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Let $f = \sum_n a_n q^n$ be a cuspidal newform of some weight and level. Here I want to view the $a_n$ of $p$-adic numbers (by embedding $\overline{{\bf Q}}_p$ in ${\bf C}$ in some way).

Let $k_f$ denote the finite field generated by the reduction mod $p$ of all of the $a_q$ as $q$ varies over all primes. Let $k'_f$ denote the same field except that we omit a single $a_q$ for one prime $q$.

My question: can $k'_f$ be strictly smaller than $k_f$?

(Compare to the characteristic 0 analogue in:

Field generated by the Fourier coefficients of a modular form


A sentence in Shimura's "On The Periods of Modular Forms"

where strong multiplicity one guarantees that the field is unchanged if finitely many $a_q$ are deleted.)

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@Dror: could you elaborate? I don't see the argument. – MF1 May 25 '13 at 18:35
@Dror: I think one has to be careful about which rings (as opposed to fields) are being generated by these Fourier coefficients in the char 0 case. Indeed, I'm worried about the possibility where say all of the $a_q$ generate the full ring of integers, but if one removes a single $a_q$ then you get a smaller order which has a smaller residue field. – MF1 May 26 '13 at 14:33
Looking at the mod p representation attached by Shimura, Deligne, Deligne-Serre, corresponding to the inclusion $K_f\rightarrow\bar{\mathbb{Q}}_p$, we see that for any $q\not | pN$, we have, from the Eichler-Shimura relation, that $a_q\ \text{mod p}$ depends only on the conjugacy class of Frobenius of $q$. By Chebotarev's theorem, there are infinitely many $q$'s in every such conjugacy class. Hence, if you delete such $q$'s, the residue field doesn't change. I'm not sure what happens at the ramified primes of the finite representation. – Dror Speiser May 26 '13 at 15:43
@Dror: I agree -- dropping a single $a_q$ for $q$ an unramified prime for the residual representation won't change the residue field. But I don't know how to deal with ramified primes -- especially the prime $p$. – MF1 May 26 '13 at 21:13
The unique (up to conjugation) normalized eigenform in $S_2(\Gamma_0(23))$ has coefficients in $\mathbf{Q}(\sqrt{5})$, and the corresponding mod-$2$ representation lands in $S_3$ (the representation comes from the Hilbert class field of $\mathbf{Q}(\sqrt{-23})$). It follows that all the $a_p$ with $p$ prime lie in $\mathbf{Z}[\sqrt{5}]$ and their mod $2$ reduction lands in $\mathbf{F}_2$. But $a_2 = (-1 + \sqrt{5})/2$, so $a_2 \mod 2$ generates $\mathbf{F}_4$. – Orac Jun 3 '13 at 4:07

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