CoCalc Shared Fileswww / papers / discheckediv / _region_.tex
Author: William A. Stein
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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{\makeatletter\gdef\[email protected]#1[#2]#3{[#3#1#2]}\gdef\cite{\@ifnextchar[{\[email protected]{, }}{\[email protected]{}[]}}}
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19
20\maketitle
21\begin{abstract}
22  We study $p$-divisibility of discriminant of Hecke algebras
23  associated to spaces of cusp forms of prime level.  By considering
24  cusp forms of weight bigger than~$2$, we are are led to make a
25  conjecture about indexes of Hecke algebras in their normalization
26  which, if true, implies that there are no mod~$p$ congruences
27  between non-conjugate newforms in $S_2(\Gamma_0(p))$.
28\end{abstract}
29
30\section{Introduction}
32discriminants of Hecke algebras at prime level.  I've recently
33revisited this question and, with the help of Frank Calegari, have
35
36\section{Discriminants of Hecke Algebras}
37Let~$R$ be a ring and let~$A$ be an~$R$ algebra that is free as an~$R$
38module.  The trace of an element of~$A$ is the trace, in the sense of
39linear algebra, of left multiplication by that element on~$A$.
40
41\begin{definition}[Discriminant]
42  Let $\omega_1,\ldots,\omega_n$ is a~$R$-basis for~$A$.  Then the
43  {\em discriminant} of~$A$, denoted $\disc(A)$, is the determinant of
44  the $n\times n$ matrix $(\tr(\omega_i\omega_j))$, which is well
45  defined modulo squares of units in~$A$.
46\end{definition}
47When $R=\Z$ the discriminant is well defined, since the only units are
48$\pm 1$.
49
50\begin{proposition}\label{prop:separable}
51  Suppose~$R$ is a field.  Then~$A$ has discriminant~$0$ if and only
52  if~$A$ is separable over~$R$, i.e., for every extension $R'$ of $R$,
53  the ring $A\tensor R'$ contains no nilpotents.
54\end{proposition}
55The following proof is summarized from Section~26 of Matsumura.
56If~$A$ contains a nilpotent then that nilpotent is in the kernel of
57the trace pairing.  If~$A$ is separable then we may assume that~$R$ is
58algebraically closed.  Then~$A$ is an Artinian reduced ring, hence
59isomorphic as a ring to a finite product of copies of~$R$, since~$R$
60is algebraically closed.  Thus the trace form on~$A$ is 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 $\Gamma_1(p)$.  For any integer $k\geq 1$, let
65$S_k(\Gamma)$ denote the space of holomorphic weight-$k$ cusp forms
66for $\Gamma$.  Let
67$$68\T = \Z[\ldots,T_n,\ldots] \subset \End(S_k(\Gamma)) 69$$
70be the associated Hecke algebra.  Then~$\T$ is a commutative ring
71that is free and of finite rank as a $\Z$-module.  Also of interest is
72the image $\T^{\new}$ of~$\T$ in $\End(S_k(\Gamma)^{\new})$.
73\begin{example}
74  Let $\Gamma=\Gamma_0(243)$, which is illustrated on my T-shirt.
75  Since $243=3^5$, experts will immediately deduce that $\disc(\T) = 76 0$.  A computation shows that
77  $$78 \disc(\T^{\new}) = 2^{13} \cdot 3^{40}, 79$$
80  which reflects the mod-$2$ and mod-$3$ intersections all over my
81  shirt.
82\end{example}
83
84
85\begin{definition}[Discriminant Valuation]
86  Let~$p$ be a prime and suppose that $\Gamma=\Gamma_0(p)$ or
87  $\Gamma_1(p)$.  The {\em discriminant valuation} is
88  $$89 d_k(\Gamma) = \ord_p(\text{the discriminant of \T}). 90$$
91%When the discriminant of $\T$ is~$0$ we define $d_k(\Gamma)$ to
92%be~$+\infty$.
