The Stacks project

Proof. Denote $\mathcal{F}$ the Zariski sheafification of the functor

For quasi-compact and quasi-separated schemes $X$ we have $\mathcal{F}(X) = \mathop{\mathrm{colim}}\nolimits _ F \Gamma (X, \mathcal{O}_ X)$. by Sheaves, Lemma 6.29.1 and the fact that $\mathcal{O}$ is a sheaf for the Zariski topology. Thus it suffices to show that $\mathcal{F}$ is a h sheaf. To prove this we check conditions (1), (2), (3), and (4) of Lemma 38.37.12. Condition (1) holds because we performed an (almost unnecessary) Zariski sheafification. 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 $\mathop{\mathrm{colim}}\nolimits _ F A$ is the equalizer of the two maps $\mathop{\mathrm{colim}}\nolimits _ F B \to \mathop{\mathrm{colim}}\nolimits _ F C$.

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

This reduces to the following algebra question: suppose $a', a'' \in A$ are such that $F^ n(a' - a'') \in fA + J$. Find $a \in A$ and $m \geq 0$ such that $a - F^ m(a') \in J$ and $a - F^ m(a'') \in fA$ and show that the pair $(a, m)$ is uniquely determined up to a replacement of the form $(a, m) \mapsto (F(a), m + 1)$. To do this just write $F^ n(a' - a'') = f h + g$ with $h \in A$ and $g \in J$ and set $a = F^ n(a') - g = F^ n(a'') + fh$ and set $m = n$. To see uniqueness, suppose $(a_1, m_1)$ is a second solution. By a replacement of the form given above we may assume $m = m_1$. Then we see that $a - a_1 \in J$ and $a - a_1 \in fA$. Since $J$ is annihilated by a power of $f$ we see that $a - a_1$ is a nilpotent element. Hence $F^ k(a - a_1)$ is zero for some large $k$. Thus after doing more replacements we get $a = a_1$.

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

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. $\square$