Author: William A. Stein
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8\title{Discriminants of Hecke Algebras at Prime Level}
9\author{William A. Stein\footnote{This will eventually be a joint paper with Frank Calegari.}}
10\date{September 24, 2002}
11\DeclareMathOperator{\charpoly}{charpoly}
12\DeclareMathOperator{\splittemp}{split}
13\renewcommand{\split}{\splittemp}
14\DeclareMathOperator{\deriv}{deriv}
15\begin{document}
16\maketitle
17\begin{abstract}
18We study $p$-divisibility of discriminant of Hecke algebras associated
19to spaces of cusp forms of prime level.  By considering cusp forms of
20weight bigger than~$2$, we are are led to make a conjecture about
21indexes of Hecke algebras in their normalization which, if true,
22implies that there are no mod~$p$ congruences between
23non-conjugate newforms in $S_2(\Gamma_0(p))$.
24\end{abstract}
25
26\section{Introduction}
27I started working in modular forms when Ken Ribet asked about discriminants of
28Hecke algebras at prime level.  I've recently revisited this question and, with
29the help of Frank Calegari, have made some interesting discoveries.
30
31
32\section{Discriminants of Hecke Algebras}
33Let~$R$ be a ring and let~$A$ be an~$R$ algebra that
34is free as an~$R$ module.
35The trace of an element of~$A$ is the trace, in the
36sense of linear algebra, of left
37multiplication by that element on~$A$.
38
39\begin{definition}[Discriminant]
40Let $\omega_1,\ldots,\omega_n$ is a~$R$-basis
41for~$A$.  Then the {\em discriminant} of~$A$, denoted $\disc(A)$,
42is the determinant of the $n\times n$ matrix
43$(\tr(\omega_i\omega_j))$, which is well defined
44modulo squares of units in~$A$.
45\end{definition}
46When $R=\Z$ the discriminant is well defined, since
47the only units are $\pm 1$.
48
49\begin{proposition}\label{prop:separable}
50Suppose~$R$ is a field.  Then~$A$ has discriminant~$0$ if and only
51if~$A$ is separable over~$R$, i.e., for every extension
52$R'$ of $R$, the ring $A\tensor R'$ contains no nilpotents.
53\end{proposition}
54The following proof is summarized from Section~26 of Matsumura.
55 If~$A$ contains a nilpotent then that nilpotent is in the kernel of
56 the trace pairing.  If~$A$ is separable then we may assume that~$R$ is
57 algebraically closed.  Then~$A$ is an Artinian reduced ring, hence
58 isomorphic as a ring to a finite product of copies of~$R$, since~$R$
59 is algebraically closed.  Thus the trace form on~$A$ is
60 nondegenerate.
61
62\subsection{The Discriminant Valuation}
63Let $\Gamma$ be a congruence subgroup of $\SL_2(\Z)$, e.g.,
64$\Gamma=\Gamma_0(p)$ or $\Gamma_1(p)$.
65For any integer $k\geq 1$, let $S_k(\Gamma)$ denote
66the space of holomorphic weight-$k$ cusp forms for $\Gamma$.  Let
67$$68\T = \Z[\ldots,T_n,\ldots] \subset \End(S_k(\Gamma)) 69$$
70be the associated Hecke algebra.
71Then~$\T$ is a commutative ring that is free and of finite rank
72as a $\Z$-module.  Also of interest is the image $\T^{\new}$
73of~$\T$ in $\End(S_k(\Gamma)^{\new})$.
74\begin{example}
75Let $\Gamma=\Gamma_0(243)$, which is illustrated on
76my T-shirt.  Since $243=3^5$, experts will immediately
77deduce that $\disc(\T) = 0$.   A computation shows that
78$$79 \disc(\T^{\new}) = 2^{13} \cdot 3^{40}, 80$$
81which reflects the mod-$2$ and mod-$3$ intersections all
82over my shirt.
83\end{example}
84
85
86\begin{definition}[Discriminant Valuation]
87Let~$p$ be a prime and suppose that $\Gamma=\Gamma_0(p)$ or
88$\Gamma_1(p)$.
89The {\em discriminant valuation} is
90$$91 d_k(\Gamma) = \ord_p(\text{the discriminant of \T}). 92$$
93%When the discriminant of $\T$ is~$0$ we define $d_k(\Gamma)$ to
94%be~$+\infty$.
95\end{definition}
96
97\section{Motivation and Applications}
98Let~$p$ be a prime and suppose that $\Gamma=\Gamma_0(p)$ or
99$\Gamma_1(p)$.
