of \( B \) that are integral over \( A \), which is an \( A \)-algebra. @@ -554,7 +556,7 @@ \subsection{Integrality over ideals} \end{proof} \begin{proposition} Let \( A \) be an integrally closed integral domain (in its field of fractions). - Let \( A \subseteq B \) be an extension of rings, let \( \mathfrak a \) be an ideal in \( A \), and let \( a \in B \). + Let \( A \subseteq B \) be an extension of rings, let \( \mathfrak a \) be an ideal in \( A \), and let \( b \in B \). The following are equivalent: \begin{enumerate} \item \( b \) is integral over \( \mathfrak a \); @@ -611,7 +613,7 @@ \subsection{Cohen--Seidenberg theorems} Now, \( \mathfrak q B_{\mathfrak p} \subseteq \mathfrak q' B_{\mathfrak p} \) are maximal ideals of \( B_{\mathfrak p} \), so they must coincide. \end{proof} \begin{proposition}[lying over] - Let \( A \subseteq B \) be an integral exxtension of rings, and let \( \mathfrak p \in \Spec A \). + Let \( A \subseteq B \) be an integral extension of rings, and let \( \mathfrak p \in \Spec A \). Then there is a prime ideal \( \mathfrak q \in \Spec B \) such that \( \mathfrak q \cap A = \mathfrak p \). In other words, \( \iota^\star : \Spec B \to \Spec A \) is surjective. \end{proposition} diff --git a/iii/commalg/06_direct_and_inverse_limits.tex b/iii/commalg/06_direct_and_inverse_limits.tex index e2d94b3..6309753 100644 --- a/iii/commalg/06_direct_and_inverse_limits.tex +++ b/iii/commalg/06_direct_and_inverse_limits.tex @@ -92,7 +92,7 @@ \subsection{Limits and completions} \begin{enumerate} \item the \( \mathfrak a \)-adic completion \( \hat R \) is Noetherian; \item the functor \( \hat R \otimes_R (-) \) is exact; - \item if \( M \) is a finitely generated \( R \)-module, then the natural map \( \hat R \otimes_R M \to \hat M \) is an \( \hat R \)-linear isomorphism. + \item if \( M \) is a finitely generated \( R \)-module, then the natural map \( \hat R \otimes_R M \to \hat M \) is an \( \hat R \)-linear isomorphism. \end{enumerate} \end{theorem} Thus \( \mathfrak a \)-adic completion is an exact functor from the category of finitely generated \( R \)-modules if \( R \) is Noetherian. @@ -204,3 +204,77 @@ \subsection{Graded rings and modules} Let \( M \) be an \( R \)-module. We say that filtrations \( (M_n), (M_n') \) of \( M \) are \emph{equivalent} if there exists \( n_0 \) such that for all \( n \geq 0 \), we have \( M_{n + n_0} \subseteq M_n' \) and \( M'_{n + n_0} \subseteq M_n \). \end{definition} +\begin{lemma} + Let \( \mathfrak a \) be an ideal of \( R \). + Let \( M \) be an \( R \)-module, and let \( (M_n)_{n \geq 0} \) be a stable \( \mathfrak a \)-filtration of \( M \). + Then \( (M_n)_{n \geq 0} \) is equivalent to \( (\mathfrak a^n M)_{n \geq 0} \). +\end{lemma} +\begin{proof} + As \( (M_n)_{n \geq 0} \) is an \( \mathfrak a \)-filtration, for all \( n \geq 0 \), we have + \[ M_n \supseteq \mathfrak a M_{n-1} \supseteq \mathfrak a^2 M_{n-2} \supseteq \dots \supseteq \mathfrak a^n M \supseteq \mathfrak a^{n + n_0} M \] + For the other direction, as the filtration is stable, there exists \( n_0 \) such that for each \( n \geq n_0 \), we have \( \mathfrak a M_n = M_{n+1} \). + Then \( M_{m + n_0} = \mathfrak a^n M_{n_0} \subseteq \mathfrak a^n M \) as required. +\end{proof} + +\subsection{Artin--Rees lemma} +\begin{definition} + Let \( \mathfrak a \) be an ideal of \( R \). + Let \( M \) be an \( R \)-module, and let \( (M_n)_{n \geq 0} \) be an \( \mathfrak a \)-filtration of \( M \). + Then we define + \[ R^\star = \bigoplus_{n \geq 0} \mathfrak a^n;\quad M^\star = \bigoplus_{n \geq 0} M_n \] +\end{definition} +Note that \( R^\star \) is a graded ring, as for \( x \in \mathfrak a^n, y \in \mathfrak a^\ell \), we have \( xy \in \mathfrak a^{n + \ell} \). +As \( (M_n)_{n \geq 0} \) is an \( \mathfrak a \)-filtration, \( M^\star \) is a graded \( R^\star \)-module. +Indeed, for \( x \in \mathfrak a^n \) and \( m \in M_\ell \), we have \( xm \in M_{n + \ell} \) as required. + +If \( R \) is Noetherian, the ideal \( \mathfrak a \) is finitely generated, say by \( x_1, \dots, x_r \). +Then \( R^\star \) is generated as an \( R \)-algebra by \( x_1, \dots, x_r \) by taking sums and products. +By Hilbert's basis theorem, \( R^\star \) is a Noetherian ring. +\begin{lemma} + Let \( R \) be a Noetherian ring, and let \( \mathfrak a \) be an ideal of \( R \). + Let \( M \) be a finitely generated \( R \)-module, and let \( (M_n)_{n \geq 0} \) be an \( \mathfrak a \)-filtration of \( M \). + Then, the following are equivalent: + \begin{enumerate} + \item \( M^\star \) is finitely generated as an \( R^\star \)-module; + \item the \( \mathfrak a \)-filtration \( (M_n)_{n \geq 0} \) is stable. + \end{enumerate} +\end{lemma} +\begin{proof} + First, note that each \( M_n \) is a finitely generated \( R \)-module. + Indeed, \( R \) is a Noetherian ring and \( M \) is finitely generated, so \( M \) is a Noetherian module, or equivalently, every submodule is finitely generated. + Now, consider + \[ M^\star_n = M_0 \oplus \dots \oplus M_n \oplus \mathfrak a M_n \oplus \mathfrak a^2 M_n \oplus \cdots \] + This is an \( R^\star \)-submodule of \( M^\star \). + Note that \( (M^\star_n)_{n \geq 0} \) is an ascending chain of \( R^\star \)-submodules of \( M^\star \), and this chain stabilises if and only if the \( \mathfrak a \)-filtration \( (M_n)_{n \geq 0} \) is stable. + + \emph{(i) implies (ii).} + As \( R \) is Noetherian, so is \( R^\star \) by the discussion above. + By assumption, \( M^\star \) is finitely generated as a module over a Noetherian ring, so it is Noetherian. + Hence the ascending chain \( (M^\star_n)_{n \geq 0} \) stabilises, giving the result. + + \emph{(ii) implies (i).} + Suppose \( (M_n)_{n \geq 0} \) is stable. + Then \( (M^\star_n)_{n \geq 0} \) stabilises at some \( n_0 \geq 0 \), so + \[ M^\star = \bigcup_{n \geq 0} M_n^\star = M_{n_0}^\star \] + Now, note that \( M_0 \oplus \dots \oplus M_{n_0} \) + generatees \( M^\star_{n_0} \) as an \( R^\star \)-module. + Each \( M_n \) is a finitely generated \( R \)-module, so \( M_0 \oplus \dots \oplus M_{n_0} \) is also finitely generated as an \( R \)-module. + So these generators span \( M^\star_{n_0} = M^\star \) as an \( R^\star \)-module, as required. +\end{proof} +\begin{proposition}[Artin--Rees] + Let \( R \) be a Noetherian ring, and let \( \mathfrak a \) be an ideal of \( R \). + Let \( M \) be a finitely generated \( R \)-module, and let \( (M_n)_{n \geq 0} \) be a stable \( \mathfrak a \)-filtration of \( M \). + Then for any submodule \( N \leq M \), \( (N \cap M_n)_{n \geq 0} \) is a stable \( \mathfrak a \)-filtration of \( N \). +\end{proposition} +Thus, stable filtrations pass to submodules. +\begin{proof} + First, we show that \( (N \cap M_n)_{n \geq 0} \) is indeed an \( \mathfrak a \)-filtration. + \[ \mathfrak a(N \cap M_n) \subseteq N \cap \mathfrak a M_n \subseteq N \cap M_{n + 1} \] + Now, define + \[ M^\star = \bigoplus_{n \geq 0} M_n;\quad N^\star = \bigoplus_{n \geq 0} (N \cap M_n) \] + Note that \( M^\star \) is an \( R^\star \)-submodule of \( N^\star \). + As \( R \) is Noetherian, so is \( R^\star \). + Then as \( (M_n)_{n \geq 0} \) is stable, \( M^\star \) is a finitely generated \( R^\star \)-module by the previous lemma. + Thus \( M^\star \) is a Noetherian \( R^\star \)-module. + Its submodule \( N^\star \) is then finitely generated, so \( (N \cap M_n)_{n \geq 0} \) is stable. +\end{proof} diff --git a/iii/commalg/07_dimension_theory.tex b/iii/commalg/07_dimension_theory.tex new file mode 100644 index 0000000..3a9374d --- /dev/null +++ b/iii/commalg/07_dimension_theory.tex @@ -0,0 +1,45 @@ +\subsection{???