MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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....)

share|cite|improve this question
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
up vote 4 down vote accepted

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 $[\zeta^\sigma]\circ\sigma$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.