Let $ K $ be a $ p $-adic field. Suppose we have an isogeny of elliptic curves $ \phi : E \to E' $ defined over $ K $, where $ E $ and $ E' $ both have multiplicative reduction.

1) Is there anything we can say about the structure of the induced map on the Tate modules $ V_l(E) \to V_l(E') $? Mostly I'm interested in the eigenvalues.

I should note that for question 1) it is actually easy to determine the determinant of this linear map using the Weil pairing. I'm only interested in additional results, such as a way to compute the trace.

Note that by possibly enlarging the field K we may assume the multiplicative reduction for both curves is split, hence $ E $ and $ E' $ are isomorphic to Tate curves $ E_q $ and $ E_{q'} $ with $ q, q' \in K^{*} $ of positive valuation. Note that for any finite extension $ L / K $ we have that $ E_q( L ) \cong L^{*} / q^{\mathbb{Z}} $. For a suitably large choice of $ L $ the $ m $-torsion points of the latter have a basis $ \{ \zeta_m , q^{1/m} \} $, where $ \zeta_m $ is an $ m $-th root of unity.

2) Can we say anything about the structure of the induced map $ \phi : E[m] \to E'[m] $ with respect to the bases given by $ \{ \zeta_m, q^{1/m} \} $ and $ \{ \zeta_m, (q')^{1/m} \} $?

Although I'm interested in general results, I would be happy getting a result in case E' is a Galois conjugate of E, so in case q' is a Galois conjugate of q.


1 Answer 1


I believe this is the answer in the split case: Let $E$ be the Tate curve with parameter $q$. Let $n>1$. We look for isogenies with cyclic kernel of order $n$. We may suppose that $n$ is prime.

First, there is the isogeny to the Tate curve $E'$ with parameter $q' = q^n$ and the map is induced from $K\to K$ sending $x$ to $x^n$. With respect to the basis of $\ell^{n}$ torsion where the first element is an $\ell^n$-th root of unity $\zeta_{\ell^n}$ and the second element is a choice of an $\ell^n$-th root of $q$ (and correpondingly of $q'$), the matrix for this isogeny on $V_{\ell}(E) \to V_{\ell}(E')$ is diagonal with entries $n$, $1$.

All other cyclic isogenies of degree $n$ leaving $E$ have the kernel generated by an $n$-th root $q'$ of $q$ (and the isogeny is only defined over $K$ if this $q'$ belongs to it). The corresponding map $K/q^{\mathbb{Z}}\to K/(q')^{\mathbb{Z}}$ is induced by the identity map. This time the matrix is also diagonal by with diagonal entries first $1$ then $n$.

  • $\begingroup$ You seem to use that every isogeny is uniquely determined by its kernel, but I can multiply any isogeny with a degree 1 endomorphism to get an isogeny with the same kernel, but maybe different eigenvalues. For example replace $ \phi $ by $ - \phi $. $\endgroup$ Mar 31, 2020 at 8:43
  • $\begingroup$ You are absolutely right. This is up to automorphisms that fix the kernel. Luckily our curves only have $\pm 1$ as automorphisms. $\endgroup$ Mar 31, 2020 at 10:22

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