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Let $X = \text{Spec }B$ be a smooth affine curve over $\mathbb{C}$ with finitely many automorphisms. Ie, $B$ is a smooth $\mathbb{C}$-algebra with finitely many $\mathbb{C}$-algebra automorphisms.

Let $A\subset B$ be a $\overline{\mathbb{Q}}$-subalgebra such that the inclusion $A\hookrightarrow B$ induces an isomorphism $A\otimes_{\overline{\mathbb{Q}}}\mathbb{C}\cong B$. Ie, $A$ generates $B$ as a $\mathbb{C}$-algebra.

Given that such an $A$ exists, is it unique? (I don't mean unique up to isomorphisms - I literally mean that it is the only $\overline{\mathbb{Q}}$-subalgebra of $B$ satisfying the above properties.)

Note that for $X = \text{Spec }\mathbb{C}[t]$, the statement is false - there are infinitely many $\overline{\mathbb{Q}}$-subalgebras which generate $\mathbb{C}[t]$. For example, you can take $\overline{\mathbb{Q}}[t+n\pi]$ for $n\in\mathbb{Z}$. However, this doesn't satisfy the hypotheses since $\mathbb{C}[t]$ has the whole affine group as automorphisms.

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    $\begingroup$ Uniqueness up to isomorphism follows from "spreading-out and specialization", so it remains to show if $K/k$ is an extension of algebraically closed fields, $X$ is smooth proper connected curve over $k$, and $U\subset X$ is a non-empty affine open with ${\rm{Aut}}_k(U)$ finite, then ${\rm{Aut}}_k(U)\to {\rm{Aut}}_K(U_K)$ is bijective. For $D=X-U\hookrightarrow X$ with reduced structure, you're studying the preimage $G$ of ${\rm{Aut}}_{D/k}$ under the map ${\rm{Aut}}_{X/k}\to {\rm{Hom}}(D,X)$ of locally finite type $k$-schemes. Since $G(k)$ is finite, so $G$ is $k$-finite, clearly $G(K)=G(k)$. $\endgroup$
    – nfdc23
    Commented Oct 14, 2017 at 13:09
  • $\begingroup$ @nfdc23 You seem to be proving that every automorphism of $B$ as a $\mathbb{C}$ algebra is the base change of an automorphism of $A$ as a $\overline{\mathbb{Q}}$-algebra. I agree with this statement, though I don't see how it relates to my question... $\endgroup$
    – Will Chen
    Commented Oct 16, 2017 at 17:45
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    $\begingroup$ It completely solves your question in the affirmative in view of my first sentence. Say $A, A'$ are coordinate rings of smooth connected affine curves over $k$ and we're given isomorphisms $f:A\otimes_k K \simeq B$ and $f':A'\otimes_k K \simeq B$ for some $K$-algebra $B$. We want the $K$-algebra isomorphism $f^{-1} \circ f'$ to carry $A'$ to $A$. But abstractly there exists $i:A \simeq A'$ as $k$-algebras (by my first sentence), so harmlessly composing with $i_K$ reduces you to exactly what I do in my first comment. OK? $\endgroup$
    – nfdc23
    Commented Oct 16, 2017 at 21:48
  • $\begingroup$ @nfdc23 Ah, okay I was being dumb. Sorry. $\endgroup$
    – Will Chen
    Commented Oct 16, 2017 at 22:14

1 Answer 1

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This question has already been answered in the comments, but I'm in a cab and have nothing better to do but give a different answer. Namely, if we could determine a priori an infinite set of points $x_n$ which have to be defined over the smaller field, then we can characterize $A$ to be those regular functions on $X$ whose value at $x_n$ is in the smaller field for all $n$. Hence uniqueness of $A$.

For example, if the genus of a smooth projective model of $X$ is $2$ or more, then you can use Weierstrass points and higher Weierstrass points. There's​ an infinite number of these. Ok, got to my destination! Bye

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