% compgroup.tex
\documentclass{article}
\usepackage[all]{xy}
\include{macros}
\DeclareMathOperator{\numer}{numer}
\DeclareMathOperator{\order}{order}
\newcommand{\eisen}{\numer\left(\frac{p-1}{12}\right)}
\title{The refined Eisenstein conjecture}
\author{William A.~Stein}
\begin{document}
\maketitle
\section{Introduction}
Let~$N$ be a prime, and consider the
the curve~$X_0(N)$ whose complex points
are the isomorphism classes of pairs consisting of a
(generalized) elliptic curve and a cyclic subgroup of order~$N$.
Let~$J_0(N)$ denote the Jacobian of $X_0(N)$; this is an abelian
variety of dimension equal to the genus of~$X_0(N)$ whose points
correspond to the degree~$0$ divisor classes on~$X_0(N)$.
In~\cite{mazur:eisenstein} B.~Mazur gave a complete description
of the torsion subgroup and component group of~$J_0(N)$.
Consider a normalized weight-two eigenform $f = \sum a_n q^n$
on~$X_0(N)$, and let~$I$ be the kernel of the natural
map $\T\ra\Z[\ldots,a_n,\ldots]$ that sends a Hecke
operator~$T_n$ to~$a_n$. Denote by $A_f$ the optimal abelian variety
quotient $J_0(N)/IJ_0(N)$. In this paper we give evidence for
a conjectural refinement of Mazur's theorem to quotients~$A_f$
of~$J_0(N)$. Let $C_f$ denote the cyclic subgroup of $A_f(\Q)$
generated by the image of the point $0-\infty\in J_0(N)$.
\begin{conjecture}[Refined Eisenstein conjecture]\label{conj:main}\mbox{}
\vspace{-1ex}
\begin{itemize}
\item[(a)] Let $\pi:J_0(N)\ra A_f$ be the optimal quotient corresponding
to the newform~$f$. Then
$$A_f(\Q)_{\tor}\isom \Adual_f(\Q)_{\tor}
\isom \Phi_{A_f}(\Fp)
\isom \Phi_{A_f}(\Fpbar)
\isom C_f.$$
\vspace{-5ex}
\item[(b)] Assume conjecture (a), and
let~$n_f$ denote the common order.
Then
$\prod n_f = \numer\left(\frac{N-1}{12}\right),$
where the product is over the Galois conjugacy classes
of normalized eigenforms.
\end{itemize}
\end{conjecture}
The conjecture is false when the level is composite.
For example, let $E=${\bf 33A}; then
$E(\Q)=\Z/2\cross\Z/2$ but $\Phi_E(\Fbar_3)=\Z/6\Z$.
\section{Computational evidence}
There are $305$ factors $A_f$ of level up to $631$;
there are $860$ factors up to level $2113$.
\begin{proposition}
Conjecture~\ref{conj:main} is true
for $N\leq 631$; it is true for $N\leq 2113$, up to a $2$-power.
(Except possibly for the statement about $\Adual_f(\Q)$.)
\end{proposition}
\begin{proof}
We verified this using a computer program written in~\cite{magma}.
The computation was done as follows.
Viewed as an abelian variety over the complex number,
$J_0(N)$ is the quotient
$$J_0(N)(\C)=\Hom(S_2(N),\C)/H_1(X_0(N),\Z).$$
Here $S_2(N)$ is
the space of weight two cusp forms of level~$N$ and~$H_1(X_0(N),\Z)$
is the first integral homology of the Riemann surface $X_0(N)$;
both of these spaces, and the action of the
Hecke operators~$T_n$, can be computed using the
algorithms in~\cite{cremona:algs}.
Using the Hecke operators, we list each optimal quotient~$A_f$.
As a complex abelian variety, $A_f$ is the quotient
$$A_f(\C) = \Hom(S_2(N)[I],\C)/\pi_* H_1(X_0(N),\Z).$$
The torsion subgroup $A_f(\Q)_{\tor}$ can be computed by
combining the following upper and lower bound.
Manin proved in~\cite[p.~28 and Thm.~2.7b]{manin:parabolic}
that $\xi = 0-\infty$ lies in $J_0(N)(\Q)_{\tor}$.
The point~$\xi$ corresponds to the element
$\{0,\infty\}\in H_1(X_0(N),\Z)\tensor\Q$, so we can
compute~$\xi$ and its image in $A_f(\C)$.
Combining the Eichler-Shimura relation with the
injectivity of rational torsion under reduction modulo
an odd prime~$p$ (see~\cite[p.~70]{cassels-flynn}),
we obtain an upper bound on the order of the torsion
subgroup of any abelian variety isogeneous to~$A_f$;
this upper bound is computed from the characteristic
polynomials of Hecke operators.
