# Automorphism groups of Elliptic curves as Galois module

Let $E/k$ be an elliptic curve over a field of characteristic $\neq$ 2, 3. Then we have an isomorphism $[ \ \ ] :\mu_n \rightarrow\mathrm{Aut}_{\overline{k}}(E)$, $[ \zeta ] : (x,y) \rightarrow (\zeta^2x, \zeta^3y)$, here $n=2, 4,6$, depending on the $j$-invariant $j(E)$. See Corollary 10.2 on Ch3 in "The arithmetic of Elliptic Curves" by Silverman. There it mentioned that this isomorphism commutes with the Galois action, but I am confused. For example, let $\sigma \in G=\mathrm{Gal}(\overline{k}/k)$, then $[\zeta^\sigma] : (x,y) \rightarrow ( (\zeta^\sigma)^2x, (\zeta^\sigma)^3y)$, but $\sigma( [\zeta])$ is $(x,y) \rightarrow (\zeta^2x, \zeta^3y) \rightarrow ( (\zeta^2x)^\sigma, (\zeta^3y)^\sigma)$, hence they are different. Am I thinking something in the wrong way? ( Sorry about such level of question....)

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You have to consider the Galois action on the points of $E$ too! The correct formula is $\sigma([\zeta](x,y)) = [\sigma(\zeta)]\sigma(x,y).$ – Emerton Aug 10 '11 at 22:54

It's as Matt indicates. You might find it easier to look at the commutative diagram, which is what it means for two maps to commute. Thus \begin{aligned} E(\bar{k}) &\qquad\xrightarrow{[\zeta]} & E(\bar{k}) \cr \downarrow\sigma && \downarrow\sigma \cr E(\bar{k}) &\qquad\xrightarrow{[\zeta^\sigma]} & E(\bar{k}) \cr \end{aligned} The map $\sigma([\zeta])$ is the composition $\sigma\circ[\zeta]$, which from the diagram is equal to $[\zeta^\sigma]\circ\sigma$.
And in cooordinates, with $P=(x,y)\in E$, this reads $([\zeta](x,y))^\sigma=(\zeta^2x,\zeta^3y)^\sigma=((\zeta^2x)^\sigma,(\zeta^3y)^‌​\sigma)=((\zeta^\sigma)^2x^\sigma,(\zeta^\sigma)^3y^\sigma)=[\zeta^\sigma](x,y)^\‌​sigma$. – Álvaro Lozano-Robledo Aug 11 '11 at 15:06