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For Noetherian schemes this follows from Serre's criterion for affineness by a filtration argument.

flag	asked Oct 4 2009 at 21:33 David Zureick-Brown♦ 6,064●10●41

David Rydh · Accepted Answer · 2009-10-05 00:19:28Z UTC

No, if X is any algebraic space such that X_red is an affine scheme, then X is an affine scheme. This follows from Chevalley's theorem. For X noetherian scheme/alg. space this theorem is in EGA/Knutson. As you noted, this can also be showed using Serre's criterion for affineness or by an even simpler argument (see EGA I 5.1.9, first edition).

For X non-noetherian, the following general version of Chevalley's theorem is proved in my paper "Noetherian approximation of algebraic spaces and stacks" (arXiv:0904.0227):

Theorem: Let W->X be an integral and surjective morphism of algebraic spaces. If W is an affine scheme, then so is X.

Recall that any finite morphism is integral, in particular X_red -> X. As a corollary, it follows that under the same assumptions, if W is a scheme then so is X.

Ilya Nikokoshev · Answer 2 · 2009-10-04 21:50:48Z UTC

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I'm not sure this is a correct answer, but I can't post comments yet -- consider this a comment.

How about you take a limit over lots of nilpotent schemes with the same X_red?

answered Oct 4 2009 at 21:50

Ilya Nikokoshev
8,989●5●30●73

Is there an example of a scheme X whose reduction X_red is affine but X is not affine?

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