Lemma 48.25.10. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. Let $x \in X$. If $f$ is flat, then the following are equivalent

$f$ is Gorenstein at $x$,

$f^!\mathcal{O}_ X$ is isomorphic to an invertible object in a neighbourhood of $x$.

In particular, the set of points where $f$ is Gorenstein is open in $X$.

**Proof.**
Set $\omega ^\bullet = f^!\mathcal{O}_ Y$. By Lemma 48.18.4 this is a bounded complex with coherent cohomology sheaves whose derived restriction $Lh^*\omega ^\bullet $ to the fibre $X_ y$ is a dualizing complex on $X_ y$. Denote $i : x \to X_ y$ the inclusion of a point. Then the following are equivalent

$f$ is Gorenstein at $x$,

$\mathcal{O}_{X_ y, x}$ is Gorenstein,

$Lh^*\omega ^\bullet $ is invertible in a neighbourhood of $x$,

$Li^* Lh^* \omega ^\bullet $ has exactly one nonzero cohomology of dimension $1$ over $\kappa (x)$,

$L(h \circ i)^* \omega ^\bullet $ has exactly one nonzero cohomology of dimension $1$ over $\kappa (x)$,

$\omega ^\bullet $ is invertible in a neighbourhood of $x$.

The equivalence of (1) and (2) is by definition (as $f$ is flat). The equivalence of (2) and (3) follows from Lemma 48.24.4. The equivalence of (3) and (4) follows from More on Algebra, Lemma 15.76.1. The equivalence of (4) and (5) holds because $Li^* Lh^* = L(h \circ i)^*$. The equivalence of (5) and (6) holds by More on Algebra, Lemma 15.76.1. Thus the lemma is clear.
$\square$

## Comments (1)

Comment #6225 by Stefano on