Let $\underline{\mathbb{A}}^r$ be the functor from $\textbf{Schemes}$ to $\textbf{Sets}$ which associates to each scheme $S$ the set of morphisms $\bigoplus_{k=1}^rO_S \to O_S$ ($O_S$ is the structure sheaf of $S$). Let $\mathbb{A}^r$ be the scheme $Spec(\mathbb{Z}[X_1, \dots, X_r])$.

Assume it is given that whenever $X$ is a scheme the restriction of $\underline{\mathbb{A}}^r$ to $Top(X)$ (which I presume is the subcategory of $\textbf{Schemes}$ consisting of the open subschemes of $X$ and the open immersions between them) is a sheaf of sets on $X$.

Assume further that the restriction of the representable functor $h_{\mathbb{A}^r} := Mor(\_, \mathbb{A}^r)$ to Top(X) is also a sheaf of sets on $X$.

If we wish to show that $\underline{\mathbb{A}}^r$ is representable by $\mathbb{A}^r$, why does it suffice to show that there is an isomorphism between $\mathbb{A}^r$ restricted to the category $\textbf{Aff}$ of affine schemes and $\underline{\mathbb{A}}^r$ restricted also to $\textbf{Aff}$? This is just stated at some point in a proof in my Schemes course.

I vaguely see that perhaps the fact that every scheme is a bunch of affine schemes glued together, and that 'sheaves allow us to glue', might help - but given that $Top(X)$ does not appear to be a full subcategory of $\textbf{Schemes}$ in general I do not quite follow the logic. It is stated that this fact follows from the two facts stated above about restrictions to $Top(X)$.

What am I misunderstanding, here?

  • $\begingroup$ Your gluing idea is correct. A hint is that every scheme is a colimit of affines and Hom preserves colimits $\endgroup$ – Samir Canning Feb 22 at 16:52
  • $\begingroup$ @SamirCanning I have thought about this but I still cannot see how it follows. You can decompose any Hom(X, Y) perhaps into a double colimit of Hom(Spec(R), Spec(S))'s, but unless each morphism between affine schemes is an open embedding I cannot really see how that helps. $\endgroup$ – Nethesis Mar 7 at 16:56
  • $\begingroup$ Oh wait, it is just because any scheme has an open affine cover, and then you use the sheaf property to make the deduction. $\endgroup$ – Nethesis Mar 7 at 18:59
  • $\begingroup$ Yes, that is it. $\endgroup$ – Samir Canning Mar 7 at 19:40

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