TITLE: Rational eigenforms
QUESTION [12 upvotes]: Probably an easy question: let $f$ be an eigenform in $S_k^{\text{new}}(\Gamma_0(N),\chi)$ assumed to be with rational fourier coefficients.
Then $\chi$ is necessarily trivial or quadratic. But more precisely, $\chi$ seems to
be unique: if $k$ is even, then $\chi$ must be the trivial character, and
if $k$ is odd then $N$ must be of the form $N=-Df^2$ with $D$ a negative
discriminant, and then $\chi(n)=(D/n)$. Can someone explain why this is true?
More generally, in even weight and nontrivial quadratic character the irreducible Galois orbits all have even
dimension (equivalently, the irreducible factors of the characteristic polynomial of Hecke operators all have even degree). Is this true, and why?

REPLY [2 votes]: I only answer the first part of the question. As it is well-known, $\chi$ and $k$ have the same parity. In particular, if $\chi$ is trivial then $k$ must be even. Conversely, if $\chi$ is non trivial then $f$ has CM by its own Nebentypus $\chi$ because $f$ is assumed to have rational (hence real) coefficients. Therefore $\chi$ is the quadratic character associated with an imaginary quadratic field $K$ of (negative) discriminant $D$ (see Ribet, Galois representations attached to eigenforms with Nebentypus, Modular Functions of One Variable V, LNM Vol. 601). In particular, we have $\chi(-1)=-1$ and $k$ is odd.
Let us now assume that we are in that latter case, i.e. $k$ is odd. By a result of Hecke and Shimura (loc. cit. pp. 34-36), $f$ is the newform attached to a Hecke character $\psi$ of $K$ with conductor, say, $\mathfrak{m}$ and infinite type $k-1$. (That is, if we view $\psi$ as a homomorphism from the group of prime-to-$\mathfrak{m}$ fractional ideals of $K$ into $\mathbf{C}^*$, then $\psi(\alpha \mathcal{O}_K)=\alpha^{k-1}$ for all $\alpha\in K^*$ with $\alpha\equiv 1\pmod{\mathfrak{m}}$; here $\mathcal{O}_K$ denotes the integer ring of $K$.) Moreover, in the notation of the question, we have $N=-\mathcal{N}(\mathfrak{m})D$ where $\mathcal{N}$ denotes the norm map from $K$ to $\mathbf{Q}$.
If $\mathfrak{p}$ is a prime of $\mathcal{O}_K$, denote by $e_{\mathfrak{p}}$ the exponent of the conductor $\mathfrak{m}$ of $\psi$ at $\mathfrak{p}$. By Proposition 6.1 and Table 1 of Schütt's paper CM newforms with rational coefficients (Ramanujan Journal 19 (2009), 187-205), for all prime ideals $\mathfrak{p}$, we have $e_{\mathfrak{p}}=e_{\overline{\mathfrak{p}}}$  and $e_{\mathfrak{p}}$ even if $\mathfrak{p}$ ramifies in $K$. (For a more detailed proof of these results, see chapter II of the author's dissertation available on his homepage.) This in turn implies that $\mathcal{N}(\mathfrak{m})=N/(-D)$ is a square, as desired.