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Author: William A. Stein
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\documentclass[11pt]{article} \usepackage{fancybox}
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\hoffset=-0.1\textwidth \textwidth=1.2\textwidth
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\voffset=-0.1\textheight \textheight=1.2\textheight
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\include{macros}
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\title{Discriminants of Hecke Algebras at Prime Level} \author{William
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A. Stein\footnote{This will probably eventually be a joint paper
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with Frank Calegari and Romyar Sharifi (?)}} \date{September 24,
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2002} \DeclareMathOperator{\charpoly}{charpoly}
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\DeclareMathOperator{\splittemp}{split}
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\renewcommand{\split}{\splittemp} \DeclareMathOperator{\deriv}{deriv}
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\newcommand{\Zpbar}{\overline{\Z}_p}
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\newcommand{\kr}[2]{\left(\frac{#1}{#2}\right)}
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\begin{document}
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\maketitle
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\begin{abstract}
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We study $p$-divisibility of discriminant of Hecke algebras
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associated to spaces of cusp forms of prime level. By considering
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cusp forms of weight bigger than~$2$, we are are led to make a
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conjecture about indexes of Hecke algebras in their normalization
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which, if true, implies that there are no mod~$p$ congruences
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between non-conjugate newforms in $S_2(\Gamma_0(p))$.
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\end{abstract}
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\section{Introduction}
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I started working in modular forms when Ken Ribet asked about
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discriminants of Hecke algebras at prime level. I've recently
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revisited this question and, with the help of Frank Calegari, have
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made some interesting discoveries.
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\section{Discriminants of Hecke Algebras}
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Let~$R$ be a ring and let~$A$ be an~$R$ algebra that is free as an~$R$
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module. The trace of an element of~$A$ is the trace, in the sense of
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linear algebra, of left multiplication by that element on~$A$.
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\begin{definition}[Discriminant]
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Let $\omega_1,\ldots,\omega_n$ is a~$R$-basis for~$A$. Then the
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{\em discriminant} of~$A$, denoted $\disc(A)$, is the determinant of
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the $n\times n$ matrix $(\tr(\omega_i\omega_j))$, which is well
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defined modulo squares of units in~$A$.
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\end{definition}
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When $R=\Z$ the discriminant is well defined, since the only units are
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$\pm 1$.
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\begin{proposition}\label{prop:separable}
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Suppose~$R$ is a field. Then~$A$ has discriminant~$0$ if and only
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if~$A$ is separable over~$R$, i.e., for every extension $R'$ of $R$,
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the ring $A\tensor R'$ contains no nilpotents.
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\end{proposition}
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The following proof is summarized from Section~26 of Matsumura.
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If~$A$ contains a nilpotent then that nilpotent is in the kernel of
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the trace pairing. If~$A$ is separable then we may assume that~$R$ is
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algebraically closed. Then~$A$ is an Artinian reduced ring, hence
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isomorphic as a ring to a finite product of copies of~$R$, since~$R$
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is algebraically closed. Thus the trace form on~$A$ is nondegenerate.
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\subsection{The Discriminant Valuation}
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Let $\Gamma$ be a congruence subgroup of $\SL_2(\Z)$, e.g.,
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$\Gamma=\Gamma_0(p)$ or $\Gamma_1(p)$. For any integer $k\geq 1$, let
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$S_k(\Gamma)$ denote the space of holomorphic weight-$k$ cusp forms
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for $\Gamma$. Let
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$$
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\T = \Z[\ldots,T_n,\ldots] \subset \End(S_k(\Gamma))
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$$
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be the associated Hecke algebra. Then~$\T$ is a commutative ring
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that is free and of finite rank as a $\Z$-module. Also of interest is
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the image $\T^{\new}$ of~$\T$ in $\End(S_k(\Gamma)^{\new})$.
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\begin{example}
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Let $\Gamma=\Gamma_0(243)$, which is illustrated on my T-shirt.
