Lemma 38.18.2. Let $f : X \to S$ be a morphism of schemes of finite type. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module of finite type. Let $s \in S$. Let $(S', s') \to (S, s)$ be an elementary étale neighbourhood and let
\[ \xymatrix{ X \ar[d] & X' \ar[l]^ g \ar[d] \\ S & S' \ar[l] } \]
be a commutative diagram of morphisms of schemes. Assume
$\mathcal{F}$ is flat over $S$ at all points of $X_ s$,
$X' \to S'$ is of finite type,
$g^*\mathcal{F}$ is pure along $X'_{s'}$,
$g : X' \to X$ is étale, and
$g(X')$ contains $\text{Ass}_{X_ s}(\mathcal{F}_ s)$.
In this situation $\mathcal{F}$ is pure along $X_ s$ if and only if the image of $X' \to X \times _ S S'$ contains the points of $\text{Ass}_{X \times _ S S'/S'}(\mathcal{F} \times _ S S')$ lying over points in $S'$ which specialize to $s'$.
Proof.
Since the morphism $S' \to S$ is étale, we see that if $\mathcal{F}$ is pure along $X_ s$, then $\mathcal{F} \times _ S S'$ is pure along $X_ s$, see Lemma 38.16.4. Since purity satisfies flat descent, see Lemma 38.16.6, we see that if $\mathcal{F} \times _ S S'$ is pure along $X_{s'}$, then $\mathcal{F}$ is pure along $X_ s$. Hence we may replace $S$ by $S'$ and assume that $S = S'$ so that $g : X' \to X$ is an étale morphism between schemes of finite type over $S$. Moreover, we may replace $S$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ and assume that $S$ is local.
First, assume that $\mathcal{F}$ is pure along $X_ s$. In this case every point of $\text{Ass}_{X/S}(\mathcal{F})$ specializes to a point of $X_ s$ by purity. Hence by Lemma 38.18.1 we see that every point of $\text{Ass}_{X/S}(\mathcal{F})$ specializes to a point of $\text{Ass}_{X_ s}(\mathcal{F}_ s)$. Thus every point of $\text{Ass}_{X/S}(\mathcal{F})$ is in the image of $g$ (as the image is open and contains $\text{Ass}_{X_ s}(\mathcal{F}_ s)$).
Conversely, assume that $g(X')$ contains $\text{Ass}_{X/S}(\mathcal{F})$. Let $S^ h = \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}^ h)$ be the henselization of $S$ at $s$. Denote $g^ h : (X')^ h \to X^ h$ the base change of $g$ by $S^ h \to S$, and denote $\mathcal{F}^ h$ the pullback of $\mathcal{F}$ to $X^ h$. By Divisors, Lemma 31.7.3 and Remark 31.7.4 the relative assassin $\text{Ass}_{X^ h/S^ h}(\mathcal{F}^ h)$ is the inverse image of $\text{Ass}_{X/S}(\mathcal{F})$ via the projection $X^ h \to X$. As we have assumed that $g(X')$ contains $\text{Ass}_{X/S}(\mathcal{F})$ we conclude that the base change $g^ h((X')^ h) = g(X') \times _ S S^ h$ contains $\text{Ass}_{X^ h/S^ h}(\mathcal{F}^ h)$. In this way we reduce to the case where $S$ is the spectrum of a henselian local ring. Let $x \in \text{Ass}_{X/S}(\mathcal{F})$. To finish the proof of the lemma we have to show that $x$ specializes to a point of $X_ s$, see criterion (4) for purity in discussion following Definition 38.16.1. By assumption there exists a $x' \in X'$ such that $g(x') = x$. As $g : X' \to X$ is étale, we see that $x' \in \text{Ass}_{X'/S}(g^*\mathcal{F})$, see Lemma 38.2.8 (applied to the morphism of fibres $X'_ w \to X_ w$ where $w \in S$ is the image of $x'$). Since $g^*\mathcal{F}$ is pure along $X'_ s$ we see that $x' \leadsto y$ for some $y \in X'_ s$. Hence $x = g(x') \leadsto g(y)$ and $g(y) \in X_ s$ as desired.
$\square$
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