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Let $X$ be a curve over $\overline{\mathbf{Q}}$.

I can prove that $X$ can be defined over some number field. (Take two equations defining $X$ in $\mathbf{P}^3$ and consider the number field containing the coefficients of this equation.)

Suppose that $X$ can be defined over a number field $K$.

It seems to be a basic fact that there are infinitely many non-isomorphic curves $Y$ over $K$, called twists I believe, such that $Y$ is isomorphic to $X$ over $\overline{\mathbf{Q}}$. Why is this?

Now, this actually bothers me a bit. Because I was hoping to have only a finite number of such "twists".

How can I guarantee that there are only a finite number of twists of a given curve $X/K$?

Let me be more precise.

For example, if I also want all the twists of $X$ over $K$ to have semi-stable reduction over $K$, is the number then finite?

Another example, if I also want all the twists of $X$ over $K$ to have good reduction over $K$, is the number then finite?

I might be asking for something that's impossible.

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Let $A$ be the automorphism group of $X$ over $\overline{K}$. Note that $\mathrm{Gal}(\overline{K}/K)$ acts on $A$. The twists of $X$ are classified by the cohomology group pointed set $H^1(\mathrm{Gal}(\overline{K}/K), A)$. If the action of the Galois group on $A$ is trivial (i.e. all automorphisms are defined over $K$), then this is $\mathrm{Hom}(\mathrm{Gal}(\overline{K}/K), A)/\sim$ where $\sim$ is the equivalence relation of conjugation by elements of $A$.

If the automorphism group of $X$ is trivial, this will force the $H^1$ to likewise be trivial. I don't know any other condition that forces $H^1$ to be finite.

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Your $H^1$ is generally a only a set, no? ($A$ is usually not abelian) – Mariano Suárez-Alvarez Oct 27 '11 at 19:37
Right. Plus some more characters. (Perhaps what you are asking is how $A$ acts on $\mathrm{Hom}(\mathrm{Gal}, A)$ by conjugation? Given $\psi : \mathrm{Gal} \to A$, we can conjugate it to $g \mapsto a \psi(g) a^{-1}$.) – David Speyer Oct 27 '11 at 19:41
I was only very midly objecting to $H^1$ being called a cohomology group :) – Mariano Suárez-Alvarez Oct 27 '11 at 20:27
Fair enough, fixed. – David Speyer Oct 27 '11 at 20:44

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