100The quantity $d_k(\Gamma)$ is of interest because it measures mod~$p$
101congruences between eigenforms in $S_k(\Gamma)$.
102\begin{proposition}
103Suppose that $d_k(\Gamma)$ is finite.
104Then the discriminant  valuation $d_k(\Gamma)$ is nonzero
105if and only if there is a mod-$p$ congruence
106between two Hecke eigenforms in $S_k(\Gamma)$
107(note that the two congruent eigenforms might
108be Galois conjugate).
109\end{proposition}
110\begin{proof}
111It follows from Proposition~\ref{prop:separable} that
112$d_k(\Gamma)>0$ if and only if $\T\tensor \Fpbar$ is not
113separable.  The Artinian ring $\T\tensor\Fpbar$ is
114not separable if and only if the number of ring
115homomorphisms $\T\tensor\Fpbar \ra \Fpbar$ is
116less than
117$$118\dim_{\Fpbar} \T\tensor\Fpbar = \dim_\C S_k(\Gamma). 119$$
120Since $d_k(\Gamma)$ is finite, the number of ring
121homomorphisms $\T\tensor\Qpbar \ra \Qpbar$ equals
122$\dim_\C S_k(\Gamma)$.  Using the standard bijection between
123congruences and normalized eigenforms, we see that
124$\T\tensor\Fpbar$ is not separable if and only
125if there is a mod-$p$ congruence between two eigenforms.
126\end{proof}
127
128\begin{example}
129If $\Gamma=\Gamma_0(389)$ and $k=2$,
130then $\dim_\C S_2(\Gamma) = 32$.
131Let~$f$ be the characteristic polynomial of $T_2$.
132One can check that~$f$ is square free and $389$ exactly
133divides the discriminant of~$f$, so $T_2$ generated
134$\T\tensor \Z_{389}$ as a ring. (If it generated a subring of $\T\tensor\Z_{389}$
135of finite index, then the discriminant of~$f$ would be divisible
136by $389^2$.)
137
138Modulo~$389$ the polynomial~$f$ is congruent to
139$$\begin{array}{l} 140 (x+2)(x+56)(x+135)(x+158)(x+175)^2(x+315)(x+342)(x^2+387)\\ 141 (x^2+97x+164)(x^2 + 231x + 64)(x^2 + 286x + 63)(x^5 + 88x^4 +196x^3 + \\ 142 113x^2 +168x + 349)(x^{11} + 276x^{10} + 182x^9 + 13x^8 + 298x^7 + 316x^6 +\\ 143 213x^5 + 248x^4 + 108x^3 + 283x^2 + x + 101) 144 \end{array} 145$$
146The factor $(x+175)^2$ indicates that
147$\T\tensor \Fbar_{389}$ is not separable
148since the image of $T_2+175$ is nilpotent
149(its square is~$0$).  There are $32$ eigenforms over~$\Q_2$
150but only $31$ mod-$389$ eigenforms, so there must be a congruence.
151Let~$F$ be the $389$-adic newform  whose $a_2$ term is a root of
152$$153x^2 + (-39 + 190\cdot 389 + 96\cdot 389^2 +\cdots) x + (-106 + 43\cdot 389 + 15419\cdot 389^2 + \cdots). 155$$
156Then the congruence is between~$F$
157and its $\Gal(\Qbar_{389}/\Q_{389})$-conjugate.
158\end{example}
159
160\begin{example}
161The discriminant of the Hecke algebra $\T$ associated
162to $S_2(\Gamma_0(389))$ is
163$$164 2^{53} \!\cdot\! 3^{4} \!\cdot\! 5^{6} \!\cdot\! 31^{2} \!\cdot\! 37 \!\cdot\! 389 \!\cdot\! 3881 \!\cdot\! 215517113148241 \!\cdot\! 477439237737571441 165$$
166I computed this using the following algorithm, which was suggested
167by Hendrik Lenstra.  Using the Sturm bound I found a~$b$
168such that $T_1,\ldots,T_b$ generate $\T$ as a $\Z$-module.  I then
169found a subset~$B$ of the $T_i$ that form a $\Q$-basis for
170$\T\tensor_\Z\Q$.
171Next, viewing $\T$ as a ring of matrices acting on $\Q^{32}$,
172I found a random vector $v\in\Q^{32}$ such that the set of
173vectors $C=\{T(v) : T \in B\}$ is linearly independent.  Then
174I wrote each of $T_1(v),\ldots, T_b(v)$ as $\Q$-linear combinations
175of the elements of~$C$.   Next I found a $\Z$-basis~$D$ for the
176$\Z$-span of these $\Q$-linear combinations of elements of~$C$.