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Since $243=3^5$, experts will immediately deduce that $\disc(\T) =
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0$. A computation shows that
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$$
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\disc(\T^{\new}) = 2^{13} \cdot 3^{40},
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$$
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which reflects the mod-$2$ and mod-$3$ intersections all over my
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shirt.
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\end{example}
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\begin{definition}[Discriminant Valuation]
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Let~$p$ be a prime and suppose that $\Gamma=\Gamma_0(p)$ or
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$\Gamma_1(p)$. The {\em discriminant valuation} is
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$$
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d_k(\Gamma) = \ord_p(\text{the discriminant of $\T$}).
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$$
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%When the discriminant of $\T$ is~$0$ we define $d_k(\Gamma)$ to
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%be~$+\infty$.
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\end{definition}
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\section{Motivation and Applications}
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Let~$p$ be a prime and suppose that $\Gamma=\Gamma_0(p)$ or
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$\Gamma_1(p)$. The quantity $d_k(\Gamma)$ is of interest because it
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measures mod~$p$ congruences between eigenforms in $S_k(\Gamma)$.
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\begin{proposition}
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Suppose that $d_k(\Gamma)$ is finite. Then the discriminant
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valuation $d_k(\Gamma)$ is nonzero if and only if there is a mod-$p$
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congruence between two Hecke eigenforms in $S_k(\Gamma)$ (note that
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the two congruent eigenforms might be Galois conjugate).
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\end{proposition}
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\begin{proof}
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It follows from Proposition~\ref{prop:separable} that
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$d_k(\Gamma)>0$ if and only if $\T\tensor \Fpbar$ is not separable.
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The Artinian ring $\T\tensor\Fpbar$ is not separable if and only if
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the number of ring homomorphisms $\T\tensor\Fpbar \ra \Fpbar$ is
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less than
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$$
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\dim_{\Fpbar} \T\tensor\Fpbar = \dim_\C S_k(\Gamma).
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$$
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Since $d_k(\Gamma)$ is finite, the number of ring homomorphisms
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$\T\tensor\Qpbar \ra \Qpbar$ equals $\dim_\C S_k(\Gamma)$. Using
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the standard bijection between homomorphisms and normalized
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eigenforms, we see that $\T\tensor\Fpbar$ is not separable if and
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only if there is a mod-$p$ congruence between two eigenforms.
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\end{proof}
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\begin{example}
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If $\Gamma=\Gamma_0(389)$ and $k=2$, then $\dim_\C S_2(\Gamma) =
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32$. Let~$f$ be the characteristic polynomial of $T_2$. One can
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check that~$f$ is square free and $389$ exactly divides the
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discriminant of~$f$, so $T_2$ generated $\T\tensor \Z_{389}$ as a
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ring. (If it generated a subring of $\T\tensor\Z_{389}$ of finite
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index, then the discriminant of~$f$ would be divisible by $389^2$.)
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Modulo~$389$ the polynomial~$f$ is congruent to
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$$\begin{array}{l}
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(x+2)(x+56)(x+135)(x+158)(x+175)^2(x+315)(x+342)(x^2+387)\\
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(x^2+97x+164)(x^2 + 231x + 64)(x^2 + 286x + 63)(x^5 + 88x^4 +196x^3 + \\
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113x^2 +168x + 349)(x^{11} + 276x^{10} + 182x^9 + 13x^8 + 298x^7 + 316x^6 +\\
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213x^5 + 248x^4 + 108x^3 + 283x^2 + x + 101)
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\end{array}
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$$
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The factor $(x+175)^2$ indicates that $\T\tensor \Fbar_{389}$ is
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not separable since the image of $T_2+175$ is nilpotent (its square
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is~$0$). There are $32$ eigenforms over~$\Q_2$ but only $31$
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mod-$389$ eigenforms, so there must be a congruence. Let~$F$ be the
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$389$-adic newform whose $a_2$ term is a root of
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$$
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x^2 + (-39 + 190\cdot 389 + 96\cdot 389^2 +\cdots) x + (-106 +
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43\cdot 389 + 19\cdot 389^2 + \cdots).
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$$
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Then the congruence is between~$F$ and its
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$\Gal(\Qbar_{389}/\Q_{389})$-conjugate.
