% 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} \bibliographystyle{amsplain} \bibliography{biblio} \end{document}