CoCalc Shared Fileswww / papers / discheckediv / a.tex
Author: William A. Stein
1\documentclass[11pt]{article} \usepackage{fancybox}
2\hoffset=-0.1\textwidth \textwidth=1.2\textwidth
3\voffset=-0.1\textheight \textheight=1.2\textheight
4\include{macros}
5\title{Discriminants of Hecke Algebras at Prime Level} \author{William
6  A. Stein\footnote{This will probably eventually be a joint paper
7    with Frank Calegari and Romyar Sharifi (?)}}  \date{September 24,
8  2002} \DeclareMathOperator{\charpoly}{charpoly}
9\DeclareMathOperator{\splittemp}{split}
10\renewcommand{\split}{\splittemp} \DeclareMathOperator{\deriv}{deriv}
11\newcommand{\Zpbar}{\overline{\Z}_p}
12\newcommand{\kr}[2]{\left(\frac{#1}{#2}\right)}
13
14
15\begin{document}
16\maketitle
17\begin{abstract}
18  We study $p$-divisibility of discriminant of Hecke algebras
19  associated to spaces of cusp forms of prime level.  By considering
20  cusp forms of weight bigger than~$2$, we are are led to make a
21  conjecture about indexes of Hecke algebras in their normalization
22  which, if true, implies that there are no mod~$p$ congruences
23  between non-conjugate newforms in $S_2(\Gamma_0(p))$.
24\end{abstract}
25
26\section{Introduction}
28discriminants of Hecke algebras at prime level.  I've recently
29revisited this question and, with the help of Frank Calegari, have
31
32\section{Discriminants of Hecke Algebras}
33Let~$R$ be a ring and let~$A$ be an~$R$ algebra that is free as an~$R$
34module.  The trace of an element of~$A$ is the trace, in the sense of
35linear algebra, of left multiplication by that element on~$A$.
36
37\begin{definition}[Discriminant]
38  Let $\omega_1,\ldots,\omega_n$ is a~$R$-basis for~$A$.  Then the
39  {\em discriminant} of~$A$, denoted $\disc(A)$, is the determinant of
40  the $n\times n$ matrix $(\tr(\omega_i\omega_j))$, which is well
41  defined modulo squares of units in~$A$.
42\end{definition}
43When $R=\Z$ the discriminant is well defined, since the only units are
44$\pm 1$.
45
46\begin{proposition}\label{prop:separable}
47  Suppose~$R$ is a field.  Then~$A$ has discriminant~$0$ if and only
48  if~$A$ is separable over~$R$, i.e., for every extension $R'$ of $R$,
49  the ring $A\tensor R'$ contains no nilpotents.
50\end{proposition}
51The following proof is summarized from Section~26 of Matsumura.
52If~$A$ contains a nilpotent then that nilpotent is in the kernel of
53the trace pairing.  If~$A$ is separable then we may assume that~$R$ is
54algebraically closed.  Then~$A$ is an Artinian reduced ring, hence
55isomorphic as a ring to a finite product of copies of~$R$, since~$R$
56is algebraically closed.  Thus the trace form on~$A$ is nondegenerate.
57
58\subsection{The Discriminant Valuation}
59Let $\Gamma$ be a congruence subgroup of $\SL_2(\Z)$, e.g.,
60$\Gamma=\Gamma_0(p)$ or $\Gamma_1(p)$.  For any integer $k\geq 1$, let
61$S_k(\Gamma)$ denote the space of holomorphic weight-$k$ cusp forms
62for $\Gamma$.  Let
63$$64\T = \Z[\ldots,T_n,\ldots] \subset \End(S_k(\Gamma)) 65$$
66be the associated Hecke algebra.  Then~$\T$ is a commutative ring
67that is free and of finite rank as a $\Z$-module.  Also of interest is
68the image $\T^{\new}$ of~$\T$ in $\End(S_k(\Gamma)^{\new})$.
