Notation : $M_k(SL_2(\mathbb{Z}))$ denotes the set of all modular forms of weight k on $SL_2(\mathbb{Z})$
$E_k=1-\frac{2k}{B_k}\sum_{n=1}^\infty \sigma_{k-1}(n)q^n$
Now the exercise :
Let f be a modular form of weight k for the full modular group whose Fourier expansion at $i\infty$ is given by $$f(z)=\sum_{n=0}^\infty a_ne^{2\pi inz} \ \ , \ \ a_n \in \mathbb{C}$$ There are unique $c_{ab}$'s such that $$f(z)=\sum_{4a+6b=k} c_{ab}E_4^aE_6^b \ \ , \ \ c_{ab} \in \mathbb{C}$$ Let F be an algebraic extension of $\mathbb{Q}$ . Suppose that $$ \# \lbrace n:a_n \in F\rbrace \ge dim M_k(SL_2(\mathbb{Z})).$$ Then , show that $c_{ab}$ belong to F , whence all the Fourier coefficients belong to F.
Solution :
We begin by noting that the Fourier coefficients of $E_4$ and $E_6$ are rational (in fact, integral). Inserting the q-expansions of f and $E_4^aE_6^b$ in our expression, and comparing the coefficients of $q^n$ on both sides for exactly $dimM_k(SL_2(\mathbb{Z}))$ many values of n such that $a_n \in F$ , we get a system of linear equations satisfied by the $c_{ab}$ say , $A(c_{ab})=v$ , for some matrix A with rational entries and vector v with entries from F.
Now the uniqueness (and the existence) of complex numbers $c_{ab}$ imply the invertibility of A. Since A has rational entries, so does $A^{-1}$ . But then , the $c_{ab}$'s belong to F .
I do not really understand how to get the system of linear equations .