Is a torsor over a variety a variety?

Question

Let $X$ be an algebraic variety over some field $k$ of characteristic 0. Let $g : Y \to X$ be a $X$-torsor under some linear algebraic $k$-group $G$.

Is $Y$ also an algebraic variety over $k$?

Answer 1

I'm not entirely sure what you mean by 'is a variety', but I would think the below result is somewhat close:

Fix a scheme $X$, and a group scheme $G/X$. A pair $(\pi:P\to X,\rho)$, where $\rho:G\times_X P\to P$ is claled a principal $G$-bundle if

1. $\pi$ is faithfully flat and finitely presented.
2. $\rho$ is an action of $G$ on $P$ over $X$ (i.e. the obvious diagrams must commute, saying that the identity section of $G/X$ acts trivially, and that the action is associative).
3. The map $(\rho,\text{pr}_2):G\times_X P\to P\times_X P$ is an isomorphism.

So, this is precisely what you would define if you were analogizing the definition of principal $G$-bundle from topology.

We then have the following theorem:

Theorem: If $G/X$ is affine, then there is an equivalence of categories $$\left\{G-\text{torsors on }X\right\}\leftrightarrow\left\{\text{principal }G-\text{bundles on }X\right\}$$

This follows almost immediately from the fact that $\mathsf{fppf}$-locally on $X$, any torsor $T$ is isomorphic to a pullback of $G$, and the fact that this implies (by standard descent arguments) that $T$ itself must be represented by a relatively affine scheme $T/X$.

In particular, the above tells you that if $X=\text{Spec}(k)$, and $G$ is a linear algebraic group, then $G$-torsors are just principal $G$-bundles which 'are schemes', in the finest sense of the word.