Corollary 1

Let $\rho : A \to B$ be a homomorphism of noetherian rings making $B$ into a flat $A$-module. The following conditions are equivalent:

(i) $B$ is a Gorenstein ring; (ii) (resp. (iii)) for every prime ideal (resp. maximal ideal) $q$ of $B$, the rings $A_{\rho^{-1}(q)}$ and $\kappa(\rho^{-1}(q)) \otimes_A B$ are Gorenstein.

If moreover $B$ is a faithfully flat $A$-module, these conditions imply that $A$ is a Gorenstein ring.

Let $q$ be a prime ideal of $B$; put $p = \rho^{-1}(q)$. The ring $B_q$, isomorphic to a ring of fractions of $B_p$, is flat over $A_p$. In order that $B_q$ be a Gorenstein ring, it is necessary and sufficient that the rings $A_p$ and $\kappa(p) \otimes_{A_p} B_q$ should be so (Prop. 12).

(i) $\Rightarrow$ (ii): let $q$ be a prime ideal of $B$ and $p = \rho^{-1}(q)$. If $B$ is a Gorenstein ring, the same is true of $B_q$, hence of $A_p$ and $\kappa(p) \otimes_{A_p} B_q$ by what precedes. By the remark of § 2, No. 6, the local ring of $\kappa(p) \otimes_A B$ at any prime ideal whatever is then a Gorenstein ring, which implies that $\kappa(p) \otimes_A B$ is a Gorenstein ring, whence (ii).

(ii) $\Rightarrow$ (iii): this is clear.

(iii) $\Rightarrow$ (i): for every maximal ideal $n$ of $B$, it follows from the beginning of the proof (applied with $q = n$) that $B_n$ is a Gorenstein ring, whence (i).

If $B$ is a faithfully flat $A$-module, the mapping $^a\rho : \mathrm{Spec}(B) \longrightarrow \mathrm{Spec}(A)$ is surjective (II, § 2, No. 5, cor. 4 of prop. 11), whence the last assertion.