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Author: William A. Stein
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\documentclass{article}
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\include{macros}
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\begin{document}
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\myheadauth{Generating the Hecke algebra}{0.1}
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{{\large A. Agashe}\vspace{1ex}\\
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{\large W.A. Stein}\vspace{1ex}\\
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{\small Department of Mathematics, University of California, Berkeley,
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CA 94720, USA}}
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\begin{abstract}
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Let $\T$ be the Hecke algebra associate to
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weight $k$ modular forms for $X_0(N)$.
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We give a bound for the number of Hecke operators $T_n$
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needed to generate $\T$ as a $\Z$-module.
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\end{abstract}
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\section*{Introduction}
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In this note we apply a theorem of Sturm \cite{sturm}
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to prove a bound on the number of Hecke operators
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needed to generate the Hecke algebra as a $\Z$-module.
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This bound was observed by to Ken Ribet, but has not
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been written down.
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In section 2 we record our notation and some standard theorems.
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In section 3 we state Sturm's theorem and use it to deduce
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a bound on the number of generators of the Hecke algebra.
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\section{Modular forms and Hecke operators}
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Let $N$ and $k$ be positive integers and let
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$M_k(N)=M_k(\Gamma_0(N))$ be the $\C$-vector space of weight $k$
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modular
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forms on $X_0(N)$. This space can be viewed as the set of
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functions $f(z)$, holomorphic on the upper half-plane, such that
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$$f(z)=f|[\gamma]_k(z):=(cz+d)^{-k}f\left(\frac{az+b}{cz+d}\right)$$
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for all $\gamma\in\Gamma_0(N)$,
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and such that $f$ satisfies a certain holomorphic condition at
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the cusps.
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Any $f\in M_k(N)$ has a Fourier expansion
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$$f = a_0(f) + a_1(f) q + a_2(f)q^2 + \cdots =\sum a_n q^n \in \C[[q]]$$
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where $q=e^{2\pi i z}$.
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The map sending $f$ to its $q$-expansion is an injective
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map
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$M_k(N)\hookrightarrow\C[[q]]$
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called the $q$-expansion map.
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Define $M_k(N;\Z)$ to be the inverse image
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of $\Z[[q]]$ under this map. It is known (see \S12.3, \cite{diamondim}) that
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$$M_k(N)=M_k(N;\Z)\tensor \C.$$
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For any ring $R$, define
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$M_k(N;R):=M_k(N;\Z)\tensor_{\Z} R.$
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Let $p$ be a prime. Define two operators on $\C[[q]]$:
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$$V_p(\sum a_n q^n) = \sum a_n q^{np}$$
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and
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$$U_p(\sum a_n q^n) = \sum a_{np} q^n.$$
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The Hecke operator $T_p$
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acts on $q$-expansions by
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$$T_p = U_p + \eps(p) p^{k-1} V_p$$
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where $\eps(p) = 1$, unless $p|N$ in which case $\eps(p)=0$.
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If $m$ and $n$ are coprime, the Hecke operators satisfy
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$T_{nm}=T_n T_m = T_m T_n$. If $p$ is a prime and $r\geq 2$,
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$$T_{p^r}=T_{p^{r-1}}T_p - \eps(p) p^{k-1} T_{p^{r-2}}.$$
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The $T_n$ are linear maps which
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preserves $M_k(N;\Z)$.
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The Hecke algebra $\T=\T(N)=\Z[T_1,T_2,T_3,\ldots]$, which is
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viewed as a subring of the ring of linear endomorphisms of $M_k(N)$,
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is a finite commutative $\Z$-algebra.
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\begin{proposition}\label{prop1}
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Let $\sum a_n q^n$ be the $q$-expansion of
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$f\in M_k(N)$ and let $\sum b_n q^n$ be
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the $q$-expansion of $T_m f$. Then the coefficients
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$b_n$ are given by
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$$b_n = \sum_{d|(m,n)} \eps(d) d^{k-1} a_{mn/d^2}.$$
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Note in particular that $a_1(T_m f) = a_m(f)$.
