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Is the ring of global functions on an integral scheme an integral domain (if the scheme is affine then it is try by definition so we are interested in non-affine schemes)? It is necessarily a reduced ring (https://stacks.math.columbia.edu/tag/01OL) so the question is whether it is irreducible.

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If $X$ is any reduced scheme and $U$ is a dense open subset, then the restriction map $\mathcal{O}_X(X)\to\mathcal{O}_X(U)$ is injective. Indeed, an element of the kernel vanishes in the fiber at every point of $U$, but the set where a section of $\mathcal{O}_X$ vanishes in the fiber is closed, so it must vanish in the fiber everywhere. Since $X$ is reduced, a function which vanishes in the fiber everywhere must be $0$.

In particular, if $X$ is integral, we can consider $\mathcal{O}_X(X)$ as a subring of the domain $\mathcal{O}_X(U)$ for any nonempty affine open $U$, and so $\mathcal{O}_X(X)$ is also a domain.

Eric Wofsey
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  • Is the "fiber" the residue field of the local ring $\mathcal{O}{X,x}$ ? So, do you mean $U\subset X\setminus D(f)$ for any $f\in ker(\mathcal{O}{X}(X)\longrightarrow\mathcal{O}{X}(U))$, where $D(f) = {x\in X : f\in\mathcal{O}{X, x}^{\times}}$? – Sem Jun 24 '22 at 02:21
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    That's correct. – Eric Wofsey Jun 24 '22 at 02:50
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This is an alternative answer to your question.

Question: "Q1. Is the ring of global functions on an integral scheme an integral domain (if the scheme is affine then it is try by definition so we are interested in non-affine schemes)?

Q2. It is necessarily a reduced ring?"

Answer: If you use the definition in Hartshorne (page 82) you define a scheme $(X,\mathcal{O}_X)$ to be "integral" iff for any open set $U\subseteq X$ it follows $\mathcal{O}_X(U)$ is an integral domain. In particular, since $X$ is an open set, it follows $\Gamma(X,\mathcal{O}_X):=\mathcal{O}_X(X)$ is an integral domain. Hence it seems to me the answer to Q1 is "yes, this is true by definition". Since an integral domain is a reduced ring it follows $\mathcal{O}_X(X)$ is a reduced ring. Hence the answer to Q2 is also "yes".

hm2020
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