## 46.24 Gorenstein schemes

This section is the continuation of Dualizing Complexes, Section 45.21.

Definition 46.24.1. Let $X$ be a scheme. We say $X$ is Gorenstein if $X$ is locally Noetherian and $\mathcal{O}_{X, x}$ is Gorenstein for all $x \in X$.

This definition makes sense because a Noetherian ring is said to be Gorenstein if and only if all of its local rings are Gorenstein, see Dualizing Complexes, Definition 45.21.1.

Proof. Looking affine locally this follows from the corresponding result in algebra, namely Dualizing Complexes, Lemma 45.21.2. $\square$

Proof. Looking affine locally this follows from the corresponding result in algebra, namely Dualizing Complexes, Lemma 45.21.3. $\square$

Lemma 46.24.4. Let $X$ be a locally Noetherian scheme.

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

1. $X$ is Gorenstein $\Leftrightarrow$ $\omega _ X^\bullet$ is an invertible object of $D(\mathcal{O}_ X)$,

2. $\mathcal{O}_{X, x}$ is Gorenstein $\Leftrightarrow$ $\omega _{X, x}^\bullet$ is an invertible object of $D(\mathcal{O}_{X, x})$,

3. $U = \{ x \in X \mid \mathcal{O}_{X, x}\text{ is Gorenstein}\}$ is an open Gorenstein subscheme.

2. If $X$ is Gorenstein, then $X$ has a dualizing complex if and only if $\mathcal{O}_ X$ is a dualizing complex.

Proof. Looking affine locally this follows from the corresponding result in algebra, namely Dualizing Complexes, Lemma 45.21.4. $\square$

Lemma 46.24.5. If $f : Y \to X$ is a local complete intersection morphism with $X$ a Gorenstein scheme, then $Y$ is Gorenstein.

Proof. By More on Morphisms, Lemma 36.54.5 it suffices to prove the corresponding statement about ring maps. This is Dualizing Complexes, Lemma 45.21.7. $\square$

Lemma 46.24.6. The property $\mathcal{P}(S) =$“$S$ is Gorenstein” is local in the syntomic topology.

Proof. Let $\{ S_ i \to S\}$ be a syntomic covering. The scheme $S$ is locally Noetherian if and only if each $S_ i$ is Noetherian, see Descent, Lemma 34.13.1. Thus we may now assume $S$ and $S_ i$ are locally Noetherian. If $S$ is Gorenstein, then each $S_ i$ is Gorenstein by Lemma 46.24.5. Conversely, if each $S_ i$ is Gorenstein, then for each point $s \in S$ we can pick $i$ and $t \in S_ i$ mapping to $s$. Then $\mathcal{O}_{S, s} \to \mathcal{O}_{S_ i, t}$ is a flat local ring homomorphism with $\mathcal{O}_{S_ i, t}$ Gorenstein. Hence $\mathcal{O}_{S, s}$ is Gorenstein by Dualizing Complexes, Lemma 45.21.8. $\square$

## Comments (4)

Comment #1314 by Liran Shaul on

Re "Nonetheless many rings in algebraic geometry have dualizing complexes simply because they are quotients of Gorenstein rings, which are defined as follows.".

It seems natural to mention here that this condition is in fact both necessary and sufficient. That is: a noetherian ring has dualizing complexes if and only if it is a quotient of a finite dimensional Gorenstein ring. This was proved in Corollary 1.4 of

Kawasaki, T. (2002). On arithmetic Macaulayfication of Noetherian rings. Transactions of the American Mathematical Society, 354(1), 123-149

Comment #1316 by on

OK, thanks. I've added this [here]{https://github.com/stacks/stacks-project/commit/38254971765fad7f779e6f0453aa986274828f5b}. Of course, it would be even better if somebody had a, already written (?), write up of the history of dualizing complexes and their existence, which we could then add as a section to this chapter.

The initial goal for this chapter was to write just enough so it can be used for: (a) the relative dualizing sheaf of a family of curves used in dealing with moduli of curves, (b) the proof of resolution of surfaces, and (c) the finiteness theorem in local cohomology.

Comment #2256 by David Hansen on

First sentence: "we seen" --> "we've seen".

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