69\begin{example}
70  Let $\Gamma=\Gamma_0(243)$, which is illustrated on my T-shirt.
71  Since $243=3^5$, experts will immediately deduce that $\disc(\T) = 72 0$.  A computation shows that
73  $$74 \disc(\T^{\new}) = 2^{13} \cdot 3^{40}, 75$$
76  which reflects the mod-$2$ and mod-$3$ intersections all over my
77  shirt.
78\end{example}
79
80
81\begin{definition}[Discriminant Valuation]
82  Let~$p$ be a prime and suppose that $\Gamma=\Gamma_0(p)$ or
83  $\Gamma_1(p)$.  The {\em discriminant valuation} is
84  $$85 d_k(\Gamma) = \ord_p(\text{the discriminant of \T}). 86$$
87%When the discriminant of $\T$ is~$0$ we define $d_k(\Gamma)$ to
88%be~$+\infty$.
89\end{definition}
90
91\section{Motivation and Applications}
92Let~$p$ be a prime and suppose that $\Gamma=\Gamma_0(p)$ or
93$\Gamma_1(p)$.  The quantity $d_k(\Gamma)$ is of interest because it
94measures mod~$p$ congruences between eigenforms in $S_k(\Gamma)$.
95\begin{proposition}
96  Suppose that $d_k(\Gamma)$ is finite.  Then the discriminant
97  valuation $d_k(\Gamma)$ is nonzero if and only if there is a mod-$p$
98  congruence between two Hecke eigenforms in $S_k(\Gamma)$ (note that
99  the two congruent eigenforms might be Galois conjugate).
100\end{proposition}
101\begin{proof}
102  It follows from Proposition~\ref{prop:separable} that
103  $d_k(\Gamma)>0$ if and only if $\T\tensor \Fpbar$ is not separable.
104  The Artinian ring $\T\tensor\Fpbar$ is not separable if and only if
105  the number of ring homomorphisms $\T\tensor\Fpbar \ra \Fpbar$ is
106  less than
107  $$108 \dim_{\Fpbar} \T\tensor\Fpbar = \dim_\C S_k(\Gamma). 109$$
110  Since $d_k(\Gamma)$ is finite, the number of ring homomorphisms
111  $\T\tensor\Qpbar \ra \Qpbar$ equals $\dim_\C S_k(\Gamma)$.  Using
112  the standard bijection between homomorphisms and normalized
113  eigenforms, we see that $\T\tensor\Fpbar$ is not separable if and
114  only if there is a mod-$p$ congruence between two eigenforms.
115\end{proof}
116
117\begin{example}
118  If $\Gamma=\Gamma_0(389)$ and $k=2$, then $\dim_\C S_2(\Gamma) = 119 32$.  Let~$f$ be the characteristic polynomial of $T_2$.  One can
120  check that~$f$ is square free and $389$ exactly divides the
121  discriminant of~$f$, so $T_2$ generated $\T\tensor \Z_{389}$ as a
122  ring. (If it generated a subring of $\T\tensor\Z_{389}$ of finite
123  index, then the discriminant of~$f$ would be divisible by $389^2$.)
124
125  Modulo~$389$ the polynomial~$f$ is congruent to
126  $$\begin{array}{l} 127 (x+2)(x+56)(x+135)(x+158)(x+175)^2(x+315)(x+342)(x^2+387)\\ 128 (x^2+97x+164)(x^2 + 231x + 64)(x^2 + 286x + 63)(x^5 + 88x^4 +196x^3 + \\ 129 113x^2 +168x + 349)(x^{11} + 276x^{10} + 182x^9 + 13x^8 + 298x^7 + 316x^6 +\\ 130 213x^5 + 248x^4 + 108x^3 + 283x^2 + x + 101) 131 \end{array} 132$$
133  The factor $(x+175)^2$ indicates that $\T\tensor \Fbar_{389}$ is
134  not separable since the image of $T_2+175$ is nilpotent (its square
135  is~$0$).  There are $32$ eigenforms over~$\Q_2$ but only $31$
136  mod-$389$ eigenforms, so there must be a congruence.  Let~$F$ be the
137  $389$-adic newform whose $a_2$ term is a root of
138  $$139 x^2 + (-39 + 190\cdot 389 + 96\cdot 389^2 +\cdots) x + (-106 + 140 43\cdot 389 + 19\cdot 389^2 + \cdots). 141$$
142  Then the congruence is between~$F$ and its
143  $\Gal(\Qbar_{389}/\Q_{389})$-conjugate.
