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 $\overline{\mathbb{Q}}^{\ker\varepsilon}=F$?



Fixed field of the Nebentypus of a newform for $\Gamma_1(N)$
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 $\varepsilon$.


