# Torsor whose ring of function is a field

Let $G$ be an affine group scheme over $\mathbb{Q}$. Then it is easy to see that if the ring of regular functions $H^0(G,\mathcal{O}_G)$ is a field then $G$ is the trivial group.

Let $P$ be a $G$-torsor (for the etale or fpqc topology). Is it possible for $H^0(P,\mathcal{O}_P)$ to be a field of transcendance degree $>0$?

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Not even a comment or an up vote? Is something wrong with the question? – AFK Aug 12 '12 at 11:06

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I guess by $O(P)$ you mean the regular functions on $P$ ? As $P$ is affine, then surely $P$ is trivial, hence $G$ is also trivial.

Edit As I don't have enough reputation to comment, I have to answer here: I thought your schemes are of finite type over $\mathbb Q$, in which cas, as $P$ is affine, $P=\mathrm{Spec}(K)$ where $K$ is the field $O(P)$, so $P$ is one point, hence $K$ is finite over $\mathbb Q$ and $O(G)$ must also be finite over $\mathbb Q$. As $G$ has a rational point, this forces $G$ to be trivial and $K=\mathbb Q$.

If $G$ is not necessarily of finite type over the base field, I don't know. I have to think about it.

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I don't understand "As $P$ is affine then surely $P$ is trivial". $P(\mathbb{Q})$ may be empty. – AFK Aug 4 '12 at 15:58
Also I slightly changed the question – AFK Aug 4 '12 at 16:01
Even if $P$ is of finite type I think your reasoning is false for the same reason: $P(\mathbb{Q})$ still may be empty. For example, take $P = Spec(K)$ the spectrum of a finite Galois extension and $G$ the corresponding constant group. – AFK Aug 4 '12 at 21:32
@YBL, you are right. – Cantlog Aug 5 '12 at 1:45