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\end{example}
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\begin{example}
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The discriminant of the Hecke algebra $\T$ associated to
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$S_2(\Gamma_0(389))$ is
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$$
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2^{53} \!\cdot\! 3^{4} \!\cdot\! 5^{6} \!\cdot\! 31^{2} \!\cdot\!
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37 \!\cdot\! 389 \!\cdot\! 3881 \!\cdot\! 215517113148241 \!\cdot\!
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477439237737571441
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$$
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I computed this using the following algorithm, which was
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suggested by Hendrik Lenstra. Using the Sturm bound I found a~$b$
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such that $T_1,\ldots,T_b$ generate $\T$ as a $\Z$-module. I then
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found a subset~$B$ of the $T_i$ that form a $\Q$-basis for
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$\T\tensor_\Z\Q$. Next, viewing $\T$ as a ring of matrices acting
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on $\Q^{32}$, I found a random vector $v\in\Q^{32}$ such that the
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set of vectors $C=\{T(v) : T \in B\}$ is linearly independent. Then
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I wrote each of $T_1(v),\ldots, T_b(v)$ as $\Q$-linear combinations
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of the elements of~$C$. Next I found a $\Z$-basis~$D$ for the
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$\Z$-span of these $\Q$-linear combinations of elements of~$C$.
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Tracing everything back, I find the trace pairing on the elements
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of~$D$, and deduce the discriminant by computing the determinant of
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the trace pairing matrix. The most difficult step is computing~$D$
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from $T_1(v),\ldots,T_b(v)$ expressed in terms of~$C$, and this
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explains why we embed $\T$ in $\Q^{32}$ instead of viewing the
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elements of $\T$ as vectors in $\Q^{32^2}$. This whole computation
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takes one second on an Athlon 2000 processor.
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\end{example}
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\subsection{Literature}
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I've seen a version of Theorem~\ref{thm:disc} referred to in the
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following papers:
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\begin{enumerate}
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\item Ribet: {\em Torsion points on $J_0(N)$ and Galois
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representations}
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\item Lo\"\i{}c Merel and William Stein: {\em The field generated by
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the points of small prime order on an elliptic curve}
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\item Ken Ono and William McGraw: {\em Modular form Congruences and
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Selmer groups} (McGraw will speak about this next week in this
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seminar!)
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\item Momose and Ozawa: {\em Rational points of modular curves
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$X_{\split}(p)$}
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\end{enumerate}
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\section{Data About Discriminant Valuations}
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\subsection{Weight Two}
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\begin{theorem}\label{thm:disc}
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The only prime $p<60000$ such that $d_2(\Gamma_0(p))>0$ is $p=389$.
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(Except possibly $50923$ and $51437$, which I haven't finished
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checking yet.)
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\end{theorem}
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\begin{proof}
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This is the result of a large computer computation, and perhaps
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couldn't be verified any other way, since I know of no general
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theorems about $d_2(\Gamma_0(p))$. The rest of this proof describes
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how I did the computation, so you can be convinced that there is
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valid mathematics behind my computation, and that you could verify
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the computation given sufficient time. The computation described
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below took about one week using $12$ Athlon 2000MP processors. In
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1999 I had checked the result stated above but only for $p<14000$
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using a completely different implementation of the algorithm and a
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200Mhz Pentium computer. These computations are nontrivial; we
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compute spaces of modular symbols, supersingular points, and Hecke
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operators on spaces of dimensions up to~$5000$.
