TITLE: Varieties which become isomorphic to algebraic groups over an algebraic closure
QUESTION [9 upvotes]: My question is as follows:

Let $k$ be a field of characteristic zero and let $\overline{k}$ be an algebraic closure. Let $V$ be an algebraic variety over $k$ and let $\overline{V}=V \times_k \overline{k}$. Suppose that $\overline{V}$ admits the structure of an algebraic group. Then is $V$ itself a principal homogeneous space for some algebraic group?

Note that the answer to my question is yes when $V$ is projective; in this case it is well-known that such a $V$ is a principal homogeneous space for its Albanese variety.
So I am really interested in the case where $V$ is affine. Here I am not aware of an analogue of the Albanese variety for linear algebraic groups.
Even if the answer is no in general, I would still be interested in some positive results for special cases, e.g. for reductive groups or semisimple groups.

REPLY [7 votes]: Here is a counterexample with $V$ finite.
Suppose that $k'$ is a separable extension of degree $5$ of $k$, and set $V = \mathop{\rm Spec} k \sqcup \mathop{\rm Spec} k'$. Suppose that $V$ is a homogenous space; then it is a group scheme, since $V(k) \neq \emptyset$.
If $\overline k$ is the separable closure of $k$, then $V_{\overline k}$ is the disjoint union of six copies of $\mathop{\rm Spec} \overline k$, and these copies form a group of order $6$. The Galois group of $k$ acts on $V_{\overline k}$ by permuting the non-identity component transitively. But a group of order $6$ contains elements of order $2$ and $3$, so this is impossible.
I am convinced that one can get geometrically connected examples using forms of $\mathbb G_{\rm a}^n$.