Proposition 5

Let $\rho : A \to B$ be a homomorphism of noetherian rings, making $B$ into a flat $A$-module. Assume that for every maximal ideal $n$ of $B$, the ring $\kappa(\rho^{-1}(n)) \otimes_A B$ is a Gorenstein ring. Let $\Omega$ be a dualizing $A$-module ; the $B$-module $\Omega_{(B)}$ is dualizing.

Let $n$ be a maximal ideal of $B$, and $p$ its inverse image in $A$. The $A_p$-module $B_n$ is flat, the $A_p$-module $\Omega_p$ is dualizing, $\Omega_{(B)} \otimes_B B_n$ is identified with $\Omega_p \otimes_{A_p} B_n$ and $\kappa_{A_p} \otimes_{A_p} B_n$, which is identified with a ring of fractions of $\kappa(p) \otimes_A B$, is a Gorenstein ring. It is therefore enough to prove the proposition when $\rho$ is a local homomorphism of local rings, which we shall henceforth assume.

Let us first treat the case where the rings $A$ and $B$ are artinian. Put $C = B/\mathfrak{m}_A B$. Since $B$ is flat over $A$, the $B$-module $\mathrm{Hom}_B(C, \Omega_{(B)})$ is isomorphic to $\mathrm{Hom}_A(\kappa_A, \Omega) \otimes_A B$ (I, § 2, n° 10, prop. 11), hence to $\mathrm{Hom}_A(\kappa_A, \Omega) \otimes_{\kappa_A} C$. One deduces a sequence of isomorphisms

The $\kappa_A$-vector space $\mathrm{Hom}_A(\kappa_A, \Omega)$ is of dimension 1 since $\Omega$ is dualizing, and the same holds for the $\kappa_C$-vector space $\mathrm{Hom}_C(\kappa_C, C)$ since $C$ is a Gorenstein ring; consequently the $\kappa_B$-vector space $\mathrm{Hom}_B(\kappa_B, \Omega_{(B)})$ is of dimension 1.

Let $M$ be a $B$-module of finite length; let us prove by induction on $\mathrm{long}_B(M)$ that one has $\mathrm{long}_B(\mathrm{Hom}_B(M, \Omega_{(B)})) \leqslant \mathrm{long}_B(M)$. The assertion is clear if $M = 0$, and it follows from the preceding if $M = \kappa_B$. Suppose $\mathrm{long}_B(M) \geqslant 2$. There exists an exact sequence of $B$-modules

with $\mathrm{long}_B(M') < \mathrm{long}_B(M)$. One deduces from this an exact sequence and one concludes by applying the induction hypothesis to $M'$.

Let $N$ be the kernel of the canonical surjection of $\kappa_A \otimes_A B$ onto $\kappa_B$. Put $m = \mathrm{long}_B(\kappa_A \otimes_A B)$; one has $\mathrm{long}_B(N) = m - 1$. Consider the exact sequence of $B$-modules

The $B$-modules $\mathrm{Hom}_B(\kappa_A \otimes_A B, \Omega_{(B)})$ and $\mathrm{Ext}_B^1(\kappa_A \otimes_A B, \Omega_{(B)})$ are respectively isomorphic to $\mathrm{Hom}_A(\kappa_A, \Omega) \otimes_A B$ and $\mathrm{Ext}_A^1(\kappa_A, \Omega) \otimes_A B$, that is to say to $\kappa_A \otimes_A B$ and to 0. The lengths of the $B$-modules $\mathrm{Hom}_B(\kappa_B, \Omega_{(B)})$ and $\mathrm{Hom}_B(\kappa_A \otimes_A B, \Omega_{(B)})$ are 1 and $m$, and that of $\mathrm{Hom}_B(N, \Omega_{(B)})$) is $\leqslant m - 1$; one deduces from this that the $B$-module $\mathrm{Ext}_B^1(\kappa_B, \Omega_{(B)})$ is zero. By Prop. 6 of § 3, No. 3, the $B$-module $\Omega_{(B)}$ is injective; consequently it is a dualizing module (No. 1, Example 1).

Let us pass to the general case. Put $C = \kappa_A \otimes_A B$; by hypothesis this is a Gorenstein ring, hence a Macaulay ring ($§ 3, n° 7$, prop. 10). By prop. 1 of No. 1, $A$ is a Macaulay ring, and the $A$-module $\Omega$ is Macaulay. Hence $B$ is a Macaulay ring, and the $B$-module $\Omega_{(B)}$ is Macaulay ($§ 2, n° 7$, cor. 1 of prop. 9). Put $r = \dim(A)$, $s = \dim(C)$. There exists a sequence $(x_1, \ldots, x_r)$ of elements of $\mathfrak{m}_A$ regular for the $A$-modules $A$ and $\Omega$, and a sequence $(y_1, \ldots, y_s)$ of elements of $\mathfrak{m}_B$ regular for the $B$-module $C$; let $x$ denote the ideal of $A$ and $\mathfrak{g}$ the ideal of $B$ generated by them respectively. The sequence $(y_1, \ldots, y_s, \rho(x_1), \ldots, \rho(x_r))$ is regular for the $B$-modules $B$ and $\Omega_{(B)}$ ($§ 1, n° 6$, prop. 11), and the $A$-module $B/\mathfrak{g}$ is flat (loc. cit., prop. 10). Put $A' = A/x$, $B' = B/(xB + \mathfrak{g})$ and let $\rho' : A' \to B'$ denote the homomorphism deduced from $\rho$ by passing to quotients. The rings $A'$ and $B'$ are artinian, the $A'$-module $B'$ is flat, the ring $\kappa_{A'} \otimes_{A'} B'$, which is identified with $C/\mathfrak{g}$, is a Gorenstein ring ($§ 3, n° 7$, example 2) and the $A'$-module $\Omega_{(A')}$ is dualizing (No. 2, prop. 4). By the first part of the proof, the $B'$-module $\Omega_{(B')}$ is dualizing. It then follows from loc. cit. that the $B$-module $\Omega_{(B)}$ is dualizing.