Let $f=\sum\_{n=1}^\infty a\_nq^n$ be a newform of level $N$ and weight $k\ge 2$. Suppose that $f$ is a CM modular form in the sense of §3 of Ribet's paper [Galois representations attached to eigenforms with nebentypus](http://link.springer.com/chapter/10.1007%2FBFb0063943): i.e. there exists a quadratic character $\varphi$ such that $$a\_p = \varphi(p)a\_p$$for all primes $p$ in a set of primes of density $1$. Let $K$ be the quadratic field cut out by $\varphi$.



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> Is there a down-to-earth explanation of why $K$ must be an imaginary quadratic field?
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The proof given in theorem 4.5 of the above paper depends quite heavily on the properties of the Serre group $S\_{\mathfrak m/K}$.



Field cut out by a CM modular form is imaginary
As Joel says, $a\_p=0$ if $p$ is inert in $K$, so this means that the Galois representation attached to $f$ must be induced from a 1-dimensional representation of the absolute Galois group of $K$. If $K$ were real quadratic then the grossencharacter corresponding to this 1-dimensional representation would have to be equal to $|.|^n$ on both the $\mathbf{R}\_{>0}$'s at the infinite places. But this means that the associated Galois representation has finite order up to twist (both Hodge-Tate weights are equal) so the weight would have to be 1.