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The aim is to determine whether or not~$p$ divides the discriminant
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of the Hecke algegra of level~$p$ for each $p < 60000$. If~$T$ is
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an operator with integral characteristic polynomial, we write
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$\disc(T)$ for $\disc(\charpoly(T))$, which also equals
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$\disc(\Z[T])$. We will often use that
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$$\disc(T)\!\!\!\!\mod{p} = \disc(\charpoly(T)\!\!\!\!\mod p).$$
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Most levels~$p<60000$ were ruled out by computing characteristic
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polynomials of Hecke operators using an algorithm that David Kohel
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and I implemented in MAGMA, which is based on the Mestre-Oesterle
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method of graphs (our implementation is ``The Modular of
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Supersingular Points'' package that comes with MAGMA). I computed
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$\disc(T_q)$ modulo~$p$ for several primes~$q$, and in most cases
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found a~$q$ such that this discriminant is nonzero. The following
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table summarizes how often we used each prime~$q$ (note that there
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are $6057$ primes up to $60000$):
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\begin{center}
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\begin{tabular}{|l|l|}\hline
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$q$ & number of $p< 60000$ where~$q$ smallest
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s.t. $\disc(T_q)\neq 0$ mod~$p$\\\hline
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2& 5809 times\\
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3& 161 (largest: 59471)\\
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5& 43 (largest: 57793)\\
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7& 15 (largest: 58699)\\
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11& 15 (the smallest is 307; the largest 50971)\\
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13& 2 (they are 577 and 5417)\\
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17& 3 (they are 17209, 24533, and 47387)\\
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19& 1 (it is 15661 )\\\hline
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\end{tabular}
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\end{center}
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The numbers in the right column sum to 6049, so 8 levels are missing.
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These are
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$$
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389,487,2341,7057,15641,28279, 50923, \text{ and } 51437.
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$$
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(The last two are still being processed. $51437$ has the property
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that $\disc(T_q)=0$ for $q=2,3,\ldots,17$.) We determined the
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situation with the remaining 6 levels using Hecke operators $T_n$
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with~$n$ composite.
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\begin{center}
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\begin{tabular}{|l|l|}\hline
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$p$ & How we rule level~$p$ out, if possible\\\hline
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389& $p$ does divide discriminant\\
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487& using charpoly($T_{12}$)\\
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2341& using charpoly($T_6$)\\
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7057& using charpoly($T_{18}$)\\
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15641& using charpoly($T_6$)\\
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28279& using charpoly($T_{34}$)\\\hline
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\end{tabular}
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\end{center}
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Computing $T_n$ with~$n$ composite is very time consuming when~$p$ is
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large, so it is important to choose the right $T_n$ quickly. For
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$p=28279$, here is the trick I used to quickly find an~$n$ such that
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$\disc(T_n)$ is not divisible by~$p$. This trick might be used to
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speed up the computation for some other levels. The key idea is to
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efficiently discover which $T_n$ to compute. Though computing $T_n$
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on the full space of modular symbols is quite hard, it turns out that
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there is an algorithm that quickly computes $T_n$ on subspaces of
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modular symbols with small dimension (see \S3.5.2 of my Ph.D. thesis).
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Let~$M$ be the space of mod~$p$ modular symbols of level $p=28279$,
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and let $f=\gcd(\charpoly(T_2),\deriv(\charpoly(T_2)))$. Let~$V$ be
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the kernel of $f(T_2)$ (this takes 7 minutes to compute). If $V=0$,
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we would be done, since then $\disc(T_2)\neq 0\in\F_p$. In fact,~$V$
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has dimension~$7$. We find the first few integers~$n$ so that the
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charpoly of $T_n$ on $V_1$ has distinct roots, and they are $n=34$,
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$47$, $53$, and $89$. I then computed $\charpoly(T_{34})$ directly on
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the whole space and found that it has distinct roots modulo~$p$.
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\end{proof}
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\subsection{Higher Weight Data}
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\begin{enumerate}
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\item The following are the valuations $d=d_4(\Gamma_0(p))$ at~$p$ of
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the discriminant of the Hecke algebras associated to
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$S_4(\Gamma_0(p))$ for $p<500$.
