Lemma 46.6.1. Let $S$ be a scheme. Let $\mathcal{F}$ be an adequate $\mathcal{O}$-module on $(\mathit{Sch}/S)_\tau $. The following are equivalent:
$v\mathcal{F} = 0$,
$\mathcal{F}$ is parasitic,
$\mathcal{F}$ is parasitic for the $\tau $-topology,
$\mathcal{F}(U) = 0$ for all $U \subset S$ open, and
there exists an affine open covering $S = \bigcup U_ i$ such that $\mathcal{F}(U_ i) = 0$ for all $i$.
Proof. The implications (2) $\Rightarrow $ (3) $\Rightarrow $ (4) $\Rightarrow $ (5) are immediate from the definitions. Assume (5). Suppose that $S = \bigcup U_ i$ is an affine open covering such that $\mathcal{F}(U_ i) = 0$ for all $i$. Let $V \to S$ be a flat morphism. There exists an affine open covering $V = \bigcup V_ j$ such that each $V_ j$ maps into some $U_ i$. As the morphism $V_ j \to S$ is flat, also $V_ j \to U_ i$ is flat. Hence the corresponding ring map $A_ i = \mathcal{O}(U_ i) \to \mathcal{O}(V_ j) = B_ j$ is flat. Thus by Lemma 46.5.2 and Lemma 46.3.5 we see that $\mathcal{F}(U_ i) \otimes _{A_ i} B_ j \to \mathcal{F}(V_ j)$ is an isomorphism. Hence $\mathcal{F}(V_ j) = 0$. Since $\mathcal{F}$ is a sheaf for the Zariski topology we conclude that $\mathcal{F}(V) = 0$. In this way we see that (5) implies (2).
This proves the equivalence of (2), (3), (4), and (5). As (1) is equivalent to (3) (see Remark 46.5.9) we conclude that all five conditions are equivalent. $\square$
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