The Stacks project

Theorem 76.4.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite type. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module of finite type. Let $x \in |X|$ with image $y \in |Y|$. Set $F = f^{-1}(\{ y\} ) \subset |X|$. Consider the conditions

  1. $\mathcal{F}$ is flat at $x$ over $Y$, and

  2. for every $x' \in F \cap \text{Ass}_{X/Y}(\mathcal{F})$ which specializes to $x$ we have that $\mathcal{F}$ is flat at $x'$ over $Y$.

Then we always have (2) $\Rightarrow $ (1). If $X$ and $Y$ are decent, then (1) $\Rightarrow $ (2).

Proof. Assume (2). Choose a scheme $V$ and a surjective ├ętale morphism $V \to Y$. Choose a scheme $U$ and a surjective ├ętale morphism $U \to V \times _ Y X$. Choose a point $u \in U$ mapping to $x$. Let $v \in V$ be the image of $u$. We will deduce the result from the corresponding result for $\mathcal{F}|_ U = (U \to X)^*\mathcal{F}$ and the point $u$. $U_ v$. This works because $\text{Ass}_{U/V}(\mathcal{F}|_ U) \cap |U_ v|$ is equal to $\text{Ass}_{U_ v}(\mathcal{F}|_{U_ v})$ and equal to the inverse image of $F \cap \text{Ass}_{X/Y}(\mathcal{F})$. Since the map $|U_ v| \to F$ is continuous we see that specializations in $|U_ v|$ map to specializations in $F$, hence condition (2) is inherited by $U \to V$, $\mathcal{F}|_ U$, and the point $u$. Thus More on Flatness, Theorem 38.26.1 applies and we conclude that (1) holds.

If $Y$ is decent, then we can represent $y$ by a quasi-compact monomorphism $\mathop{\mathrm{Spec}}(k) \to Y$ (by definition of decent spaces, see Decent Spaces, Definition 67.6.1). Then $F = |X_ k|$, see Decent Spaces, Lemma 67.18.6. If in addition $X$ is decent (or more generally if $f$ is decent, see Decent Spaces, Definition 67.17.1 and Decent Spaces, Lemma 67.17.3), then $X_ y$ is a decent space too. Furthermore, specializations in $F$ can be lifted to specializations in $U_ v \to X_ y$, see Decent Spaces, Lemma 67.12.2. Having said this it is clear that the reverse implication holds, because it holds in the case of schemes. $\square$


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