\documentclass[11pt]{article} \usepackage{fancybox} \hoffset=-0.1\textwidth \textwidth=1.2\textwidth \voffset=-0.1\textheight \textheight=1.2\textheight \include{macros} \title{Discriminants of Hecke Algebras at Prime Level} \author{William A. Stein\footnote{This will eventually be a joint paper with Frank Calegari.}} \date{September 24, 2002} \DeclareMathOperator{\charpoly}{charpoly} \DeclareMathOperator{\splittemp}{split} \renewcommand{\split}{\splittemp} \DeclareMathOperator{\deriv}{deriv} \begin{document} \maketitle \begin{abstract} We study $p$-divisibility of discriminant of Hecke algebras associated to spaces of cusp forms of prime level. By considering cusp forms of weight bigger than~$2$, we are are led to make a conjecture about indexes of Hecke algebras in their normalization which, if true, implies that there are no mod~$p$ congruences between non-conjugate newforms in $S_2(\Gamma_0(p))$. \end{abstract} \section{Introduction} I started working in modular forms when Ken Ribet asked about discriminants of Hecke algebras at prime level. I've recently revisited this question and, with the help of Frank Calegari, have made some interesting discoveries. \section{Discriminants of Hecke Algebras} Let~$R$ be a ring and let~$A$ be an~$R$ algebra that is free as an~$R$ module. The trace of an element of~$A$ is the trace, in the sense of linear algebra, of left multiplication by that element on~$A$. \begin{definition}[Discriminant] Let $\omega_1,\ldots,\omega_n$ is a~$R$-basis for~$A$. Then the {\em discriminant} of~$A$, denoted $\disc(A)$, is the determinant of the $n\times n$ matrix $(\tr(\omega_i\omega_j))$, which is well defined modulo squares of units in~$A$. \end{definition} When $R=\Z$ the discriminant is well defined, since the only units are $\pm 1$. \begin{proposition}\label{prop:separable} Suppose~$R$ is a field. Then~$A$ has discriminant~$0$ if and only if~$A$ is separable over~$R$, i.e., for every extension $R'$ of $R$, the ring $A\tensor R'$ contains no nilpotents. \end{proposition} The following proof is summarized from Section~26 of Matsumura. If~$A$ contains a nilpotent then that nilpotent is in the kernel of the trace pairing. If~$A$ is separable then we may assume that~$R$ is algebraically closed. Then~$A$ is an Artinian reduced ring, hence isomorphic as a ring to a finite product of copies of~$R$, since~$R$ is algebraically closed. Thus the trace form on~$A$ is nondegenerate. \subsection{The Discriminant Valuation} Let $\Gamma$ be a congruence subgroup of $\SL_2(\Z)$, e.g., $\Gamma=\Gamma_0(p)$ or $\Gamma_1(p)$. For any integer $k\geq 1$, let $S_k(\Gamma)$ denote the space of holomorphic weight-$k$ cusp forms for $\Gamma$. Let $$ \T = \Z[\ldots,T_n,\ldots] \subset \End(S_k(\Gamma)) $$ be the associated Hecke algebra. Then~$\T$ is a commutative ring that is free and of finite rank as a $\Z$-module. Also of interest is the image $\T^{\new}$ of~$\T$ in $\End(S_k(\Gamma)^{\new})$. \begin{example} Let $\Gamma=\Gamma_0(243)$, which is illustrated on my T-shirt. Since $243=3^5$, experts will immediately deduce that $\disc(\T) = 0$. A computation shows that $$ \disc(\T^{\new}) = 2^{13} \cdot 3^{40}, $$ which reflects the mod-$2$ and mod-$3$ intersections all over my shirt. \end{example} \begin{definition}[Discriminant Valuation] Let~$p$ be a prime and suppose that $\Gamma=\Gamma_0(p)$ or $\Gamma_1(p)$. The {\em discriminant valuation} is $$ d_k(\Gamma) = \ord_p(\text{the discriminant of $\T$}). $$ %When the discriminant of $\T$ is~$0$ we define $d_k(\Gamma)$ to %be~$+\infty$. \end{definition} \section{Motivation and Applications} Let~$p$ be a prime and suppose that $\Gamma=\Gamma_0(p)$ or $\Gamma_1(p)$. The quantity $d_k(\Gamma)$ is of interest because it measures mod~$p$ congruences between eigenforms in $S_k(\Gamma)$. \begin{proposition} Suppose that $d_k(\Gamma)$ is finite. Then the discriminant valuation $d_k(\Gamma)$ is nonzero if and only if there is a mod-$p$ congruence between two Hecke eigenforms in $S_k(\Gamma)$ (note that the two congruent eigenforms might be Galois conjugate). \end{proposition} \begin{proof} It follows from Proposition~\ref{prop:separable} that $d_k(\Gamma)>0$ if and only if $\T\tensor \Fpbar$ is not separable. The Artinian ring $\T\tensor\Fpbar$ is not separable if and only if the number of ring homomorphisms $\T\tensor\Fpbar \ra \Fpbar$ is less than $$ \dim_{\Fpbar} \T\tensor\Fpbar = \dim_\C S_k(\Gamma). $$ Since $d_k(\Gamma)$ is finite, the number of ring homomorphisms $\T\tensor\Qpbar \ra \Qpbar$ equals $\dim_\C S_k(\Gamma)$. Using the standard bijection between congruences and normalized eigenforms, we see that $\T\tensor\Fpbar$ is not separable if and only if there is a mod-$p$ congruence between two eigenforms. \end{proof} \begin{example} If $\Gamma=\Gamma_0(389)$ and $k=2$, then $\dim_\C S_2(\Gamma) = 32$. Let~$f$ be the characteristic polynomial of $T_2$. One can check that~$f$ is square free and $389$ exactly divides the discriminant of~$f$, so $T_2$ generated $\T\tensor \Z_{389}$ as a ring. (If it generated a subring of $\T\tensor\Z_{389}$ of finite index, then the discriminant of~$f$ would be divisible by $389^2$.) Modulo~$389$ the polynomial~$f$ is congruent to $$\begin{array}{l} (x+2)(x+56)(x+135)(x+158)(x+175)^2(x+315)(x+342)(x^2+387)\\ (x^2+97x+164)(x^2 + 231x + 64)(x^2 + 286x + 63)(x^5 + 88x^4 +196x^3 + \\ 113x^2 +168x + 349)(x^{11} + 276x^{10} + 182x^9 + 13x^8 + 298x^7 + 316x^6 +\\ 213x^5 + 248x^4 + 108x^3 + 283x^2 + x + 101) \end{array} $$ The factor $(x+175)^2$ indicates that $\T\tensor \Fbar_{389}$ is not separable since the image of $T_2+175$ is nilpotent (its square is~$0$). There are $32$ eigenforms over~$\Q_2$ but only $31$ mod-$389$ eigenforms, so there must be a congruence. Let~$F$ be the $389$-adic newform whose $a_2$ term is a root of $$ x^2 + (-39 + 190\cdot 389 + 96\cdot 389^2 +\cdots) x + (-106 + 43\cdot 389 + 19\cdot 389^2 + \cdots). $$ Then the congruence is between~$F$ and its $\Gal(\Qbar_{389}/\Q_{389})$-conjugate. \end{example} \begin{example} The discriminant of the Hecke algebra $\T$ associated to $S_2(\Gamma_0(389))$ is $$ 2^{53} \!\cdot\! 3^{4} \!\cdot\! 5^{6} \!\cdot\! 31^{2} \!\cdot\! 37 \!\cdot\! 389 \!\cdot\! 3881 \!\cdot\! 215517113148241 \!\cdot\! 