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\bigskip\bigskip
\beginsection Appendix by A.~Agashe and W.~Stein.

In this appendix, we apply a result of J.~Sturm\footnote*{J. Sturm,
{\sl On the Congruence of Modular Forms}.  Number theory (New York,
1984--1985), 275--280, Lecture Notes in Math., 1240, Springer,
Berlin-New York, 1987.}  to obtain a bound on the number of Hecke
operators needed to generate the Hecke algebra as an abelian group.
This bound was suggested to the authors of this appendix by
Lo\"\i{}c Merel and Ken Ribet.

\bigskip
\proclaim Theorem. 
The ring ${\bf T}$ of Hecke operators acting on the space of cusp forms 
of weight~$k$ and level~$N$ is generated as an abelian group by the Hecke
operators $T_n$ with
$$
n\le {{kN}\over{12}}\cdot \prod_{p|N}\left(1+{1\over p}\right).
$$

\smallskip\noindent{\bf Proof.} 
For any ring~$R$, let $S_k(N;R) = S_k(N;{\bf Z})\otimes{}R$,
where $S_k(N;{\bf Z})$ is the subgroup of cusp forms with integer Fourier
expansion at the cusp~$\infty$, and
let ${\bf T}_R = {\bf T}\otimes_{\bf Z} R$.
There is a perfect pairing
$S_k(N;R) \otimes_R {\bf T}_R \rightarrow R$
given by
    $\langle f,T\rangle\mapsto a_1(T(f))$.

Let~$M$ be the submodule of ${\bf T}$ generated by
   $T_1,T_2,\ldots,T_r$, where~$r$ is the largest integer
   $\leq {{kN}\over{12}}\cdot \prod_{p|N}\left(1+{1\over p}\right)$.
 Consider the exact sequence of
additive abelian groups
  $$0\rightarrow M \mathop{\rightarrow}\limits^{i}\ {\bf T} 
         \rightarrow {\bf T}/M \rightarrow 0.$$
Let~$p$ be a prime and use that tensor product is right exact to obtain
an exact sequence
  $$M\otimes {\bf F}_p \mathop{\rightarrow}\limits^{\overline{i}}\ 
      {\bf T}\otimes{\bf F}_p \rightarrow 
     ({\bf T}/M)\otimes{\bf F}_p\rightarrow 0.$$
Suppose that
$f\in S_k(N;{\bf F}_p)$ pairs to~$0$ with each of $T_1,\ldots, T_r$.
Then $a_m(f)=a_1(T_m f)=\langle f, T_m\rangle = 0$ in 
${\bf F}_p$ for each $m\leq r$.  By Theorem~1 of
Sturm's paper, it follows that $f = 0$. Thus the pairing restricted
to the image of $M\otimes{\bf F}_p$ in ${\bf T}\otimes{\bf F}_p$ is 
nondegenerate, so 
$$\dim_{{\bf F}_p} \overline{i}(M\otimes{\bf F}_p) 
   = \dim_{{\bf F}_p} S_k(N,{\bf F}_p)= 
       \dim_{{\bf F}_p} {\bf T}\otimes{\bf F}_p.$$
It follows that $({\bf T}/M) \otimes {\bf F}_p = 0$; repeating the
argument for  all primes~$p$ shows that ${\bf T}/M=0$, as claimed.

\bigskip
\noindent{\bf Remark.}
In general, the theorem is not true if one considers only
 $T_n$ where~$n$ runs over the {\sl primes} less than the bound.
Consider, for example, $S_2(11)$, where the bound is~$2$ and
$T_2$ is the $1\times 1$ matrix $[2]$, which
does not generate the full Hecke algebra as a ${\bf Z}$-submodule of
${\rm End}(S_2(\Gamma_0(N),{\bf Z}))$.  One needs, in addition, 
the matrix $[1]$.
\bye