In all cases that we computed, the computed upper bound agrees with
the lower bound, so we obtain $A_f(\Q)_{\tor}$.
Similar methods can be used to compute $\Adual_f(\Q)$, but we have
not carried out this computation.
The component group $\Phi_{A_f}(\Fbar_N)$
can be computed using the algorithm
described in~\cite{stein:compgroup}, which involves
the monodromy pairing on the character groups of certain tori.
In all cases
in which the $\Gal(\Fbar_N/\F_N)$-action is nontrivial, the component
group $\Phi_{A_f}(\Fbar_N)$ is trivial, so it suffices
to know $\Phi_{A_f}(\Fbar_N)$.
\end{proof}
\subsection{Examples}
In this section we give two examples. We decompose
$J_0(N)$ as a product of $A_d$ where each
$A_d$ is a $d$-dimensional factor corresponding to an eigenform.
\begin{example} $N=113$ and $\numer\left(\frac{113-1}{12}\right) = 28$:
$$\begin{array}{lrrrr}
J_0(113) \sim&A_1 &\cross A_2&\cross A_3 & \cross B_3\\
\#A_f(\Q)_{\tor}&2&2&1&7\\
\#\Phi_f(\F_N)&2&2&1&7\\
\frac{L(1)}{\Omega}&\frac{1}{2}&\frac{1}{2}&0&\frac{8}{7}
\end{array}$$
Remark: The Shafarevich-Tate group $\text{Sha}(B_3)$ probably
has order $8$, and is explained by the Mordell-Weil
group of $A_3$.
\end{example}
\begin{example} $N=389$ and $\numer\left(\frac{389-1}{12}\right) = 97$:
$$\begin{array}{lrrrrr}
J_0(389) \sim& A_1 &\cross A_2&\cross A_3 &\cross A_6 &\cross A_{20}\\
\#A_f(\Q)_{\tor}& 1 & 1& 1&1&97\\
\#\Phi_f(\F_N)&1&1&1&1&97\\
\frac{L(1)}{\Omega}&0&0&0&0&\frac{2^{11}\cdot 5^2}{97}
\end{array}$$
\end{example}
\section{Theoretical evidence}
Mestre and Oesterl\'{e} proved in~\cite{mestre-oesterle:crelle}
that if $A_f$ has dimension~1 then Conjecture~\ref{conj:main}(a)
is true.
Mazur proved in~\cite{mazur:eisenstein} that %page 141?
if $\ell \mid\# A_f(\Q)$, then $\ell \mid \eisen$; furthermore,
$$\# J_0(p)(\Q)_{\tor} = \# \Phi_{J_0(p)}(\Fpbar)
= \numer\left(\frac{p-1}{12}\right).$$
Upon hearing of Conjecture~\ref{conj:main} Mazur said:
\begin{quote}
What you are conjecturing is something I don't think I ever thought
of---at least in those terms---It has some implications which are interesting
(to me). For example, suppose that a prime~$q$ divides the numerator of
$(p-1)/12$ but~$q^2$ does not; let~$A$ be the $q$-Eisenstein
quotient. Then only
one optimal quotient of~$A$ would be allowed to have nontrivial $q$-torsion,
and the other(s), if there are any, would, I guess, then have to settle for
a~$\mu_q$ subgroup. I suppose, though, that it is very often the case that
(under the above hypotheses of~$q$) the $q$-Eisenstein quotient is simple.
\end{quote}
\section{Application of Ribet's level lowering theorem}
It is almost possible to obtain a partial consequence of
Conjecture~\ref{conj:main} by applying the level lowering
theorem proved by Ribet in~\cite{ribet:modreps}.
If $\ell\mid \#\Phi_{A_f}(\Fpbar)$ then $\rho_{f,\m}$ tends
to be finite for some~$\m$ containing~$\ell$.
If so, and $\rho_{f,\m}$ is irreducible (and~$\ell$ is odd),
then Ribet's level
lowering theorem leads to a contradiction.
We now record the present state of our argument.
If $\Phi_J\ra \Phi_{A_f}$ is surjective, then
the primes on the left hand side of Conjecture~\ref{conj:main}(b)
divide the right hand side, which prompts us to make
the following weaker conjecture.
\begin{conjecture}\label{conj:surjective}
The natural map $\Phi_J \into \Phi_{A_f}$ is surjective.
\end{conjecture}
The component group~$\Phi_A$ of~$A$ tends to force some~$A[\m]$
to arise from a finite flat group scheme.
Ribet's theorem implies such~$\m$ are reducible, and then
Mazur's theorem implies that such~$\m$ are Eisenstein.
Suppose $\ell\mid\#\Phi_A$; does $\ell\mid \numer(\frac{N-1}{12})$?