477439237737571441 $$ I computed this using the following algorithm, which was suggested by Hendrik Lenstra. Using the Sturm bound I found a~$b$ such that $T_1,\ldots,T_b$ generate $\T$ as a $\Z$-module. I then found a subset~$B$ of the $T_i$ that form a $\Q$-basis for $\T\tensor_\Z\Q$. Next, viewing $\T$ as a ring of matrices acting on $\Q^{32}$, I found a random vector $v\in\Q^{32}$ such that the set of vectors $C=\{T(v) : T \in B\}$ is linearly independent. Then I wrote each of $T_1(v),\ldots, T_b(v)$ as $\Q$-linear combinations of the elements of~$C$. Next I found a $\Z$-basis~$D$ for the $\Z$-span of these $\Q$-linear combinations of elements of~$C$. Tracing everything back, I find the trace pairing on the elements of~$D$, and deduce the discriminant by computing the determinant of the trace pairing matrix. The most difficult step is computing~$D$ from $T_1(v),\ldots,T_b(v)$ expressed in terms of~$C$, and this explains why we embed $\T$ in $\Q^{32}$ instead of viewing the elements of $\T$ as vectors in $\Q^{32^2}$. This whole computation takes one second on an Athlon 2000 processor. \end{example} \subsection{Literature} I've seen a version of Theorem~\ref{thm:disc} referred to in the following papers: \begin{enumerate} \item Ribet: {\em Torsion points on $J_0(N)$ and Galois representations} \item Lo\"\i{}c Merel and William Stein: {\em The field generated by the points of small prime order on an elliptic curve} \item Ken Ono and William McGraw: {\em Modular form Congruences and Selmer groups} (McGraw will speak about this next week in this seminar!) \item Momose and Ozawa: {\em Rational points of modular curves $X_{\split}(p)$} \end{enumerate} \section{Data About Discriminant Valuations} \subsection{Weight Two} \begin{theorem}\label{thm:disc} The only prime $p<60000$ such that $d_2(\Gamma_0(p))>0$ is $p=389$. (Except possibly $50923$ and $51437$, which I haven't finished checking yet.) \end{theorem} \begin{proof} This is the result of a large computer computation, and perhaps couldn't be verified any other way, since I know of no general theorems about $d_2(\Gamma_0(p))$. The rest of this proof describes how I did the computation, so you can be convinced that there is valid mathematics behind my computation, and that you could verify the computation given sufficient time. The computation described below took about one week using $12$ Athlon 2000MP processors. In 1999 I had checked the result stated above but only for $p<14000$ using a completely different implementation of the algorithm and a 200Mhz Pentium computer. These computations are nontrivial; we compute spaces of modular symbols, supersingular points, and Hecke operators on spaces of dimensions up to~$5000$. The aim is to determine whether or not~$p$ divides the discriminant of the Hecke algegra of level~$p$ for each $p < 60000$. If~$T$ is an operator with integral characteristic polynomial, we write $\disc(T)$ for $\disc(\charpoly(T))$, which also equals $\disc(\Z[T])$. We will often use that $$\disc(T)\!\!\!\!\mod{p} = \disc(\charpoly(T)\!\!\!\!\mod p).$$ Most levels~$p<60000$ were ruled out by computing characteristic polynomials of Hecke operators using an algorithm that David Kohel and I implemented in MAGMA, which is based on the Mestre-Oesterle method of graphs (our implementation is ``The Modular of Supersingular Points'' package that comes with MAGMA). I computed $\disc(T_q)$ modulo~$p$ for several primes~$q$, and in most cases found a~$q$ such that this discriminant is nonzero. The following table summarizes how often we used each prime~$q$ (note that there are $6057$ primes up to $60000$): \begin{center} \begin{tabular}{|l|l|}\hline $q$ & number of $p< 60000$ where~$q$ smallest s.t. $\disc(T_q)\neq 0$ mod~$p$\\\hline 2& 5809 times\\ 3& 161 (largest: 59471)\\ 5& 43 (largest: 57793)\\ 7& 15 (largest: 58699)\\ 11& 15 (the smallest is 307; the largest 50971)\\ 13& 2 (they are 577 and 5417)\\ 17& 3 (they are 17209, 24533, and 47387)\\ 19& 1 (it is 15661 )\\\hline \end{tabular} \end{center} The numbers in the right column sum to 6049, so 8 levels are missing. These are $$ 389,487,2341,7057,15641,28279, 50923, \text{ and } 51437. $$ (The last two are still being processed. $51437$ has the property that $\disc(T_q)=0$ for $q=2,3,\ldots,17$.) We determined the situation with the remaining 6 levels using Hecke operators $T_n$ with~$n$ composite. \begin{center} \begin{tabular}{|l|l|}\hline $p$ & How we rule level~$p$ out, if possible\\\hline 389& $p$ does divide discriminant\\ 487& using charpoly($T_{12}$)\\ 2341& using charpoly($T_6$)\\ 7057& using charpoly($T_{18}$)\\ 15641& using charpoly($T_6$)\\ 28279& using charpoly($T_{34}$)\\\hline \end{tabular} \end{center} Computing $T_n$ with~$n$ composite is very time consuming when~$p$ is large, so it is important to choose the right $T_n$ quickly. For $p=28279$, here is the trick I used to quickly find an~$n$ such that $\disc(T_n)$ is not divisible by~$p$. This trick might be used to speed up the computation for some other levels. The key idea is to efficiently discover which $T_n$ to compute. Though computing $T_n$ on the full space of modular symbols is quite hard, it turns out that there is an algorithm that quickly computes $T_n$ on subspaces of modular symbols with small dimension (see \S3.5.2 of my Ph.D. thesis). Let~$M$ be the space of mod~$p$ modular symbols of level $p=28279$, and let $f=\gcd(\charpoly(T_2),\deriv(\charpoly(T_2)))$. Let~$V$ be the kernel of $f(T_2)$ (this takes 7 minutes to compute). If $V=0$, we would be done, since then $\disc(T_2)\neq 0\in\F_p$. In fact,~$V$ has dimension~$7$. We find the first few integers~$n$ so that the charpoly of $T_n$ on $V_1$ has distinct roots, and they are $n=34$, $47$, $53$, and $89$. I then computed $\charpoly(T_{34})$ directly on the whole space and found that it has distinct roots modulo~$p$. \end{proof} \subsection{Higher Weight Data} \begin{enumerate} \item The following are the valuations $d=d_4(\Gamma_0(p))$ at~$p$ of the discriminant of the Hecke algebras associated to $S_4(\Gamma_0(p))$ for $p<500$. \hspace{-4em}\shadowbox{\begin{minipage}[b]{1.15\textwidth} \begin{tabular}{|c|ccccccccccccccccc|}\hline $p$ &2& 3& 5& 7& 11& 13& 17& 19& 23& 29& 31& 37& 41& 43& 47& 53& 59\\ $d$ &0& 0& 0& 0& 0& 2& 2& 2& 2& 4& 4& 6& 6& 6& 6& 8& 8\\\hline $p$&61& 67& 71& 73& 79& 83& 89& 97& 101& 103& 107& 109& 113& 127& 131& 137& 139\\ $d$ & 10& 10& 10& 12& 12& 12& 14& 16& 16& 16& 16& 18& 18& 20& 20& 22&24\\\hline $p$ & 149& 151& 157& 163& 167& 173& 179& 181& 191& 193& 197& 199& 211& 223& 227& 229& 233\\ $d$ & 24& 24& 26& 26& 26&28& 28& 30& 30& 32& 32& 32& 34& 36& 36& 38& 38\\ \hline $p$ & 239& 241& 251& 257& 263& 269& 271& 277& 281& 283& 293& 307& 311& 313& 317& 331& 337\\ $d$ & 38& 40& 40& 42& 42&44& 44& 46& 46& 46& 48& 50& 50& 52& 52& 54& 56\\\hline $p$ & 347& 349& 353& 359& 367& 373& 379& 383& 389&397& 401& 409& 419& 421& 431& 433& 439 \\ $d$ & 56& 58& 58& 58& 60&62& 62& 62& 65 &66& 66& 68& 68& 70& 70& 72& 72\\\hline $p$ & 443& 449& 457& 461& 463& 467& 479& 487& 491& 499 &&&&&&&\\ $d$ & 72& 74& 76& 76& 76& 76& 78& 80& 80& 82 &&&&&&&\\\hline \end{tabular} \end{minipage}} \comment{\item For each prime~$p$, let $$ \delta(p) = \dim