## 47.21 Gorenstein rings

So far, the only explicit dualizing complex we've seen is $\kappa$ on $\kappa$ for a field $\kappa$, see proof of Lemma 47.15.12. By Proposition 47.15.11 this means that any finite type algebra over a field has a dualizing complex. However, it turns out that there are Noetherian (local) rings which do not have a dualizing complex. Namely, we have seen that a ring which has a dualizing complex is universally catenary (Lemma 47.17.4) but there are examples of Noetherian local rings which are not catenary, see Examples, Section 110.19.

Nonetheless many rings in algebraic geometry have dualizing complexes simply because they are quotients of Gorenstein rings. This condition is in fact both necessary and sufficient. That is: a Noetherian ring has a dualizing complex if and only if it is a quotient of a finite dimensional Gorenstein ring. This is Sharp's conjecture ([Sharp]) which can be found as [Corollary 1.4, Kawasaki] in the literature. Returning to our current topic, here is the definition of Gorenstein rings.

Definition 47.21.1. Gorenstein rings.

1. Let $A$ be a Noetherian local ring. We say $A$ is Gorenstein if $A[0]$ is a dualizing complex for $A$.

2. Let $A$ be a Noetherian ring. We say $A$ is Gorenstein if $A_\mathfrak p$ is Gorenstein for every prime $\mathfrak p$ of $A$.

This definition makes sense, because if $A[0]$ is a dualizing complex for $A$, then $S^{-1}A[0]$ is a dualizing complex for $S^{-1}A$ by Lemma 47.15.6. We will see later that a finite dimensional Noetherian ring is Gorenstein if it has finite injective dimension as a module over itself.

Proof. Follows from Lemma 47.20.2. $\square$

An example of a Gorenstein ring is a regular ring.

Lemma 47.21.3. A regular local ring is Gorenstein. A regular ring is Gorenstein.

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. $\square$

Lemma 47.21.4. Let $A$ be a Noetherian ring.

1. If $A$ has a dualizing complex $\omega _ A^\bullet$, then

1. $A$ is Gorenstein $\Leftrightarrow$ $\omega _ A^\bullet$ is an invertible object of $D(A)$,

2. $A_\mathfrak p$ is Gorenstein $\Leftrightarrow$ $(\omega _ A^\bullet )_\mathfrak p$ is an invertible object of $D(A_\mathfrak p)$,

3. $\{ \mathfrak p \in \mathop{\mathrm{Spec}}(A) \mid A_\mathfrak p\text{ is Gorenstein}\}$ is an open subset.

2. If $A$ is Gorenstein, then $A$ has a dualizing complex if and only if $A[0]$ is a dualizing complex.

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. $\square$

Lemma 47.21.5. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring. Then $A$ is Gorenstein if and only if $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(\kappa , A)$ is zero for $i \gg 0$.

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. $\square$

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). $\square$

Lemma 47.21.7. If $A \to B$ is a local complete intersection homomorphism of rings and $A$ is a Noetherian Gorenstein ring, then $B$ is a Gorenstein ring.

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). $\square$

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. $\square$

Lemma 47.21.9. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local Gorenstein ring of dimension $d$. Let $E$ be the injective hull of $\kappa$. Then $\text{Tor}_ i^ A(E, \kappa )$ is zero for $i \not= d$ and $\text{Tor}_ d^ A(E, \kappa ) = \kappa$.

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

$E \otimes _ A^\mathbf {L} \kappa = R\Gamma _\mathfrak m(\omega _ A^\bullet ) \otimes _ A^\mathbf {L} \kappa = R\Gamma _\mathfrak m(\omega _ A^\bullet \otimes _ A^\mathbf {L} \kappa ) = R\Gamma _\mathfrak m(\kappa [d]) = \kappa [d]$

and the lemma follows. $\square$

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