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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.

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up vote 6 down vote accepted

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.

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Thanks. Why does $\phi^2 = \mathrm{id}$ imply $\phi = \pm \mathrm{id}$? Is it because those are the only elements of order 1 and 2 in $\mathrm{Aut}(E,O)$? (Why is this true in arbitrary characteristic?) Or is there a simpler reason? –  Ricardo Buring Aug 16 '13 at 12:37
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Well, at least over $\mathbb{C}$, the only automorphisms that can occur are multiplication by a complex number. This means that $\phi$ is induced by multiplication by some complex number $\lambda$ such that $\lambda^2=1$, and so $\lambda=\pm1$. –  Robert Auffarth Aug 16 '13 at 12:51
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For characteristic 2 and $j = 0$, we have $\mathrm{Aut}(E,O) \cong \mathrm{SL}_2(\mathbb{F}_3)$ (SL(2,3)). For characteristic 3 and $j=0$, we have $\mathrm{Aut}(E,O) \cong \mathbb{Z}/4\mathbb{Z} \rtimes \mathbb{Z}/3\mathbb{Z} \cong \mathrm{Dic}_{12}$ (Dic12). These also have only one element of order 2, so it's true in general. –  Ricardo Buring Aug 16 '13 at 17:42
    
Thanks for your comment, I forgot how the situation was in positive characteristic. –  Robert Auffarth Aug 16 '13 at 23:52

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