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Let $(X_i)_{i\in I}$ be a family of affine schemes, where $I$ is an infinite set and $X_i = Spec(A_i)$ for each $i \in I$. Let $X$ be a coproduct of $(X_i)_{i\in I}$ in the category of schemes. Let $\Gamma(X, \mathcal{O}_X)$ be the ring of global sections.

(1) Is $X$ affine?

(2) Can we deterimine the struture of $\Gamma(X, \mathcal{O}_X)$ by $(A_i)_{i\in I}$?

Makoto Kato
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2 Answers2

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$\newcommand{\Spec}{\operatorname{Spec}} \newcommand{\Sch}{\text{Sch}} \newcommand{\Affsch}{\text{Affsch}}$(1) No, the scheme $X=\bigsqcup X_i$ is not affine (unless almost all $X_i$ are empty!) because its underlying topological space $|X|=\bigsqcup |X_i|$ is not quasi-compact.

(2) Yes, $\Gamma (X,\mathcal O_X)$ is determined by the formula $$\Gamma (X,\mathcal O_X)=\prod \Gamma (X_i,\mathcal O_{X_i})=\prod A_i$$

Remarks
a) The scheme $X$ has as underlying topological space $| X|=\bigsqcup | X_i|$, as already mentioned, and its structure sheaf is the unique sheaf of rings $\mathcal O_X$ satisfying $\mathcal O_X|_{ X_i}=\mathcal O_{X_i}$.
[O my dear nitpicking brothers , notice that "unique" here really means unique, and not unique up to isomorphism!]

b) In order to prevent any misunderstanding, let me emphasize that the scheme $X$ is the coproduct of the schemes $X_i$ in the category of all schemes (not in the category of affine schemes!).
In other words, the open immersions $u_i :X_i\hookrightarrow X$ produce bijections $\operatorname{Hom}_{\Sch}(X,Y)=\prod \operatorname{Hom}_{\Sch}(X_i,Y): f\mapsto f\circ u_i$ which are functorial in the scheme $Y$.

Edit
c) Beware the subtle fact that the family of affine schemes $X_i=\Spec(A_i)$ also has a coproduct in the category $\Affsch$ of affine schemes, namely $X'=\Spec(\prod A_i)$.
There is a canonical morphism of schemes $$\alpha: X=\bigsqcup_{\Sch} X_i \to X'=\bigsqcup_{\Affsch} X_i=\Spec(\prod A_i)$$ of the coproduct of the $X_i$'s in the category of all schemes to the coproduct of the $X_i$'s in the category of affine schemes.
This morphism $\alpha$ is determined by its restrictions $\alpha|_{X_j}:X_j=\Spec(A_j)\to \Spec(\prod A_i)$, which are dual to the ring projections $\prod A_i\to A_j$.
And finally, let me insist: this canonical morphism $\alpha$ is an isomorphism of schemes if and only the family of schemes $(X_i)$ is a finite family.
[This edit is the consequence of a pleasant discussion with my friend and very competent colleague Dehon: thanks François-Xavier!]

Elías Guisado Villalgordo
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Georges Elencwajg
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  • Dear Georges, Could you explain why (2) follows from b)? – Makoto Kato Jan 04 '13 at 00:39
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    Dear @Makoto, The first equality in (2) has nothing to do with schemes or even locally ringed spaces. If $X$ is a topological space with an open cover $X=\bigcup_iX_i$ such that $X_i\cap X_j=\emptyset$ for $i\neq j$, and $F$ is any sheaf on $X$ (of modules, sets, rings, etc.), then the sheaf axioms stipulate that natural map $F(X)\rightarrow\prod_iF(X_i)$ is an isomorphism. Explicitly, separatedness of $F$ states that the map is injective, and the "gluability" axiom says that an arbitrary element of the RHS, i.e., a tuple $(s_i)_i$ lifts to to an element of the LHS (because the condition – Keenan Kidwell Jan 04 '13 at 02:23
  • $s_i\vert_{X_i\cap X_j}=s_j\vert_{X_i\cap X_j}$ is vacuous here). The second equality in (2) follows from the definition of the structure sheaf of $\mathrm{Spec}(A)$. – Keenan Kidwell Jan 04 '13 at 02:26
  • @KeenanKidwell I think you are right. I was thinking about $Y = Spec(\mathbb{Z}[X])$. I would like to accept the both answers(this and yours), but I can only accept one. – Makoto Kato Jan 04 '13 at 04:08
  • Dear @Keenan, thanks for your perfect comment :+1. And while I'm at it , +1 for your answer too ! – Georges Elencwajg Jan 04 '13 at 08:15
  • Dear @Georges, Thank you! I was starting to doubt my answer until I saw yours and was reassured :) – Keenan Kidwell Jan 05 '13 at 00:41
  • Dear @Georges, sorry to be one of those nitpicking brothers, but could you explain why "unique" really meant "unique" and not "unique up to unique isomorphism"? After all, "\mathcal O_X" is the kernel of the map $\prod_i \mathcal O_{X_i} \rightarrow\prod_{ij} \mathcal O_{X_i \cap X_j} = 0$ and kernels are defined up to unique isomorphism. – Rodrigo Jan 21 '14 at 12:01
  • Dear @Rodrigo, it is true that kernels are defined up to isomorphism in general categories, but for sheaves there is a concrete description not involving any isomorphism: $(ker (F\to G))(U)=ker(F(U)\to G(U))$ – Georges Elencwajg Jan 21 '14 at 12:25
  • Is it correct that infinite coproduct of schemes in the category of all schemes is disjoint union? – Lao-tzu Jul 15 '21 at 07:44
  • @Lao-tzu : Yes. – Georges Elencwajg Jul 27 '21 at 08:42
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No, $X$ is not affine if $I$ is infinite because an infinite disjoint union of schemes is not quasi-compact. The natural map $\mathscr{O}_X(X)\rightarrow\prod_i\mathscr{O}_X(X_i)=\prod_iA_i$ is an isomorphism by the definition of a sheaf together with the fact that $X_i\cap X_j=\emptyset$ inside $X$ for all $i\neq j$.

Keenan Kidwell
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