Let~$X$ be the character group of the torus attached to~$A$,
a let $Y$ be character group attached to $\Adual$.
The monodromy pairing expresses the component group
of~$A$ by exactness of the sequence
$$0 \ra Y \ra \Hom(X,\Z) \ra \Phi_A \ra 0.$$
By tensoring this sequence with $\Z/\ell\Z$, we obtain the exact sequence
\begin{equation}\label{eqn:comptor}
0 \ra \Phi_A[\ell] \ra Y\tensor \Z/\ell\Z \ra \Hom(X,\Z/\ell\Z)
\ra \Phi_A \tensor\Z/\ell\Z\ra 0.
\end{equation}
The Mumford-Tate uniformizations of~$A$ and~$\Adual$
yield the following diagrams:
$$\xymatrix{
& 0\ar[r]\ar[d]& Y\ar[d]\ar[rr]^{\ell}& & Y\ar[d]\ar[r]& Y/\ell Y\ar[r]\ar[d]& 0\\
0\ar[r]& {\Hom(X/\ell X,\mu_\ell)}\ar[d]\ar[r]
& T\ar[d]\ar[rr]^{\cdot^\ell}&& T\ar[d]\ar[r] & 0\ar[d]\\
0\ar[r]& {A[\ell]}\ar[r]& A\ar[rr]^{\ell}&& A\ar[r]& 0.}$$
The snake lemma then gives an exact sequence
$$0 \ra \Hom(X/\ell X,\mu_{\ell})
\ra A[\ell] \xrightarrow{\,f\,} Y/\ell Y \ra 0$$
which, when combined with~(\ref{eqn:comptor}), produces the following diagram:
$$\xymatrix{
& & f^{-1}(\Phi_A[\ell])\ar@{^(->}[d]
& {\Phi_A[\ell]} \ar@{^(->}[d]\\
0\ar[r]&{\Hom(X/\ell X,\mu_{\ell})}
\ar[r]& {A[\ell]}\ar[r]^{f} & {Y/\ell Y}\ar[r]&0.
}$$
The maximal finite part of $A[\ell]$ is $f^{-1}(\Phi_A[\ell])$,
which is {\em bigger} than $\Hom(X/\ell X,\mu_{\ell})$.
Because~$N$ is cube-free, the Hecke algebra~$\T$ is semisimple,
so {\em it is almost certainly possible to show}\footnote{I will
do this soon.} that $f^{-1}(\Phi_A[\ell])$
in $A[\ell]$ forces some $A[\m]$ to be finite.
\begin{example}[Dimension~$1$]
If $\dim A=1$ then $Y/\ell Y$ has dimension one
and we have
$$0 \ra \Hom(X/\ell X,\mu_{\ell})
\ra A[\ell] \ra \Phi_A[\ell] \ra 0.$$
This forces $A[\ell]$ to be finite, so $\ell\mid\eisen$.
\end{example}
\section{Further computations (informal)}
Barry's next email:
\begin{quote}
The tough cases to try your conjecture on are the prime levels~$p$ for which
there is a prime $q>3$ dividing $n=$ numerator of $(p-1)/12$ such
that (you wouldn't believe this!)\\
{\em Condition($p;q$):}
$\displaystyle \prod_{k=1}^{\frac{p-1}{2}}k^k$
is a $q$-th power in $\F_p.$\\
There are also conditions Condition($p;2$), Condition($p;3$),
slightly different to state, which cover $q=2,3$ when they divide $n$.
Condition($p;3$) is just that $({\frac{p-1}{3}})!$ is a cube mod~$p$.
I make these conditions because I think that
your conjecture should be, very likely, true unless
CONDITION($p;q$) holds for some prime~$q$ dividing~$n$.
\end{quote}
I used the following \magma{} program to list all tough pairs
$p,q$, with $p<2113$ such that condition Condition($p;q$) is satisfied:
\begin{verbatim}
function PowerProd(p)
F := GF(p);
return &*[(F!k)^k : k in [1..Integers()!((p-1) div 2)]];
end function;
function Condition(p,q)
if not IsPrime(p) or not IsPrime(q) then
error "p and q must be prime.";
end if;
n := Numerator((p-1)/12);
if q eq 2 then
error "I don't know condition 2.";
end if;
if n mod q ne 0 then
error "q must divide the numerator of (p-1)/12.";
end if;
F := GF(p);
if q eq 3 then
a := &*[F!i : i in [1..Integers()!((p-1)/3)]];
else
a := PowerProd(p);
end if;
// return true iff a is a q-th power, i.e., iff p-1 divided by
// the order of a in the cyclic group Fp^* is divisible by q.
