6
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ @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 '16 at 11:42
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.