177Tracing everything back, I find the trace pairing on
178the elements of~$D$, and deduce the discriminant by computing
179the determinant of the trace pairing matrix.  The most difficult
180step is computing~$D$ from $T_1(v),\ldots,T_b(v)$ expressed
181in terms of~$C$, and this explains why we embed $\T$ in $\Q^{32}$
182instead of viewing the elements of $\T$ as vectors in $\Q^{32^2}$.
183This whole computation takes one
184second on an Athlon 2000 processor.
185\end{example}
186
187\subsection{Literature}
188I've seen a version of Theorem~\ref{thm:disc} referred to in the
189following papers:
190\begin{enumerate}
191\item Ribet: {\em Torsion points on $J_0(N)$ and Galois representations}
192\item Lo\"\i{}c Merel and William Stein: {\em The field generated by the points of small prime
193order on an elliptic curve}
194\item Ken Ono and William McGraw:
195{\em Modular form Congruences and Selmer groups}
197\item Momose and Ozawa: {\em Rational points of
198modular curves $X_{\split}(p)$}
199\end{enumerate}
200
201
203
204\subsection{Weight Two}
205\begin{theorem}\label{thm:disc}
206The only prime $p<60000$ such that $d_2(\Gamma_0(p))>0$ is $p=389$.
207(Except possibly $50923$ and $51437$, which I
208haven't finished checking yet.)
209\end{theorem}
210\begin{proof}
211This is the result of a large computer computation, and perhaps
212couldn't be verified any other way, since I know of no general
213theorems about $d_2(\Gamma_0(p))$.  The rest of this proof describes
214how I did the computation, so you can be convinced that there is valid
215mathematics behind my computation, and that you could verify the
216computation given sufficient time.  The computation described below
217took about one week using $12$ Athlon 2000MP processors.  In 1999 I
218had checked the result stated above but only for $p<14000$ using a
219completely different implementation of the algorithm and a 200Mhz
220Pentium computer.  These computations are nontrivial; we compute
221spaces of modular symbols, supersingular points, and Hecke operators
222on spaces of dimensions up to~$5000$.
223
224The aim is to determine whether or not~$p$ divides the discriminant of
225the Hecke algegra of level~$p$ for each $p < 60000$.  If~$T$ is an
226operator with integral characteristic polynomial, we write $\disc(T)$
227for $\disc(\charpoly(T))$, which also equals $\disc(\Z[T])$. We will
228often use that
229$$\disc(T)\!\!\!\!\mod{p} = \disc(\charpoly(T)\!\!\!\!\mod p).$$
230
231Most levels~$p<60000$ were ruled out by
232computing characteristic polynomials of Hecke operators using an
233algorithm that David Kohel and I implemented in MAGMA, which is based
234on the Mestre-Oesterle method of graphs (our implementation is The Modular of
235Supersingular Points'' package that comes with MAGMA).  I computed
236$\disc(T_q)$ modulo~$p$ for several primes~$q$, and in most
237cases found a~$q$ such that this discriminant is nonzero.  The
238following table summarizes how often we used each prime~$q$ (note
239that there are $6057$ primes up to $60000$):
240\begin{center}
241\begin{tabular}{|l|l|}\hline
242$q$  & number of $p< 60000$ where~$q$ smallest
243  s.t. $\disc(T_q)\neq 0$ mod~$p$\\\hline
2442&             5809 times\\
2453&             161   (largest: 59471)\\
2465&             43    (largest: 57793)\\
2477&             15    (largest: 58699)\\
24811&            15    (the smallest is 307; the largest 50971)\\
24913&            2     (they are 577 and 5417)\\
25017&            3     (they are 17209, 24533, and 47387)\\
25119&            1     (it is 15661 )\\\hline
252\end{tabular}
253\end{center}
254
255The numbers in the right column sum to 6049, so 8 levels are
256missing.  These are
257$$258 389,487,2341,7057,15641,28279, 50923, \text{ and } 51437. 259$$
260(The last two are still being processed.  $51437$ has the property
261that $\disc(T_q)=0$ for $q=2,3,\ldots,17$.)
262We determined the situation with the remaining 6 levels
263using Hecke operators $T_n$ with~$n$ composite.
264\begin{center}
265\begin{tabular}{|l|l|}\hline
266$p$ & How we rule level~$p$ out, if possible\\\hline
267389&   $p$ does divide discriminant\\
268487&   using charpoly($T_{12}$)\\
2692341&  using charpoly($T_6$)\\
2707057&  using charpoly($T_{18}$)\\
27115641& using charpoly($T_6$)\\
27228279& using charpoly($T_{34}$)\\\hline
273\end{tabular}
274\end{center}
275
276Computing $T_n$ with~$n$ composite is very time consuming when~$p$ is
277large, so it is important to choose the right $T_n$ quickly.
