
Lemma 10.103.8. Let $R$ be a Noetherian local Cohen-Macaulay ring of dimension $d$. Let $0 \to K \to R^{\oplus n} \to M \to 0$ be an exact sequence of $R$-modules. Then either $M = 0$, or $\text{depth}(K) > \text{depth}(M)$, or $\text{depth}(K) = \text{depth}(M) = d$.

Proof. If $d = 0$, then every nonzero $R$-module has depth $0$ and the lemma is true. Assume $d > 0$. Then $\text{depth}(K) > 0$ as $K$ is a submodule of a module of depth $> 0$. Hence the lemma holds if $\text{depth}(M) = 0$. Assume both $\text{depth}(M) > 0$ and $d > 0$. Then we choose $x \in \mathfrak m$ which is a nonzerodivisor on $M$ and on $R$. Then $x$ is a nonzerodivisor on $M$ and on $K$ and it follows by an easy diagram chase that $0 \to K/xK \to (R/xR)^{\oplus n} \to M/xM \to 0$ is exact. Using Lemmas 10.71.7 and 10.103.2 we find the result follows from the result for $K/xK$ over $R/xR$ which has smaller dimension. $\square$

Comment #2966 by Dario Weißmann on

So what if the depth of $R$ is zero?

Also if $M$ is the zero module then the conventions in section 10.71 say it has infinite depth.

Comment #2967 by Dario Weißmann on

Then every (nonzero) module has depth zero, right? So maybe the first line of the proof should be "If $\text{depth}(M)=0$ or $\text{depth}(R)=0$ then the lemma is clear."

Comment #3092 by on

This is an annoying but important lemma. I tried to fix it so it is actually true! Hope I succeeded this time. The fix is here.

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