Proposition 3

*Let A be a local noetherian ring, $\Omega$ a dualizing A-module and M a finitely generated A-module.

a) One has $\operatorname{Ext}_A^i(M, \Omega) = 0$ for $i < \dim(A) - \dim_A(M)$.

b) Put $c = \dim(A) - \dim_A(M)$. If M is nonzero, the A-module $\operatorname{Ext}_A^c(M, \Omega)$ is nonzero.

c) One has $\operatorname{Ext}_A^i(M, \Omega) = 0$ for $i > \dim(A) - \operatorname{prof}_A(M)$.

Assume M nonzero and denote by F its support. By proposition 9 of § 1, No. 5, the conjunction of assertions a) and b) is equivalent to $\operatorname{prof}_F(\Omega) = c$. But since $\Omega$ is Cohen-Macaulay and its support is equal to Spec(A) (proposition 1 and corollary 2 of proposition 2), one has

(§ 2, No. 1, corollary of proposition 1 and No. 2, corollary of proposition 2).

Let us prove c) by induction on the depth of M. If $\operatorname{prof}_A(M) = 0$, one indeed has $\operatorname{Ext}_A^i(M, \Omega) = 0$ for $i > \dim(A)$, since $\operatorname{di}_A(\Omega) = \dim(A)$ (proposition 1). Assume $\operatorname{prof}_A(M) > 0$; then there exists an element $x$ of $\mathfrak{m}_A$ such that the homothety of ratio $x$ is injective in M. One has $\operatorname{prof}_A(M/xM) = \operatorname{prof}_A(M) - 1$ (§ 1, No. 4, proposition 7).

Consider the exact sequence of extension modules

associated with the exact sequence

For $i > \dim(A) - \operatorname{prof}_A(M)$, the $A$-module $\operatorname{Ext}_A^{i+1}(M/xM, \Omega)$ is zero by the induction hypothesis; therefore the homothety with ratio $x$ is surjective in $\operatorname{Ext}_A^i(M, \Omega)$, which implies that this $A$-module is zero (Nakayama's lemma). This proves c).