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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?

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    $\begingroup$ 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. $\endgroup$
    – Daniel Loughran
    Commented May 22, 2016 at 16:12
  • $\begingroup$ 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). $\endgroup$
    – Watson
    Commented Oct 31, 2018 at 19:09
  • $\begingroup$ This is also partially proved in universiteitleiden.nl/binaries/content/assets/science/mi/… (Arno Kret, Galois Representations), theorem 8, p. 13-18. $\endgroup$
    – Watson
    Commented Oct 19, 2022 at 18:33

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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.

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    $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Commented May 22, 2016 at 17:14
  • $\begingroup$ 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. $\endgroup$
    – Jeff H
    Commented May 22, 2016 at 17:23
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    $\begingroup$ @WillSawin Maybe it makes sense to define the trace of any linear map on a 0-dimensional vector space to be 0? Then there are no special cases. Ditto for the trace of a 0-by-0 square matrix! $\endgroup$
    – Joe Silverman
    Commented May 22, 2016 at 17:30
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    $\begingroup$ The trace is well-defined (at least) for endomorphisms of finite-type projective modules over commutative rings (e.g, Bourbaki, Algebra, Ch. 2, \S 4, no 3), so the trace interpretation always works. In fact, so does the "sum of eigenvalues" interpretation, since an empty sum is zero. $\endgroup$
    – Denis Chaperon de Lauzières
    Commented May 22, 2016 at 18:45
  • $\begingroup$ @JoeSilverman Yes, I agree. I was just trying to answer Jeff's question about the eigenvalue. $\endgroup$
    – Will Sawin
    Commented May 22, 2016 at 20:09

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