Lemma 10.104.8. 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. This is a special case of Lemma 10.72.6. $\square$

## Comments (5)

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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• 7 comment(s) on Section 10.104: Cohen-Macaulay rings

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