return ((p-1) div Order(a)) mod q eq 0;
end function;
function Toughness(p)
n := Numerator((p-1)/12);
Q := [q : q in PrimeDivisors(n) | q ne 2 and Condition(p,q)];
return Q;
end function;
\end{verbatim}
There are~$52$ {\em tough} primes~$p\leq 2113$, in the sense that
Condition($p,q$) is satisfied for some odd~$q$. In the notation
$(p,\text{tough $q$})$ they are:\\
(31;5), (103;17), (127;7), (131;5), (181;5), (199;3), (211;5),
(271;3), (281;5), (401;5), (409;17), (487;3), (523;3), (541;3,5),
(571;5), (661;11), (683;11),
(691;5), (701;5), (733;61), (751;5), (761;5,19), (911;7), (919;3), (941;5),
(971;5), (1021;17), (1091;5), (1279;3), (1289;7), (1291;5), (1297;3),
(1321;11),
(1381;5,23), (1447;241), (1471;5), (1483;13), (1511;5), (1531;3,5), (1571;5),
(1621;3), (1693;3), (1697;53), (1747;3), (1789;149), (1831;5), (1861;5),
(1871;5), (1999;3), (2003;11), (2017;3), (2081;5).
There are $19$ more tough~$p<3000$:\\
(2143;3), (2161;3), (2269;3), (2281;5), (2339;7), (2351;5), (2371;5),
(2377;3,11),
(2411;5), (2467;3), (2521;5), (2531;5), (2551;5), (2593;3), (2621;5),
(2707;11), (2861;5), (2887;13,37), (2917;3).
Barry's response:
\begin{quote}
Great. I'll try to send you a proof that, in the ``non-tough''
cases, your conjecture is safe, maybe even true.
But in the first $52$ of the $(p,\text{tough $q$})$
cases can you tell me which have the property that the $q$-Eisenstein
quotient of the jacobian of $X_0(p)$ is a simple abelian variety, and which
not? When they split, can you given the simple abelian variety factors?
\end{quote}
The $q$-Eisenstein quotient $J^{(q)}$ is the abelian variety
quotient of $J_0(p)$ corresponding to the union of the
irreducible components of $\Spec(\T)$ containing
$\mathcal{I}+q$ where $\mathcal{I}$ is the Eisenstein ideal,
which is generated by $1+\ell-T_{\ell}$, for all~$\ell\neq p$,
and $1-T_p$.
\begin{verbatim}
Barry,
Great. I'll try to send you a proof that in the "non-tough" cases, your
conjecture is safe, maybe even true.
Please do, if possible; even a rough sketch will be very much appreciated.
But in the first 52
cases can you tell me which have the property that the q-Eisenstein
quotient of the jacobian of X_0(p) is a simple abelian variety, and which
not? When they split can you given the simple abelian variety factors?
It appears that there are exactly two non-simple q-Eisenstein quotients
(q odd) at level p<=2113. These just happen to be "tough". They occur at
levels 487 and 1999; in both cases q=3. The details follow.
p=487
numer((487-1)/12) = 3^4.
g_+ = 17
g_- = 23 = 2 + 2 + 3 + 16
A + B + C + D
#A(Q)=1
#B(Q)=3
#C(Q)=1
#D(Q)=3^3
3-Eisenstein quotient = B+D.
p=1999
numer((1999-1)/12) = 3^2*37.
g_+ = 70
g_- = 96 = 2 + 94
A + B
#A(Q) = 3,
#B(Q) = 3*37.
3-Eisenstein quotient = A+B.
I just looked at Armand Brumer's published table of splittings of
J_0(N)^- for N<10000. As far as I can tell, he seems to claim that
only the piece of largest dimension is Eisenstein in both the 487
and 1999 cases; this is in direct contradiction with my computation.
(I've found another mistake in this table, so this doesn't mean I'm wrong;
I'll email Armand.) The only example he lists in which the q-Eisenstein
quotient (q odd) is not simple is at level 3001. Here q=5 and n = 2*5^3,
so again q^2 divides n.
Here is how I computed the torsion subgroup of each optimal quotient
A=A_f. I computed a lower bound on #A(Q) by computing
the order of the image of 0-oo using modular symbols. Then I computed
the upper bound
G_f = gcd{ #A(Fq) : q does not divide 2p, q<=37 }.
In every case considered, I was lucky and the two bounds coincided.
I think that if A_f is q-Eisenstein then q must divide G_f.
In general, q dividing G_f probably doesn't imply that A_f is
q-Eisenstein; however, it probably does within the range of conductors
I am considering.
I wish I could test an example in which the q-Eisenstein quotient
is not simple and q exactly divides numer((p-1)/12)...
Did you ever prove that such quotient exists when q is odd?
-- William
\end{verbatim}
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\bibliography{biblio}
\end{document}