93\end{definition}
94
95\section{Motivation and Applications}
96Let~$p$ be a prime and suppose that $\Gamma=\Gamma_0(p)$ or
97$\Gamma_1(p)$.  The quantity $d_k(\Gamma)$ is of interest because it
98measures mod~$p$ congruences between eigenforms in $S_k(\Gamma)$.
99\begin{proposition}
100  Suppose that $d_k(\Gamma)$ is finite.  Then the discriminant
101  valuation $d_k(\Gamma)$ is nonzero if and only if there is a mod-$p$
102  congruence between two Hecke eigenforms in $S_k(\Gamma)$ (note that
103  the two congruent eigenforms might be Galois conjugate).
104\end{proposition}
105\begin{proof}
106  It follows from Proposition~\ref{prop:separable} that
107  $d_k(\Gamma)>0$ if and only if $\T\tensor \Fpbar$ is not separable.
108  The Artinian ring $\T\tensor\Fpbar$ is not separable if and only if
109  the number of ring homomorphisms $\T\tensor\Fpbar \ra \Fpbar$ is
110  less than
111  $$112 \dim_{\Fpbar} \T\tensor\Fpbar = \dim_\C S_k(\Gamma). 113$$
114  Since $d_k(\Gamma)$ is finite, the number of ring homomorphisms
115  $\T\tensor\Qpbar \ra \Qpbar$ equals $\dim_\C S_k(\Gamma)$.  Using
116  the standard bijection between homomorphisms and normalized
117  eigenforms, we see that $\T\tensor\Fpbar$ is not separable if and
118  only if there is a mod-$p$ congruence between two eigenforms.
119\end{proof}
120
121\begin{example}
122  If $\Gamma=\Gamma_0(389)$ and $k=2$, then $\dim_\C S_2(\Gamma) = 123 32$.  Let~$f$ be the characteristic polynomial of $T_2$.  One can
124  check that~$f$ is square free and $389$ exactly divides the
125  discriminant of~$f$, so $T_2$ generated $\T\tensor \Z_{389}$ as a
126  ring. (If it generated a subring of $\T\tensor\Z_{389}$ of finite
127  index, then the discriminant of~$f$ would be divisible by $389^2$.)
128
129  Modulo~$389$ the polynomial~$f$ is congruent to
130  $$\begin{array}{l} 131 (x+2)(x+56)(x+135)(x+158)(x+175)^2(x+315)(x+342)(x^2+387)\\ 132 (x^2+97x+164)(x^2 + 231x + 64)(x^2 + 286x + 63)(x^5 + 88x^4 +196x^3 + \\ 133 113x^2 +168x + 349)(x^{11} + 276x^{10} + 182x^9 + 13x^8 + 298x^7 + 316x^6 +\\ 134 213x^5 + 248x^4 + 108x^3 + 283x^2 + x + 101) 135 \end{array} 136$$
137  The factor $(x+175)^2$ indicates that $\T\tensor \Fbar_{389}$ is
138  not separable since the image of $T_2+175$ is nilpotent (its square
139  is~$0$).  There are $32$ eigenforms over~$\Q_2$ but only $31$
140  mod-$389$ eigenforms, so there must be a congruence.  Let~$F$ be the
141  $389$-adic newform whose $a_2$ term is a root of
142  $$143 x^2 + (-39 + 190\cdot 389 + 96\cdot 389^2 +\cdots) x + (-106 + 144 43\cdot 389 + 19\cdot 389^2 + \cdots). 145$$
146  Then the congruence is between~$F$ and its
147  $\Gal(\Qbar_{389}/\Q_{389})$-conjugate.
