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$X$ is projective and reduced over a field $k$ (not necessarily algebraically closed). Why is $H^0(X,\mathcal{O}_X)$ a field?

Are there any good lecture notes on this (valuative criteria, properness, projectiveness, completeness)? I really don't have time to go through EGA/SGA.

HLC
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1 Answers1

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This is not true. The scheme $X=\mathrm{Spec}(k\times k)$ is projective and reduced over $k$ but the global sections of the structure sheaf do not form a field. You need to assume $X$ is connected too. In any case, by properness, $H^0(X,\mathscr{O}_X)$ is a finite-dimensional $k$-algebra (this is the hard part), hence a finite product of Artin local rings. Since $X$ is connected, $H^0(X,\mathscr{O}_X)$ is a connected ring, meaning that it must be a single Artin local ring. The unique maximal ideal of such a ring coincides with its nilradical, which is zero because $X$ is reduced, so $H^0(X,\mathscr{O}_X)$ must be a field.

Keenan Kidwell
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  • Why does properness imply finite-dimensional property? I know for example theorem 5.19 Hartshorne says projectiveness implies finite-dimensional but this statement is stronger, right? – HLC Nov 20 '15 at 16:51
  • Yes, the statement is stronger. I don't know a precise reference off hand. It follows from coherence of higher direct images of coherent modules under proper morphisms of locally Noetherian schemes, but that's serious overkill. I think there is a place in Liu's book where he proves that $H^0(X,\mathscr{O}_X)$ is integral over $k$...maybe something slightly more general. But integral plus finite type is finite, so that's adequate. – Keenan Kidwell Nov 20 '15 at 20:52
  • nice proof amigo! – Juan Fran Aug 03 '16 at 13:57