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Let $\ell$ be a prime number, denote by $K_\ell$ the maximal algebraic extension of $\Bbb{Q}$ ramified only at $\ell$. Let $f = \sum a_n q^n$ be a Hecke eigenform of level $1$ with integer coefficients. By a theorem of Serre and Deligne, there is a continuous homomorphism $$ \rho_\ell: \mathrm{Gal}(K_\ell / \Bbb{Q}) \to GL_2(\Bbb{Z}_\ell) $$ such that $\rho_\ell(\mathrm{Frob}_p)$ has characteristic polynomial $$ x^2 - a_p x + p^{k-1} $$ for each $p \neq \ell$. This induces the mod $\ell$ representation $\tilde{\rho}_\ell: \mathrm{Gal}(K_\ell / \Bbb{Q}) \to GL_2(\Bbb{F}_\ell)$.

Question: How to find the conductor of $\tilde{\rho}_\ell$?

We know that the conductor is a power of $\ell$ since $K_\ell$ is ramified only at $\ell$. But I don't know if we have control on the exponent.

We've looked at section 5.5 of the notes by Bryden Cais on Serre's Conjectures (, which seems to be related to my question, but I'm not familiar with the topic enough to understand the relation.

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Did you mean to ask about the conductor of the mod $\ell$ representation, instead of $\ell$--adic? – Jesse Silliman Jul 18 '13 at 13:14
Yes, I mean the mod $\ell$ representation. Thanks. – Ping Ngai Chung Jul 18 '13 at 13:28
The $\ell$-part of the conductor of the mod $\ell$-representation comes out of the recipe for the weight in Serre's conjecture. A special case is that the representation is unramified at $\ell$ if and only if the weight of the representation is $1$. This should be in Cais's notes, or in Edixhoven's paper Inv. Math. 109 (1992) 563-594. – Felipe Voloch Jul 18 '13 at 14:54

It is difficult to even define the conductor at $\ell$. The problem is that rho is infinitely wildly ramified at $\ell$ (it must be, since its determinant is a power of the $\ell$-adic cyclotomic character); so the naive definition of the conductor would be infinity.

The morally right way to define the conductor at p of a p-adic Galois representation is to use Fontaine's p-adic Hodge theory. In your case the representation is crystalline at $\ell$, so the conductor is 1, matching the level of the modular form.

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Oh I'm sorry. I mean the mod $\ell$ representation (edited above) instead of the $\ell$-adic one. Sorry for the typo. – Ping Ngai Chung Jul 18 '13 at 13:30

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