Lemma 47.11.1. Let $A$ be a Noetherian ring, let $I \subset A$ be an ideal, and let $M$ be a finite $A$-module such that $IM \not= M$. Then the following integers are equal:

$\text{depth}_ I(M)$,

the smallest integer $i$ such that $\mathop{\mathrm{Ext}}\nolimits _ A^ i(A/I, M)$ is nonzero, and

the smallest integer $i$ such that $H^ i_ I(M)$ is nonzero.

Moreover, we have $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(N, M) = 0$ for $i < \text{depth}_ I(M)$ for any finite $A$-module $N$ annihilated by a power of $I$.

**Proof.**
We prove the equality of (1) and (2) by induction on $\text{depth}_ I(M)$ which is allowed by Algebra, Lemma 10.71.4.

Base case. If $\text{depth}_ I(M) = 0$, then $I$ is contained in the union of the associated primes of $M$ (Algebra, Lemma 10.62.9). By prime avoidance (Algebra, Lemma 10.14.2) we see that $I \subset \mathfrak p$ for some associated prime $\mathfrak p$. Hence $\mathop{\mathrm{Hom}}\nolimits _ A(A/I, M)$ is nonzero. Thus equality holds in this case.

Assume that $\text{depth}_ I(M) > 0$. Let $f \in I$ be a nonzerodivisor on $M$ such that $\text{depth}_ I(M/fM) = \text{depth}_ I(M) - 1$. Consider the short exact sequence

\[ 0 \to M \to M \to M/fM \to 0 \]

and the associated long exact sequence for $\mathop{\mathrm{Ext}}\nolimits ^*_ A(A/I, -)$. Note that $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(A/I, M)$ is a finite $A/I$-module (Algebra, Lemmas 10.70.9 and 10.70.8). Hence we obtain

\[ \mathop{\mathrm{Hom}}\nolimits _ A(A/I, M/fM) = \mathop{\mathrm{Ext}}\nolimits ^1_ A(A/I, M) \]

and short exact sequences

\[ 0 \to \mathop{\mathrm{Ext}}\nolimits ^ i_ A(A/I, M) \to \text{Ext}^ i_ A(A/I, M/fM) \to \mathop{\mathrm{Ext}}\nolimits ^{i + 1}_ A(A/I, M) \to 0 \]

Thus the equality of (1) and (2) by induction.

Observe that $\text{depth}_ I(M) = \text{depth}_{I^ n}(M)$ for all $n \geq 1$ for example by Algebra, Lemma 10.67.9. Hence by the equality of (1) and (2) we see that $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(A/I^ n, M) = 0$ for all $n$ and $i < \text{depth}_ I(M)$. Let $N$ be a finite $A$-module annihilated by a power of $I$. Then we can choose a short exact sequence

\[ 0 \to N' \to (A/I^ n)^{\oplus m} \to N \to 0 \]

for some $n, m \geq 0$. Then $\mathop{\mathrm{Hom}}\nolimits _ A(N, M) \subset \mathop{\mathrm{Hom}}\nolimits _ A((A/I^ n)^{\oplus m}, M)$ and $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(N, M) \subset \text{Ext}^{i - 1}_ A(N', M)$ for $i < \text{depth}_ I(M)$. Thus a simply induction argument shows that the final statement of the lemma holds.

Finally, we prove that (3) is equal to (1) and (2). We have $H^ p_ I(M) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^ p_ A(A/I^ n, M)$ by Lemma 47.8.2. Thus we see that $H^ i_ I(M) = 0$ for $i < \text{depth}_ I(M)$. For $i = \text{depth}_ I(M)$, using the vanishing of $\mathop{\mathrm{Ext}}\nolimits _ A^{i - 1}(I/I^ n, M)$ we see that the map $\mathop{\mathrm{Ext}}\nolimits _ A^ i(A/I, M) \to H_ I^ i(M)$ is injective which proves nonvanishing in the correct degree.
$\square$

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