I want to show that if a normalised modular form of level $1$, say $f$, is a simultaneous eigenform of the Hecke operators $T(n)$ for all $n$, then the corresponding eigenvalues are all algebraic integers. My approach goes like this:

Let $M_k(SL_2(\mathbb{Z}))$ be the space of modular forms of level $1$ and weight $k$. Let $M_k(SL_2(\mathbb{Z}))_{\mathbb{Z}}$ be the subset of those modular forms all of whose Fourier coefficients are integers. Then, one can check that the action of $T(n)$ descends to an action on $M_k(SL_2(\mathbb{Z}))_{\mathbb{Z}}$. Hence, if we can find a $\mathbb{Z}$-basis for $M_k(SL_2(\mathbb{Z}))_{\mathbb{Z}}$ which is also a $\mathbb{C}$-basis for $M_k(SL_2(\mathbb{Z}))$, then, the matrix of the action of $T(n)$ will have integer entries, and so, the eigenvalues will be algebraic integers. So, I only need to find such a basis.

I feel that some combinations of $E_4$ and $E_6$ (the normalised Weierstrass forms) should work, but I haven't been able to work out the details.

  • $\begingroup$ @Mathmo123 I found what I was looking for. Actually, I was referring to something on the lines of Serre's 'A Course in Arithmetic', Page 105, Section 5.6.2 on Integrality Properties. $\endgroup$
    – MathManiac
    Nov 21, 2016 at 11:42

1 Answer 1


You are correct. For a given $k\geqslant 12$ a basis for $M_k$ is given by $E_4^aE_6^b$ with $4a+6b=k$.


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