Lemma 38.38.5. Let $S$ be a scheme. Choose a site $(\mathit{Sch}/S)_ h$ as in Definition 38.34.13. The rule

is the sheafification of the “structure sheaf” $\mathcal{O}$ on $(\mathit{Sch}/S)_ h$. Similarly for the ph topology.

Lemma 38.38.5. Let $S$ be a scheme. Choose a site $(\mathit{Sch}/S)_ h$ as in Definition 38.34.13. The rule

\[ X \longmapsto \Gamma (X^{awn}, \mathcal{O}_{X^{awn}}) \]

is the sheafification of the “structure sheaf” $\mathcal{O}$ on $(\mathit{Sch}/S)_ h$. Similarly for the ph topology.

**Proof.**
In Lemma 38.38.4 we have seen that the rule $\mathcal{F}$ of the lemma defines a sheaf in the ph topology and hence a fortiori a sheaf for the h topology. Clearly, there is a canonical map of presheaves of rings $\mathcal{O} \to \mathcal{F}$. To finish the proof, it suffices to show

if $f \in \mathcal{O}(X)$ maps to zero in $\mathcal{F}(X)$, then there is a h covering $\{ X_ i \to X\} $ such that $f|_{X_ i} = 0$, and

given $f \in \mathcal{F}(X)$ there is a h covering $\{ X_ i \to X\} $ such that $f|_{X_ i}$ is the image of $f_ i \in \mathcal{O}(X_ i)$.

Let $f$ be as in (1). Then $f|_{X^{awn}} = 0$. This means that $f$ is locally nilpotent. Thus if $X' \subset X$ is the closed subscheme cut out by $f$, then $X' \to X$ is a surjective closed immersion of finite presentation. Hence $\{ X' \to X\} $ is the desired h covering. Let $f$ be as in (2). After replacing $X$ by the members of an affine open covering we may assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. Then $f \in A^{awn}$, see Morphisms, Lemma 29.47.6. By Morphisms, Lemma 29.46.11 we can find a ring map $A \to B$ of finite presentation such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism and such that $f$ is the image of an element $b \in B$ under the canonical map $B \to A^{awn}$. Then $\{ \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)\} $ is an h covering and we conclude. The statement about the ph topology follows in the same manner (or it can be deduced from the statement for the h topology). $\square$

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