$\DeclareMathOperator{\colim}{colim} \DeclareMathOperator{\Spec}{Spec}$

[Edit1] I should point out that the colimits below are in the category of schemes, since the statements are trivially false for colimits in the category of affine schemes. [/Edit1]

In this answer Martin points out that $\coprod_i \Spec R_i \ne \Spec \prod_i R_i$ in general. This also proves for $$ \colim_{i \in I} \Spec R_i \ne \Spec \lim_{i \in I} R_i. $$ (Though taking a non-affine scheme, and writing it as colimit of affines might even be a more 'natural' proof.) Now I wondered if what happens if we take global sections on both sides, i.e., $$ \mathcal{O}(\colim_{i \in I} \Spec R_i) \stackrel{?}{=} \lim_{i \in I} R_i. $$ For several (co)limits, I verified this is true. And actually there is a natural map from right to left.

But is this map

$$ \lim_{i \in I} R_i \to \mathcal{O}(\colim_{i \in I} \Spec R_i) $$

an isomorphism:

  • when $I$ is finite?
  • if so, when $I$ is small?
  • 5
    $\begingroup$ Elements of ${\cal O}(X)$ are maps $X \to \mathbb{A}^1$. $\endgroup$ – Tyler Lawson Apr 2 '12 at 19:29
  • $\begingroup$ @Tyler and what does that imply in this situation? $\endgroup$ – jmc Apr 2 '12 at 19:45
  • 4
    $\begingroup$ I think what Tyler had in mind was: $\mathcal{O}(colim_i Spec R_i) = Hom(colim_i Spec R_i, A^1) = lim_i Hom(Spec R_i, A^1) = lim_i R_i$ $\endgroup$ – Yosemite Sam Apr 2 '12 at 20:27
  • 1
    $\begingroup$ where the second identity is the defining property of a colimit and the last identity follows from $Sch(Spec R, A^1) = Ring(Z[x],R) = R$. $\endgroup$ – Yosemite Sam Apr 2 '12 at 23:49

We have an equivalence of categories $Aff\simeq Ring^{op}$ and a pair of adjoint functors $$\mathcal{O}:Sch\rightleftharpoons Ring^{op} : Spec$$ $$\mathcal{O} \dashv Spec$$ The category of affine schemes is clearly a full subcategory of $Sch$. This leads us to reformulation of the previous statement: the inclusion of full subcategory $J:Aff\hookrightarrow Sch$ has a left adjoint $L=Spec\circ \mathcal{O}$. So $Aff$ is a reflective subcategory of $Sch$. Your question is then about the isomorphism $$\mathrm{colim}_i A_i \simeq L(\mathrm{colim}_i J(A_i))$$ This is true, because for a reflective subcategory $L\circ J \simeq 1$ (see, for example, Maclane's book) and left adjoints ($L$) commute with colimits. $$\mathrm{colim}_i A_i \simeq \mathrm{colim}_i LJ(A_i) \simeq L(\mathrm{colim}_i J(A_i))$$

It is an isomorphism whenever any mentioned colimit exists, it also defines then all other mentioned colimits.

  • $\begingroup$ Thans, this really helps. I guess I can now figure out myself whether things like $ \lim_{i \in I} \Spec R_i \ne \Spec \colim_{i \in I} R_i $ are true. $\endgroup$ – jmc Apr 3 '12 at 6:05

Some additions:

It is rather rare that a diagram of schemes has a colimit in the category of schemes - even if all schemes in the diagram are affine. For a nice counterexample, see here. Positive examples include: Coproducts, as well as pushouts along open immersions (this is the usual gluing procedure), but also along closed immersions (see Karl Schwede's paper, and for more results in this direction see this MO discussion). In general, it is quite hard to prove that some diagram of schemes has no colimit (because it is not enough to show that the colimit in the larger categry of locally ringed spaces (or algebraic spaces or alike) is no scheme); but in any event, the functor $\mathcal{O} : \mathrm{Sch} \to \mathrm{Ring}^{op}$ is left adjoint to $\mathrm{Spec}$ and therefore preserves all colimits which exist in $\mathrm{Sch}$, no matter how they look like.

For exmaple, if the colimit $X$ of the sequence of closed immersions $\mathbb{A}^0 \hookrightarrow \mathbb{A}^1 \hookrightarrow \mathbb{A}^2 \hookrightarrow \dotsc$ exists, then we have $\mathcal{O}(X) = \mathrm{lim}_n ~ k[x_1,\dotsc,x_n]$. It is still an open question whether $X$ exists (but not really relevant for algebraic geometers because as soon as no colimit is available in first sight, they use ind-schemes instead).

Other MO-questions about colimits of schemes which illustrate the problems:

  • $\begingroup$ Thanks for your reply. Just to make sure I am understanding stuff correctly: Every scheme can be viewed as colimit of affine schemes, right? (The gluing data forms a colimit if I am correct.) $\endgroup$ – jmc Apr 3 '12 at 8:55
  • $\begingroup$ @johan: before martin writes a better comment perhaps I'll just say that a scheme is the gluing of an open affine cover along intersections, but the intersections needn't be affine. It works if your schemes are geometric (i.e. with affine diagonal), in general you would have to refine each step of the Cech diagram, which I believe is called a(n affine) hypercover. $\endgroup$ – Yosemite Sam Apr 3 '12 at 9:24
  • $\begingroup$ @martin: IMHO, you can always take colimits of schemes inside the (cocomplete) category of sheaves on (say the etale site of) affine schemes. If that colimit isn't a scheme then you should interpret this fact as meaning that the colimit doesn't exist, regardless of whether the categorical colim exists in the category of schemes. I know it makes no sense from a categorical perspective, but I am under the impression that the `correct' limits and colimits should take place in the topos of affine schemes, not in schemes (e.g. quotients via a free action as algebraic spaces, not schemes) $\endgroup$ – Yosemite Sam Apr 3 '12 at 9:28
  • $\begingroup$ @Johan: Yes. In fact, every scheme is the canonical colimit of all affine schemes mapping into it. This means that the restricted Yoneda embeding $\mathrm{Sch} \to [\mathrm{Ring},\mathrm{Set}]$ is fully faithful, which justifies the functor of points approach to AG. $\endgroup$ – Martin Brandenburg Apr 3 '12 at 9:47
  • $\begingroup$ @Yosemite: I agree with everything you have said, except for "you can always take colimits of schemes inside the (cocomplete) category of sheaves on (say the etale site of) affine schemes.". You rather mean (which you also have explained in your comment) that it is not interesting for most algebraic geometers to look at colimits in the category $\mathrm{Sch}$. $\endgroup$ – Martin Brandenburg Apr 3 '12 at 9:47

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