The Stacks project

Lemma 102.9.2. Let $\mathcal{X}$ be an algebraic stack. Let $\mathcal{F}$ be a presheaf of $\mathcal{O}_\mathcal {X}$-modules.

  1. If $\mathcal{F}$ is parasitic and $g : \mathcal{Y} \to \mathcal{X}$ is a flat morphism of algebraic stacks, then $g^*\mathcal{F}$ is parasitic.

  2. For $\tau \in \{ Zariski, {\acute{e}tale}, smooth, syntomic, fppf\} $ we have

    1. the $\tau $ sheafification of a parasitic presheaf of modules is parasitic, and

    2. the full subcategory of $\textit{Mod}(\mathcal{X}_\tau , \mathcal{O}_\mathcal {X})$ consisting of parasitic modules is a Serre subcategory.

  3. Suppose $\mathcal{F}$ is a sheaf for the étale topology. Let $f_ i : \mathcal{X}_ i \to \mathcal{X}$ be a family of smooth morphisms of algebraic stacks such that $|\mathcal{X}| = \bigcup _ i |f_ i|(|\mathcal{X}_ i|)$. If each $f_ i^*\mathcal{F}$ is parasitic then so is $\mathcal{F}$.

  4. Suppose $\mathcal{F}$ is a sheaf for the fppf topology. Let $f_ i : \mathcal{X}_ i \to \mathcal{X}$ be a family of flat and locally finitely presented morphisms of algebraic stacks such that $|\mathcal{X}| = \bigcup _ i |f_ i|(|\mathcal{X}_ i|)$. If each $f_ i^*\mathcal{F}$ is parasitic then so is $\mathcal{F}$.

Proof. To see part (1) let $y$ be an object of $\mathcal{Y}$ which lies over a scheme $V$ such that the corresponding morphism $y : V \to \mathcal{Y}$ is flat. Then $g(y) : V \to \mathcal{Y} \to \mathcal{X}$ is flat as a composition of flat morphisms (see Morphisms of Stacks, Lemma 100.25.2) hence $\mathcal{F}(g(y))$ is zero by assumption. Since $g^*\mathcal{F} = g^{-1}\mathcal{F}(y) = \mathcal{F}(g(y))$ we conclude $g^*\mathcal{F}$ is parasitic.

To see part (2)(a) note that if $\{ x_ i \to x\} $ is a $\tau $-covering of $\mathcal{X}$, then each of the morphisms $x_ i \to x$ lies over a flat morphism of schemes. Hence if $x$ lies over a scheme $U$ such that $x : U \to \mathcal{X}$ is flat, so do all of the objects $x_ i$. Hence the presheaf $\mathcal{F}^+$ (see Sites, Section 7.10) is parasitic if the presheaf $\mathcal{F}$ is parasitic. This proves (2)(a) as the sheafification of $\mathcal{F}$ is $(\mathcal{F}^+)^+$.

Let $\mathcal{F}$ be a parasitic $\tau $-module. It is immediate from the definitions that any submodule of $\mathcal{F}$ is parasitic. On the other hand, if $\mathcal{F}' \subset \mathcal{F}$ is a submodule, then it is equally clear that the presheaf $x \mapsto \mathcal{F}(x)/\mathcal{F}'(x)$ is parasitic. Hence the quotient $\mathcal{F}/\mathcal{F}'$ is a parasitic module by (2)(a). Finally, we have to show that given a short exact sequence $0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0$ with $\mathcal{F}_1$ and $\mathcal{F}_3$ parasitic, then $\mathcal{F}_2$ is parasitic. This follows immediately on evaluating on $x$ lying over a scheme flat over $\mathcal{X}$. This proves (2)(b), see Homology, Lemma 12.10.2.

Let $f_ i : \mathcal{X}_ i \to \mathcal{X}$ be a jointly surjective family of smooth morphisms of algebraic stacks and assume each $f_ i^*\mathcal{F}$ is parasitic. Let $x$ be an object of $\mathcal{X}$ which lies over a scheme $U$ such that $x : U \to \mathcal{X}$ is flat. Consider a surjective smooth covering $W_ i \to U \times _{x, \mathcal{X}} \mathcal{X}_ i$. Denote $y_ i : W_ i \to \mathcal{X}_ i$ the projection. It follows that $\{ f_ i(y_ i) \to x\} $ is a covering for the smooth topology on $\mathcal{X}$. Since a composition of flat morphisms is flat we see that $f_ i^*\mathcal{F}(y_ i) = 0$. On the other hand, as we saw in the proof of (1), we have $f_ i^*\mathcal{F}(y_ i) = \mathcal{F}(f_ i(y_ i))$. Hence we see that for some smooth covering $\{ x_ i \to x\} _{i \in I}$ in $\mathcal{X}$ we have $\mathcal{F}(x_ i) = 0$. This implies $\mathcal{F}(x) = 0$ because the smooth topology is the same as the étale topology, see More on Morphisms, Lemma 37.37.7. Namely, $\{ x_ i \to x\} _{i \in I}$ lies over a smooth covering $\{ U_ i \to U\} _{i \in I}$ of schemes. By the lemma just referenced there exists an étale covering $\{ V_ j \to U\} _{j \in J}$ which refines $\{ U_ i \to U\} _{i \in I}$. Denote $x'_ j = x|_{V_ j}$. Then $\{ x'_ j \to x\} $ is an étale covering in $\mathcal{X}$ refining $\{ x_ i \to x\} _{i \in I}$. This means the map $\mathcal{F}(x) \to \prod _{j \in J} \mathcal{F}(x'_ j)$, which is injective as $\mathcal{F}$ is a sheaf in the étale topology, factors through $\mathcal{F}(x) \to \prod _{i \in I} \mathcal{F}(x_ i)$ which is zero. Hence $\mathcal{F}(x) = 0$ as desired.

Proof of (4): omitted. Hint: similar, but simpler, than the proof of (3). $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0774. Beware of the difference between the letter 'O' and the digit '0'.