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Let $k$ be a field with a fixed separable closure $k_s$ and $G$ a finite type $k$-group scheme. Assume $F:(\mathrm{Sch}/k)^{opp}\rightarrow\mathrm{Set}$ is a contravariant functor whose restriction $F_{k_s}$ to schemes over $k_s$ is representable by a closed $k_s$-subscheme $Z_{k_s}\hookrightarrow G_{k_s}$. I want to prove that $F$ itself is representable by a closed $k$-subscheme $Z\hookrightarrow G$ via descent theory and approximation results, and I'm not sure I'm going about it the right way.

Since $Z_{k_s}$ is of finite presentation over $k_s$, results on inductive limits show that there is a finite separable extension $k^\prime$ and a closed subscheme $Z_{k^\prime}\hookrightarrow G_{k^\prime}$ that becomes isomorphic to $Z_{k_s}$ upon base change to $k_s$.

In this generality, is there any reason to believe that $Z_{k^\prime}$ represents the restriction $F_{k^\prime}$ of $F$ to $k^\prime$-schemes?

If $G$ is affine (and so all other schemes under consideration are affine), then my understanding is that faithfully flat descent relative to the extension $k_s/k$ is effective (because the condition about coverings by open affine subschemes stable under the descent datum is trivially satisfied), and in this case one can descend $Z_{k_s}$ uniquely to $k$ (without the intermediate field $k^\prime$), and I think that the descended scheme will represent $F$ (based on Stacks Tag 02W5, which admittedly I don't fully understand.

If the schemes under consideration are not affine, then I don't think descent relative to the extension $k_s/k$ is necessarily effective, and so my thought was that one must first descend to a finite separable extension $k^\prime$ (which can be taken Galois) using considerations with inductive limits, and then try to apply ``Galois descent" relative to $k^\prime/k$ by establishing the relevant Galois equivariance to ensure effectivity of descent. But again, I don't understand why the object constructed over $k^\prime$ should represent $F_{k^\prime}$.

I apologize for this somewhat imprecise question. In reality I'm only interested in the affine case, but I'm worried that even my understanding of that is flawed.

Keenan Kidwell
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1 Answers1

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Suppose that $Z_1$ and $Z_2$ are two different $k'$-schemes which become isomorphic over $k_s$ (to some common $Z_{k_s}$). Suppose that $Z_1$ even descends to some $Z$ over $k$ (e.g. maybe $k = k'$). If we choose $F$ to be the functor rep'd by $Z$, and but we choose $Z_{k'}$ to be $Z_2$, then we see that $Z_{k'}$ doesn't represented $F_{k'}$.

So the answer to your question in bold is no.

However, suppose that $F$ is loc. of fin. pres. (which it had better be if it is going to representable by some loc. f.t. $k$-scheme $Z$).

Then since $Z_{k_s}$ is q.c., writing $Z_{k_s}$ as the limit of $Z_{k'}$ for $k'$ finite over $k$, the isomorphism $Z_{k,s} \to F$ is induced by $Z_{k'} \to F$ for some $k'$ finite over $k$.

So you have a morphism $Z_{k'} \to F$ which becomes an isomorphism after pulling back to $k_s$. If $F$ has some sheaf properties too (say is an fpqc sheaf) then this map will have to be an isomorphism, since it becomes an isomorphism on the fpqc covers $Z_{k_s}$ and $F_{k_s}$ of its source and target.

Anyway, this may not be the best way to argue in your context, but it gives some indication of of what you should do if you want to make these sorts of arguments, which is to use properties of $F$.

In your context, it might be easiest just to follow the argument you already ouline: note that $F_{k_s}$ has descent data to $k$, and if the isomorphism between $F_{k_s}$ and $Z_{k_s}$ is going to descend to an isomorphism between $F$ and some $k$-scheme underlying $Z_{k_s}$, then $Z_{k_s}$ will have to be equipped with descent data to $k$ in a manner that is compatible with its identification with $F_{k_s}$.

But, since $F_{k_s} \cong Z_{k_s}$, you can transfer the descent data from $F_{k_s}$ to $Z_{k_s}$, and apply fpqc descent. (Incidentally, if this descent data is not effective, you are not going to be able to get around that by descending to something over a finite extension first. After all, descending through a finite Galois action is analogous to descending through a finite fixed-point free group action, and there are famous examples, due to Hironaka, of quotients by a group of order two which are not schemes.)

Matt E
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  • Dear @Matt, Thank you for this explanation. My specific context was just trying to understand how to prove representability of some basic subgroup functors of linear algebraic groups over fields (like centralizers). I think this is more than enough to understand that particular application now. – Keenan Kidwell Jan 12 '14 at 17:16
  • Dear Keenan, As a side remark, note that the centralizer of an element (or collection of elements) is defined by equations (the equations $z a z^{-1} = a$, where $a$ runs over the set of elements to be cenralized), and so is manifestly representable by a scheme. Regards, – Matt E Jan 12 '14 at 17:29
  • Dear @Matt, Right. I guess what I had in mind was something like the following (which I guess is actually a normalizer): if $G$ is a linear group over a field $k$ and $T$ is a closed subgroup scheme, I want to prove (under appropriate hypotheses) representability of the functor on $k$-algebras $R$ whose $R$-points are those $g\in G(R)$ such that the conjugation $g:G_R\rightarrow G_R$ has $g(T_R)=T_R$ scheme-theoretically, or equivalently, $g(T(A))=T(A)$ for all $R$-algebras $A$. – Keenan Kidwell Jan 12 '14 at 17:49