0
$\begingroup$

Let $E$ be an elliptic curve defined over an algebraic number field $K$ and $X$ be a cyclic subgroup of $E(K)$ of order $p^n$, then we have the isogeny $E\rightarrow E/X$ and the dual isogeny $ E/X\rightarrow E$.

I want to ask why the kernel of the dual isogeny $ E/X\rightarrow E$ is also a cyclic group of order $p^n$. I have read most of Silverman's book(arithmetic of elliptic curves), any reference or proof is appreciated. Thanks in advance.

$\endgroup$
1
  • 2
    $\begingroup$ Even better. The Weil pairing shows that the kernel of the dual isogeny is a Galois module isomorphic to $\mu_{p^n}$. $\endgroup$ – Chris Wuthrich Dec 16 '20 at 10:41
1
$\begingroup$

Let $\phi$ be the map $$\phi:E \rightarrow E/X$$ then we have to use $\phi\circ \hat{\phi} = [p^{n}]$, and so there's an exact sequence given by $$0\rightarrow X \rightarrow E[p^{n}] \rightarrow \text{ker}\hat{\phi} \rightarrow 0 .$$ This gives an exact sequence $$0\rightarrow \mathbb{Z}_{p^n} \rightarrow \mathbb{Z}_{p^{n}}^2\rightarrow \text{ker}\hat{\phi} \rightarrow 0. $$

Now just pick a generator of $\mathbb{Z}_{p^n}^2$ that isn't in $\mathbb{Z}_{p^n}$ and its image should generate $\text{ker}\hat{\phi}$.

$\endgroup$
1
  • $\begingroup$ Thanks, the first exact sequence should be $0\rightarrow\ker\phi$... $\endgroup$ – user169802 Dec 15 '20 at 23:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy