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Let $f=\sum_{n\geq 1}\in S_2(\Gamma_1(N),\varepsilon)$ be a normalized newform without CM and with Nebentypus $\varepsilon$. Let $L=\mathbb Q(a_n\colon n\in \mathbb N)$ be the number field generated by the coefficients of $f$. This is either a totally real field or a CM field, namely a totally imaginary quadratic extension of a totally real field. Let $F\subseteq L$ be the biggest totally real subfield of $L$, so that $[L\colon F]$ is either $1$ or $2$.

Now assume that $\varepsilon$ is nontrivial, which implies that $L$ is a CM field. If we consider $\varepsilon$ a character of $\text{Gal}(\overline{\mathbb{Q}}/\mathbb Q)$, is it true that $\overline{\mathbb{Q}}^{\ker\varepsilon}=F$?

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No, this is not the case. In general these fields will have very little to do with each other.

For instance, let $\varepsilon$ be the unique character of $(\mathbf{Z} / 17 \mathbf{Z})^\times$ mapping 3 to $i$. Then there is a modular form of level $34$, weight $2$ and nebentypus $\varepsilon$ whose $q$-expansion is

q - i*q^2 + (-i - 1)*q^3 - q^4 + (2*i + 2)*q^5 + O(q^6)

(thanks, Sage). For this form we evidently have $L = \mathbf{Q}(i)$ and $F = \mathbf{Q}$, but $(\overline{\mathbf{Q}})^{\ker \varepsilon}$ is the unique degree 4 subfield of the cyclotomic field $\mathbf{Q}(\zeta_{17})$ (generated by a root of $x^4 + x^3 - 6x^2 - x + 1$) which is linearly disjoint from $\mathbf{Q}(i)$.

Note that $(\overline{\mathbf{Q}})^{\ker \varepsilon}$ is always abelian, and in particular Galois over $\mathbf{Q}$, while there is no reason why the fields $L$ or $F$ should have either property.

EDIT: I couldn't resist giving another example. There is a form of weight 2, character the same $\varepsilon$ as before, and level 51; this form is unique up to Galois conjugacy, and its q-expansion is given by

q + a0*q^2 + ((-1/4*i + 1/4)*a0^3 + (-1/4*i - 1/4)*a0^2 + (-i + 1)*a0 - 1/2*i - 1/2)*q^3 + (a0^2 + 2)*q^4 + ((1/2*i - 1/2)*a0^3 + (5/2*i - 5/2)*a0 - i - 1)*q^5 + O(q^6)

where $a_0$ is a root of $x^4 - 2ix^3 + 5x^2 - 8ix + 2$. Then the field $F$ is generated by $a_0^2$, which has minimal polynomial over $\mathbf{Q}$ given by $x^4 + 14 x^3 + 61 x^2 + 84x + 4$. Neither $L$ nor $F$ is Galois over $\mathbf{Q}$, and $L$ is not Galois over $\mathbf{Q}(i)$ either.

I think it's reasonable to expect that the degrees of the fields $L$ and $F$ can get arbitrarily large, as $f$ varies over newforms of weight 2 and nebentypus $\varepsilon$.

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  • $\begingroup$ David, just one question: when you say that $F$ is evidently $\mathbf{Q}(i)$, do you really mean that or simply that it is very much reasonable to guess from the first five coefficients? Of course, this does not affect your argument, but just to know. $\endgroup$ Jan 23, 2015 at 7:09
  • $\begingroup$ You mean $L = Q(i)$; $F$ is totally real in the notation of the question. Anyway, I didn't mean to say that the five coefficients above were enough for a proof; to be completely sure I guess you'd have to compute the first 10 coefficients (10 is the Sturm bound for this level and weight). $\endgroup$ Jan 23, 2015 at 7:26
  • $\begingroup$ Oh, yes, I was sloppy with $L$ and $F$; thanks anyway. $\endgroup$ Jan 23, 2015 at 7:53

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