} +\begin{definition} + Let \( \mathfrak p \) be a prime ideal of \( R \). + The \emph{height} of \( \mathfrak p \), denoted \( \operatorname{ht}(p) \), is + \[ \operatorname{ht}(\mathfrak p) = \sup \qty{ d \mid \mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \dots \subsetneq \mathfrak p_d = \mathfrak p; \mathfrak p_i \in \Spec R } \] + The \emph{(Krull) dimension} of \( R \) is + \[ \dim R = \sup \qty{\operatorname{ht}(\mathfrak p) \mid \mathfrak p \in \Spec R} = \sup \qty{\operatorname{ht}(\mathfrak m) \mid \mathfrak m \in \mSpec R} \] +\end{definition} +\begin{remark} + The height of a prime ideal \( \mathfrak p \) is the Krull dimension of the localisation \( R_{\mathfrak p} \). + In particular, + \[ \dim R = \sup \qty{\dim R_{\mathfrak p} \mid \mathfrak p \in \Spec R} = \sup \qty{\dim R_{\mathfrak m} \mid \mathfrak m \in \mSpec R} \] + So the problem of computing dimension can be reduced to computing dimension of local rings. +\end{remark} +\begin{definition} + Let \( I \) be a proper ideal of \( R \). + Then the \emph{height} of \( I \) is + \[ \operatorname{ht}(I) = \inf \qty{\operatorname{ht}(\mathfrak p) \mid I \subseteq \mathfrak p} \] +\end{definition} +\begin{proposition} + Let \( A \subseteq B \) be an integral extension of rings. + Then, + \begin{enumerate} + \item \( \dim A = \dim B \); and + \item if \( A, B \) are integral domains and \( k \)-algebras for some field \( k \), they have the same transcendence degree over \( k \). + \end{enumerate} +\end{proposition} +We prove part (i); the second part is not particularly relevant for this course. +\begin{proof} + First, we show that \( \dim A \leq \dim B \). + Consider a chain of prime ideals \( \mathfrak p_0 \subsetneq \dots \subsetneq \mathfrak p_d \) in \( \Spec A \). + By the lying over theirem and the going up theorem, we obtain a chain of prime ideals \( \mathfrak q_0 \subseteq \dots \subseteq \mathfrak q_d \) in \( \Spec B \). + As \( \mathfrak p_i = \mathfrak q_i \cap A \) and \( \mathfrak p_i \neq \mathfrak p_{i+1} \), we must have \( \mathfrak q_i \neq \mathfrak q_{i+1} \). + So this produces a chain of length \( d \) in \( B \), as required. + + Now consider a chain \( \mathfrak q_0 \subsetneq \dots \subsetneq \mathfrak q_d \) in \( \Spec B \). + Contracting each ideal, we produce a chain \( \mathfrak p_0 \subseteq \dots \subseteq \mathfrak p_d \) in \( \Spec A \). + Suppose that \( \mathfrak q_i \) and \( \mathfrak q_{i+1} \) contract to the same prime ideal \( \mathfrak p_i \) in \( \Spec A \). + Note that \( \mathfrak q_i \subseteq \mathfrak q_{i+1} \), so by incomparability, they must be equal, but this is a contradiction. +\end{proof} +\begin{remark} + If \( A \) is a finitely generated \( k \)-algebra for some field \( k \), then by Noether normalisation, we obtain a \( k \)-algebra embedding \( k[T_1, \dots, T_d] \to A \), and the extension is integral. + Thus \( \dim A = \dim k[T_1, \dots, T_d] \). + One can show that \( \dim k[T_1, \dots, T_d] = d \), and hence that the integer \( d \) obtained by Noether normalisation is uniquely determined by \( A \) and \( k \). +\end{remark} diff --git a/iii/commalg/main.tex b/iii/commalg/main.tex index 04af5b1..a488d0c 100644 --- a/iii/commalg/main.tex +++ b/iii/commalg/main.tex @@ -22,5 +22,7 @@ \section{Primary decomposition} \input{05_primary_decomposition.tex} \section{Direct and inverse limits} \input{06_direct_and_inverse_limits.tex} +\section{Dimension theory} +\input{07_dimension_theory.tex} \end{document}