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\end{proposition}
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\begin{proof}
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Proposition 3.4.3, \cite{diamondim}.
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\end{proof}
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\begin{proposition}\label{prop2}
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For any ring $R$, there is a perfect pairing
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$$ \T_R\tensor_RM_k(N;R) \ra R,\qquad (T,f)\mapsto a_1(Tf),$$
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where $\T_R = \T\tensor_{\Z} R$.
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\end{proposition}
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\begin{proof}
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Proposition 12.4.13, \cite{diamondim}.
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\end{proof}
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\section{Bounding the number of generators}
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Let $\mu(N)=N\prod_{p|N}(1+\frac{1}{p})$ be the
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index of $\Gamma_0(N)$ in $\sltwoz$.
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\begin{theorem}\label{sturm}
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Let $\lambda$ be a prime ideal in the ring
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of integers $\O$ of some number field. Suppose $f\in M_k(N;\O)$
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is such that $a_n(f)\con 0\pmod{\lambda}$ for $n\leq \frac{k}{12}\mu(N)$.
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Then $f\con 0\pmod{\lambda}$.
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\end{theorem}
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\begin{proof}
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Theorem 1, \cite{sturm}.
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\end{proof}
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Denote by $\lceil{}x\rceil$ the smallest integer $\geq x$.
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\begin{proposition}\label{determine}
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Suppose $f\in M_k(N)$ and $$a_n(f)=0 \quad\text{for}\quad
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n\leq r=\left\lceil\frac{k}{12}\mu(N)\right\rceil.$$ Then $f=0$.
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\end{proposition}
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\begin{proof}
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We must show that the composite map
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$$M_k(N)\hookrightarrow\C[[q]]\into\C[[q]]/(q^{r+1})$$
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is injective. Because $\C$ is a flat $\Z$-module, it suffices
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to show that the map $\Phi:M_k(N;\Z)\into\Z[[q]]/(q^{r+1})$ is injective.
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Suppose $\Phi(f)=0$, and let $p$ be a prime number.
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Then $a_n(f)=0$ for $n\leq r$, hence plainly
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$a_n(f)\con 0\pmod{p}$ for any such $n$.
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By Theorem~\ref{sturm}, it follows that $f\con 0\pmod{p}$.
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Repeating this argument shows that
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the coefficients of $f$ are divisible by all primes $p$,
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i.e., they are $0$.
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\end{proof}
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\begin{theorem}\label{bound}
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The Hecke algebra is generated as a $\Z$-module by
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$T_1,\ldots,T_r$
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where $r=\lceil \frac{k}{12}\mu(N)\rceil $.
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\comment{Thus $$\T=\Z[T_1,T_2,T_3,\ldots] = \Z\{T_1,\ldots,T_r\}.$$}
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\end{theorem}
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\begin{proof}
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Let $A$ be the submodule of $\T$ generated by
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$T_1,T_2,\ldots,T_r$. Consider the exact sequence of
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additive abelian groups
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$$0\into A \xrightarrow{i} \T \into \T/A \into 0.$$
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Let $p$ be a prime and tensor with $\F_p$ to obtain
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$$A\tensor \F_p\xrightarrow{\overline{i}} \T\tensor\F_p \into (\T/A)\tensor\F_p\into 0$$
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(tensor product is right exact).
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Put $R=\Fp$ in Proposition~\ref{prop2}, and suppose
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$f\in M_k(N,\Fp)$ pairs to $0$ with each of $T_1,\ldots, T_r$.
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Then by Proposition~\ref{prop1}, $a_m(f)=a_1(T_m f)=0$ in
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$\Fp$ for each $m$, $1\leq m\leq r$. By Theorem~\ref{sturm}
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it follows that $f = 0$. Thus the pairing, when restricted
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to the image of $A\tensor\Fp$ in $\T\tensor\Fp$, is also perfect and so
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$$\dim_{\Fp} \overline{i}(A\tensor\Fp)
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= \dim_{\Fp} M_k(N,\Fp)= \dim_{\Fp} \T\tensor\Fp.$$
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We see that $(\T/A) \tensor \F_p = 0$; repeating the argument for
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all $p$ shows that the finitely generated abelian group
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$\T/A$ must be trivial.
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\end{proof}
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\comment{
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Let $S_k^{\new}(N)$ be the new subspace of $S_k(N)$. It is the
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orthogonal complement, with respect to the Peterson pairing
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(VII, \S5, \cite{lang}), of the subspace spanned
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by the images of $S_k(M)$ for $M|N$ under the natural
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inclusion maps. Let $\T^{\new}$ be the
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image of the Hecke algebra in the ring of endomorphisms of
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$S_k^{\new}(N)$. By (VIII, \S3, \cite{lang}), $S_k^{\new}(N)$
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is a direct sum of distinct one dimensional eigenspaces.
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We call $f\in S_k^{\new}(N)$ a {\em newform}
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if it is an eigenform for all Hecke operators $T_p$
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and if it is normalized so that $a_1(f)=1$.
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\begin{proposition}\label{sign}
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If $f$ is a newform level $N$ and $p|N$, then
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$$a_p(f) = \begin{cases}\pm p^{k/2 -1}&\text{ if $p||N$}\\
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0 & \text{ if $p^2|N$}\end{cases}.$$
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\end{proposition}
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\begin{proof}
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See the end of \S6 in \cite{diamondim}.
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\end{proof}
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Fix a square free positive integer $N$.
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Let $\{p_1,p_2,\ldots p_s\}$ be a subset (possibly empty, in which
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case $s=0$) of the prime divisors of $N$
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and set $$r:=\left[\frac{k\mu(N)}{(12\cdot 2^s)}\right].$$
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\begin{theorem}
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Let $\lambda$ be a prime ideal in the ring
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of integers $\O$ of some number field.
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Suppose $f$ and $g$ are newforms in $S_k^{\new}(N;\O)$.
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Assume
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\begin{enumerate}
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\item $a_n(f-g) \con 0 \pmod{\lambda}$ for $n\leq r$ and
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\item $a_{p_i}(f) = b_{p_i}(g)$ for each $i=1,\ldots s$.
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\end{enumerate}
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Then $f\con g\pmod{\lambda}$.
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\end{theorem}
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\begin{proof}
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Theorem 2 of \cite{sturm}.
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\end{proof}
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Note that by Proposition~\ref{sign} $a_p(f) = \pm b_p(f)$.
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I wonder:
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is $$\T^{\new}=\Z\{ T_1,\ldots, T_r, T_{p_1},\ldots, T_{p_s}\}?$$
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I DON'T see that it does because theorem 3.5 says that,
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essentially, the first vectors of first $r$ entries of
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a basis of eigenforms are all different. But, there's no
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reason I can see that they have to be linearly independent.
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}
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\begin{thebibliography}{HHHHHHH}
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\bibitem[A]{agashe} A. Agashe, {\em On the Calculation of Eisenstein quotient}.
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Preprint, 1998.
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\bibitem[DI]{diamondim} F. Diamond, J. Im, {\em Modular forms and
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modular curves}. Seminar on Fermat's Last Theorem (Toronto, ON,
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1993--1994), 39--133, CMS Conf. Proc., 17, Amer. Math. Soc.,
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Providence, RI, 1995.
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\bibitem[L]{lang} S. Lang, {\em Introduction to modular forms.}
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Grundlehren der Mathematischen Wissenschaften, 222.
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Springer-Verlag, Berlin, 1995.
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\bibitem[S]{sturm} J. Sturm, {\em On the Congruence of Modular Forms}.
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Number theory (New York, 1984--1985), 275--280, Lecture Notes in
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Math., 1240, Springer, Berlin-New York, 1987.
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\end{thebibliography} \normalsize\vspace*{1 cm}
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\end{document}