What if the base change of an algebraic space is representable 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?
 A: 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 $Y_\overline{k}\times_\overline{k}Y_\overline{k}$).
