(This was originally asked on math.stackexchange, but didn't get any responses. I figured it might be worthwhile to move it here and try again.)

This paper gives a proof that the underlying topological spaces of affine schemes are precisely the spectral spaces (compact, T0, sober, and the compact open subsets are closed under finite intersection and generate the topology; equivalently, they are the inverse limits of families of finite T0 spaces).

Instead of starting with a bare topological space, suppose we have a locally ringed space. If it is an affine scheme, then the underlying space is spectral, the supports of sections generate the topology, every restriction map of the structure sheaf to a distinguished open set is a localization of rings, etc.

Does anyone know whether we can impose restrictions (such as the ones I have just listed, and probably together with others) to guarantee that a given locally ringed space is actually an affine scheme?

This poster asked something similar - whether locally ringed spaces had been classified in some way. This is not exactly what I'm looking for, but it would be interesting to know in any case.

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    $\begingroup$ This question should be relevant mathoverflow.net/questions/89263/… $\endgroup$
    – Mark Grant
    Dec 10, 2012 at 21:56
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    $\begingroup$ It's actually not true that "every restriction map of the [structure] sheaf is a localization of ring"; this is only guaranteed for restrictions to distinguished open sets. $\endgroup$ Dec 11, 2012 at 0:37
  • $\begingroup$ Thanks for the correction @Charles! It's been fixed. @Mark: In fact, that question was identical to mine. Your response there (quoting Eisenbud) was actually exactly the kind of characterization I was looking for. It seems closely related to Ricky's response below, since it states that Ricky's map is a homeomorphism, and then applies some less obvious algebraic conditions. $\endgroup$ Dec 11, 2012 at 1:35

1 Answer 1


I think the easiest condition is the fact that the natural morphism $$ (X, \mathcal O_X) \to \operatorname{Spec}(\mathcal O_X(X)) $$ is an isomorphism.

  • $\begingroup$ I struggle with this: it does give a "deterministic" criterion, in the sense that we don't need to guess at the ring. However, the fact that it essentially states "an l.r.s is an affine scheme if it is isomorphic to (this) scheme" makes it difficult for me to decide if I'm satisfied by it. $\endgroup$ Dec 11, 2012 at 1:26
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    $\begingroup$ It gives a little more, you do not have to guess the isomorphism, just check that morphism. $\endgroup$
    – Ricky
    Dec 11, 2012 at 10:49

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