For Noetherian schemes this follows from Serre's criterion for affineness by a filtration argument.
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asked Oct 4, 2009 at 21:33
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$\begingroup$ 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? $\endgroup$– Ilya NikokoshevCommented Oct 4, 2009 at 21:50
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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.
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1$\begingroup$ Note that for schemes this is already in Conrad's Deligne's notes on Nagata compactification, Corollary A.2, which you cite in your paper. $\endgroup$ Commented Jan 23, 2020 at 4:49