148\end{example}
149
150\begin{example}
151  The discriminant of the Hecke algebra $\T$ associated to
152  $S_2(\Gamma_0(389))$ is
153  $$154 2^{53} \!\cdot\! 3^{4} \!\cdot\! 5^{6} \!\cdot\! 31^{2} \!\cdot\! 155 37 \!\cdot\! 389 \!\cdot\! 3881 \!\cdot\! 215517113148241 \!\cdot\! 156 477439237737571441 157$$
158  I computed this using the following algorithm, which was
159  suggested by Hendrik Lenstra.  Using the Sturm bound I found a~$b$
160  such that $T_1,\ldots,T_b$ generate $\T$ as a $\Z$-module.  I then
161  found a subset~$B$ of the $T_i$ that form a $\Q$-basis for
162  $\T\tensor_\Z\Q$.  Next, viewing $\T$ as a ring of matrices acting
163  on $\Q^{32}$, I found a random vector $v\in\Q^{32}$ such that the
164  set of vectors $C=\{T(v) : T \in B\}$ is linearly independent.  Then
165  I wrote each of $T_1(v),\ldots, T_b(v)$ as $\Q$-linear combinations
166  of the elements of~$C$.  Next I found a $\Z$-basis~$D$ for the
167  $\Z$-span of these $\Q$-linear combinations of elements of~$C$.
168  Tracing everything back, I find the trace pairing on the elements
169  of~$D$, and deduce the discriminant by computing the determinant of
170  the trace pairing matrix.  The most difficult step is computing~$D$
171  from $T_1(v),\ldots,T_b(v)$ expressed in terms of~$C$, and this
172  explains why we embed $\T$ in $\Q^{32}$ instead of viewing the
173  elements of $\T$ as vectors in $\Q^{32^2}$.  This whole computation
174  takes one second on an Athlon 2000 processor.
175\end{example}
176
177\subsection{Literature}
178I've seen a version of Theorem~\ref{thm:disc} referred to in the
179following papers:
180\begin{enumerate}
181\item Ribet: {\em Torsion points on $J_0(N)$ and Galois
182    representations}
183\item Lo\"\i{}c Merel and William Stein: {\em The field generated by
184    the points of small prime order on an elliptic curve}
185\item Ken Ono and William McGraw: {\em Modular form Congruences and
187  seminar!)
188\item Momose and Ozawa: {\em Rational points of modular curves
189    $X_{\split}(p)$}
190\end{enumerate}
191
192
194
195\subsection{Weight Two}
196\begin{theorem}\label{thm:disc}
197  The only prime $p<60000$ such that $d_2(\Gamma_0(p))>0$ is $p=389$.
198  (Except possibly $50923$ and $51437$, which I haven't finished
199  checking yet.)
200\end{theorem}
201\begin{proof}
202  This is the result of a large computer computation, and perhaps
203  couldn't be verified any other way, since I know of no general
204  theorems about $d_2(\Gamma_0(p))$.  The rest of this proof describes
205  how I did the computation, so you can be convinced that there is
206  valid mathematics behind my computation, and that you could verify
207  the computation given sufficient time.  The computation described
208  below took about one week using $12$ Athlon 2000MP processors.  In
209  1999 I had checked the result stated above but only for $p<14000$
210  using a completely different implementation of the algorithm and a
211  200Mhz Pentium computer.  These computations are nontrivial; we
212  compute spaces of modular symbols, supersingular points, and Hecke
213  operators on spaces of dimensions up to~$5000$.
214
215  The aim is to determine whether or not~$p$ divides the discriminant
216  of the Hecke algegra of level~$p$ for each $p < 60000$.  If~$T$ is
217  an operator with integral characteristic polynomial, we write
218  $\disc(T)$ for $\disc(\charpoly(T))$, which also equals
219  $\disc(\Z[T])$. We will often use that
220  $$\disc(T)\!\!\!\!\mod{p} = \disc(\charpoly(T)\!\!\!\!\mod p).$$
221
222  Most levels~$p<60000$ were ruled out by computing characteristic
223  polynomials of Hecke operators using an algorithm that David Kohel
224  and I implemented in MAGMA, which is based on the Mestre-Oesterle
225  method of graphs (our implementation is The Modular of
226  Supersingular Points'' package that comes with MAGMA).  I computed
227  $\disc(T_q)$ modulo~$p$ for several primes~$q$, and in most cases
228  found a~$q$ such that this discriminant is nonzero.  The following
229  table summarizes how often we used each prime~$q$ (note that there
230  are $6057$ primes up to $60000$):
231\begin{center}
232\begin{tabular}{|l|l|}\hline
233$q$  & number of $p< 60000$ where~$q$ smallest
234  s.t. $\disc(T_q)\neq 0$ mod~$p$\\\hline
2352&             5809 times\\
2363&             161   (largest: 59471)\\
2375&             43    (largest: 57793)\\
2387&             15    (largest: 58699)\\
23911&            15    (the smallest is 307; the largest 50971)\\
24013&            2     (they are 577 and 5417)\\
24117&            3     (they are 17209, 24533, and 47387)\\
24219&            1     (it is 15661 )\\\hline
243\end{tabular}
244\end{center}
245
246The numbers in the right column sum to 6049, so 8 levels are missing.
