Proposition 9

Let $A$ be a noetherian local ring and $M$ a nonzero finitely generated $A$-module of finite injective dimension. Then $\operatorname{di}_A(M) = \operatorname{prof}(A)$.

Put $r = \operatorname{di}_A(M)$. We have $\operatorname{Ext}^i_A(\kappa_A, M) = 0$ for $i > r$, hence $\operatorname{Ext}^r_A(\kappa_A, M) \neq 0$ by Proposition 6 of No. 3, (iv) $\Rightarrow$ (i). Let $s$ be the depth of $A$ and let $(x_1, \ldots, x_s)$ be an $A$-regular sequence of elements of $\mathfrak{m}_A$ (§ 1, No. 4, Theorem 2); put $N = A/(x_1A + \ldots + x_sA)$. By the example of No. 1, we have $\operatorname{dp}_A(N) = s$ and $\operatorname{Ext}^s_A(N, M) \neq 0$, hence $s \leq \operatorname{di}_A(M) = r$. But $N$ is of depth 0 (§ 1, No. 4, Proposition 7), hence there exists an exact sequence of $A$-modules

From this one deduces an exact sequence of extension modules

since $\operatorname{Ext}^{r+1}_A(N', M) = 0$ and $\operatorname{Ext}^r_A(\kappa_A, M) \neq 0$, one obtains $\operatorname{Ext}^r_A(N, M) \neq 0$, whence $r \leq \operatorname{dp}_A(N) = s$. In the end, we have $r = s$, which completes the proof.