The Stacks project

Proof. Follows from Lemma 47.20.2. $\square$

Proof. Let $A$ be a regular ring of finite dimension $d$. Then $A$ has finite global dimension $d$, see Algebra, Lemma 10.110.8. Hence $\mathop{\mathrm{Ext}}\nolimits ^{d + 1}_ A(M, A) = 0$ for all $A$-modules $M$, see Algebra, Lemma 10.109.8. Thus $A$ has finite injective dimension as an $A$-module by More on Algebra, Lemma 15.69.2. It follows that $A[0]$ is a dualizing complex, hence $A$ is Gorenstein by the remark following the definition. $\square$

Proof. For invertible objects of $D(A)$, see More on Algebra, Lemma 15.126.4 and the discussion in Section 47.15.

By Lemma 47.15.6 for every $\mathfrak p$ the complex $(\omega _ A^\bullet )_\mathfrak p$ is a dualizing complex over $A_\mathfrak p$. By definition and uniqueness of dualizing complexes (Lemma 47.15.5) we see that (1)(b) holds.

To see (1)(c) assume that $A_\mathfrak p$ is Gorenstein. Let $n_ x$ be the unique integer such that $H^{n_{x}}((\omega _ A^\bullet )_\mathfrak p)$ is nonzero and isomorphic to $A_\mathfrak p$. Since $\omega _ A^\bullet $ is in $D^ b_{\textit{Coh}}(A)$ there are finitely many nonzero finite $A$-modules $H^ i(\omega _ A^\bullet )$. Thus there exists some $f \in A$, $f \not\in \mathfrak p$ such that only $H^{n_ x}((\omega _ A^\bullet )_ f)$ is nonzero and generated by $1$ element over $A_ f$. Since dualizing complexes are faithful (by definition) we conclude that $A_ f \cong H^{n_ x}((\omega _ A^\bullet )_ f)$. In this way we see that $A_\mathfrak q$ is Gorenstein for every $\mathfrak q \in D(f)$. This proves that the set in (1)(c) is open.

Proof of (1)(a). The implication $\Leftarrow $ follows from (1)(b). The implication $\Rightarrow $ follows from the discussion in the previous paragraph, where we showed that if $A_\mathfrak p$ is Gorenstein, then for some $f \in A$, $f \not\in \mathfrak p$ the complex $(\omega _ A^\bullet )_ f$ has only one nonzero cohomology module which is invertible.

If $A[0]$ is a dualizing complex then $A$ is Gorenstein by part (1). Conversely, we see that part (1) shows that $\omega _ A^\bullet $ is locally isomorphic to a shift of $A$. Since being a dualizing complex is local (Lemma 47.15.7) the result is clear. $\square$

Proof. Observe that $A[0]$ is a dualizing complex for $A$ if and only if $A$ has finite injective dimension as an $A$-module (follows immediately from Definition 47.15.1). Thus the lemma follows from More on Algebra, Lemma 15.69.7. $\square$

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

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). $\square$

Proof. By More on Algebra, Definition 15.33.2 we can write $B = A[x_1, \ldots , x_ n]/I$ where $I$ is a Koszul-regular ideal. Observe that a polynomial ring over a Gorenstein ring $A$ is Gorenstein: reduce to $A$ local and then use Lemmas 47.15.10 and 47.21.4. A Koszul-regular ideal is by definition locally generated by a Koszul-regular sequence, see More on Algebra, Section 15.32. Looking at local rings of $A[x_1, \ldots , x_ n]$ we see it suffices to show: if $R$ is a Noetherian local Gorenstein ring and $f_1, \ldots , f_ c \in \mathfrak m_ R$ is a Koszul regular sequence, then $R/(f_1, \ldots , f_ c)$ is Gorenstein. This follows from Lemma 47.21.6 and the fact that a Koszul regular sequence in $R$ is just a regular sequence (More on Algebra, Lemma 15.30.7). $\square$

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

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

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. $\square$

Proof. Since $A$ is Gorenstein $\omega _ A^\bullet = A[d]$ is a normalized dualizing complex for $A$. Also $E$ is the only nonzero cohomology module of $R\Gamma _\mathfrak m(\omega _ A^\bullet )$ sitting in degree $0$, see Lemma 47.18.1. By Lemma 47.9.5 we have

and the lemma follows. $\square$