278For $p=28279$, here is the trick I used to quickly find an~$n$ such
279that $\disc(T_n)$ is not divisible by~$p$.  This trick might be used
280to speed up the computation for some other levels.  The key idea is to
281efficiently discover which $T_n$ to compute.  Though computing $T_n$
282on the full space of modular symbols is quite hard, it turns out that
283there is an algorithm that quickly computes $T_n$ on subspaces of
284modular symbols with small dimension (see \S3.5.2 of my Ph.D. thesis).
285Let~$M$ be the space of mod~$p$ modular symbols of level $p=28279$,
286and let $f=\gcd(\charpoly(T_2),\deriv(\charpoly(T_2)))$.  Let~$V$ be the
287kernel of $f(T_2)$ (this takes 7 minutes to compute).  If $V=0$, we
288would be done, since then $\disc(T_2)\neq 0\in\F_p$.  In fact,~$V$ has
289dimension~$7$.  We find the first few integers~$n$ so that the
290charpoly of $T_n$ on $V_1$ has distinct roots, and they are
291$n=34$, $47$, $53$,  and $89$.  I then computed
292$\charpoly(T_{34})$ directly on the whole space and found that it has
293distinct roots modulo~$p$.
294\end{proof}
295
296\subsection{Higher Weight Data}
297\begin{enumerate}
298\item The following are the valuations $d=d_4(\Gamma_0(p))$ at~$p$ of the discriminant
299of the Hecke algebras associated to $S_4(\Gamma_0(p))$ for $p<500$.
300
302\begin{tabular}{|c|ccccccccccccccccc|}\hline
303$p$ &2& 3& 5& 7& 11& 13& 17& 19& 23& 29& 31& 37& 41& 43& 47& 53& 59\\
304$d$ &0& 0& 0& 0& 0& 2& 2& 2& 2& 4& 4& 6& 6& 6& 6& 8& 8\\\hline
305$p$&61& 67& 71& 73& 79& 83& 89& 97& 101& 103& 107& 109& 113& 127& 131&  137& 139\\
306$d$ & 10& 10& 10& 12& 12& 12& 14& 16& 16& 16& 16& 18& 18& 20& 20& 22&24\\\hline
307$p$ & 149& 151& 157& 163& 167& 173& 179& 181& 191& 193& 197& 199&
308  211& 223& 227& 229& 233\\
309$d$ & 24& 24& 26& 26& 26&28& 28& 30& 30& 32& 32& 32& 34& 36& 36& 38& 38\\ \hline
310$p$ & 239& 241& 251& 257& 263& 269& 271& 277&
311  281& 283& 293& 307& 311& 313& 317& 331& 337\\
312$d$ & 38& 40& 40& 42& 42&44& 44& 46& 46& 46& 48& 50& 50& 52& 52& 54& 56\\\hline
313$p$ & 347& 349& 353& 359& 367& 373& 379& 383& 389&397& 401& 409& 419& 421& 431& 433& 439 \\
314$d$ & 56& 58& 58& 58& 60&62& 62& 62& 65  &66& 66& 68& 68& 70& 70& 72& 72\\\hline
315$p$ &  443& 449& 457& 461& 463& 467& 479& 487& 491& 499 &&&&&&&\\
316$d$ &  72& 74& 76& 76& 76& 76& 78& 80& 80& 82 &&&&&&&\\\hline
317\end{tabular}
318\end{minipage}}
319
320\comment{\item
321For each prime~$p$, let
322$$323 \delta(p) = \dim S_4(\Gamma_0(p)) - \dim S_{p+3}(\Gamma_0(1)). 324$$
325Then $|\delta(p) - d_4(\Gamma_0(p))| \leq 2$ for each $p<500$.
326Moreover, for every $p\neq 139$ we have that $\delta(p)\geq 327d_4(\Gamma_0(p))$, but for $p=139$, $\delta(p)=23$ but
328$d_4(\Gamma_0(p))=24$.
329}
330\end{enumerate}
331
332
333\section{The Conjecture}
334\newcommand{\tT}{\tilde{\T}}
335
336Let~$k=2m$ be an even integer and~$p$ a prime.
337Let $\T$ be the Hecke algebra associated to $S_k(\Gamma_0(p))$ and
338let $\tT$ be the normalization of $\tT$ in $\T\tensor\Q$.
