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I have question regarding exercise 3.12 of J. Silverman "The arithmetic of Elliptic curves". It states the following:

Let $m \geq 2$ be an integer, prime to $\text{char}(K) > 0$. Prove that the natural map $\text{Aut}(E) \longrightarrow \text{Aut}(E[m])$ is injective except for $m=2$, where the kernel is $[\pm1]$.

I want to prove this without too much prior knowledge of the structure of $\text{Aut}(E)$ (to be more precise, without the use of theorem III.10.1 in Silverman).

($E[m] =\{ P \in E : [m]P=O\}$).

Anyone got any ideas?

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Suppose that $\sigma$ is an automorphism of $E$ whose image is trivial in $Aut(E[m])$.

The assumption about the characteristic implies that the endomorphism $[m] \in E$, defined by $[m]P = P$, is separable: this is precisely corollary III.5.4 in Silverman's Arithmetic of Elliptic Curves.

Corollary III.4.11 in Silverman says that if $\phi:E_1 \rightarrow E_2$ and $\psi:E_1 \rightarrow E_3$ are non-constant isogenies with $\phi$ separable and $ker\phi \subset ker\psi$, then there exists a unique isogeny $\lambda:E_2 \rightarrow E_3$ such that $\psi = \lambda \circ \phi$. By the assumption on $\sigma$, one has $ker[m] \subset ker(Id-\sigma)$, so since $[m]$ is separable we obtain an isogeny $\lambda$ with $\lambda \circ [m] = Id - \sigma$.

Taking duals of both sides of the equation, we obtain $\hat{[m]} \circ \hat{\lambda} = \hat{Id} - \hat{\sigma}$. Since $\sigma$ is an isomorphism, its dual is simply its inverse, so recalling that $\hat{[m]} = [m]$ this equation simplifies to $\hat{\lambda} \circ [m] = Id - \sigma^{-1}$. Composing the two equations gives $[2]- \sigma - \sigma^{-1} = (1-\sigma) \circ (1-\sigma^{-1}) = [m^2] \circ \lambda \circ \hat{\lambda} = [deg\lambda m^2]$, hence $\sigma + \sigma^{-1} = [2-deg\lambda m^2]$.

Now put $\tau = 2\sigma - [2-deg\lambda m^2]$. Obviously $deg\tau \geq 0$. On the other hand we have $[deg\tau] = \tau \circ \hat{\tau}$, which is equal to $4\sigma \circ \sigma^{-1} - 2(\sigma + \sigma^{-1})[2-deg\lambda m^2] + [2-deg\lambda m^2]^2 = [4-(2-deg\lambda m^2)^2].$

It follows that $|2-deg\lambda m^2| \leq 2$. Since $deg\lambda$ is a non-negative integer, it is easy to see there are two cases:

  1. $deg\lambda = 0$; then $\lambda$ must be $0$ so $Id = \sigma$, hence $\sigma$ is indeed trivial.

  2. $deg\lambda = 1, m =2$. But in this case one sees that $\sigma^2 + 2\sigma + Id = 0$, hence $(\sigma + Id)^2 = 0$. Since $End(E)$ is a domain it follows that $\sigma = [-1]$ like we wanted to show. (Notice that we can always use the two relations of $\sigma,\sigma^{-1}$ to get a quadratic equation, but without knowing the structure of $End(E)$ this is only useful if the discriminant is zero, which is precisely the current case).

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