# isogenies between elliptic curves with multiplicative reduction

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.

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 $$n$$.
• 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 $- \phi$. – Joey van Langen Mar 31 '20 at 8:43
• You are absolutely right. This is up to automorphisms that fix the kernel. Luckily our curves only have $\pm 1$ as automorphisms. – Chris Wuthrich Mar 31 '20 at 10:22