Is $Aut(K/\mathbb F_q)$ finite?

Question

I do not know whether this result is true or not. I did not find any reference about it. Let $K/\mathbb F_q$ be a function field defined over a finite field. Is $Aut(K/\mathbb F_q)$ finite where $Aut(K/\mathbb F_q)$ denotes the set of automorphisms of $K$ that are identity on $\mathbb F_q$. All I found is about function field defined over an algebraically closed field. So when the genus is $\ge2$ one deduces that $Aut(K/\mathbb F_q)$ is finite. The genus $0$ case is easy. But that says nothing about the genus $1$ case. Any reference or proof would be welcome for the genus $1$ case.

Answer 1

$\newcommand{\End}{\mathrm{End}}$$\newcommand{\Aut}{\mathrm{Aut}}$Let $C/\mathbb{F}_q$ be a genus $1$ curve. Then, the Hasse-Weil bound implies that $C(\mathbb{F}_q)\ne \varnothing$. So, we can turn $C$ into an elliptic curve $E=(C,e)$ (i.e. I'm choosing the point $e\in C(\mathbb{F}_q)$).

Note then that

$$\End(C)=C(\mathbb{F}_q)\rtimes\End(E)$$

where $C(\mathbb{F}_q)$ acts by translations. Indeed, if $f:C\to C$ is any endomorphism then $f(e)\in C(\mathbb{F}_q)$ and thus $t_{-f(e)}\circ f$ takes $e$ to $e$ so is an endomorphism of $E$.

Thus, we see that

$$\Aut(C)=C(\mathbb{F}_q)\rtimes \Aut(E)$$

So your question becomes "Is $\Aut(E)$ finite?".

As is well-known, $\End(E)$ can be on of three things over $\mathbb{F}_q$:

• $\mathbb{Z}$
• An order in an imaginary quadratic extension $F/\mathbb{Q}$
• An order in a (certain) quaternion algebra $D/\mathbb{Q}$.

In the first two cases we see that its units are finite (by Dirichlet's unit theorem). In the last case we see that $\Aut(E)$ is also finite, but this requires more work. In fact, if you work through the details $|\Aut(E)|\mid 24$ (e.g. see Theorem 10.1 of Silverman's book on elliptic curves).

Of course, wheras the result you stated for genus $\geqslant 2$ is not a pecularity of finite fields, the result for genus $1$ is. In general, you have that if $C$ is a genus $1$ curve over $k$ with a $k$-point then $\Aut(C)$ looks like $C(k)\rtimes G$ where $G$ finite. In particular, if $k$ is infinite, then $C(k)$ and thus $\Aut(C)$ are infinite.