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\hspace{-4em}\shadowbox{\begin{minipage}[b]{1.15\textwidth}
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\begin{tabular}{|c|ccccccccccccccccc|}\hline
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$p$ &2& 3& 5& 7& 11& 13& 17& 19& 23& 29& 31& 37& 41& 43& 47& 53& 59\\
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$d$ &0& 0& 0& 0& 0& 2& 2& 2& 2& 4& 4& 6& 6& 6& 6& 8& 8\\\hline
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$p$&61& 67& 71& 73& 79& 83& 89& 97& 101& 103& 107& 109& 113& 127& 131& 137& 139\\
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$d$ & 10& 10& 10& 12& 12& 12& 14& 16& 16& 16& 16& 18& 18& 20& 20& 22&24\\\hline
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$p$ & 149& 151& 157& 163& 167& 173& 179& 181& 191& 193& 197& 199&
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211& 223& 227& 229& 233\\
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$d$ & 24& 24& 26& 26& 26&28& 28& 30& 30& 32& 32& 32& 34& 36& 36& 38& 38\\ \hline
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$p$ & 239& 241& 251& 257& 263& 269& 271& 277&
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281& 283& 293& 307& 311& 313& 317& 331& 337\\
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$d$ & 38& 40& 40& 42& 42&44& 44& 46& 46& 46& 48& 50& 50& 52& 52& 54& 56\\\hline
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$p$ & 347& 349& 353& 359& 367& 373& 379& 383& 389&397& 401& 409& 419& 421& 431& 433& 439 \\
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$d$ & 56& 58& 58& 58& 60&62& 62& 62& 65 &66& 66& 68& 68& 70& 70& 72& 72\\\hline
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$p$ & 443& 449& 457& 461& 463& 467& 479& 487& 491& 499 &&&&&&&\\
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$d$ & 72& 74& 76& 76& 76& 76& 78& 80& 80& 82 &&&&&&&\\\hline
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\end{tabular}
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\end{minipage}}
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\comment{\item For each prime~$p$, let
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$$
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\delta(p) = \dim S_4(\Gamma_0(p)) - \dim S_{p+3}(\Gamma_0(1)).
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$$
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Then $|\delta(p) - d_4(\Gamma_0(p))| \leq 2$ for each $p<500$.
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Moreover, for every $p\neq 139$ we have that $\delta(p)\geq
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d_4(\Gamma_0(p))$, but for $p=139$, $\delta(p)=23$ but
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$d_4(\Gamma_0(p))=24$. }
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\end{enumerate}
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\section{The Discriminant is Divisible by~$p$|}
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In this section we prove that for $k\geq 4$ the discriminant of the
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Hecke algebra associated to $S_k(\Gamma_0(p))$ is almost always
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divisible by~$p$.
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\begin{theorem}
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Suppose~$p$ is a prime and $k\geq 4$ is an even integer. Then
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$d_k(\Gamma_0(p))>0$ unless
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\begin{align*}
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(p,k) \not\in \{&(2,4),(2,6),(2,8),(2,10),\\
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&(3,4),(3,6), (3,8),\\
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&(5,4), (5,6), (7,4), (11,4)\},
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\end{align*}
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in which case $d_k(\Gamma_0(p))=0$.
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\end{theorem}
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\begin{proof}
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(Romyar and William came up with this.)
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Let~$p$ be a prime. For~$N$ and~$k$ integers, let
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$$
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S_k(\Gamma_0(N),\Zpbar) := S_k(\Gamma_0(N),\Z) \tensor \Zpbar
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$$
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and
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$$
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S_k(\Gamma_0(N),\Fpbar) := S_k(\Gamma_0(N),\Zpbar) \tensor
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\Fpbar.
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$$
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Let $\wp$ denote the maximal ideal of $\Zpbar$. In
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\cite[\S3]{serre:antwerp72} Serre defines a linear map
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$$
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t: S_k(\Gamma_0(p), \Zpbar) \ra S_{k+(p-1)}(\Gamma_0(p),\Zpbar)
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$$
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given by
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$$
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t(f) = \Tr(f\cdot{} G)
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$$
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where~$G$ is an Eisenstein series of weight~$p-1$ such that
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$G\con 1\pmod{\wp}$. Serre shows (see Lemme~9 with $m=0$) that
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$t(f) \con f \pmod{\wp}$.
