14
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 Nikokoshev Oct 4 '09 at 21:50
18
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.