Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Say $X$ is an algebraic variety and $U\subset X$ is open. Consider the natural map $U\rightarrow \operatorname{Spm}(\mathcal{O}_X(U))$ given by sending a point of $U$ to the ideal of sections over $U$ that vanish on that point. $U$ is affine if and only if this map is an isomorphism of varieties.

Is it enough to know $U$ is a bijection of sets? (It seems to me it should be! The ring of global sections on $\operatorname{Spm}(\mathcal{O}_X(U))$ is already $\mathcal{O}_X(U)$ right? So the rings already match! We just need to check that the sets match... what am I missing?)

EDIT (in response to comments by QiL):

First, let me add the assumption that $\mathcal{O}_X(U)$ is finitely generated as a $k$-algebra, so that $\operatorname{Spm}(\mathcal{O}_X(U))$ is an affine variety.

Second, let me state the definition of algebraic variety I'm working with: a separated prevariety. A prevariety is a quasicompact topological space with a sheaf of $k$-valued functions (for $k$ some algebraically closed field) such that every point is contained in an open set such that the restriction of the sheaf to that set makes it isomorphic (as a ringed space) to an affine variety. An affine variety is the $\operatorname{Spm}$ of a finitely generated reduced $k$-algebra.

share|improve this question
    
Well, in principle you have to check that the topologies match as well... –  Zhen Lin Nov 6 '12 at 21:25
    
@ZhenLin - my idea was that the topology is basically coming from the ring... –  Ben Blum-Smith Nov 12 '12 at 0:26

1 Answer 1

up vote 2 down vote accepted

This is not a complete answer.

First you should probably work with Spec rather that Spm because $O_X(U)$ in general is not a finitely generated algebra and you can't call Spm$(O_X(U))$ an algebraic variety.

Working with Spec, there is a lemma in Stack project (23.14.4 in my version) which says that if $U$ is quasi-affine, then $U\to \mathrm{Spec}(O_X(U))$ is always an open immersion (and the converse is quasi-true). So under your hypothesis it is an isomorphism and $U$ is affine.

share|improve this answer
    
Suppose I add the assumption that $\mathcal{O}_X(U)$ is finitely generated? Then $\operatorname{Spm}\mathcal{O}_X(U)$ is a variety. It seems to me that the map $U\rightarrow \mathcal{O}_X(U)$ will automatically be well-defined because the sections over $U$ already include constant functions, so the quotient of $\mathcal{O}_X(U)$ by the ideal of sections vanishing at $p\in U$ will always be the ground field. Now if it is a bijection, then I have two varieties with a correspondence between the points that pulls back to an isomorphism of global sections. Is this enough? –  Ben Blum-Smith Nov 11 '12 at 23:08
    
@BenBlum-Smith: if $O_X(U)$ is finite type over the ground field, then yes your map is a well defined morphism of varieties. If $U$ is moreover quasi-affine, then you morphism is an isomorphism. Otherwise, I don't know. Also please define "algebraic variety". –  user18119 Nov 12 '12 at 9:40
    
@QiL- I added a definition to the OP. Thanks. –  Ben Blum-Smith Nov 12 '12 at 14:15
    
@BenBlum-Smith: thanks for clarification. –  user18119 Nov 12 '12 at 23:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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