247These are
248$$249389,487,2341,7057,15641,28279, 50923, \text{ and } 51437. 250$$
251(The last two are still being processed.  $51437$ has the property
252that $\disc(T_q)=0$ for $q=2,3,\ldots,17$.)  We determined the
253situation with the remaining 6 levels using Hecke operators $T_n$
254with~$n$ composite.
255\begin{center}
256\begin{tabular}{|l|l|}\hline
257$p$ & How we rule level~$p$ out, if possible\\\hline
258389&   $p$ does divide discriminant\\
259487&   using charpoly($T_{12}$)\\
2602341&  using charpoly($T_6$)\\
2617057&  using charpoly($T_{18}$)\\
26215641& using charpoly($T_6$)\\
26328279& using charpoly($T_{34}$)\\\hline
264\end{tabular}
265\end{center}
266
267Computing $T_n$ with~$n$ composite is very time consuming when~$p$ is
268large, so it is important to choose the right $T_n$ quickly.  For
269$p=28279$, here is the trick I used to quickly find an~$n$ such that
270$\disc(T_n)$ is not divisible by~$p$.  This trick might be used to
271speed up the computation for some other levels.  The key idea is to
272efficiently discover which $T_n$ to compute.  Though computing $T_n$
273on the full space of modular symbols is quite hard, it turns out that
274there is an algorithm that quickly computes $T_n$ on subspaces of
275modular symbols with small dimension (see \S3.5.2 of my Ph.D. thesis).
276Let~$M$ be the space of mod~$p$ modular symbols of level $p=28279$,
277and let $f=\gcd(\charpoly(T_2),\deriv(\charpoly(T_2)))$.  Let~$V$ be
278the kernel of $f(T_2)$ (this takes 7 minutes to compute).  If $V=0$,
279we would be done, since then $\disc(T_2)\neq 0\in\F_p$.  In fact,~$V$
280has dimension~$7$.  We find the first few integers~$n$ so that the
281charpoly of $T_n$ on $V_1$ has distinct roots, and they are $n=34$,
282$47$, $53$, and $89$.  I then computed $\charpoly(T_{34})$ directly on
283the whole space and found that it has distinct roots modulo~$p$.
284\end{proof}
285
286\subsection{Higher Weight Data}
287\begin{enumerate}
288\item The following are the valuations $d=d_4(\Gamma_0(p))$ at~$p$ of
289  the discriminant of the Hecke algebras associated to
290  $S_4(\Gamma_0(p))$ for $p<500$.
