Skip to content

# stacks/stacks-project

Lemma about depth of localization

• Loading branch information...
aisejohan committed Aug 11, 2019
1 parent da121b7 commit ec9697a3fb8a496813dd432f175684e8f935d262
Showing with 36 additions and 32 deletions.
1. +36 −32 algebra.tex
 @@ -17299,6 +17299,31 @@ \section{Depth} $\dim(R/\mathfrak q) \geq \text{depth}(M/x^nM)$ and we win. \end{proof} \begin{lemma} \label{lemma-depth-localization} Let $R$ be a local Noetherian ring and $M$ a finite $R$-module. For a prime ideal $\mathfrak p \subset R$ we have $\text{depth}(M_\mathfrak p) + \dim(R/\mathfrak p) \geq \text{depth}(M)$. \end{lemma} \begin{proof} If $M_\mathfrak p = 0$, then $\text{depth}(M_\mathfrak p) = \infty$ and the lemma holds. If $\text{depth}(M) \leq \dim(R/\mathfrak p)$, then the lemma is true. If $\text{depth}(M) > \dim(R/\mathfrak p)$, then $\mathfrak p$ is not contained in any associated prime $\mathfrak q$ of $M$ by Lemma \ref{lemma-depth-dim-associated-primes}. Hence we can find an $x \in \mathfrak p$ not contained in any associated prime of $M$ by Lemma \ref{lemma-silly} and Lemma \ref{lemma-finite-ass}. Then $x$ is a nonzerodivisor on $M$, see Lemma \ref{lemma-ass-zero-divisors}. Hence $\text{depth}(M/xM) = \text{depth}(M) - 1$ and $\text{depth}(M_\mathfrak p / x M_\mathfrak p) = \text{depth}(M_\mathfrak p) - 1$ provided $M_\mathfrak p$ is nonzero, see Lemma \ref{lemma-depth-drops-by-one}. Thus we conclude by induction on $\text{depth}(M)$. \end{proof} \begin{lemma} \label{lemma-depth-goes-down-finite} Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $R \to S$ @@ -24145,43 +24170,22 @@ \section{Cohen-Macaulay modules} \end{lemma} \begin{proof} We may and do assume $\mathfrak p \not = \mathfrak m$ and $M$ not zero. Choose a maximal chain of primes $\mathfrak p = \mathfrak p_c \subset \mathfrak p_{c - 1} \subset \ldots \subset \mathfrak p_1 \subset \mathfrak m$. If we prove the result for $M_{\mathfrak p_1}$ over $R_{\mathfrak p_1}$, then the lemma will follow by induction on $c$. Thus we may assume that there is no prime strictly between $\mathfrak p$ and $\mathfrak m$. \medskip\noindent If $M_\mathfrak p = 0$, then the lemma holds. Assume $M_\mathfrak p \not = 0$. We have $\dim(\text{Supp}(M_\mathfrak p)) \leq \dim(\text{Supp}(M)) - 1$ as a chain of primes in the support of $M_\mathfrak p$ is a chain a primes in the support of $M$ not including $\mathfrak m$. Thus it suffices to show that the depth of $M_\mathfrak p$ is at least the depth of $M$ minus $1$. We will prove by induction on the depth of $M$ that there exists an $M$-regular sequence $f_1, \ldots, f_{\text{depth}(M) - 1}$ in $\mathfrak p$. This will prove the lemma since localization at $\mathfrak p$ is exact. Since $\text{depth}(M) = \dim((\text{Supp}(M)) \geq \dim(\text{Supp}(M_\mathfrak p)) + 1 \geq 1$ we see that the base case happens when the depth of $M$ is $1$ and this case is trivial. Assume the depth of $M$ is at least $2$. \medskip\noindent Let $I \subset R$ be the annihilator of $M$ such that $\Spec(R/I) = V(I) = \text{Supp}(M)$ (Lemma \ref{lemma-support-closed}). By Lemmas \ref{lemma-CM-over-quotient} and \ref{lemma-maximal-chain-maximal-CM} every maximal chain of primes in $V(I)$ has length $\geq 2$. Hence none of the minimal primes of $V(I)$ are equal to $\mathfrak p$. Thus we can use Lemma \ref{lemma-silly} to find a $f_1 \in \mathfrak p$ which is not contained in any of the minimal primes of $V(I)$. Then $f_1$ is a nonzerodivisor on $M$ and $M/f_1M$ has depth exactly one less by Lemma \ref{lemma-CM-one-g}. By induction we can extend to an $M$-regular sequence $f_1, \ldots, f_r \in \mathfrak p$ with $r = \text{depth}(M) - 1$ as desired. Note that $\dim(\text{Supp}(M_\mathfrak p)) \leq \dim(\text{Supp}(M)) - 1$ because any chain of primes in the support of $M_\mathfrak p$ can be extended by one more prime (namely $\mathfrak m$) in the support of $M$. On the other hand, we have $\text{depth}(M_\mathfrak p) \geq \text{depth}(M) - \dim(R/\mathfrak p) = \text{depth}(M) - 1$ by Lemma \ref{lemma-depth-localization} and our choice of $\mathfrak p$. Thus $\text{depth}(M_\mathfrak p) \geq \dim(\text{Supp}(M_\mathfrak p)$ as desired (the other inequality is Lemma \ref{lemma-bound-depth}). \end{proof} \begin{definition}

#### 0 comments on commit ec9697a

Please sign in to comment.
You can’t perform that action at this time.