Proposition 6

*Let $A$ be a noetherian local ring and $\Omega$ a dualizing $A$-module. Let $B$ be a finite $A$-algebra; suppose that the $A$-module $B$ is Macaulay. The $B$-module $\mathrm{Ext}_A^i(B, \Omega)$ is zero for $i \neq \dim(A) - \dim(B)$ and dualizing for $i = \dim(A) - \dim(B)$.

We have $\dim(B) = \dim_A(B) \leq \dim(A)$ (VIII, § 2, No. 3, Theorem 1 c)); let $c = \dim(A) - \dim(B)$. We have $\mathrm{Ext}_A^i(B, \Omega) = 0$ for $i \neq c$ since the $A$-module $B$ is Macaulay (No. 1, Corollary to Proposition 3). Let us prove that the $B$-module $\mathrm{Ext}_A^c(B, \Omega)$ is dualizing.

Suppose first that $\dim(B) = 0$. The spectrum $X$ of $B$ is finite and formed of maximal ideals (IV, § 2, No. 5, Proposition 9); the canonical mapping $B \to \prod_{n \in X} B_n$ is an isomorphism (loc. cit., Corollary 1). The $B$-module $\Omega' = \mathrm{Ext}_A^c(B, \Omega)$ is therefore the direct sum of the modules $\mathrm{Ext}_A^c(B_n, \Omega)$; since $\mathrm{Ext}_A^c(B_n, \Omega)$ has support in $\{n\}$, it identifies with $\Omega'_n$. We have $\dim(B_n) = 0$ for every $n$; to prove that the $B$-module $\Omega'$ is dualizing, it is therefore enough to prove that this is so for the $B_n$-module $\mathrm{Ext}_A^c(B_n, \Omega)$ for every $n \in X$, which brings us back to the case where the ring $B$ is local. In this case, by Example 6 of § 8, No. 5, the $B$-module $\mathrm{Ext}_A^c(B, \Omega)$ is isomorphic to $\mathrm{Hom}_A(B, I)$, where $I$ is a Matlis $A$-module; it is consequently a Matlis $B$-module (§ 8, No. 6, Corollary to Proposition 5), hence a dualizing $B$-module (No. 1, Example 1).

Now suppose that $\dim(B) > 0$ and reason by induction on $\dim(B)$. We have $\mathrm{prof}_A(B) = \dim_A(B) = \dim(B)$, whence $\mathrm{prof}_A(B) > 0$; on the other hand we have $\mathrm{prof}(A) = \dim(A) > 0$ (No. 1, Proposition 1), and consequently $\mathrm{prof}_A(A \oplus B) > 0$. There therefore exists an element $x$ of $\mathfrak{m}_A$ such that the homotheties $x_A$ and $x_B$ are injective.

Consider the exact sequence of extension modules associated with the exact sequence $0 \to B \xrightarrow{x_B} B \longrightarrow B/xB \to 0$ and with the $A$-module $\Omega$. The $A$-module $B/xB$ is Macaulay ($§ 2$, No. 1, Example 3), of dimension $\dim(B) - 1$ (VIII, $§ 3$, No. 2, Proposition 3) ; we therefore have $\mathrm{Ext}_A^i(B/xB, \Omega) = 0$ for $i \neq c + 1$ (No. 1, Corollary to Proposition 3). Since we have $\mathrm{Ext}_A^i(B, \Omega) = 0$ for $i \neq c$, we obtain an exact sequence of $B$-modules

By the induction hypothesis, the $B/xB$-module $\mathrm{Ext}_A^{c+1}(B/xB, \Omega)$ is dualizing. Since the $A$-algebra $B$ is finite, the image of $m_A$ in $B$ is contained in the radical of $B$ (V, $§ 2$, No. 1, Proposition 1) ; by Proposition 4 of No. 2, the $B$-module $\mathrm{Ext}_A^c(B, \Omega)$ is dualizing.