Let $f\in S\_k(\Gamma\_1(N))$ be an eigenform, and $K\_f$ be its number field, which is of finite degree over $\mathbb{Q}$. Consider the following statements.


1, $[K\_f:\mathbb{Q}]=\#\{$Galois conjugates of $f\}$. 


2, Any $n$-th coefficients of $f$ having degree $[K\_f:\mathbb{Q}]$ will generate $K\_f$, so could this $n$ always be $n=2$?


I guess statement 1 is true, but I have problem in proving it. 
I guess statement 2 is false, but I need a counterexample.



degree of Hecke field (number field of an eigenform)
It seems that proof of the statement 1 is quite trivial.
Indeed the $G\_\mathbb{Q}$ acts on Galois conjugates of $f$ transitively and the kernel is exactly the subgroup fixing $K\_f$ by definition. 


It has a corollary: if some Hecke operator $T\_n$ acts on a $S\_k(\Gamma\_1(N))^{new}$ with irreducible characteristic polynomial, then 
$K\_f=\mathbb{Q}(a\_n(f))$ for any newform $f$ since both sides have degree $\dim S\_k(\Gamma\_1(N))^{new}$.


I say this because some one prove this for $N=1$ using the so-called Miller basis and some computations.


