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?''.


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.

  • $\begingroup$ Sorry, the product is a good start. I'm thinking in reverse, from how to construct the stalks from the open sets. $\endgroup$ – Loki Clock Mar 13 '13 at 14:44
  • $\begingroup$ 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? $\endgroup$ – Ronald Bernard Mar 13 '13 at 15:05
  • $\begingroup$ 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. $\endgroup$ – Loki Clock Mar 13 '13 at 15:14
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    $\begingroup$ @RonaldBernard The two propositions are not equivalent. $\operatorname{Spec}$ does not always map limits to colimits. $\endgroup$ – Zhen Lin Mar 13 '13 at 15:53
  • $\begingroup$ @ZhenLin Yes it does. Spectra of commutative rings are the opposite category to commutative rings. $\endgroup$ – Loki Clock Mar 13 '13 at 15:55

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