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Serre's criterion for affineness (Hartshorne III.3.7) states that:

Let $X$ be a noetherian scheme. Suppose $H^1(X, \mathcal{F})= 0$ for every quasi-coherent sheaf on $X$. Then $X$ is affine.

There is a more general statement (EGA II.5.2.1) that reaches the same conclusion with the hypothesis that $X$ is a separated quasi-compact scheme or one where the topological space is noetherian.

I don't, however, understand why we need these hypotheses. The idea of the proof is to find a set of $g_i \in \Gamma(X, O_X)$ generating the unit ideal and such that the $X_{g_i}$ are affine. This can be done by showing the $X_{g_i}$ form a basis at each closed point using the cohomology statement (an argument which I'm pretty sure works without any conditions on the scheme). Then, you take open affine sets of the form $X_f$ containing each closed point and take their union; this is an open set whose complement is closed and must contain no closed point, hence is empty. This part of the argument relies on the fact that every closed subset contains a closed point, a fact which is true under noetherian hypotheses since a scheme is a $T_0$-space.

But isn't it true that every quasi-compact $T_0$ space has a closed point? A minimal closed set must be reduced to a point. So I'm unclear why "quasi-compactness" alone is not enough. (It shouldn't be enough. Grothendieck is not one to mince hypotheses!)

Questions:

  1. Where does the standard argument break down for schemes which are only quasi-compact?

  2. Is this true if $X$ is only quasi-separated and quasi-compact? (Are there counterexamples otherwise?)

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Hi Akhil, have you thought what happens if you take $X$ to be a quasi-compact algebraic space? (and by algebraic space I mean with no conditions on the diagonal) –  Jacob Bell Feb 29 '12 at 14:44
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1 Answer

up vote 10 down vote accepted

The short answer is that quasi-compactness is enough (for the statement you asked about): see lemma 3.1 in http://www.math.columbia.edu/algebraic_geometry/stacks-git/coherent.pdf (but its essentially the argument in Hartshorne)

this business of extra hypothesis comes up because Hartshorne and Gronthendieck are proving an iff statement; that is, these hypothesis are needed to prove that if $X$ is affine then $H^1(X,F) = 0$ for every quasicoherent sheaf. In the case of Hartshorne you need notherian hypothesis to prove

lemma II.3.3 If $I$ is an injective $A$-module and $f \in A$ then $I \to I_f$ is surjective.

with this lemma you can then show if $I$ is injective then $\tilde I$ is flasque an so you can use them to calculate cohomology: $F$ is quasicoherent $\Rightarrow F = \widetilde{M}$. If $M \to I_0 \to I_1 \to ...$ (1) is an injective resolution then $\widetilde{M} \to \widetilde{I_0} \to \widetilde{I_1} \to ...$ is flasque resolution and applying global sections just recovers (1) so all the higher cohomology vanishes.

I'm not familiar with Gronthendieck's argument and I don't know if you can replace separated with quasi-separated.

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Excellent--thanks. Grothendieck's argument for the converse (vanishing cohomology-/> affine) is the same as Hartshorne's, though for the reverse he uses the fact that $H^1$ can be interpreted as an $Ext$ and any extension of a q-c sheaf by $O_X$ is trivial (by thinking of the associated modules). For the general $H^n(X, \cal{F})=0$ he interprets the sheaf cohomology as the limit of Koszul complexes in EGA III, each of which is homotopically trivial; I haven't really grokked this one though. I guess you just need quasi-compactness then. –  Akhil Mathew Aug 24 '10 at 14:02
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