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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?2012-08-12

2 Answers 2

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See comments in this MO thread.

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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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    @YBL, you are right.2012-08-05