144\end{example}
145
146\begin{example}
147  The discriminant of the Hecke algebra $\T$ associated to
148  $S_2(\Gamma_0(389))$ is
149  $$150 2^{53} \!\cdot\! 3^{4} \!\cdot\! 5^{6} \!\cdot\! 31^{2} \!\cdot\! 151 37 \!\cdot\! 389 \!\cdot\! 3881 \!\cdot\! 215517113148241 \!\cdot\! 152 477439237737571441 153$$
154  I computed this using the following algorithm, which was
155  suggested by Hendrik Lenstra.  Using the Sturm bound I found a~$b$
156  such that $T_1,\ldots,T_b$ generate $\T$ as a $\Z$-module.  I then
157  found a subset~$B$ of the $T_i$ that form a $\Q$-basis for
158  $\T\tensor_\Z\Q$.  Next, viewing $\T$ as a ring of matrices acting
159  on $\Q^{32}$, I found a random vector $v\in\Q^{32}$ such that the
160  set of vectors $C=\{T(v) : T \in B\}$ is linearly independent.  Then
161  I wrote each of $T_1(v),\ldots, T_b(v)$ as $\Q$-linear combinations
162  of the elements of~$C$.  Next I found a $\Z$-basis~$D$ for the
163  $\Z$-span of these $\Q$-linear combinations of elements of~$C$.
164  Tracing everything back, I find the trace pairing on the elements
165  of~$D$, and deduce the discriminant by computing the determinant of
166  the trace pairing matrix.  The most difficult step is computing~$D$
167  from $T_1(v),\ldots,T_b(v)$ expressed in terms of~$C$, and this
168  explains why we embed $\T$ in $\Q^{32}$ instead of viewing the
169  elements of $\T$ as vectors in $\Q^{32^2}$.  This whole computation
170  takes one second on an Athlon 2000 processor.
171\end{example}
172
173\subsection{Literature}
174I've seen a version of Theorem~\ref{thm:disc} referred to in the
175following papers:
176\begin{enumerate}
177\item Ribet: {\em Torsion points on $J_0(N)$ and Galois
178    representations}
179\item Lo\"\i{}c Merel and William Stein: {\em The field generated by
180    the points of small prime order on an elliptic curve}
181\item Ken Ono and William McGraw: {\em Modular form Congruences and
183  seminar!)
184\item Momose and Ozawa: {\em Rational points of modular curves
185    $X_{\split}(p)$}
186\end{enumerate}
187
188
190
191\subsection{Weight Two}
192\begin{theorem}\label{thm:disc}
193  The only prime $p<60000$ such that $d_2(\Gamma_0(p))>0$ is $p=389$.
194  (Except possibly $50923$ and $51437$, which I haven't finished
195  checking yet.)
196\end{theorem}
197\begin{proof}
198  This is the result of a large computer computation, and perhaps
199  couldn't be verified any other way, since I know of no general
200  theorems about $d_2(\Gamma_0(p))$.  The rest of this proof describes
201  how I did the computation, so you can be convinced that there is
202  valid mathematics behind my computation, and that you could verify
203  the computation given sufficient time.  The computation described
204  below took about one week using $12$ Athlon 2000MP processors.  In
205  1999 I had checked the result stated above but only for $p<14000$
206  using a completely different implementation of the algorithm and a
207  200Mhz Pentium computer.  These computations are nontrivial; we
208  compute spaces of modular symbols, supersingular points, and Hecke
209  operators on spaces of dimensions up to~$5000$.
