TITLE: Field of definition of Galois representations of weight 1 modular forms
QUESTION [8 upvotes]: Let $f$ be a weight 1 modular form (let's say cuspidal, new, normalized, and a Hecke eigenform). Then there's an associated Artin representation $\rho_f: \operatorname{Gal}(\overline{\mathbf{Q}} / \mathbf{Q}) \to GL_2(\mathbf{C})$.
The character of this representation is determined by $f$, so it takes values in the number field $L = \mathbf{Q}(a_n(f): n \ge 1)$. Are there examples where $\rho_f$ is not definable over $L$, i.e. isn't conjugate to a morphism $\operatorname{Gal}(\overline{\mathbf{Q}} / \mathbf{Q}) \to GL_2(L)$? 
(There are lots of examples of even representations in the LMFDB where this happens, coming from even Artin representations with image $Q_8$ or $S_4$, but I can't find any odd ones.)

REPLY [9 votes]: Here is another proof of the result from Joel's answer, using a little more theory:
a two dimensional irrep. of a finite group $G$ corresponds to a simple algebra factor $A$ of the group ring $\mathbb Q[G]$, where $A$ is $4$-dimensional over its centre (which is some number field $F$).  Thus $A$ is either $M_2(F)$, or a non-split quat. alg. over $F$. In particular, in the second case $A$ is a division algebra. 
If $c \in G$ has order $2$, with image $\bar{c}$ in $A$, then since $c^2 = 1$,
we see that $(\bar{c} -1)(\bar{c} + 1) = 0$ in $A$.  If $A$ is a division algebra,
we conclude that either $\bar{c} = 1$ or $\bar{c} = -1$.  
Thus if the image of $c$ is not scalar, we see that $A = M_2(F)$, which is to say
that the given irrep. is defined over its trace field.