\documentclass{article}
\include{macros}
\begin{document}
\myheadauth{Generating the Hecke algebra}{1.0}
{{\large A. Agashe}\vspace{1ex}\\
{\large W.A. Stein}\vspace{1ex}\\
{\small Department of Mathematics, University of California, Berkeley,
CA 94720, USA}}
\begin{abstract}
Let $\T$ be the Hecke algebra associate to
weight $k$ modular forms for $X_0(N)$.
We give a bound for the number of Hecke operators $T_n$
needed to generate $\T$ as a $\Z$-module.
\end{abstract}
\section*{Introduction}
In this note we apply a theorem of Sturm \cite{sturm}
to prove a bound on the number of Hecke operators
needed to generate the Hecke algebra as a $\Z$-module.
This bound was observed by to Ken Ribet, but has not
been written down.
In section 2 we record our notation and some standard theorems.
In section 3 we state Sturm's theorem and use it to deduce
a bound on the number of generators of the Hecke algebra.
\section{Modular forms and Hecke operators}
Let $N$ and $k$ be positive integers and let
$M_k(N)=M_k(\Gamma_0(N))$ be the $\C$-vector space of weight $k$
modular
forms on $X_0(N)$. This space can be viewed as the set of
functions $f(z)$, holomorphic on the upper half-plane, such that
$$f(z)=f|[\gamma]_k(z):=(cz+d)^{-k}f\left(\frac{az+b}{cz+d}\right)$$
for all $\gamma\in\Gamma_0(N)$,
and such that $f$ satisfies a certain holomorphic condition at
the cusps.
Any $f\in M_k(N)$ has a Fourier expansion
$$f = a_0(f) + a_1(f) q + a_2(f)q^2 + \cdots =\sum a_n q^n \in \C[[q]]$$
where $q=e^{2\pi i z}$.
The map sending $f$ to its $q$-expansion is an injective
map
$M_k(N)\hookrightarrow\C[[q]]$
called the $q$-expansion map.
Define $M_k(N;\Z)$ to be the inverse image
of $\Z[[q]]$ under this map. It is known (see \S12.3, \cite{diamondim}) that
$$M_k(N)=M_k(N;\Z)\tensor \C.$$
For any ring $R$, define
$M_k(N;R):=M_k(N;\Z)\tensor_{\Z} R.$
Let $p$ be a prime. Define two operators on $\C[[q]]$:
$$V_p(\sum a_n q^n) = \sum a_n q^{np}$$
and
$$U_p(\sum a_n q^n) = \sum a_{np} q^n.$$
The Hecke operator $T_p$
acts on $q$-expansions by
$$T_p = U_p + \eps(p) p^{k-1} V_p$$
where $\eps(p) = 1$, unless $p|N$ in which case $\eps(p)=0$.
If $m$ and $n$ are coprime, the Hecke operators satisfy
$T_{nm}=T_n T_m = T_m T_n$. If $p$ is a prime and $r\geq 2$,
$$T_{p^r}=T_{p^{r-1}}T_p - \eps(p) p^{k-1} T_{p^{r-2}}.$$
The $T_n$ are linear maps which
preserves $M_k(N;\Z)$.
The Hecke algebra $\T=\T(N)=\Z[T_1,T_2,T_3,\ldots]$, which is
viewed as a subring of the ring of linear endomorphisms of $M_k(N)$,
is a finite commutative $\Z$-algebra.
\begin{proposition}\label{prop1}
Let $\sum a_n q^n$ be the $q$-expansion of
$f\in M_k(N)$ and let $\sum b_n q^n$ be
the $q$-expansion of $T_m f$. Then the coefficients
$b_n$ are given by
$$b_n = \sum_{d|(m,n)} \eps(d) d^{k-1} a_{mn/d^2}.$$
Note in particular that $a_1(T_m f) = a_m(f)$.
\end{proposition}
\begin{proof}
Proposition 3.4.3, \cite{diamondim}.
\end{proof}
\begin{proposition}\label{prop2}
For any ring $R$, there is a perfect pairing
$$ \T_R\tensor_RM_k(N;R) \ra R,\qquad (T,f)\mapsto a_1(Tf),$$
where $\T_R = \T\tensor_{\Z} R$.
\end{proposition}
\begin{proof}
Proposition 12.4.13, \cite{diamondim}.
\end{proof}
\section{Bounding the number of generators}
Let $\mu(N)=N\prod_{p|N}(1+\frac{1}{p})$ be the
index of $\Gamma_0(N)$ in $\sltwoz$.
\begin{theorem}\label{sturm}
Let $\lambda$ be a prime ideal in the ring
of integers $\O$ of some number field. Suppose $f\in M_k(N;\O)$
is such that $a_n(f)\con 0\pmod{\lambda}$ for $n\leq \frac{k}{12}\mu(N)$.
Then $f\con 0\pmod{\lambda}$.
\end{theorem}
\begin{proof}
Theorem 1, \cite{sturm}.
\end{proof}
Denote by $\lceil{}x\rceil$ the smallest integer $\geq x$.
\begin{proposition}\label{determine}
Suppose $f\in M_k(N)$ and $$a_n(f)=0 \quad\text{for}\quad
n\leq r=\left\lceil\frac{k}{12}\mu(N)\right\rceil.$$ Then $f=0$.
