Lemma 10.72.11. Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $R \to S$ be a finite ring map. Let $\mathfrak m_1, \ldots , \mathfrak m_ n$ be the maximal ideals of $S$. Let $N$ be a finite $S$-module. Then

$\min \nolimits _{i = 1, \ldots , n} \text{depth}(N_{\mathfrak m_ i}) = \text{depth}_\mathfrak m(N)$

Proof. By Lemmas 10.36.20, 10.36.22, and Lemma 10.36.21 the maximal ideals of $S$ are exactly the primes of $S$ lying over $\mathfrak m$ and there are finitely many of them. Hence the statement of the lemma makes sense. We will prove the lemma by induction on $k = \min \nolimits _{i = 1, \ldots , n} \text{depth}(N_{\mathfrak m_ i})$. If $k = 0$, then $\text{depth}(N_{\mathfrak m_ i}) = 0$ for some $i$. By Lemma 10.72.5 this means $\mathfrak m_ i S_{\mathfrak m_ i}$ is an associated prime of $N_{\mathfrak m_ i}$ and hence $\mathfrak m_ i$ is an associated prime of $N$ (Lemma 10.63.16). By Lemma 10.63.13 we see that $\mathfrak m$ is an associated prime of $N$ as an $R$-module. Whence $\text{depth}_\mathfrak m(N) = 0$. This proves the base case. If $k > 0$, then we see that $\mathfrak m_ i \not\in \text{Ass}_ S(N)$. Hence $\mathfrak m \not\in \text{Ass}_ R(N)$, again by Lemma 10.63.13. Thus we can find $f \in \mathfrak m$ which is not a zerodivisor on $N$, see Lemma 10.63.18. By Lemma 10.72.7 all the depths drop exactly by $1$ when passing from $N$ to $N/fN$ and the induction hypothesis does the rest. $\square$

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