Sharedwww / tables / Notes / refinedeisen.texOpen in CoCalc
Author: William A. Stein
1% compgroup.tex
2\documentclass{article}
3\usepackage[all]{xy}
4\include{macros}
5
6\DeclareMathOperator{\numer}{numer}
7\DeclareMathOperator{\order}{order}
8\newcommand{\eisen}{\numer\left(\frac{p-1}{12}\right)}
9
10\title{The refined Eisenstein conjecture}
11\author{William A.~Stein}
12\begin{document}
13\maketitle
14
15\section{Introduction}
16Let~$N$ be a prime, and consider the
17the curve~$X_0(N)$ whose complex points
18are the isomorphism classes of pairs consisting of a
19(generalized) elliptic curve and a cyclic subgroup of order~$N$.
20Let~$J_0(N)$ denote the Jacobian of $X_0(N)$; this is an abelian
21variety of dimension equal to the genus of~$X_0(N)$ whose points
22correspond to the degree~$0$ divisor classes on~$X_0(N)$.
23In~\cite{mazur:eisenstein} B.~Mazur gave a complete description
24of the torsion subgroup and component group of~$J_0(N)$.
25
26Consider a normalized weight-two eigenform $f = \sum a_n q^n$
27on~$X_0(N)$, and let~$I$ be the kernel of the natural
28map $\T\ra\Z[\ldots,a_n,\ldots]$ that sends a Hecke
29operator~$T_n$ to~$a_n$.  Denote by $A_f$ the optimal abelian variety
30quotient $J_0(N)/IJ_0(N)$.  In this paper we give evidence for
31a conjectural refinement of Mazur's theorem to quotients~$A_f$
32of~$J_0(N)$.  Let $C_f$ denote the cyclic subgroup of $A_f(\Q)$
33generated by the image of the point $0-\infty\in J_0(N)$.
34
35\begin{conjecture}[Refined Eisenstein conjecture]\label{conj:main}\mbox{}
36\vspace{-1ex}
37
38\begin{itemize}
39\item[(a)] Let $\pi:J_0(N)\ra A_f$ be the optimal quotient corresponding
40to the newform~$f$. Then
41$$A_f(\Q)_{\tor}\isom \Adual_f(\Q)_{\tor} 42\isom \Phi_{A_f}(\Fp) 43\isom \Phi_{A_f}(\Fpbar) 44\isom C_f.$$
45\vspace{-5ex}
46
47\item[(b)] Assume conjecture (a), and
48let~$n_f$ denote the common order.
49Then
50   $\prod n_f = \numer\left(\frac{N-1}{12}\right),$
51where the product is over the Galois conjugacy classes
52of normalized eigenforms.
53\end{itemize}
54\end{conjecture}
55
56The conjecture is false when the level is composite.
57For example, let $E=${\bf 33A}; then
58$E(\Q)=\Z/2\cross\Z/2$ but $\Phi_E(\Fbar_3)=\Z/6\Z$.
59
60
61\section{Computational evidence}
62There are $305$ factors $A_f$ of level up to $631$;
63there are $860$ factors up to level $2113$.
64\begin{proposition}
65Conjecture~\ref{conj:main} is true
66for $N\leq 631$; it is true for $N\leq 2113$, up to a $2$-power.
67(Except possibly for the statement about $\Adual_f(\Q)$.)
68\end{proposition}
69\begin{proof}
70We verified this using a computer program written in~\cite{magma}.
71The computation was done as follows.
72Viewed as an abelian variety over the complex number,
73$J_0(N)$ is the quotient
74$$J_0(N)(\C)=\Hom(S_2(N),\C)/H_1(X_0(N),\Z).$$
75Here $S_2(N)$ is
76the space of weight two cusp forms of level~$N$ and~$H_1(X_0(N),\Z)$
77is the first integral homology of the Riemann surface $X_0(N)$;
78both of these spaces, and the action of the
79Hecke operators~$T_n$, can be computed using the
80algorithms in~\cite{cremona:algs}.
81Using the Hecke operators, we list each optimal quotient~$A_f$.