339\begin{conjecture}\label{conj:big}
340$$341 \ord_p([\tT : \T]) 342 = \left\lfloor\frac{p}{12}\right\rfloor\cdot \binom{m}{2} + a(p,m), 343$$
344where
345$$346a(p,m) = 347\begin{cases} 348 0 & \text{if p\con 1\pmod{12},}\\ 349 3\cdot\ds\binom{\lceil \frac{m}{3}\rceil}{2} & \text{if p\con 5\pmod{12},}\\ 350 2\cdot\ds\binom{\lceil \frac{m}{2}\rceil}{2} & \text{if p\con 7\pmod{12},}\\ 351 a(5,m)+a(7,m) & \text{if p\con 11\pmod{12}.} 352\end{cases} 353$$
354In particular, when $k=2$ we conjecture that
355$[\tT:\T]$ is not divisible by~$p$.
356\end{conjecture}
357Here $\binom{x}{y}$ is the binomial coefficient $x$ choose $y$'',
358and floor and ceiling are as usual.  We have checked this conjecture
359against significant numerical data.  (Will describe here.)
360
361
362\section{Conjectures}
363\begin{conjecture}
364Suppose~$p$ is a prime and $k\geq 4$ is an even integer.
365If
366\begin{align*}
367  (p,k) \not\in \{&(2,4),(2,6),(2,8),(2,10),\\
368        &(3,4),(3,6), (3,8),\\
369        &(5,4), (5,6), (7,4), (11,4)\}
370\end{align*}
371then  $d_k(\Gamma_0(p))>0$.
372\end{conjecture}
373Frank Calegari outlined a possible strategy for proving this conjecture.
374
375\begin{conjecture}
376Suppose $p>2$ is a prime and $k\geq 3$ is an integer.
377If
378\begin{align*}
379  (p,k) \not\in \{&(3,3),(3,4),(3,5),(3,6),(3,7),(3,8),\\
380        &(5,3),(5,4), (5,5), (5,6), (5,7)\\
381        &(7,3), (7,4), (7,5), (11,3), (11,4), (11,5),\\
382        &(13,3), (17,3), (19,3)\}
383\end{align*}
384then $d_k(\Gamma_1(p))>0$.
385\end{conjecture}
386\end{document}
387
388
389
390\begin{conjecture}
391For any even $k\geq 4$,
392$$\lim_{p\ra \infty} d_k(\Gamma_0(p)) = \infty.$$
393\end{conjecture}
394
395\begin{conjecture}
396For any integer $k\geq 3$,
397$$\lim_{p\ra \infty} d_k(\Gamma_1(p)) = \infty.$$
398\end{conjecture}
399
400The sequence of the limit need not be nonincreasing.
401
402\begin{conjecture}
403For any even integer $k\geq 4$, let
404$$405 \delta_k(p) = \dim S_k(\Gamma_0(p)) - \dim S_{k+p-1}(\Gamma_0(1)). 406$$
407Then
408$$409\lim_{p\ra \infty} 410 \frac{\delta_k(p)}{d_k(\Gamma_0(p))} = 1. 411$$
412\end{conjecture}
413
414\comment{By Theorem~\ref{thm:disc} the only~$p<50923$ such that
415$d_2(\Gamma_0(p))>0$ is $p=389$.  We have
416$d_{2}(\Gamma_0(389))=1$ and $d_4(\Gamma_0(389))=65$.
417This is mysterious, because there's no (obvious) ring-theoretic
418relationship between weight-$2$ and weight-$4$ Hecke algebras.
419\begin{question}
420Is $d_{k}(\Gamma_0(389))$ odd for all even integers $k\geq 2$?
421\end{question}
422\begin{question}\label{ques:divdown}
423If $d_k(\Gamma_0(389))$ is odd for some~$k$,
424does it necessarily follow that $d_2(\Gamma_0(p))$
425is odd?  Note that $d_4(\Gamma_0(p))$ is even for all $p<500$,
426except $p=389$.
427\end{question}
429is no'' if $\Gamma_0(p)$ is replaced
430by $\Gamma_1(p)$.  For example,
431$d_3(\Gamma_1(23))=1$,
432but $d_2(\Gamma_1(23))=0$.
433There is a newform in $S_3(\Gamma_1(23))$
436$K=\Q(\alpha)$ where $\alpha^3-12\alpha+7=0$, and
437the discriminant of~$K$ is $3^3\cdot 23$.
438We have $d_4(\Gamma_1(23)) = 4$.% and $d_5(\Gamma_1(23))=???$.