Lemma 47.21.6. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring. Let $f \in \mathfrak m$ be a nonzerodivisor. Set $B = A/(f)$. Then $A$ is Gorenstein if and only if $B$ is Gorenstein.

Proof. If $A$ is Gorenstein, then $B$ is Gorenstein by Lemma 47.16.10. Conversely, suppose that $B$ is Gorenstein. Then $\mathop{\mathrm{Ext}}\nolimits ^ i_ B(\kappa , B)$ is zero for $i \gg 0$ (Lemma 47.21.5). Recall that $R\mathop{\mathrm{Hom}}\nolimits (B, -) : D(A) \to D(B)$ is a right adjoint to restriction (Lemma 47.13.1). Hence

$R\mathop{\mathrm{Hom}}\nolimits _ A(\kappa , A) = R\mathop{\mathrm{Hom}}\nolimits _ B(\kappa , R\mathop{\mathrm{Hom}}\nolimits (B, A)) = R\mathop{\mathrm{Hom}}\nolimits _ B(\kappa , B[1])$

The final equality by direct computation or by Lemma 47.13.10. Thus we see that $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(\kappa , A)$ is zero for $i \gg 0$ and $A$ is Gorenstein (Lemma 47.21.5). $\square$

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