The Stacks project

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 15.115.4 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. $\square$

Comments (2)

Comment #4360 by comment_bot on

In the statement, it would be good to refer to a tag where an invertible object is defined. Currently, this is done before without an explicit tag for that definition.

Comment #4502 by on

OK, the definition of invertible objects in the Stacks project was messed up a bit from the beginning. We should move the material on monoidal categories earlier and then we should, whenever we define an invertible object, explain which monoidal structure we are using. In particular, the definition given in Section 47.15 of an invertible object of isn't the right one. Going to leave it as is for now but I put a reminder on my calendar to look at it later this semester.

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