Let $\mathbb F_p$ be the finite field with $p$ (prime) elements, and let $i_p:\operatorname{Spec}\mathbb F_p \to \operatorname{Spec}\mathbb Z$ be the canonical morphism. A ring is of char $p$ if $p\cdot 1=0$. The following are equivalent.

(i) For every open subset $U\subset X$, the ring $\Gamma(U,\mathcal O_X)$ has char $p$.

(ii) The ring $\Gamma(X,\mathcal O_X)$ has char p.

(iii) The scheme morphism $X\to \operatorname{Spec}\mathbb Z$ factors through $i_p$.

(i) $\Rightarrow$ (ii): Clearly.

(ii) $\Rightarrow$ (i): I wonder whether $\Gamma(U,\mathcal O_X)$ can be viewed as $\Gamma(X,\mathcal O_X)$-algebra. If yes, then $p$ is also the char of $\Gamma(U,\mathcal O_X)$.

I wonder whether the statement is right. If not, how to prove ii implies i?


1 Answer 1


We have that $ii)\implies i)$ because part of the data of the sheaf $O_X$ is the restriction maps, i.e. if $U\subset V$, then we have restriction maps (of rings in this case) $$i^V_U:\Gamma(V,O_X)\to \Gamma(U,O_X).$$ Applying this with $V=X$ in your case gives all the local sections the structure of $\Gamma(X,O_X)$-algebras.


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