Let $C$ be the projective smooth genus 2 curve defined by $y^2=x^5-x$ over $\mathbb F_5.$ What is the order of its automorphism group (automorphisms over $\mathbb F_5$)?
I have seen different answers.
In Hartshorne's Algebraic Geometry, p. 306, the answer is $2p(p^2-1)=240.$ In INFORMATION
Volume 8, Number 6, pp. 837-844, Isomorphism classes of genus-2 hyperelliptic curves over finite fields $\mathbb F_{5^m},$ by L. Hernández Encinas and J. Muñoz Masqué, theorem 2, the answer is $|A_{4221}|=20$ (using notations there). A professor (let me not to mention the name for now) told me the order is 120.
Maybe I misunderstand some of the references above?
Best Answer
According to Magma it is 120, and it is an extension of $A_5$ by $C_2$ (A:=AutomorphismGroup(HyperellipticCurve(Polynomial(GF(5),[0,-1,0,0,0,1])))), and over $F_{25}$ or over $\bar F_5$ it is 240.
Edit: Hartshorne works over an algebraically closed field, so Exc. 2.5 on p.305 proves that over $\bar F_5$ the automorphism group has order 240. Explicitly, it is generated by
$\alpha: x\mapsto x+1, y\mapsto y$ of order 5,
$\beta: x\mapsto \frac{1}{x+1}, y\mapsto \frac{y}{(1+x)^3}$ of order 6,
$\gamma: x\mapsto 2x, y\mapsto \sqrt{2}y$ of order 8.
Actually, it is clear that the group they generate has order 240 and not less, because $\beta^3$ is not the hyperelliptic involution and $\gamma^4$ is. On the other hand, as Dan explains, you cannot get more than a double cover of $PGL(2,F_5)$, so this is the whole group. Over $F_5$ however, the automorphism group is generated by $\alpha, \beta$ and $\gamma^2$, and it has order 120.