Lemma 10.71.5. Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$. Let $M$ be a nonzero finite $R$-module. Then $\text{depth}(M)$ is equal to the smallest integer $i$ such that $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(R/\mathfrak m, M)$ is nonzero.

**Proof.**
Let $\delta (M)$ denote the depth of $M$ and let $i(M)$ denote the smallest integer $i$ such that $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(R/\mathfrak m, M)$ is nonzero. We will see in a moment that $i(M) < \infty $. By Lemma 10.62.18 we have $\delta (M) = 0$ if and only if $i(M) = 0$, because $\mathfrak m \in \text{Ass}(M)$ exactly means that $i(M) = 0$. Hence if $\delta (M)$ or $i(M)$ is $> 0$, then we may choose $x \in \mathfrak m$ such that (a) $x$ is a nonzerodivisor on $M$, and (b) $\text{depth}(M/xM) = \delta (M) - 1$. Consider the long exact sequence of Ext-groups associated to the short exact sequence $0 \to M \to M \to M/xM \to 0$ by Lemma 10.70.6:

Since $x \in \mathfrak m$ all the maps $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(\kappa , M) \to \mathop{\mathrm{Ext}}\nolimits ^ i_ R(\kappa , M)$ are zero, see Lemma 10.70.8. Thus it is clear that $i(M/xM) = i(M) - 1$. Induction on $\delta (M)$ finishes the proof. $\square$

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