Let $f\in S_k(\Gamma_1(N))$ be an eigenform, and $K_f$ be its number field, which is of finite degree over $\mathbb{Q}$. Consider the following statements.

1, $[K_f:\mathbb{Q}]=\#\{$Galois conjugates of $f\}$.

2, Any $n$-th coefficients of $f$ having degree $[K_f:\mathbb{Q}]$ will generate $K_f$, so could this $n$ always be $n=2$?

I guess statement 1 is true, but I have problem in proving it. I guess statement 2 is false, but I need a counterexample.

  • $\begingroup$ Have you tried looking at some tables, e.g. those on lmfdb.org? $\endgroup$ Oct 19 '15 at 6:36
  • $\begingroup$ Dear David, they only give examples of weight $\leq$ 36. And I used magma to check, but all examples the online calculator can handle are not counterexamples. $\endgroup$
    – user42690
    Oct 19 '15 at 12:51
  • 3
    $\begingroup$ You apparently didn't look very hard. After clicking the fourth or fifth space of dimension $> 1$ I ended up at lmfdb.org/ModularForm/GL2/Q/holomorphic/22/6/0/d, which is an example of an eigenform $f$ where the coefficient of $q^2$ does not generate $K_f$. $\endgroup$ Oct 19 '15 at 16:00
  • $\begingroup$ Great! I made a mistake, Indeed I got this question when I was think about Maeda's conjecture. Thus my search is only for Level 1 forms. I should be more careful. Thanks for your help again. $\endgroup$
    – user42690
    Oct 19 '15 at 17:59

It seems that proof of the statement 1 is quite trivial. Indeed the $G_\mathbb{Q}$ acts on Galois conjugates of $f$ transitively and the kernel is exactly the subgroup fixing $K_f$ by definition.

It has a corollary: if some Hecke operator $T_n$ acts on a $S_k(\Gamma_1(N))^{new}$ with irreducible characteristic polynomial, then $K_f=\mathbb{Q}(a_n(f))$ for any newform $f$ since both sides have degree $\dim S_k(\Gamma_1(N))^{new}$.

I say this because some one prove this for $N=1$ using the so-called Miller basis and some computations.


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