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.

This is a follow up question to this question.

Is every scheme over a field $K$ the colimit (over some arbitrary complicated diagram) of affine schemes $\operatorname{Spec}(R_\alpha)$ where each $R_\alpha$ is a $K$-algebra and a local ring?

Intuitively, my question is ''scheme $=$ local rings $+$ glueing information?''.

share|improve this question

1 Answer 1

EDIT: Sorry, I misread w/c category you were asking had the colimit. The schemes are at least sort of colimits of the spectra, because each $\mathscr{O}_X(X)$ is a certain limit (the inverse limit) of $O_X$'s rings at its basis sets.

You don't glue the open sets as colimits of the stalks, you glue them as inverse limits over the sets within it that are in a basis of open sets. See the gluing axiom for sheaves, in particular using B-sheaves. Since the restriction map sends $\mathscr{O}(U)$ to $\mathscr{O}(x)$ for every neighborhood $U$ of $x$, the coproduct over all the stalks or values over the basis is a good start, but the question is what's the quotient from the coproduct to the inverse limit. See Geometry of Schemes for a full explanation.

share|improve this answer
    
Sorry, the product is a good start. I'm thinking in reverse, from how to construct the stalks from the open sets. –  Loki Clock Mar 13 '13 at 14:44
    
I know that a scheme $X$ is a certain colimit of affine schemes. This is true by definition, if you like. An other formulation of the question would be: Is every affine $K$-scheme $X=\operatorname{Spec}(R)$ the colimit (over some arbitrary complicated diagram) of $\operatorname{Spec}(R_\alpha)$? Equivalently: Is every $K$-algebra a limit of $K$-algebras $R_\alpha$ where each $R_\alpha$ is a local ring? –  Ronald Bernard Mar 13 '13 at 15:05
    
Depends on whether the sheaf functor sends limits to colimits. So, that's mostly true for spectra, except I don't know if being a limit of stalks necessarily makes it a limit of the basis of open sets for the sheaf. If it's true up to sheaf isomorphism, it's true for affine schemes in general. –  Loki Clock Mar 13 '13 at 15:14
1  
@RonaldBernard The two propositions are not equivalent. $\operatorname{Spec}$ does not always map limits to colimits. –  Zhen Lin Mar 13 '13 at 15:53
    
@ZhenLin Yes it does. Spectra of commutative rings are the opposite category to commutative rings. –  Loki Clock Mar 13 '13 at 15:55

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.