\end{proposition}
\begin{proof}
We must show that the composite map
$$M_k(N)\hookrightarrow\C[[q]]\into\C[[q]]/(q^{r+1})$$
is injective. Because $\C$ is a flat $\Z$-module, it suffices
to show that the map $\Phi:M_k(N;\Z)\into\Z[[q]]/(q^{r+1})$ is injective.
Suppose $\Phi(f)=0$, and let $p$ be a prime number.
Then $a_n(f)=0$ for $n\leq r$, hence plainly
$a_n(f)\con 0\pmod{p}$ for any such $n$.
By Theorem~\ref{sturm}, it follows that $f\con 0\pmod{p}$.
Repeating this argument shows that
the coefficients of $f$ are divisible by all primes $p$,
i.e., they are $0$.
\end{proof}
\begin{theorem}\label{bound}
The Hecke algebra is generated as a $\Z$-module by
$T_1,\ldots,T_r$
where $r=\lceil \frac{k}{12}\mu(N)\rceil $.
\comment{Thus $$\T=\Z[T_1,T_2,T_3,\ldots] = \Z\{T_1,\ldots,T_r\}.$$}
\end{theorem}
\begin{proof}
Let $A$ be the submodule of $\T$ generated by
$T_1,T_2,\ldots,T_r$. Consider the exact sequence of
additive abelian groups
$$0\into A \xrightarrow{i} \T \into \T/A \into 0.$$
Let $p$ be a prime and tensor with $\F_p$ to obtain
$$A\tensor \F_p\xrightarrow{\overline{i}} \T\tensor\F_p \into (\T/A)\tensor\F_p\into 0$$
(tensor product is right exact).
Put $R=\Fp$ in Proposition~\ref{prop2}, and suppose
$f\in M_k(N,\Fp)$ pairs to $0$ with each of $T_1,\ldots, T_r$.
Then by Proposition~\ref{prop1}, $a_m(f)=a_1(T_m f)=0$ in
$\Fp$ for each $m$, $1\leq m\leq r$. By Theorem~\ref{sturm}
it follows that $f = 0$. Thus the pairing, when restricted
to the image of $A\tensor\Fp$ in $\T\tensor\Fp$, is also perfect and so
$$\dim_{\Fp} \overline{i}(A\tensor\Fp)
= \dim_{\Fp} M_k(N,\Fp)= \dim_{\Fp} \T\tensor\Fp.$$
We see that $(\T/A) \tensor \F_p = 0$; repeating the argument for
all $p$ shows that the finitely generated abelian group
$\T/A$ must be trivial.
\end{proof}
\comment{
Let $S_k^{\new}(N)$ be the new subspace of $S_k(N)$. It is the
orthogonal complement, with respect to the Peterson pairing
(VII, \S5, \cite{lang}), of the subspace spanned
by the images of $S_k(M)$ for $M|N$ under the natural
inclusion maps. Let $\T^{\new}$ be the
image of the Hecke algebra in the ring of endomorphisms of
$S_k^{\new}(N)$. By (VIII, \S3, \cite{lang}), $S_k^{\new}(N)$
is a direct sum of distinct one dimensional eigenspaces.
We call $f\in S_k^{\new}(N)$ a {\em newform}
if it is an eigenform for all Hecke operators $T_p$
and if it is normalized so that $a_1(f)=1$.
\begin{proposition}\label{sign}
If $f$ is a newform level $N$ and $p|N$, then
$$a_p(f) = \begin{cases}\pm p^{k/2 -1}&\text{ if $p||N$}\\
0 & \text{ if $p^2|N$}\end{cases}.$$
\end{proposition}
\begin{proof}
See the end of \S6 in \cite{diamondim}.
\end{proof}
Fix a square free positive integer $N$.
Let $\{p_1,p_2,\ldots p_s\}$ be a subset (possibly empty, in which
case $s=0$) of the prime divisors of $N$
and set $$r:=\left[\frac{k\mu(N)}{(12\cdot 2^s)}\right].$$
\begin{theorem}
Let $\lambda$ be a prime ideal in the ring
of integers $\O$ of some number field.
Suppose $f$ and $g$ are newforms in $S_k^{\new}(N;\O)$.
Assume
\begin{enumerate}
\item $a_n(f-g) \con 0 \pmod{\lambda}$ for $n\leq r$ and
\item $a_{p_i}(f) = b_{p_i}(g)$ for each $i=1,\ldots s$.
\end{enumerate}
Then $f\con g\pmod{\lambda}$.
\end{theorem}
\begin{proof}
Theorem 2 of \cite{sturm}.
\end{proof}
Note that by Proposition~\ref{sign} $a_p(f) = \pm b_p(f)$.
I wonder:
is $$\T^{\new}=\Z\{ T_1,\ldots, T_r, T_{p_1},\ldots, T_{p_s}\}?$$
I DON'T see that it does because theorem 3.5 says that,
essentially, the first vectors of first $r$ entries of
a basis of eigenforms are all different. But, there's no
reason I can see that they have to be linearly independent.
}
\begin{thebibliography}{HHHHHHH}
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Providence, RI, 1995.
\bibitem[L]{lang} S. Lang, {\em Introduction to modular forms.}
Grundlehren der Mathematischen Wissenschaften, 222.
Springer-Verlag, Berlin, 1995.
\bibitem[S]{sturm} J. Sturm, {\em On the Congruence of Modular Forms}.
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\end{thebibliography} \normalsize\vspace*{1 cm}
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