# Frobenius at ramified primes

Let $E$ be an elliptic curve defined over $\mathbf{Q}$, fix an odd prime $p>3$, let $T_p$ denote the $p$-adic Tate module of $E$, and let $V_p = T_p \otimes \mathbf{Q}_p$.

If the action of $G_\mathbf{Q}$ is unramified at $\ell$, it is known that the characteristic polynomial of Frobenius, $\mathrm{Frob}_\ell$, is

$x^2 - a_\ell(E)x + \ell$.

Now let $\ell$ be a ramified prime for this representation, and let $(V_p)_{I_\ell}$ be the maximal quotient of $V_p$ on which the $\ell$-inertia group $I_\ell$ acts trivially (so it is one-dimensional). Then it makes sense to ask the following question.

Is it possible to describe the eigenvalue of Frob$_\ell$ on $(V_p)_{I_\ell}$ in terms of explicit data from the elliptic curve, as in the unramified case?

• Yes: see this wikipedia page en.wikipedia.org/wiki/Hasse%E2%80%93Weil_zeta_function. Note that the quotient is only one-dimensional if the prime exactly divides the conductor; otherwise it is trivial. – Daniel Loughran May 22 '16 at 16:12
• See also Notes on the Parity Conjecture by Tim Dokchitser in Elliptic Curves, Hilbert Modular Forms and Galois Deformations, after remark 3.6. (an earlier version is available here, with a few typos). – Watson Oct 31 '18 at 19:09

Are you looking for the following sort of answer? $a_\ell(E)=1$ if $E$ has split multiplicative reduction, $a_\ell(E)=-1$ if $E$ has non-split multiplicative reduction, and $a_\ell(E)=0$ if $E$ has additive reduction. This gives the "right" local factors for the $L$-series, i.e., $L_p=1\pm p^{-s}$ for multiplicative reduction and $L_p=1$ for additive reduction.
• and $a_\ell(E)$ is also the trace of $\operatorname{Frob}_\ell$ on $(V_p)_{I_\ell}$, so it is the eigenvalue in the multiplicative case when that module is one-dimensional. However, that space is zero-dimensional in the additive case, so there is no eigenvalue. – Will Sawin May 22 '16 at 17:14
• Great, this is perfect. And thank you for clarifying, @WillSawin. I was primarily curious about whether $a_\ell(E)$ was still the trace in this case. – Jeff H May 22 '16 at 17:23