Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.