Lemma 47.21.8. Let $A \to B$ be a flat local homomorphism of Noetherian local rings. The following are equivalent

1. $B$ is Gorenstein, and

2. $A$ and $B/\mathfrak m_ A B$ are Gorenstein.

Proof. Below we will use without further mention that a local Gorenstein ring has finite injective dimension as well as Lemma 47.21.5. By More on Algebra, Lemma 15.65.4 we have

$\mathop{\mathrm{Ext}}\nolimits ^ i_ A(\kappa _ A, A) \otimes _ A B = \mathop{\mathrm{Ext}}\nolimits ^ i_ B(B/\mathfrak m_ A B, B)$

for all $i$.

Assume (2). Using that $R\mathop{\mathrm{Hom}}\nolimits (B/\mathfrak m_ A B, -) : D(B) \to D(B/\mathfrak m_ A B)$ is a right adjoint to restriction (Lemma 47.13.1) we obtain

$R\mathop{\mathrm{Hom}}\nolimits _ B(\kappa _ B, B) = R\mathop{\mathrm{Hom}}\nolimits _{B/\mathfrak m_ A B}(\kappa _ B, R\mathop{\mathrm{Hom}}\nolimits (B/\mathfrak m_ A B, B))$

The cohomology modules of $R\mathop{\mathrm{Hom}}\nolimits (B/\mathfrak m_ A B, B)$ are the modules $\mathop{\mathrm{Ext}}\nolimits ^ i_ B(B/\mathfrak m_ A B, B) = \mathop{\mathrm{Ext}}\nolimits ^ i_ A(\kappa _ A, A) \otimes _ A B$. Since $A$ is Gorenstein, we conclude only a finite number of these are nonzero and each is isomorphic to a direct sum of copies of $B/\mathfrak m_ A B$. Hence since $B/\mathfrak m_ A B$ is Gorenstein we conclude that $R\mathop{\mathrm{Hom}}\nolimits _ B(B/\mathfrak m_ B, B)$ has only a finite number of nonzero cohomology modules. Hence $B$ is Gorenstein.

Assume (1). Since $B$ has finite injective dimension, $\mathop{\mathrm{Ext}}\nolimits ^ i_ B(B/\mathfrak m_ A B, B)$ is $0$ for $i \gg 0$. Since $A \to B$ is faithfully flat we conclude that $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(\kappa _ A, A)$ is $0$ for $i \gg 0$. We conclude that $A$ is Gorenstein. This implies that $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(\kappa _ A, A)$ is nonzero for exactly one $i$, namely for $i = \dim (A)$, and $\mathop{\mathrm{Ext}}\nolimits ^{\dim (A)}_ A(\kappa _ A, A) \cong \kappa _ A$ (see Lemmas 47.16.1, 47.20.2, and 47.21.2). Thus we see that $\mathop{\mathrm{Ext}}\nolimits ^ i_ B(B/\mathfrak m_ A B, B)$ is zero except for one $i$, namely $i = \dim (A)$ and $\mathop{\mathrm{Ext}}\nolimits ^{\dim (A)}_ B(B/\mathfrak m_ A B, B) \cong B/\mathfrak m_ A B$. Thus $B/\mathfrak m_ A B$ is Gorenstein by Lemma 47.16.1. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).