Modular forms with rational and $p$-integer coefficients (mod $p$)

Question

I'm studying $p$-adic modular forms, and in particular I was trying to understand Swinnerton-Dyer's proof on the structure of the algebra of modular forms (of level 1) modulo $p\geq 5$.

In his proof, he states that given any modular form $f=\sum c_{a,b}E_4^aE_6^b$ with rational and $p$-integer coefficients in its $q$-expansion, the $c_{a,b}$ are necessarily rational and $p$-integer. This seems very natural, but getting my hands on it, I can't seem to find a proof.

In his "Formes modulaires et fonctions zêta $p$-adiques", Serre hints that this may be done by induction on the weight, using that $\Delta$ has this property; however this hasn't helped me much.

Thank you for your help!

Answer 1

I’m assuming $p\geq 5$. In weight at most $14$, I’ll leave it to you to check it.

Let $f \in \mathbb{Z}_{(p)}[[q]]$ the $q$-expansion of a modular form of weight $w >14$, and write $f=\sum_{a,b}{c_{a,b}E_4^aE_6^b}$, and we want to show that the $c_{a,b}$ are $p$-integers. Write $\gamma \in \mathbb{Z}_{(p)}$ the value at infinity of $f$, and consider $f_2=f-\gamma E_4^aE_6^b$: it has a $p$-integral and rational $q$-expansion.

It’s easy to see that $\frac{\Delta(q)}{q} \in \mathbb{Z}[[q]]$ has constant value $1$, so that it has an inverse $i(q) \in \mathbb{Z}[[q]]$. Now, let $f_3=i(q)\frac{f_2(q)}{q}=\frac{f_2}{\Delta}$, then $f_3$ is the $p$-integral $q$-expansion of a modular form of weight $w-12$ so it is a $\sum_{a,b}{d_{a,b}E_4^aE_6^b}$ with rational $p$-integral $d_{a,b}$. Now $f_2=f_3\frac{E_4^3-E_6^2}{1728}$ and $1728=12^3$ is a unit in $\mathbb{Z}_{(p)}$.

Answer 2

You need $p\nmid 6$. From that $\Delta= \frac{E_4^3-E_6^2}{2^6 3^3}$ is a weight $12$ cusp form with no zeros on the upper half-plane and a simple zero at $i\infty$,

For $k$ even $\ne 2$ taking $a_k,b_k$ such that $k=4a_k+6b_k$ we have $$M_k(SL_2(\Bbb{Z}))= \Bbb{C}E_4^{a_k} E_6^{b_k} \oplus \Delta \ M_{k-12}(SL_2(\Bbb{Z})) $$ For $f=\sum_{n\ge 0} A_n(f)e^{2i\pi nz}\in M_k(SL_2(\Bbb{Z}))$ with rational coefficients you'll have $f = A_0(f)E_4^{a_k} E_6^{b_k}+g \Delta$ where $g = \Delta^{-1}(f-A_0(f) E_4^{a_k} E_6^{b_k})$ has rational coefficients, so we can repeat with $g$ instead of $f$.

This process terminates when $k< 12$ because $g$ will have weight $<0$ and the maximum modulus principle will apply to give that $g$ is constant $=0$.

By induction you'll get that $$f=\sum_{m\le k/12}c_m E_4^{a_{k-12m}} E_6^{b_{k-12m}}\Delta^m$$ with $c_m$ rational.

Moreover $E_4,E_6,\Delta^{-1}$ have integer coefficients, whence if $f$ has its coefficients in a subring $S\subset \Bbb{Q}$ then so does $g$ and by induction you'll have that the $c_m$ are in $S$.

Expanding $\Delta^m$ in term of powers of $E_4,E_6$ you'll get that $$f = \sum_{4u+6b=k} C_{u,v} E_4^u E_6^v, \qquad C_{u,v}\in S[1/6]$$