Automorphism group of elliptic curves in characteristic 3

Question

I would show that the automorphism group of an elliptic curve E in characteristic $3$ and with J-invariant $0$ is isomorphic to the semidirect product of $Z/4Z$ and $Z/3Z$. The curve has equation of the form $Y^2Z=X^3+a_4XZ^2+a_6Z^3$ The substitutions preserving this form are: $$X=u^2X+rZ\\Y=u^3Y\\Z=Z$$ Then automorphisms of E have:$$u^4=1$$ and $$r^3+a_4r+a_6(1-u^2)=0$$ This is all I'm able to say about $u$ and $r$. How can I proceed?

Answer 1

This is a bit of a bash, but let's display how far you can get by just following your nose. First, we assume we're working over an algebraically closed field of characteristic $3$, as otherwise your result may fail (you at least need a square root of $-1$ and all the solutions of $r^3+a_4r+a_6(1\pm 1)$). Now to determine the composition law on the pairs $(u,r)$. Suppose $(u,r)$ represents one automorphism and $(v,s)$ another. Let's look at what $(u,r)\circ (v,s)$ is:

$$X= u^2(v^2X+sZ)+rZ=(uv)^2X+(r+u^2s)Z$$ $$Y= u^3(v^3Y)=(uv)^3Y$$ $$Z=Z$$

So $(u,r)\circ (v,s)=(uv,r+u^2s)$, and we see after a bit of algebra that if restrictions $u^4=v^4=1$ and $r^3+a_4r+a_6(1-u^2)=s^3+a_4s+a_6(1-v^2)=0$ are satisfied, the necessary restrictions for $(uv,r+u^2s)$ are too. So these form a group, and we can pick out a couple relevant elements:

• We see by direct computation that the automorphism $\sigma=(i,r_2)$ is of order $4$, where $i$ is a nontrivial fourth root of unity and $r_2$ is a root of $r^3+a_4r+2a_6=0$.
• Pick a nonzero root $r_1$ of $r^3+a_4r$. We see by direct computation that the automorphism $\tau=(1,r_1)$ is of order $3$.
• Now we check the product of these two elements. $(i,r_2)\circ(1,r_1)=(i,r_2-r_1)=(1,-r_1)\circ(i,r_2)$, so $\sigma\tau=\tau^{-1}\sigma$

So the group generated by these symmetries is $\langle \sigma,\tau\mid \sigma^4=\tau^3=\sigma\tau\sigma^{-1}\tau=1\rangle$, which is isomorphic to the nontrivial semidirect product $\Bbb Z/3\rtimes\Bbb Z/4$ by computing the presentation of that group.

As there are at most 12 pairs $(u,v)$ which solve the required equations, the order of our group of symmetries is at most 12. On the other hand, we found a subgroup of order 12 generated by $\sigma$ and $\tau$, so that's the whole group of symmetries.