Lemma 38.38.3. Let $p$ be a prime number. Let $S$ be a scheme over $\mathbf{F}_ p$. Let $(\mathit{Sch}/S)_ h$ be a site as in Definition 38.34.13. The rule

defines a sheaf on $(\mathit{Sch}/S)_ h$.

Lemma 38.38.3. Let $p$ be a prime number. Let $S$ be a scheme over $\mathbf{F}_ p$. Let $(\mathit{Sch}/S)_ h$ be a site as in Definition 38.34.13. The rule

\[ \mathcal{F}(X) = \mathop{\mathrm{lim}}\nolimits _ F \Gamma (X, \mathcal{O}_ X) \]

defines a sheaf on $(\mathit{Sch}/S)_ h$.

**Proof.**
To prove $\mathcal{F}$ is a sheaf, let's check conditions (1), (2), (3), and (4) of Lemma 38.37.12. Condition (1) holds because limits of sheaves are sheaves and $\mathcal{O}$ is a Zariski sheaf. Condition (2) holds because $\mathcal{O}$ is an fppf sheaf (Descent, Lemma 35.8.1) and if $A$ is the equalizer of two maps $B \to C$ of $\mathbf{F}_ p$-algebras, then $\mathop{\mathrm{lim}}\nolimits _ F A$ is the equalizer of the two maps $\mathop{\mathrm{lim}}\nolimits _ F B \to \mathop{\mathrm{lim}}\nolimits _ F C$.

We check condition (3). Let $A, f, J$ be as in Example 38.37.10. We have to show that

\begin{align*} \mathop{\mathrm{lim}}\nolimits _ F A & \to \mathop{\mathrm{lim}}\nolimits _ F A/J \times _{\mathop{\mathrm{lim}}\nolimits _ F A/fA + J} \mathop{\mathrm{lim}}\nolimits _ F A/fA \\ & = \mathop{\mathrm{lim}}\nolimits _ F (A/J \times _{A/fA + J} A/fA) \\ & = \mathop{\mathrm{lim}}\nolimits _ F A/(fA \cap J) \end{align*}

is bijective. Since $J$ is annihilated by a power of $f$ we see that $\mathfrak a = fA \cap J$ is a nilpotent ideal, i.e., there exists an $n$ such that $\mathfrak a^ n = 0$. It is straightforward to verify that in this case $\mathop{\mathrm{lim}}\nolimits _ F A \to \mathop{\mathrm{lim}}\nolimits _ F A/\mathfrak a$ is bijective.

We check condition (4). Let $X, X', Z, E$ be as in Example 38.37.11. By Lemma 38.38.1 and the same argument as above we see that

\[ \mathcal{F}(X) = \mathop{\mathrm{lim}}\nolimits _ F \Gamma (X, \mathcal{O}_ X) \longrightarrow \mathop{\mathrm{lim}}\nolimits _ F \Gamma (X', \mathcal{O}_{X'}) = \mathcal{F}(X') \]

is bijective. Since $E = \mathbf{P}^1_ Z$ in this case we also see that $\mathcal{F}(Z) \to \mathcal{F}(E)$ is bijective. Thus the conclusion holds in this case as well. $\square$

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