TITLE: Is every homogeneous G-variety of the form G/H?
QUESTION [6 upvotes]: Let $G$ be an algebraic group over an algebraically closed field $k$. Then G/H is a quasi-projective homogeneous G-variety for any closed subgroup $H$. Now, several times I have seen something like "Let $X$ be a homogeneous $G$-variety, i.e. $X = G/H$ for a closed subgroup $H$ of $G$" and I wonder if this "i.e." is correct. This would imply that any homogeneous $G$-variety is already quasi-projective. I think this is true, when $\mathrm{char}(k) = 0$, because then the canonical abstract isomorphism $\pi:X \rightarrow G/G _x$ is separable and thus an isomorphism of varieties for any $x \in X$ (is this correct?). But what about $\mathrm{char}(k) > 0$? Are there counter-examples or is any (quasi-projective) homogeneous $G$-variety up to isomorphism of the form $G/H$?

REPLY [3 votes]: An aside: BCnrd's argument above shows that, over any perfect field $k$, $X$ is isomorphic to $G/H$ if and only if $X$ has a $k$-point. Otherwise, consider, e.g., conics with no point: they are homogeneous under a twist of $G=SO_3$ but are not $G/H$. I don't know what happens for more general base schemes.