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Let $F$ be a set-valued (contravariant) functor on the category of schemes. Let $F_{\mathbb Q}$ be the associated functor on the category of schemes over $\mathbb Q$.

Suppose that $F_{\mathbb Q}$ is representable by a scheme (over $\mathbb Q$). Does it follow that $F_{\mathbb Z[1/n]}$ is representable for some large integer $n$? (No.)

I can easily imagine cases this could fail to be the case.

I'm actually wondering what the minimal hypothesis is on $F$ for the answer to be positive. For instance, what if we assume $F$ is a sheaf in the etale topology?

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Being a sheaf in the étale topology is insufficient. Begin with the locally ringed space $(\text{Spec}\mathbb{Z},\mathcal{O}_{\text{Spec}\mathbb{Z}})$. Let $$i:X\hookrightarrow \text{Spec}\mathbb{Z}$$ be the complement of the generic point $\langle 0 \rangle$. Give $X$ the sheaf of rings $\mathcal{O}_X := i^{-1}\mathcal{O}_{\text{Spec}\mathbb{Z}}$. Let $$i^\#:\mathcal{O}_{\text{Spec}\mathbb{Z}} \to i_*\mathcal{O}_X,$$
be the natural map defined by adjunction of $i^{-1}$ and $i_*$. Then $(i,i^\#)$ is a morphism of locally ringed spaces. Moreover, $(i,i^\#)$ has the following universal property: for every locally ringed space $(T,\mathcal{O}_T)$ and for every morphism of locally ringed spaces $$ (f,f^\#):(T,\mathcal{O}_T) \to (\text{Spec}\mathbb{Z},\mathcal{O}_{\text{Spec}\mathbb{Z}}),$$ (of course there is a unique such morphism!), there exists a morphism of locally ringed spaces $$ (g,g^\#):(T,\mathcal{O}_T) \to (X,\mathcal{O}_X),$$ with $(i,i^\#)\circ(g,g^\#) = (f,f^\#)$ if and only if $g(T)$ is contained in $i(X)$, in which case the morphism $(g,g^\#)$ is unique. To learn more about any of this, please read about the "max spectrum".

Now consider the Yoneda functor $h_X$ of $(X,\mathcal{O}_X)$, but restricted to the full category of schemes. This is simply the functor that associates to every scheme $(T,\mathcal{O}_T)$ the singleton set if every residue field has positive characteristic, and that associates the empty set if some residue field has characteristic $0$. This is an étale sheaf. It is representable by the empty set if you restrict to the category of $\mathbb{Q}$-schemes. However, it is not representable by a scheme or algebraic space, not even when restricted to the category of $\mathbb{Z}[1/n]$-schemes. To see this, let $p$ be any prime not dividing $n$, and consider the value of the functor restricted to the sequence of schemes $$(\text{Spec}\mathbb{Z}/p^r\mathbb{Z},\mathcal{O}_{\text{Spec}\mathbb{Z}/p^r\mathbb{Z}})_r$$ as well as the scheme $(\text{Spec}\widehat{\mathbb{Z}}_p,\mathcal{O})$. Since the functor fails to be effectively prorepresentable, it is not a scheme or algebraic space.

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  • $\begingroup$ Thank you very mcuh for this example. I have one question to reassure myself I understood correctly. The values of the functor restricted to the sequence are singletons, whereas the value on the formal scheme \wideht{Spec} $\mathbb Z_p$ is the empty set, right? This is what obstructs the functor being effetively pro-representable? $\endgroup$ – Konan Jun 1 '15 at 11:14
  • $\begingroup$ @Konan: Yes, that is what prevents the functor from being effectively prorepresentable. Thus the functor is not an algebraic space. $\endgroup$ – Jason Starr Jun 1 '15 at 11:18

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