210
211  The aim is to determine whether or not~$p$ divides the discriminant
212  of the Hecke algegra of level~$p$ for each $p < 60000$.  If~$T$ is
213  an operator with integral characteristic polynomial, we write
214  $\disc(T)$ for $\disc(\charpoly(T))$, which also equals
215  $\disc(\Z[T])$. We will often use that
216  $$\disc(T)\!\!\!\!\mod{p} = \disc(\charpoly(T)\!\!\!\!\mod p).$$
217
218  Most levels~$p<60000$ were ruled out by computing characteristic
219  polynomials of Hecke operators using an algorithm that David Kohel
220  and I implemented in MAGMA, which is based on the Mestre-Oesterle
221  method of graphs (our implementation is The Modular of
222  Supersingular Points'' package that comes with MAGMA).  I computed
223  $\disc(T_q)$ modulo~$p$ for several primes~$q$, and in most cases
224  found a~$q$ such that this discriminant is nonzero.  The following
225  table summarizes how often we used each prime~$q$ (note that there
226  are $6057$ primes up to $60000$):
227\begin{center}
228\begin{tabular}{|l|l|}\hline
229$q$  & number of $p< 60000$ where~$q$ smallest
230  s.t. $\disc(T_q)\neq 0$ mod~$p$\\\hline
2312&             5809 times\\
2323&             161   (largest: 59471)\\
2335&             43    (largest: 57793)\\
2347&             15    (largest: 58699)\\
23511&            15    (the smallest is 307; the largest 50971)\\
23613&            2     (they are 577 and 5417)\\
23717&            3     (they are 17209, 24533, and 47387)\\
23819&            1     (it is 15661 )\\\hline
239\end{tabular}
240\end{center}
241
242The numbers in the right column sum to 6049, so 8 levels are missing.
243These are
244$$245389,487,2341,7057,15641,28279, 50923, \text{ and } 51437. 246$$
247(The last two are still being processed.  $51437$ has the property
248that $\disc(T_q)=0$ for $q=2,3,\ldots,17$.)  We determined the
249situation with the remaining 6 levels using Hecke operators $T_n$
250with~$n$ composite.
251\begin{center}
252\begin{tabular}{|l|l|}\hline
253$p$ & How we rule level~$p$ out, if possible\\\hline
254389&   $p$ does divide discriminant\\
255487&   using charpoly($T_{12}$)\\
2562341&  using charpoly($T_6$)\\
2577057&  using charpoly($T_{18}$)\\
25815641& using charpoly($T_6$)\\
25928279& using charpoly($T_{34}$)\\\hline
260\end{tabular}
261\end{center}
262
263Computing $T_n$ with~$n$ composite is very time consuming when~$p$ is
264large, so it is important to choose the right $T_n$ quickly.  For
265$p=28279$, here is the trick I used to quickly find an~$n$ such that
266$\disc(T_n)$ is not divisible by~$p$.  This trick might be used to
267speed up the computation for some other levels.  The key idea is to
268efficiently discover which $T_n$ to compute.  Though computing $T_n$
269on the full space of modular symbols is quite hard, it turns out that
270there is an algorithm that quickly computes $T_n$ on subspaces of
271modular symbols with small dimension (see \S3.5.2 of my Ph.D. thesis).
272Let~$M$ be the space of mod~$p$ modular symbols of level $p=28279$,
273and let $f=\gcd(\charpoly(T_2),\deriv(\charpoly(T_2)))$.  Let~$V$ be
274the kernel of $f(T_2)$ (this takes 7 minutes to compute).  If $V=0$,
275we would be done, since then $\disc(T_2)\neq 0\in\F_p$.  In fact,~$V$
276has dimension~$7$.  We find the first few integers~$n$ so that the
277charpoly of $T_n$ on $V_1$ has distinct roots, and they are $n=34$,
278$47$, $53$, and $89$.  I then computed $\charpoly(T_{34})$ directly on
279the whole space and found that it has distinct roots modulo~$p$.
280\end{proof}
281
282\subsection{Higher Weight Data}
283\begin{enumerate}
284\item The following are the valuations $d=d_4(\Gamma_0(p))$ at~$p$ of
285  the discriminant of the Hecke algebras associated to
286  $S_4(\Gamma_0(p))$ for $p<500$.
