Every automorphism $\varphi \in \mathrm{Aut}(E)$ of an elliptic curve $E$ (with base point $O$ over a field $k$) can be written $\varphi = \tau_Q\phi$ where $\phi \in \mathrm{Aut}(E,O)$ is an isogeny and $\tau_Q$ is a translation by $Q \in E$ (Silverman, The Arithmetic of Elliptic Curves, p. 71), by putting $Q = \varphi(O)$ and $\phi = \tau_{-Q}\varphi$.
Suppose $\varphi$ is an involution, so $\varphi^{-1} = \varphi$, and $\varphi = \tau_Q\phi$ as above. What can we say about $\tau_Q$ and $\phi$?
If $\phi = \mathrm{id}_E$, then $\varphi = \tau_Q$ where $Q$ is a point of order 2.
If $\phi = [-1]$, then $\varphi(P) = Q – P$ for some $Q \in E$.
I think that these are the only possibilities, but I don't see how to prove it.
I'm most interested in the case that $k$ has characteristic zero, in which case $[-1]$ is the only isogeny of order 2, because $\mathrm{Aut}(E,O)$ is cyclic of order 2, 4, or 6 (p. 103). Can we somehow prove that $\phi$ can't have an order greater than 2? Then we would be done.
Using $\varphi^{-1} = \phi^{-1}\tau_{-Q}$, the condition for $\varphi$ to be an involution becomes $\phi^2(P) + \phi(Q) = P – Q$. This might be helpful, but I am not sure as I am quite sleepy.
Thanks in advance for your replies.
Best Answer
Well, $(\tau_Q\phi)^2=\tau_Q\phi\tau_Q\phi$, and since $\phi\tau_Q=\tau_{\phi(Q)}\phi$, we get that $(\tau_Q\phi)^2=\tau_{Q+\phi(Q)}\phi^2=\mbox{id}$. In particular, $\phi(Q)=-Q$ (since $\phi(0)=0$), and so we actually have that $\phi^2=\mbox{id}$. This means that $\phi=\pm\mbox{id}$.
Conclusion: If $\phi=-\mbox{id}$, any $Q$ will do. If $\phi=\mbox{id}$, then $Q$ must be a 2-torsion point.