TITLE: Conductor of monomial forms with trivial nebentypus
QUESTION [5 upvotes]: Is it true that the conductor of a holomoprhic or a Maass cusp form with trivial nebentypus corresponding to a two-dimensional dihedral representation (over $\mathbb{Q}$ )is non-square-free?

REPLY [6 votes]: Thinking about Idoneal's question to Emerton about what is being used about the Steinberg, it's not just standard facts about the Steinberg one needs via this approach, but also local-global compatibility. The standard facts about the Steinberg are also easily proved if one uses local Langlands (i.e. works on the Galois side). This made me realise that in fact one can pull off the entire argument on the Galois side! Let's assume you're talking about complex 2-dimensional representations of the Galois group. Emerton has already observed that they must be ramified at some prime $p$, so it suffices to prove that $p^2$ divides the conductor. But now say
$\rho:Gal(\overline{\mathbf{Q}}_p/\mathbf{Q}_p)\to GL_2(\mathbf{C})$
has trivial determinant and conductor $p$. One instantly gets a contradiction: the inertial invariants can't be 0-dimensional because already this would mean the conductor is at least $p^2$, and they can't be 1-dimensional because if $i\in I_p$ has one eigenvalue 1 then the other eigenvalue must be 1 too, and $\rho(i)$ is diagonalisable and hence trivial, so the inertial invariants must be 2-dimensional but now the conductor must be 1. This answers the question completely without recourse to the smooth representation theory side of things and is surely the easiest approach to the question.