82As a complex abelian variety, $A_f$ is the quotient
83$$A_f(\C) = \Hom(S_2(N)[I],\C)/\pi_* H_1(X_0(N),\Z).$$
84
85The torsion subgroup $A_f(\Q)_{\tor}$ can be computed by
86combining the following upper and lower bound.
87Manin proved in~\cite[p.~28 and Thm.~2.7b]{manin:parabolic}
88that $\xi = 0-\infty$ lies in $J_0(N)(\Q)_{\tor}$.
89The point~$\xi$ corresponds to the element
90$\{0,\infty\}\in H_1(X_0(N),\Z)\tensor\Q$, so we can
91compute~$\xi$ and its image in $A_f(\C)$.
92Combining the Eichler-Shimura relation with the
93injectivity of rational torsion under reduction modulo
94an odd prime~$p$ (see~\cite[p.~70]{cassels-flynn}),
95we obtain an upper bound on the order of the torsion
96subgroup of any abelian variety isogeneous to~$A_f$;
97this upper bound is computed from the characteristic
98polynomials of Hecke operators.
99In all cases that we computed, the computed upper bound agrees with
100the lower bound, so we obtain $A_f(\Q)_{\tor}$.
101Similar methods can be used to compute $\Adual_f(\Q)$, but we have
102not carried out this computation.
103
104The component group $\Phi_{A_f}(\Fbar_N)$
105can be computed using the algorithm
106described in~\cite{stein:compgroup}, which involves
107the monodromy pairing on the character groups of certain tori.
108In all cases
109in which the $\Gal(\Fbar_N/\F_N)$-action is nontrivial, the component
110group $\Phi_{A_f}(\Fbar_N)$ is trivial, so it suffices
111to know $\Phi_{A_f}(\Fbar_N)$.
112\end{proof}
113
114\subsection{Examples}
115In this section we give two examples. We decompose
116$J_0(N)$ as a product of $A_d$ where each
117$A_d$ is a $d$-dimensional factor corresponding to an eigenform.
118
119\begin{example} $N=113$ and $\numer\left(\frac{113-1}{12}\right) = 28$:
120$$\begin{array}{lrrrr} 121 J_0(113) \sim&A_1 &\cross A_2&\cross A_3 & \cross B_3\\ 122 \#A_f(\Q)_{\tor}&2&2&1&7\\ 123 \#\Phi_f(\F_N)&2&2&1&7\\ 124\frac{L(1)}{\Omega}&\frac{1}{2}&\frac{1}{2}&0&\frac{8}{7} 125\end{array}$$
126
127
128
129Remark: The Shafarevich-Tate group $\text{Sha}(B_3)$ probably
130has order $8$, and is explained by the Mordell-Weil
131group of $A_3$.
132\end{example}
133
134\begin{example} $N=389$ and $\numer\left(\frac{389-1}{12}\right) = 97$:
135$$\begin{array}{lrrrrr} 136 J_0(389) \sim& A_1 &\cross A_2&\cross A_3 &\cross A_6 &\cross A_{20}\\ 137\#A_f(\Q)_{\tor}& 1 & 1& 1&1&97\\ 138\#\Phi_f(\F_N)&1&1&1&1&97\\ 139\frac{L(1)}{\Omega}&0&0&0&0&\frac{2^{11}\cdot 5^2}{97} 140\end{array}$$
141
142\end{example}
143
144\section{Theoretical evidence}
145Mestre and Oesterl\'{e} proved in~\cite{mestre-oesterle:crelle}
146that if $A_f$ has dimension~1 then Conjecture~\ref{conj:main}(a)
147is true.
148Mazur proved in~\cite{mazur:eisenstein} that  %page 141?