287
289\begin{tabular}{|c|ccccccccccccccccc|}\hline
290$p$ &2& 3& 5& 7& 11& 13& 17& 19& 23& 29& 31& 37& 41& 43& 47& 53& 59\\
291$d$ &0& 0& 0& 0& 0& 2& 2& 2& 2& 4& 4& 6& 6& 6& 6& 8& 8\\\hline
292$p$&61& 67& 71& 73& 79& 83& 89& 97& 101& 103& 107& 109& 113& 127& 131&  137& 139\\
293$d$ & 10& 10& 10& 12& 12& 12& 14& 16& 16& 16& 16& 18& 18& 20& 20& 22&24\\\hline
294$p$ & 149& 151& 157& 163& 167& 173& 179& 181& 191& 193& 197& 199&
295  211& 223& 227& 229& 233\\
296$d$ & 24& 24& 26& 26& 26&28& 28& 30& 30& 32& 32& 32& 34& 36& 36& 38& 38\\ \hline
297$p$ & 239& 241& 251& 257& 263& 269& 271& 277&
298  281& 283& 293& 307& 311& 313& 317& 331& 337\\
299$d$ & 38& 40& 40& 42& 42&44& 44& 46& 46& 46& 48& 50& 50& 52& 52& 54& 56\\\hline
300$p$ & 347& 349& 353& 359& 367& 373& 379& 383& 389&397& 401& 409& 419& 421& 431& 433& 439 \\
301$d$ & 56& 58& 58& 58& 60&62& 62& 62& 65  &66& 66& 68& 68& 70& 70& 72& 72\\\hline
302$p$ &  443& 449& 457& 461& 463& 467& 479& 487& 491& 499 &&&&&&&\\
303$d$ &  72& 74& 76& 76& 76& 76& 78& 80& 80& 82 &&&&&&&\\\hline
304\end{tabular}
305\end{minipage}}
306
307\comment{\item For each prime~$p$, let
308  $$309 \delta(p) = \dim S_4(\Gamma_0(p)) - \dim S_{p+3}(\Gamma_0(1)). 310$$
311  Then $|\delta(p) - d_4(\Gamma_0(p))| \leq 2$ for each $p<500$.
312  Moreover, for every $p\neq 139$ we have that $\delta(p)\geq 313 d_4(\Gamma_0(p))$, but for $p=139$, $\delta(p)=23$ but
314  $d_4(\Gamma_0(p))=24$.  }
315\end{enumerate}
316
317
318\section{The Discriminant is Divisible by~$p$|}
319
320In this section we prove that for $k\geq 4$ the discriminant of the
321Hecke algebra associated to $S_k(\Gamma_0(p))$ is almost always
322divisible by~$p$.
323
324\begin{theorem}
325  Suppose~$p$ is a prime and $k\geq 4$ is an even integer.  Then
326  $d_k(\Gamma_0(p))>0$ unless
327\begin{align*}
328  (p,k) \not\in \{&(2,4),(2,6),(2,8),(2,10),\\
329  &(3,4),(3,6), (3,8),\\
330  &(5,4), (5,6), (7,4), (11,4)\},
331\end{align*}
332in which case $d_k(\Gamma_0(p))=0$.
333\end{theorem}
334\begin{proof}
335  (Romyar and William came up with this.)
336
337  Let~$p$ be a prime.  For~$N$ and~$k$ integers, let
338  $$339 S_k(\Gamma_0(N),\Zpbar) := S_k(\Gamma_0(N),\Z) \tensor \Zpbar 340$$
341  and
342  $$343 S_k(\Gamma_0(N),\Fpbar) := S_k(\Gamma_0(N),\Zpbar) \tensor 344 \Fpbar. 345$$
346
347  Let $\wp$ denote the maximal ideal of $\Zpbar$.  In
348  \cite[\S3]{serre:antwerp72} Serre defines a linear map
349  $$350 t: S_k(\Gamma_0(p), \Zpbar) \ra S_{k+(p-1)}(\Gamma_0(p),\Zpbar) 351$$
352  given by
353  $$354 t(f) = \Tr(f\cdot{} G) 355$$
356  where~$G$ is an Eisenstein series of weight~$p-1$ such that
357  $G\con 1\pmod{\wp}$.  Serre shows (see Lemme~9 with $m=0$) that
358  $t(f) \con f \pmod{\wp}$.
