Lemma 76.27.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume the fibres of $f$ are locally Noetherian. The following are equivalent
$f$ is Gorenstein,
$f$ is flat and for some surjective étale morphism $V \to Y$ where $V$ is a scheme, the fibres of $X_ V \to V$ are Gorenstein algebraic spaces, and
$f$ is flat and for any étale morphism $V \to Y$ where $V$ is a scheme, the fibres of $X_ V \to V$ are Gorenstein algebraic spaces.
Given $x \in |X|$ with image $y \in |Y|$ the following are equivalent
$f$ is Gorenstein at $x$, and
$\mathcal{O}_{Y, \overline{y}} \to \mathcal{O}_{X, \overline{x}}$ is flat and $\mathcal{O}_{X, \overline{x}}/ \mathfrak m_{\overline{y}}\mathcal{O}_{X, \overline{x}}$ is Gorenstein.
Proof.
Given an étale morphism $V \to Y$ where $V$ is a scheme choose a scheme $U$ and a surjective étale morphism $U \to X \times _ Y V$. Consider the commutative diagram
\[ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } \]
Let $u \in U$ with images $x \in |X|$, $y \in |Y|$, and $v \in V$. Then $f$ is Gorenstein at $x$ if and only if $U \to V$ is Gorenstein at $u$ (by definition). Moreover the morphism $U_ v \to X_ v = (X_ V)_ v$ is surjective étale. Hence the scheme $U_ v$ is Gorenstein if and only if the algebraic space $X_ v$ is Gorenstein. Thus the equivalence of (1), (2), and (3) follows from the corresponding equivalence for morphisms of schemes, see Duality for Schemes, Lemma 48.24.4 by a formal argument.
Proof of equivalence of (a) and (b). The corresponding equivalence for flatness is Morphisms of Spaces, Lemma 67.30.8. Thus we may assume $f$ is flat at $x$ when proving the equivalence. Consider a diagram and $x, y, u, v$ as above. Then $\mathcal{O}_{Y, \overline{y}} \to \mathcal{O}_{X, \overline{x}}$ is equal to the map $\mathcal{O}_{V, v}^{sh} \to \mathcal{O}_{U, u}^{sh}$ on strict henselizations of local rings, see Properties of Spaces, Lemma 66.22.1. Thus we have
\[ \mathcal{O}_{X, \overline{x}}/ \mathfrak m_{\overline{y}}\mathcal{O}_{X, \overline{x}} = (\mathcal{O}_{U, u}/\mathfrak m_ v \mathcal{O}_{U, u})^{sh} \]
by Algebra, Lemma 10.156.4. Thus we have to show that the Noetherian local ring $\mathcal{O}_{U, u}/\mathfrak m_ v \mathcal{O}_{U, u}$ is Gorenstein if and only if its strict henselization is. This follows immediately from Dualizing Complexes, Lemma 47.22.3 and the definition of a Gorenstein local ring as a Noetherian local ring which is a dualizing complex over itself.
$\square$
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