149if $\ell \mid\# A_f(\Q)$, then $\ell \mid \eisen$; furthermore,
150 $$\# J_0(p)(\Q)_{\tor} = \# \Phi_{J_0(p)}(\Fpbar) 151 = \numer\left(\frac{p-1}{12}\right).$$
152
153Upon hearing of Conjecture~\ref{conj:main} Mazur said:
154\begin{quote}
155  What you are conjecturing is something I don't think I ever thought
156of---at least in those terms---It has some implications which are interesting
157(to me). For example, suppose that a prime~$q$ divides the numerator of
158$(p-1)/12$  but~$q^2$ does not; let~$A$ be the $q$-Eisenstein
159quotient. Then only
160one optimal quotient of~$A$ would be allowed to have nontrivial $q$-torsion,
161and the other(s), if there are any, would, I guess, then have to settle for
162a~$\mu_q$ subgroup.  I suppose, though, that it is very often the case that
163(under the above hypotheses of~$q$) the $q$-Eisenstein quotient is simple.
164\end{quote}
165
166
167\section{Application of Ribet's level lowering theorem}
168It is almost possible to obtain a partial consequence of
169Conjecture~\ref{conj:main} by applying the level lowering
170theorem proved by Ribet in~\cite{ribet:modreps}.
171If $\ell\mid \#\Phi_{A_f}(\Fpbar)$ then $\rho_{f,\m}$ tends
172to be finite for some~$\m$ containing~$\ell$.
173If so, and $\rho_{f,\m}$ is irreducible (and~$\ell$ is odd),
174then Ribet's level
176We now record the present state of our argument.
177
178If $\Phi_J\ra \Phi_{A_f}$ is surjective, then
179the primes on the left hand side of Conjecture~\ref{conj:main}(b)
180divide the right hand side, which prompts us to make
181the following weaker conjecture.
182\begin{conjecture}\label{conj:surjective}
183The natural map $\Phi_J \into \Phi_{A_f}$ is surjective.
184\end{conjecture}
185
186The component group~$\Phi_A$  of~$A$ tends to force some~$A[\m]$
187to arise from a finite flat group scheme.
188Ribet's theorem implies such~$\m$ are reducible, and then
189Mazur's theorem implies that such~$\m$ are Eisenstein.
190
191Suppose $\ell\mid\#\Phi_A$; does $\ell\mid \numer(\frac{N-1}{12})$?
192Let~$X$ be the character group of the torus attached to~$A$,
193a let $Y$ be character group attached to $\Adual$.
194The monodromy pairing expresses the component group
195of~$A$ by exactness of the sequence
196 $$0 \ra Y \ra \Hom(X,\Z) \ra \Phi_A \ra 0.$$
197By tensoring this sequence with $\Z/\ell\Z$, we obtain the exact sequence
198\begin{equation}\label{eqn:comptor}
1990 \ra \Phi_A[\ell] \ra Y\tensor \Z/\ell\Z \ra \Hom(X,\Z/\ell\Z)
200      \ra \Phi_A \tensor\Z/\ell\Z\ra 0.
201\end{equation}
202The Mumford-Tate uniformizations of~$A$ and~$\Adual$
203yield the following diagrams:
204$$\xymatrix{ 205 & 0\ar[r]\ar[d]& Y\ar[d]\ar[rr]^{\ell}& & Y\ar[d]\ar[r]& Y/\ell Y\ar[r]\ar[d]& 0\\ 2060\ar[r]& {\Hom(X/\ell X,\mu_\ell)}\ar[d]\ar[r] 207 & T\ar[d]\ar[rr]^{\cdot^\ell}&& T\ar[d]\ar[r] & 0\ar[d]\\ 208 0\ar[r]& {A[\ell]}\ar[r]& A\ar[rr]^{\ell}&& A\ar[r]& 0.}$$
209The snake lemma then gives an exact sequence
210 $$0 \ra \Hom(X/\ell X,\mu_{\ell}) 211 \ra A[\ell] \xrightarrow{\,f\,} Y/\ell Y \ra 0$$
212which, when combined with~(\ref{eqn:comptor}), produces the following diagram:
213$$\xymatrix{ 214 & & f^{-1}(\Phi_A[\ell])\[email protected]{^(->}[d] 215 & {\Phi_A[\ell]} \[email protected]{^(->}[d]\\ 216 0\ar[r]&{\Hom(X/\ell X,\mu_{\ell})} 217 \ar[r]& {A[\ell]}\ar[r]^{f} & {Y/\ell Y}\ar[r]&0. 218}$$
219
220The maximal finite part of $A[\ell]$ is $f^{-1}(\Phi_A[\ell])$,
221which is {\em bigger} than $\Hom(X/\ell X,\mu_{\ell})$.
