TITLE: Twists of curves over number fields
QUESTION [7 upvotes]: 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.

REPLY [6 votes]: 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.