# G-torsor whose ring of regular functions is a field.

I already asked this question on stackexchange but didn't get any answer. Maybe it is better suited for mathoverflow.

Let $G$ be an affine group scheme (not necessarily of finite type) over $\mathbb{Q}$. And let $P$ be a $G$-torsor (for the etale or fpqc topology) such that $H^0(P,\mathcal{O}_P)$ is a field.

Question: Is it possible for $H^0(P,\mathcal{O}_P)$ to be a field of transcendance degree $>0$ over $\mathbb{Q}$?

If $P$ is trivial torsor, then $H^0(P,\mathcal{O}_P) \simeq H^0(G,\mathcal{O}_G) \to \mathbb{Q}$ is an isomorphism.

If $P = Spec(K)$ is a finite Galois extension, then it is a torsor under $G = Spec(\mathbb{Q}^{Gal(K/\mathbb{Q})})$ i.e. $Gal(K/\mathbb{Q})$ seen as a $\mathbb{Q}$-group scheme.

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No, over any field $k$. The torsor $P$ is affine, and $k[G]$ is exhausted by finitely generated Hopf subalgebras. An injection between Hopf $k$-algebras is faithfully flat, so we get a decreasing system $\{N_i\}$ of closed normal subgroups of $G$ such that the fpqc sheaf $G_i = G/N_i$ is affine of finite type. The quotient sheaf $P_i = P/N_i$ is a $G_i$-torsor, affine of finite type with $P \rightarrow P_i$ faithfully flat and $k[P_i]$'s exhausting $k[P]$. But $P = {\rm{Spec}}(F)$ for a field $F$, so $P_i = {\rm{Spec}}(F_i)$ for fields $F_i$, and $[F_i:k] < \infty$ (!), so $F/k$ is algebraic. – user22479 Aug 13 '12 at 0:00
Looks great. Thanks. – AFK Aug 13 '12 at 23:25