291
293\begin{tabular}{|c|ccccccccccccccccc|}\hline
294$p$ &2& 3& 5& 7& 11& 13& 17& 19& 23& 29& 31& 37& 41& 43& 47& 53& 59\\
295$d$ &0& 0& 0& 0& 0& 2& 2& 2& 2& 4& 4& 6& 6& 6& 6& 8& 8\\\hline
296$p$&61& 67& 71& 73& 79& 83& 89& 97& 101& 103& 107& 109& 113& 127& 131&  137& 139\\
297$d$ & 10& 10& 10& 12& 12& 12& 14& 16& 16& 16& 16& 18& 18& 20& 20& 22&24\\\hline
298$p$ & 149& 151& 157& 163& 167& 173& 179& 181& 191& 193& 197& 199&
299  211& 223& 227& 229& 233\\
300$d$ & 24& 24& 26& 26& 26&28& 28& 30& 30& 32& 32& 32& 34& 36& 36& 38& 38\\ \hline
301$p$ & 239& 241& 251& 257& 263& 269& 271& 277&
302  281& 283& 293& 307& 311& 313& 317& 331& 337\\
303$d$ & 38& 40& 40& 42& 42&44& 44& 46& 46& 46& 48& 50& 50& 52& 52& 54& 56\\\hline
304$p$ & 347& 349& 353& 359& 367& 373& 379& 383& 389&397& 401& 409& 419& 421& 431& 433& 439 \\
305$d$ & 56& 58& 58& 58& 60&62& 62& 62& 65  &66& 66& 68& 68& 70& 70& 72& 72\\\hline
306$p$ &  443& 449& 457& 461& 463& 467& 479& 487& 491& 499 &&&&&&&\\
307$d$ &  72& 74& 76& 76& 76& 76& 78& 80& 80& 82 &&&&&&&\\\hline
308\end{tabular}
309\end{minipage}}
310
311\comment{\item For each prime~$p$, let
312  $$313 \delta(p) = \dim S_4(\Gamma_0(p)) - \dim S_{p+3}(\Gamma_0(1)). 314$$
315  Then $|\delta(p) - d_4(\Gamma_0(p))| \leq 2$ for each $p<500$.
316  Moreover, for every $p\neq 139$ we have that $\delta(p)\geq 317 d_4(\Gamma_0(p))$, but for $p=139$, $\delta(p)=23$ but
318  $d_4(\Gamma_0(p))=24$.  }
319\end{enumerate}
320
321
322\section{The Discriminant is Divisible by~$p$}
323In this section we prove that for $k\geq 4$ the discriminant of the
324Hecke algebra associated to $S_k(\Gamma_0(p))$ is almost always
325divisible by~$p$.
326
327\begin{theorem}
328  Suppose~$p$ is a prime and $k\geq 4$ is an even integer.  Then
329  $d_k(\Gamma_0(p))>0$ unless
330\begin{align*}
331  (p,k) \not\in \{&(2,4),(2,6),(2,8),(2,10),\\
332  &(3,4),(3,6), (3,8),\\
333  &(5,4), (5,6), (7,4), (11,4)\},
334\end{align*}
335in which case $d_k(\Gamma_0(p))=0$.
336\end{theorem}
337\begin{proof}
338  (Romyar and William came up with this.)
339
340  Let~$p$ be a prime.  For~$N$ and~$k$ integers, let
341  $$342 S_k(\Gamma_0(N),\Zpbar) := S_k(\Gamma_0(N),\Z) \tensor \Zpbar 343$$
344  and
345  $$346 S_k(\Gamma_0(N),\Fpbar) := S_k(\Gamma_0(N),\Zpbar) \tensor 347 \Fpbar. 348$$
349
350  Let $\wp$ denote the maximal ideal of $\Zpbar$.  In
351  \cite[\S3]{serre:antwerp72} Serre defines a linear map
352  $$353 t: S_k(\Gamma_0(p), \Zpbar) \ra S_{k+(p-1)}(\Gamma_0(p),\Zpbar) 354$$
355  given by
356  $$357 t(f) = \Tr(f\cdot{} G) 358$$
359  where~$G$ is an Eisenstein series of weight~$p-1$ such that
360  $G\con 1\pmod{\wp}$.  Serre shows (see Lemme~9 with $m=0$) that
361  $t(f) \con f \pmod{\wp}$.
