Lemma 76.6.3. Let $Y$ be an algebraic space over a scheme $S$. Let $g : X' \to X$ be an étale morphism of algebraic spaces over $Y$. Assume the structure morphisms $X' \to Y$ and $X \to Y$ are decent and of finite type. Let $\mathcal{F}$ be a finite type, quasi-coherent $\mathcal{O}_ X$-module. Let $y \in |Y|$. Set $F = f^{-1}(\{ y\} ) \subset |X|$.

If $\text{Ass}_{X/Y}(\mathcal{F}) \subset g(|X'|)$ and $g^*\mathcal{F}$ is (universally) pure above $y$, then $\mathcal{F}$ is (universally) pure above $y$.

If $\mathcal{F}$ is pure above $y$, $g(|X'|)$ contains $F$, and $Y$ is affine local with closed point $y$, then $\text{Ass}_{X/Y}(\mathcal{F}) \subset g(|X'|)$.

If $\mathcal{F}$ is pure above $y$, $\mathcal{F}$ is flat at all points of $F$, $g(|X'|)$ contains $\text{Ass}_{X/Y}(\mathcal{F}) \cap F$, and $Y$ is affine local with closed point $y$, then $\text{Ass}_{X/Y}(\mathcal{F}) \subset g(|X'|)$.

Add more here.

**Proof.**
The assumptions on $X \to Y$ and $X' \to Y$ guarantee that we may apply the material in Sections 76.2 and 76.3 to these morphisms and the sheaves $\mathcal{F}$ and $g^*\mathcal{F}$. Since $g$ is étale we see that $\text{Ass}_{X'/Y}(g^*\mathcal{F})$ is the inverse image of $\text{Ass}_{X/Y}(\mathcal{F})$ and the same remains true after base change.

Proof of (1). Assume $\text{Ass}_{X/Y}(\mathcal{F}) \subset g(|X'|)$. Suppose that $(T \to Y, t' \leadsto t, \xi )$ is an impurity of $\mathcal{F}$ above $y$. Since $\text{Ass}_{X_ T/T}(\mathcal{F}_ T) \subset g_ T(|X'_ T|)$ by Lemma 76.6.2 we can choose a point $\xi ' \in |X'_ T|$ mapping to $\xi $. By the above we see that $(T \to Y, t' \leadsto t, \xi ')$ is an impurity of $g^*\mathcal{F}$ above $y'$. This implies (1) is true.

Proof of (2). This follows from the fact that $g(|X'|)$ is open in $|X|$ and the fact that by purity every point of $\text{Ass}_{X/Y}(\mathcal{F})$ specializes to a point of $F$.

Proof of (3). This follows from the fact that $g(|X'|)$ is open in $|X|$ and the fact that by purity combined with Lemma 76.6.1 every point of $\text{Ass}_{X/Y}(\mathcal{F})$ specializes to a point of $\text{Ass}_{X/Y}(\mathcal{F}) \cap F$.
$\square$

## Comments (0)