\documentclass{article} \include{macros} \begin{document} \myheadauth{Generating the Hecke algebra}{0.1} {{\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} \bibitem[A]{agashe} A. Agashe, {\em On the Calculation of Eisenstein quotient}. Preprint, 1998. \bibitem[DI]{diamondim} F. Diamond, J. Im, {\em Modular forms and modular curves}. Seminar on Fermat's Last Theorem (Toronto, ON, 1993--1994), 39--133, CMS Conf. Proc., 17, Amer. Math. Soc., 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}. Number theory (New York, 1984--1985), 275--280, Lecture Notes in Math., 1240, Springer, Berlin-New York, 1987. \end{thebibliography} \normalsize\vspace*{1 cm} \end{document}