359
360  By ??, there is a basis $f_1,\ldots,f_n$ of Hecke eigenforms for
361  $S_k(\Gamma_0(p),\Qpbar)$, and we may assume these $f_i$ are
362  normalized so that the leading coefficient of each~$q$-expansion
363  is~$1$.  If
364  $$\dim S_{k+(p-1)}(\Gamma_0(p),\Zpbar) < \dim S_k(\Gamma_0(p), 365 \Zpbar)$$
366  then the set of $q$-expansions
367  $$368 t(f_1) \!\!\!\!\pmod{\wp}, \,\ldots, \,t(f_n)\!\!\!\!\pmod{\wp} 369$$
370  can not be linearly independent, which implies by Lemma~?? that
371  $d_k(\Gamma_0(p))>0$.
372
373  It follows from ?? that
374\begin{align*}
375  \dim S_k(\Gamma_0(p))
376  &= \frac{(k-1)(p+1)}{12}\\
377  &+ \left( 1 + \kr{-4}{p}\right) \cdot \left( \frac{1-k}{4} +
378    \left\lfloor\frac{k}{4}\right\rfloor \right) + \left( 1 +
379    \kr{-3}{p}\right) \cdot \left(\frac{1-k}{3} +
380    \left\lfloor\frac{k}{3}\right\rfloor \right) - 1.
381\end{align*}
382Thus
383$$384\dim S_k(\Gamma_0(p)) \geq \frac{(k-1)(p+1)}{12} - 3. 385$$
386By ??,
387$$388\dim S_{k+p-1}(\Gamma_0(1)) = \begin{cases} 389 \lfloor \frac{k+p-1}{12}\rfloor - 1 & \text{ k+p-1\con 2\pmod{12}},\\ 390 \lfloor \frac{k+p-1}{12}\rfloor & \text{otherwise.} 391 \end{cases} 392$$
393  If
394  $$395 \dim S_k(\Gamma_0(p)) \leq \dim S_{k+p-1}(\Gamma_0(1)) 396$$
397  then $(k-2)p\leq 36$.  This reduces the assertion of the theorem
398  to a very small finite computation, which we did using the
399  algorithms described earlier in this paper.
400\end{proof}
401
402
403
404\begin{conjecture}
405  Suppose $p>2$ is a prime and $k\geq 3$ is an integer.  If
406\begin{align*}
407  (p,k) \not\in \{&(3,3),(3,4),(3,5),(3,6),(3,7),(3,8),\\
408  &(5,3),(5,4), (5,5), (5,6), (5,7)\\
409  &(7,3), (7,4), (7,5), (11,3), (11,4), (11,5),\\
410  &(13,3), (17,3), (19,3)\}
411\end{align*}
412then $d_k(\Gamma_1(p))>0$.
413\end{conjecture}
414
415
416\section{The Conjecture}
417\newcommand{\tT}{\tilde{\T}}
418
419Let~$k=2m$ be an even integer and~$p$ a prime.  Let $\T$ be the Hecke
420algebra associated to $S_k(\Gamma_0(p))$ and let $\tT$ be the
421normalization 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  $$429 a(p,m) = 430\begin{cases} 431 0 & \text{if p\con 1\pmod{12},}\\ 432 3\cdot\ds\binom{\lceil \frac{m}{3}\rceil}{2} & \text{if p\con 5\pmod{12},}\\ 433 2\cdot\ds\binom{\lceil \frac{m}{2}\rceil}{2} & \text{if p\con 7\pmod{12},}\\ 434 a(5,m)+a(7,m) & \text{if p\con 11\pmod{12}.} 435\end{cases} 436$$
437In particular, when $k=2$ we conjecture that $[\tT:\T]$ is not
438divisible by~$p$.
439\end{conjecture}
440Here $\binom{x}{y}$ is the binomial coefficient $x$ choose $y$'',
441and floor and ceiling are as usual.  We have checked this conjecture
442against significant numerical data.  (Will describe here.)
443
444
445\end{document}
446