## 35.9 Parasitic modules

Parasitic modules are those which are zero when restricted to schemes flat over the base scheme. Here is the formal definition.

Definition 35.9.1. Let $S$ be a scheme. Let $\tau \in \{ Zar, {\acute{e}tale}, smooth, syntomic, fppf\} $. Let $\mathcal{F}$ be a presheaf of $\mathcal{O}$-modules on $(\mathit{Sch}/S)_\tau $.

$\mathcal{F}$ is called *parasitic*^{1} if for every flat morphism $U \to S$ we have $\mathcal{F}(U) = 0$.

$\mathcal{F}$ is called *parasitic for the $\tau $-topology* if for every $\tau $-covering $\{ U_ i \to S\} _{i \in I}$ we have $\mathcal{F}(U_ i) = 0$ for all $i$.

If $\tau = fppf$ this means that $\mathcal{F}|_{U_{Zar}} = 0$ whenever $U \to S$ is flat and locally of finite presentation; similar for the other cases.

Lemma 35.9.2. Let $S$ be a scheme. Let $\tau \in \{ Zar, {\acute{e}tale}, smooth, syntomic, fppf\} $. Let $\mathcal{G}$ be a presheaf of $\mathcal{O}$-modules on $(\mathit{Sch}/S)_\tau $.

If $\mathcal{G}$ is parasitic for the $\tau $-topology, then $H^ p_\tau (U, \mathcal{G}) = 0$ for every $U$ open in $S$, resp. étale over $S$, resp. smooth over $S$, resp. syntomic over $S$, resp. flat and locally of finite presentation over $S$.

If $\mathcal{G}$ is parasitic then $H^ p_\tau (U, \mathcal{G}) = 0$ for every $U$ flat over $S$.

**Proof.**
Proof in case $\tau = fppf$; the other cases are proved in the exact same way. The assumption means that $\mathcal{G}(U) = 0$ for any $U \to S$ flat and locally of finite presentation. Apply Cohomology on Sites, Lemma 21.10.9 to the subset $\mathcal{B} \subset \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ consisting of $U \to S$ flat and locally of finite presentation and the collection $\text{Cov}$ of all fppf coverings of elements of $\mathcal{B}$.
$\square$

Lemma 35.9.3. Let $f : T \to S$ be a morphism of schemes. For any parasitic $\mathcal{O}$-module on $(\mathit{Sch}/T)_\tau $ the pushforward $f_*\mathcal{F}$ and the higher direct images $R^ if_*\mathcal{F}$ are parasitic $\mathcal{O}$-modules on $(\mathit{Sch}/S)_\tau $.

**Proof.**
Recall that $R^ if_*\mathcal{F}$ is the sheaf associated to the presheaf

\[ U \mapsto H^ i((\mathit{Sch}/U \times _ S T)_\tau , \mathcal{F}) \]

see Cohomology on Sites, Lemma 21.7.4. If $U \to S$ is flat, then $U \times _ S T \to T$ is flat as a base change. Hence the displayed group is zero by Lemma 35.9.2. If $\{ U_ i \to U\} $ is a $\tau $-covering then $U_ i \times _ S T \to T$ is also flat. Hence it is clear that the sheafification of the displayed presheaf is zero on schemes $U$ flat over $S$.
$\square$

Lemma 35.9.4. Let $S$ be a scheme. Let $\tau \in \{ Zar, {\acute{e}tale}\} $. Let $\mathcal{G}$ be a sheaf of $\mathcal{O}$-modules on $(\mathit{Sch}/S)_{fppf}$ such that

$\mathcal{G}|_{S_\tau }$ is quasi-coherent, and

for every flat, locally finitely presented morphism $g : U \to S$ the canonical map $g_{\tau , small}^*(\mathcal{G}|_{S_\tau }) \to \mathcal{G}|_{U_\tau }$ is an isomorphism.

Then $H^ p(U, \mathcal{G}) = H^ p(U, \mathcal{G}|_{U_\tau })$ for every $U$ flat and locally of finite presentation over $S$.

**Proof.**
Let $\mathcal{F}$ be the pullback of $\mathcal{G}|_{S_\tau }$ to the big fppf site $(\mathit{Sch}/S)_{fppf}$. Note that $\mathcal{F}$ is quasi-coherent. There is a canonical comparison map $\varphi : \mathcal{F} \to \mathcal{G}$ which by assumptions (1) and (2) induces an isomorphism $\mathcal{F}|_{U_\tau } \to \mathcal{G}|_{U_\tau }$ for all $g : U \to S$ flat and locally of finite presentation. Hence in the short exact sequences

\[ 0 \to \mathop{\mathrm{Ker}}(\varphi ) \to \mathcal{F} \to \mathop{\mathrm{Im}}(\varphi ) \to 0 \]

and

\[ 0 \to \mathop{\mathrm{Im}}(\varphi ) \to \mathcal{G} \to \mathop{\mathrm{Coker}}(\varphi ) \to 0 \]

the sheaves $\mathop{\mathrm{Ker}}(\varphi )$ and $\mathop{\mathrm{Coker}}(\varphi )$ are parasitic for the fppf topology. By Lemma 35.9.2 we conclude that $H^ p(U, \mathcal{F}) \to H^ p(U, \mathcal{G})$ is an isomorphism for $g : U \to S$ flat and locally of finite presentation. Since the result holds for $\mathcal{F}$ by Proposition 35.8.10 we win.
$\square$

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