Suppose that a co-finite subset of the $a_p$'s generate a field $L$. Applying any element $\sigma$ of the absolute Galois group of $\mathrm{Gal}(K_f/L)$ L$ to $f$ yields another form $f'$ g$ whose Fourier coefficients $b_p$ are equal to $a_p$ for all but finitely many $p$. By strong multiplicity one, $f'=f$ g=f$ and $b_p = a_p$ for all $p$. Therefore $\sigma$ is trivial fixes $K_f$ as well, and since $L = \subset K_f$ .it follows that $L=K_f$.
|
2 | added 53 characters in body; edited body | ||
|
|
||||
|
|
1 |
|
||
|
Suppose that a co-finite subset of the $a_p$'s generate a field $L$. Applying any element $\sigma$ of $\mathrm{Gal}(K_f/L)$ to $f$ yields another form $f'$ whose Fourier coefficients $b_p$ are equal to $a_p$ for all but finitely many $p$. By strong multiplicity one, $f'=f$ and $b_p = a_p$ for all $p$. Therefore $\sigma$ is trivial and $L = K_f$. |
||||