222Because~$N$ is cube-free, the Hecke algebra~$\T$ is semisimple,
223so {\em it is almost certainly possible to show}\footnote{I will
224do this soon.} that $f^{-1}(\Phi_A[\ell])$
225in $A[\ell]$ forces some $A[\m]$ to be finite.
226
227\begin{example}[Dimension~$1$]
228If $\dim A=1$ then $Y/\ell Y$ has dimension one
229and we have
230$$0 \ra \Hom(X/\ell X,\mu_{\ell}) 231 \ra A[\ell] \ra \Phi_A[\ell] \ra 0.$$
232This forces $A[\ell]$ to be finite, so $\ell\mid\eisen$.
233\end{example}
234
235\section{Further computations (informal)}
236Barry's next email:
237\begin{quote}
238The tough cases to try your conjecture on are the prime levels~$p$ for which
239there is a prime $q>3$ dividing $n=$ numerator of $(p-1)/12$ such
240that (you wouldn't believe this!)\\
241{\em Condition($p;q$):}
242     $\displaystyle \prod_{k=1}^{\frac{p-1}{2}}k^k$
243  is a $q$-th power in $\F_p.$\\
244There are also conditions Condition($p;2$), Condition($p;3$),
245slightly different to state, which cover $q=2,3$ when they divide $n$.
246Condition($p;3$) is just that $({\frac{p-1}{3}})!$ is a cube mod~$p$.
247I make these conditions because I think that
248your conjecture should be, very likely, true unless
249CONDITION($p;q$) holds for some prime~$q$ dividing~$n$.
250\end{quote}
251
252I used the following \magma{} program to list all tough pairs
253$p,q$, with $p<2113$ such that condition Condition($p;q$) is satisfied:
254\begin{verbatim}
255function PowerProd(p)
256   F := GF(p);
257   return &*[(F!k)^k : k in [1..Integers()!((p-1) div 2)]];
258end function;
259
260function Condition(p,q)
261   if not IsPrime(p) or not IsPrime(q) then
262      error "p and q must be prime.";
263   end if;
264   n := Numerator((p-1)/12);
265   if q eq 2 then
266      error "I don't know condition 2.";
267   end if;
268   if n mod q ne 0 then
269      error "q must divide the numerator of (p-1)/12.";
270   end if;
271   F := GF(p);
272   if q eq 3 then
273      a := &*[F!i : i in [1..Integers()!((p-1)/3)]];
274   else
275      a := PowerProd(p);
276   end if;
277   // return true iff a is a q-th power, i.e., iff p-1 divided by
278   // the order of a in the cyclic group Fp^* is divisible by q.
279   return ((p-1) div Order(a)) mod q eq 0;
280end function;
281
282function Toughness(p)
283   n := Numerator((p-1)/12);
284   Q := [q : q in PrimeDivisors(n) | q ne 2 and Condition(p,q)];
285   return Q;
286end function;
287\end{verbatim}
288
289There are~$52$ {\em tough} primes~$p\leq 2113$, in the sense that
290Condition($p,q$) is satisfied for some odd~$q$.  In the notation
291$(p,\text{tough$q$})$ they are:\\
292(31;5), (103;17), (127;7), (131;5), (181;5), (199;3), (211;5),
293(271;3), (281;5), (401;5), (409;17), (487;3), (523;3), (541;3,5),
294(571;5), (661;11), (683;11),
295(691;5), (701;5), (733;61), (751;5), (761;5,19), (911;7), (919;3), (941;5),
296(971;5), (1021;17), (1091;5), (1279;3), (1289;7), (1291;5), (1297;3),
297(1321;11),
298(1381;5,23), (1447;241), (1471;5), (1483;13), (1511;5), (1531;3,5), (1571;5),
299(1621;3), (1693;3), (1697;53), (1747;3), (1789;149), (1831;5), (1861;5),
300(1871;5), (1999;3), (2003;11), (2017;3), (2081;5).
