Lemma 10.72.5. 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 10.63.18 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 10.71.6:
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 10.71.8. Thus it is clear that i(M/xM) = i(M) - 1. Induction on \delta (M) finishes the proof. \square
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