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By ??, there is a basis $f_1,\ldots,f_n$ of Hecke eigenforms for
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$S_k(\Gamma_0(p),\Qpbar)$, and we may assume these $f_i$ are
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normalized so that the leading coefficient of each~$q$-expansion
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is~$1$. If
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$$\dim S_{k+(p-1)}(\Gamma_0(p),\Zpbar) < \dim S_k(\Gamma_0(p),
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\Zpbar)$$
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then the set of $q$-expansions
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$$
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t(f_1) \!\!\!\!\pmod{\wp}, \,\ldots, \,t(f_n)\!\!\!\!\pmod{\wp}
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$$
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can not be linearly independent, which implies by Lemma~?? that
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$d_k(\Gamma_0(p))>0$.
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It follows from ?? that
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\begin{align*}
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\dim S_k(\Gamma_0(p))
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&= \frac{(k-1)(p+1)}{12}\\
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&+ \left( 1 + \kr{-4}{p}\right) \cdot \left( \frac{1-k}{4} +
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\left\lfloor\frac{k}{4}\right\rfloor \right) + \left( 1 +
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\kr{-3}{p}\right) \cdot \left(\frac{1-k}{3} +
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\left\lfloor\frac{k}{3}\right\rfloor \right) - 1.
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\end{align*}
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Thus
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$$
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\dim S_k(\Gamma_0(p)) \geq \frac{(k-1)(p+1)}{12} - 3.
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$$
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By ??,
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$$
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\dim S_{k+p-1}(\Gamma_0(1)) = \begin{cases}
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\lfloor \frac{k+p-1}{12}\rfloor - 1 & \text{ $k+p-1\con 2\pmod{12}$},\\
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\lfloor \frac{k+p-1}{12}\rfloor & \text{otherwise.}
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\end{cases}
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$$
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If
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$$
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\dim S_k(\Gamma_0(p)) \leq \dim S_{k+p-1}(\Gamma_0(1))
396
$$
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then $(k-2)p\leq 36$. This reduces the assertion of the theorem
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to a very small finite computation, which we did using the
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algorithms described earlier in this paper.
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\end{proof}
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403
404
\begin{conjecture}
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Suppose $p>2$ is a prime and $k\geq 3$ is an integer. If
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\begin{align*}
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(p,k) \not\in \{&(3,3),(3,4),(3,5),(3,6),(3,7),(3,8),\\
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&(5,3),(5,4), (5,5), (5,6), (5,7)\\
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&(7,3), (7,4), (7,5), (11,3), (11,4), (11,5),\\
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&(13,3), (17,3), (19,3)\}
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\end{align*}
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then $d_k(\Gamma_1(p))>0$.
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\end{conjecture}
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415
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\section{The Conjecture}
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\newcommand{\tT}{\tilde{\T}}
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419
Let~$k=2m$ be an even integer and~$p$ a prime. Let $\T$ be the Hecke
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algebra associated to $S_k(\Gamma_0(p))$ and let $\tT$ be the
421
normalization of $\tT$ in $\T\tensor\Q$.
422
\begin{conjecture}\label{conj:big}
423
$$
424
\ord_p([\tT : \T]) = \left\lfloor\frac{p}{12}\right\rfloor\cdot
425
\binom{m}{2} + a(p,m),
426
$$
427
where
428
$$
429
a(p,m) =
430
\begin{cases}
431
0 & \text{if $p\con 1\pmod{12}$,}\\
432
3\cdot\ds\binom{\lceil \frac{m}{3}\rceil}{2} & \text{if $p\con 5\pmod{12}$,}\\
433
2\cdot\ds\binom{\lceil \frac{m}{2}\rceil}{2} & \text{if $p\con 7\pmod{12}$,}\\
434
a(5,m)+a(7,m) & \text{if $p\con 11\pmod{12}$.}
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\end{cases}
436
$$
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In particular, when $k=2$ we conjecture that $[\tT:\T]$ is not
438
divisible by~$p$.
439
\end{conjecture}
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Here $\binom{x}{y}$ is the binomial coefficient ``$x$ choose $y$'',
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and floor and ceiling are as usual. We have checked this conjecture
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against significant numerical data. (Will describe here.)
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444
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\end{document}
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