Proposition 2

Let $A$ be a ring, $J$ an ideal of $A$, $M$ an $A$-module and $N$ an $A$-module annihilated by a power of $J$. One has $\operatorname{Ext}_A^i(N, M) = 0$ for every integer $i < \operatorname{prof}_A(J; M)$.

Suppose first that $JN = 0$ and reason by induction on the integer $i < \operatorname{prof}_A(J; M)$. The assertion is obvious for $i < 0$. Consider $N$ as an $(A/J)$-module and choose an exact sequence of $(A/J)$-modules

One deduces from this an exact sequence of extension modules

The $A$-module $\operatorname{Ext}_A^{i-1}(K, M)$ is zero by the induction hypothesis, and the $A$-module $\operatorname{Ext}_A^i((A/J)^{(I)}, M)$ is isomorphic to $\operatorname{Ext}_A^i(A/J, M)^I$ (A, X, p. 89, prop. 7), which is zero by definition of depth. Hence one has $\operatorname{Ext}_A^i(N, M) = 0$.

Pass to the general case, and reason by induction on the smallest integer $m > 0$ such that $J^m N = 0$. We have just treated the case $m = 1$. Suppose $m > 1$ and let $i < \operatorname{prof}_A(J; M)$ be an integer. Consider the exact sequence

deduced from the exact sequence $0 \to JN \to N \to N/JN \to 0$. The two extreme modules are zero by the induction hypothesis, since $N/JN$ and $JN$ are annihilated by $J^{m-1}$. Therefore one has $\operatorname{Ext}_A^i(N, M) = 0$, which was to be proved.