3
$\begingroup$

Let $k\subset L$ be an extension of fields of characteristic zero.

Suppose that $X/k$ is an algebraic space such that $X\otimes_k L$ is representable by a finite type $L$-scheme.

I am sure there are examples where $X/k$ is not representable. I think Mumford's example of the Picard scheme of a degenerating conic over $\mathbb R$ can be used to give an example.

What kind of conditions imply $X/k$ to be representable as well? For instance, what if we impose separatedness or properness?

$\endgroup$

1 Answer 1

10
$\begingroup$

In general $X$ is not a scheme, even if it is proper. But if $X\otimes_k L$ is a quasiprojective scheme, so is $X$.

First, assume $X\otimes_k L$ quasiprojective. By standard techniques, this still holds for some finite $L/k$. Then the projection $X\otimes_k L\to X$ identifies $X$ with the quotient (as fppf sheaf) of $X\otimes_k L$ by a finite locally free equivalence relation. The conclusion then follows from SGA3, V, Th. 4.1.

Now here is a counterexample where $X$ is proper, assuming $k$ has a separable quadratic extension $L$. We fix an algebraic closure $\overline{k}$ of $k$.
Hironaka has constructed a smooth proper $k$-variety $Y$ with a pair $\{y_1,y_2\}$ of $k$-points which is not contained in any affine open subscheme of $Y$ (not even after extension to $\overline{k}$). I claim that the Weil restriction $X:=R_{L/k} Y_L$ is not a scheme. Indeed, we have by definition $X(\overline{k})=Y(L\otimes_k\overline{k})=Y(\overline{k}\times\overline{k})=Y(\overline{k})\times Y(\overline{k})$. In particular we have a $\overline{k}$-point of $X$ corresponding to the pair $(y_1,y_2)$. If $x\in X$ is the corresponding closed point, one checks that $x$ has no affine neighborhood since this would give rise to an affine neighborhood of $\{y_1,y_2\}$.
On the other hand, $X$ is an algebraic space, proper over $k$ (in fact $X_\overline{k}$ is isomorphic to $Y_\overline{k}\times_\overline{k}Y_\overline{k}$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.