S_4(\Gamma_0(p)) - \dim S_{p+3}(\Gamma_0(1)). $$ Then $|\delta(p) - d_4(\Gamma_0(p))| \leq 2$ for each $p<500$. Moreover, for every $p\neq 139$ we have that $\delta(p)\geq d_4(\Gamma_0(p))$, but for $p=139$, $\delta(p)=23$ but $d_4(\Gamma_0(p))=24$. } \end{enumerate} \section{The Conjecture} \newcommand{\tT}{\tilde{\T}} Let~$k=2m$ be an even integer and~$p$ a prime. Let $\T$ be the Hecke algebra associated to $S_k(\Gamma_0(p))$ and let $\tT$ be the normalization of $\tT$ in $\T\tensor\Q$. \begin{conjecture}\label{conj:big} $$ \ord_p([\tT : \T]) = \left\lfloor\frac{p}{12}\right\rfloor\cdot \binom{m}{2} + a(p,m), $$ where $$ a(p,m) = \begin{cases} 0 & \text{if $p\con 1\pmod{12}$,}\\ 3\cdot\ds\binom{\lceil \frac{m}{3}\rceil}{2} & \text{if $p\con 5\pmod{12}$,}\\ 2\cdot\ds\binom{\lceil \frac{m}{2}\rceil}{2} & \text{if $p\con 7\pmod{12}$,}\\ a(5,m)+a(7,m) & \text{if $p\con 11\pmod{12}$.} \end{cases} $$ In particular, when $k=2$ we conjecture that $[\tT:\T]$ is not divisible by~$p$. \end{conjecture} Here $\binom{x}{y}$ is the binomial coefficient ``$x$ choose $y$'', and floor and ceiling are as usual. We have checked this conjecture against significant numerical data. (Will describe here.) \section{Conjectures} \begin{conjecture} Suppose~$p$ is a prime and $k\geq 4$ is an even integer. If \begin{align*} (p,k) \not\in \{&(2,4),(2,6),(2,8),(2,10),\\ &(3,4),(3,6), (3,8),\\ &(5,4), (5,6), (7,4), (11,4)\} \end{align*} then $d_k(\Gamma_0(p))>0$. \end{conjecture} Frank Calegari outlined a possible strategy for proving this conjecture. \begin{conjecture} Suppose $p>2$ is a prime and $k\geq 3$ is an integer. If \begin{align*} (p,k) \not\in \{&(3,3),(3,4),(3,5),(3,6),(3,7),(3,8),\\ &(5,3),(5,4), (5,5), (5,6), (5,7)\\ &(7,3), (7,4), (7,5), (11,3), (11,4), (11,5),\\ &(13,3), (17,3), (19,3)\} \end{align*} then $d_k(\Gamma_1(p))>0$. \end{conjecture} \end{document} \begin{conjecture} For any even $k\geq 4$, $$\lim_{p\ra \infty} d_k(\Gamma_0(p)) = \infty.$$ \end{conjecture} \begin{conjecture} For any integer $k\geq 3$, $$\lim_{p\ra \infty} d_k(\Gamma_1(p)) = \infty.$$ \end{conjecture} The sequence of the limit need not be nonincreasing. \begin{conjecture} For any even integer $k\geq 4$, let $$ \delta_k(p) = \dim S_k(\Gamma_0(p)) - \dim S_{k+p-1}(\Gamma_0(1)). $$ Then $$ \lim_{p\ra \infty} \frac{\delta_k(p)}{d_k(\Gamma_0(p))} = 1. $$ \end{conjecture} \comment{By Theorem~\ref{thm:disc} the only~$p<50923$ such that $d_2(\Gamma_0(p))>0$ is $p=389$. We have $d_{2}(\Gamma_0(389))=1$ and $d_4(\Gamma_0(389))=65$. This is mysterious, because there's no (obvious) ring-theoretic relationship between weight-$2$ and weight-$4$ Hecke algebras. \begin{question} Is $d_{k}(\Gamma_0(389))$ odd for all even integers $k\geq 2$? \end{question} \begin{question}\label{ques:divdown} If $d_k(\Gamma_0(389))$ is odd for some~$k$, does it necessarily follow that $d_2(\Gamma_0(p))$ is odd? Note that $d_4(\Gamma_0(p))$ is even for all $p<500$, except $p=389$. \end{question} The answer to Question~\ref{ques:divdown} is ``no'' if $\Gamma_0(p)$ is replaced by $\Gamma_1(p)$. For example, $d_3(\Gamma_1(23))=1$, but $d_2(\Gamma_1(23))=0$. There is a newform in $S_3(\Gamma_1(23))$ with quadratic character that is defined over $K=\Q(\alpha)$ where $\alpha^3-12\alpha+7=0$, and the discriminant of~$K$ is $3^3\cdot 23$. We have $d_4(\Gamma_1(23)) = 4$.% and $d_5(\Gamma_1(23))=???$. }