Edit: the original question was imprecise, I'm sorry about that. I hope this is better.


Let $k$ be a field, $S/k$ any scheme, $G/k$ be an algebraic group and $X$ an $S$-torsor under $G$ (so $G$ acts simply transitively on the fibers of $G \to S$). Under what (sufficient/necessary/whatever) conditions on the schemes $X$, $G$, $S$, ... do we get an isomorphism $\overline{X} \cong \overline{G} \times\_{\overline{k}} \overline{S}$, where the bar denotes base change to the algebraic closure of $\overline{k}$? For example, do we always have this isomorphism if $\text{Pic}\ \overline{S}$ is trivial?



(corrected) When does a torsor trivialize over the algebraic closure?
-The following is a general condition for trivialization:


Let $\pi:X\to S$ be the structure map and let $T\_G$ be the sheaf of trivialization maps where for $U\subset X$ open we have $T\_G(U) := \lbrace \psi: \pi^{-1}(U)\to U\times G, \mbox{an isomorphism such that } \pi = p\_1\circ\psi\rbrace$ (you can replace $G$ with any scheme you want) The collection of such trivialization is an $\underline{\mathrm{Aut}}(G)$ torsor. If there exists some $\lbrace U\_i\rbrace\_{i\in I}$ an open cover of $S$ with $t\_i\in T\_G(U\_i)$ let's define $t\_{ij}=t\_i\circ t\_j^{-1}\in \underline{\mathrm{Aut}}(G)(U\_{ij})$ so that we get a (well-defined) cohomology class
$$\eta\in \check{H}^1(S,\underline{\mathrm{Aut}}(G))$$
One can show that $\eta$ is trivial if and only if $X\cong S\times G$.


-The statement as you have it will never work since the dimensions don't match if $\dim\_{\bar{k}}(S\_{\bar k})>0$ since the relative dimension of a torsor and its group scheme are the same. Am I missing something here?


