I have read in several texts that in order to show that a given functor from the category of schemes (over, say an algebraically closed field $k$) to $\mathsf{Set}$ is representable, it suffices to check that it is representable in the subcategory of affine schemes. How is this the case? I have not been able to find a proof of this fact, and it may be rather trivial, but I do not see the connection.
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1$\begingroup$ @Fred do you mean github.com/ryankeleti/ega ? $\endgroup$– David Roberts ♦Commented Sep 25, 2019 at 7:21
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1$\begingroup$ @David: No, that's (unfortunately) a translation of the first edition. I mean the 1971 Springer edition (MR3075000). $\endgroup$– Fred RohrerCommented Sep 25, 2019 at 7:35
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1$\begingroup$ @Fred sadly, I had a look for this on SpringerLink, and it doesn't seem to be there! Other 1971 volumes of the Grundlehren der mathematischen Wissenschaften are available digitally (modulo paywalls)... $\endgroup$– David Roberts ♦Commented Sep 25, 2019 at 10:04
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1$\begingroup$ @David: Yes, it is indeed a bit hard to find (but far superior to the first edition!). If you wish, you may send me an email. $\endgroup$– Fred RohrerCommented Sep 25, 2019 at 10:47
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1$\begingroup$ An electronic version of EGA I (2nd edition) can be found here: gen.lib.rus.ec/book/… $\endgroup$– Dmitri PavlovCommented Sep 25, 2019 at 18:28
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The statement
to show that a given functor [$F\colon \mathsf{Sch}^{\mathit{op}}\to\mathsf{Set}$] is representable, it suffices to check that it is representable in the subcategory of affine schemes
isn't exactly true: for example, if $X$ is a scheme that's not affine, $\mathrm{Hom}(-, X)$ is representable in $\mathsf{Sch}$, but not in the subcategory $\mathsf{AffSch}$ of affine schemes.
What is true, though, is that any representable functor is determined (up to natural isomorphism) by its restriction to $\mathsf{AffSch}$. This is because, for any schemes $X$ and $Y$, $U\mapsto \mathrm{Hom}(U,X)$ (for $U\subset Y$ open) is a sheaf on $Y$ in the Zariski topology, and every scheme $Y$ admits an affine open cover. Hence we can determine $\mathrm{Hom}(Y, X)$ from the collection of $\mathrm{Hom}(U, X)$ as $U$ ranges over the affine open subschemes of $Y$.
This leads to the statement that you might have seen: if you're defining a scheme by first defining a functor $F\colon\mathsf{Sch}^{\mathit{op}}\to\mathsf{Set}$, then checking that $F$ is representable, you actually only need to define $F$ on affine schemes. This is particularly useful for defining group schemes, e.g. defining $\mathrm{GL}_n$ by checking that the functor $\mathrm{Spec} R\mapsto \mathrm{GL}_n(R)$ is representable.
(Note: I might have misinterpreted you when you wrote “it suffices to check that it is representable in the subcategory of affine schemes” – if you didn't mean the incorrect statement, then sorry about that!)
answered Sep 24, 2019 at 22:54
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1$\begingroup$ We don't know that a priori. But if $F$ is represented by some scheme $X$, then $F$ is determined by its restriction to affines. Let $Y$ be another scheme and $\mathfrak U$ be an affine open cover of $Y$; then the set $F(Y) = \mathrm{Hom}(Y, X)$ is the subset of $\prod_{U\in\mathfrak U}\mathrm{Hom}(U, X)$ consisting of tuples of maps which agree on all pairwise intersections. $\endgroup$ Commented Sep 24, 2019 at 23:40
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1$\begingroup$ @JoakimFærgeman Maybe this helps: the restriction functor induces an equivalences of categories between Zariski sheaves on $\mathrm{Sch}$ and Zariski sheaves on $\mathrm{AffSch}$, so to give one is the same thing as to give the other. $\endgroup$ Commented Sep 25, 2019 at 6:14
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1$\begingroup$ Maybe I was unclear in my question, or maybe you have already answered it, and I dont understand. In any case, I am dealing with the following situation: I have a contravariant functor from Schemes to Sets. I know that it is representable when restricted to affine schemes. Can I conclude that it is also representable on Schemes? $\endgroup$ Commented Sep 25, 2019 at 11:24
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1$\begingroup$ @Joakim: What precisely do you mean by "it is representable when restricted to affine schemes"? $\endgroup$ Commented Sep 25, 2019 at 13:25
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1$\begingroup$ I think (without checking all the details) that it follows from the result in EGA I mentioned above that the answer to your question is yes if your functor is a Zariski sheaf on the category of all schemes. I have no idea about the general case. $\endgroup$ Commented Sep 25, 2019 at 15:51