301
302
303There are $19$ more tough~$p<3000$:\\
304(2143;3), (2161;3), (2269;3), (2281;5), (2339;7), (2351;5), (2371;5),
305(2377;3,11),
306(2411;5), (2467;3), (2521;5), (2531;5), (2551;5), (2593;3), (2621;5),
307(2707;11), (2861;5), (2887;13,37), (2917;3).
308
309Barry's response:
310
311\begin{quote}
312   Great.  I'll try to send you a proof that, in the non-tough''
313cases, your conjecture is safe, maybe even true.
314But  in the first $52$ of the $(p,\text{tough$q$})$
315cases can you tell me which have the property that the $q$-Eisenstein
316quotient of the jacobian of $X_0(p)$ is a simple abelian variety, and which
317not?  When they split, can you given the simple abelian variety factors?
318\end{quote}
319
320The $q$-Eisenstein quotient $J^{(q)}$ is the abelian variety
321quotient of $J_0(p)$ corresponding to the union of the
322irreducible components of $\Spec(\T)$ containing
323$\mathcal{I}+q$ where $\mathcal{I}$ is the Eisenstein ideal,
324which is generated by $1+\ell-T_{\ell}$, for all~$\ell\neq p$,
325and $1-T_p$.
326
327\begin{verbatim}
328Barry,
329
330   Great.  I'll try to send you a proof that in the "non-tough" cases, your
331   conjecture is safe, maybe even true.
332
333Please do, if possible; even a rough sketch will be very much appreciated.
334
335   But in the first 52  <p, [tough q]>
336   cases can you tell me which have the property that the q-Eisenstein
337   quotient of the jacobian of X_0(p) is a simple abelian variety, and which
338   not?  When they split can you given the simple abelian variety factors?
339
340It appears that there are exactly two non-simple q-Eisenstein quotients
341(q odd) at level p<=2113.  These just happen to be "tough". They occur at
342levels 487 and 1999; in both cases q=3.   The details follow.
343
344p=487
345numer((487-1)/12) = 3^4.
346g_+ = 17
347g_- = 23 = 2 + 2 + 3 + 16
348           A + B + C + D
349#A(Q)=1
350#B(Q)=3
351#C(Q)=1
352#D(Q)=3^3
3533-Eisenstein quotient = B+D.
354
355
356p=1999
357numer((1999-1)/12) = 3^2*37.
358g_+ = 70
359g_- = 96 = 2 + 94
360           A +  B
361#A(Q) = 3,
362#B(Q) = 3*37.
3633-Eisenstein quotient = A+B.
364
365I just looked at Armand Brumer's published table of splittings of
366J_0(N)^- for N<10000.  As far as I can tell, he seems to claim that
367only the piece of largest dimension is Eisenstein in both the 487
368and 1999 cases; this is in direct contradiction with my computation.
369(I've found another mistake in this table, so this doesn't mean I'm wrong;
370I'll email Armand.)  The only example he lists in which the q-Eisenstein
371quotient (q odd) is not simple is at level 3001.  Here q=5 and n = 2*5^3,
372so again q^2 divides n.
373
374Here is how I computed the torsion subgroup of each optimal quotient
375A=A_f.  I computed a lower bound on #A(Q) by computing
376the order of the image of 0-oo using modular symbols.   Then I computed
377the upper bound
378     G_f = gcd{ #A(Fq) : q does not divide 2p, q<=37 }.
379In every case considered, I was lucky and the two bounds coincided.
380I think that if A_f is q-Eisenstein then q must divide G_f.
381In general, q dividing G_f probably doesn't imply that A_f is
382q-Eisenstein; however, it probably does within the range of conductors
383I am considering.
384
385I wish I could test an example in which the q-Eisenstein quotient
386is not simple and q exactly divides numer((p-1)/12)...
387Did you ever prove that such quotient exists when q is odd?
388
389 -- William
390\end{verbatim}
391
392\bibliographystyle{amsplain}
393\bibliography{biblio}
394
395\end{document}
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