Corollary

If $M$ is Macaulay, then $\operatorname{Ext}_A^i(M, \Omega) = 0$ for $i \neq c$; the $A$-module $\operatorname{Ext}_A^c(M, \Omega)$ is Macaulay, and its support is equal to that of $M$.

The first assertion follows from Prop. 3, a) and c). Let $p \in \operatorname{Supp}(M)$; by Prop. 1 of § 2, No. 1, applied to $M$ and to $A$, we have

since the $A_p$-module $\Omega_p$ is dualizing (Prop. 2), it follows from Prop. 3, b) that the $A_p$-module $\operatorname{Ext}_A^c(M_p, \Omega_p)$ is not zero. Hence the support of $\operatorname{Ext}_A^c(M, \Omega)$ is equal to that of $M$.

Finally, let us prove, by induction on $\dim(M)$, that the $A$-module $\operatorname{Ext}_A^c(M, \Omega)$ is Macaulay. The assertion is true when $\dim(M) = 0$, since every module of finite length is Macaulay. Suppose $\dim(M) > 0$ and choose an element $x$ of $\mathfrak{m}_A$ such that the homothety $x_M$ is injective. The $A$-module $M/xM$ is Macaulay ($§ 2$, No. 3, Prop. 4), of dimension $\dim(M) - 1$. In view of the foregoing, the exact sequence of extension modules associated with the exact sequence $0 \to M \xrightarrow{x} M \longrightarrow M/xM \to 0$ reduces to

Prop. 4 of § 2, No. 3 and the induction hypothesis then imply that $\operatorname{Ext}_A^c(M, \Omega)$ is Macaulay, whence the corollary.