362
363  By ??, there is a basis $f_1,\ldots,f_n$ of Hecke eigenforms for
364  $S_k(\Gamma_0(p),\Qpbar)$, and we may assume these $f_i$ are
365  normalized so that the leading coefficient of each~$q$-expansion
366  is~$1$.  If
367  $$\dim S_{k+(p-1)}(\Gamma_0(p),\Zpbar) < \dim S_k(\Gamma_0(p), 368 \Zpbar)$$
369  then the set of $q$-expansions
370  $$371 t(f_1) \!\!\!\!\pmod{\wp}, \,\ldots, \,t(f_n)\!\!\!\!\pmod{\wp} 372$$
373  can not be linearly independent, which implies by Lemma~?? that
374  $d_k(\Gamma_0(p))>0$.
375
376  It follows from ?? that
377\begin{align*}
378  \dim S_k(\Gamma_0(p))
379  &= \frac{(k-1)(p+1)}{12}\\
380  &+ \left( 1 + \kr{-4}{p}\right) \cdot \left( \frac{1-k}{4} +
381    \left\lfloor\frac{k}{4}\right\rfloor \right) + \left( 1 +
382    \kr{-3}{p}\right) \cdot \left(\frac{1-k}{3} +
383    \left\lfloor\frac{k}{3}\right\rfloor \right) - 1.
384\end{align*}
385Thus
386$$387\dim S_k(\Gamma_0(p)) \geq \frac{(k-1)(p+1)}{12} - 3. 388$$
389By ??,
390$$391\dim S_{k+p-1}(\Gamma_0(1)) = \begin{cases} 392 \lfloor \frac{k+p-1}{12}\rfloor - 1 & \text{ k+p-1\con 2\pmod{12}},\\ 393 \lfloor \frac{k+p-1}{12}\rfloor & \text{otherwise.} 394 \end{cases} 395$$
396  If
397  $$398 \dim S_k(\Gamma_0(p)) \leq \dim S_{k+p-1}(\Gamma_0(1)) 399$$
400  then $(k-2)p\leq 36$.  This reduces the assertion of the theorem
401  to a very small finite computation, which we did using the
402  algorithms described earlier in this paper.
403\end{proof}
404
405
406
407\begin{conjecture}
408  Suppose $p>2$ is a prime and $k\geq 3$ is an integer.  If
409\begin{align*}
410  (p,k) \not\in \{&(3,3),(3,4),(3,5),(3,6),(3,7),(3,8),\\
411  &(5,3),(5,4), (5,5), (5,6), (5,7)\\
412  &(7,3), (7,4), (7,5), (11,3), (11,4), (11,5),\\
413  &(13,3), (17,3), (19,3)\}
414\end{align*}
415then $d_k(\Gamma_1(p))>0$.
416\end{conjecture}
417
418
419\section{The Conjecture}
420\newcommand{\tT}{\tilde{\T}}
421
422Let~$k=2m$ be an even integer and~$p$ a prime.  Let $\T$ be the Hecke
423algebra associated to $S_k(\Gamma_0(p))$ and let $\tT$ be the
424normalization of $\tT$ in $\T\tensor\Q$.
425\begin{conjecture}\label{conj:big}
426  $$427 \ord_p([\tT : \T]) = \left\lfloor\frac{p}{12}\right\rfloor\cdot 428 \binom{m}{2} + a(p,m), 429$$
430  where
431  $$432 a(p,m) = 433\begin{cases} 434 0 & \text{if p\con 1\pmod{12},}\\ 435 3\cdot\ds\binom{\lceil \frac{m}{3}\rceil}{2} & \text{if p\con 5\pmod{12},}\\ 436 2\cdot\ds\binom{\lceil \frac{m}{2}\rceil}{2} & \text{if p\con 7\pmod{12},}\\ 437 a(5,m)+a(7,m) & \text{if p\con 11\pmod{12}.} 438\end{cases} 439$$
440In particular, when $k=2$ we conjecture that $[\tT:\T]$ is not
441divisible by~$p$.
442\end{conjecture}
443Here $\binom{x}{y}$ is the binomial coefficient $x$ choose $y$'',
444and floor and ceiling are as usual.  We have checked this conjecture
445against significant numerical data.  (Will describe here.)
446
447
448\end{document}
449
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