The refined Eisenstein conjecture

William A. Stein

October 10, 1999

1 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 [5] B. Mazur gave a complete description of the torsion subgroup and component group of J_{0}(N).

Consider a normalized weight-two eigenform f = ∑ a_{n}q^{n} on X_{0}(N), and let I be the kernel of the natural map T → Z[…, a_{n}, …] 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 − ∞ ∈ J_{0}(N).

Conjecture 1.1 (Refined Eisenstein conjecture).

(a) Let π : J_{0}(N) → A_{f} be the optimal quotient corresponding to the newform f. Then

A_{f}(Q)_{tor} ≅ Â_{f}(Q)_{tor} ≅ Φ_{A_{f}} (F_{p}) ≅ Φ_{A_{f}} (\overline{F}_{p}) ≅ C_{f}.

(b) Assume conjecture (a), and let n_{f} denote the common order. Then ∏ n_{f} = numer (\frac{N−1}{12}) , where the product is over the Galois conjugacy classes of normalized eigenforms.

The conjecture is false when the level is composite. For example, let E =33A; then E(Q) = Z/2 × Z/2 but Φ_{E}(\overline{F}_{3}) = Z/6Z.

2 Computational evidence

There are 305 factors A_{f} of level up to 631; there are 860 factors up to level 2113.

Proposition 2.1. Conjecture 1.1 is true for N ≤ 631; it is true for N ≤ 2113, up to a 2-power. (Except possibly for the statement about Â_{f}(Q).)

Proof. We verified this using a computer program written in [1]. 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 [3]. 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)/π_{∗}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 [4, p. 28 and Thm. 2.7b] that ξ = 0 − ∞ lies in J_{0}(N)(Q)_{tor}. The point ξ corresponds to the element {0, ∞} ∈ H_{1}(X_{0}(N), Z)⊗Q, so we can compute ξ 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 [2, p. 70]), 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 Â_{f}(Q), but we have not carried out this computation.

The component group Φ_{A_{f}} (\overline{F}_{N}) can be computed using the algorithm described in [8], which involves the monodromy pairing on the character groups of certain tori. In all cases in which the Gal(\overline{F}_{N}/F_{N})-action is nontrivial, the component group Φ_{A_{f}} (\overline{F}_{N}) is trivial, so it suffices to know Φ_{A_{f}} (\overline{F}_{N}).

2.1 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. Example 2.2. N = 113 and numer (\frac{113−1}{12}) = 28:

\begin{matrix}J_{0}(113) ∼ & A_{1} & ×A_{2} & ×A_{3} & ×B_{3} \\ #A_{f}(Q)_{tor} & 2 & 2 & 1 & 7 \\ #Φ_{f}(F_{N}) & 2 & 2 & 1 & 7 \\ \frac{L(1)}{Ω} & \frac{1}{2} & \frac{1}{2} & 0 & \frac{8}{7} \end{matrix}

Remark: The Shafarevich-Tate group Sha(B_{3}) probably has order 8, and is explained by the Mordell-Weil group of A_{3}. Example 2.3. N = 389 and numer (\frac{389−1}{12}) = 97:

\begin{matrix}J_{0}(389) ∼ & A_{1} & ×A_{2} & ×A_{3} & ×A_{6} & ×A_{20} \\ #A_{f}(Q)_{tor} & 1 & 1 & 1 & 1 & 97 \\ #Φ_{f}(F_{N}) & 1 & 1 & 1 & 1 & 97 \\ \frac{L(1)}{Ω} & 0 & 0 & 0 & 0 & \frac{2^{11}·5^{2}}{97} \end{matrix}

3 Theoretical evidence

Mestre and Oesterlé proved in [6] that if A_{f} has dimension 1 then Conjecture 1.1(a) is true. Mazur proved in [5] that if ℓ | #A_{f}(Q), then ℓ | numer (\frac{p−1}{12}); furthermore,

#J_{0}(p)(Q)_{tor} = #Φ_{J_{0}(p)}(\overline{F}_{p}) = numer (\frac{p − 1}{12}) .

Upon hearing of Conjecture 1.1 Mazur said:

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 μ_{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.

4 Application of Ribet’s level lowering theorem

It is almost possible to obtain a partial consequence of Conjecture 1.1 by applying the level lowering theorem proved by Ribet in [7]. If ℓ | #Φ_{A_{f}} (\overline{F}_{p}) then ρ_{f,𝔪} tends to be finite for some 𝔪 containing ℓ. If so, and ρ_{f,𝔪} is irreducible (and ℓ is odd), then Ribet’s level lowering theorem leads to a contradiction. We now record the present state of our argument.

If Φ_{J} → Φ_{A_{f}} is surjective, then the primes on the left hand side of Conjecture 1.1(b) divide the right hand side, which prompts us to make the following weaker conjecture.

Conjecture 4.1. The natural map Φ_{J} → Φ_{A_{f}} is surjective.

The component group Φ_{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 ℓ | #Φ_{A}; does ℓ | 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 Â. The monodromy pairing expresses the component group of A by exactness of the sequence

0 → Y → Hom(X, Z) → Φ_{A} → 0.

By tensoring this sequence with Z/ℓZ, we obtain the exact sequence

0 → Φ_{A}[ℓ] → Y ⊗ Z/ℓZ → Hom(X, Z/ℓZ) → Φ_{A} ⊗ Z/ℓZ → 0.  (1)

The Mumford-Tate uniformizations of A and Â yield the following diagrams:

\xymatrix{ & 0 \ar[r] \ar[d] & Y \ar[d] \ar[r]^{ℓ} & Y \ar[d] \ar[r] & Y/ℓY \ar[r] \ar[d] & 0 \\ 0 \ar[r] & Hom(X/ℓX, μ_{ℓ}) \ar[d] \ar[r] & T \ar[d] \ar[r]^{·^{ℓ}} & T \ar[d] \ar[r] & 0 \ar[d] \\ 0 \ar[r] & A[ℓ] \ar[r] & A \ar[r]^{ℓ} & A \ar[r] & 0.}

The snake lemma then gives an exact sequence

0 → Hom(X/ℓX, μ_{ℓ}) → A[ℓ] ⟶^{f} Y/ℓY → 0

which, when combined with (1), produces the following diagram:

\xymatrix{ &  & f^{−1}(Φ_{A}[ℓ]) \ar@{^{(}->}[d] & Φ_{A}[ℓ] \ar@{^{(}->}[d] \\ 0 \ar[r] & Hom(X/ℓX, μ_{ℓ}) \ar[r] & A[ℓ] \ar[r]^{f} & Y/ℓY \ar[r] & 0.}

The maximal finite part of A[ℓ] is f^{−1}(Φ_{A}[ℓ]), which is bigger than Hom(X/ℓX, μ_{ℓ}). Because N is cube-free, the Hecke algebra T is semisimple, so it is almost certainly possible to show^{1} that f^{−1}(Φ_{A}[ℓ]) in A[ℓ] forces some A[𝔪] to be finite. Example 4.2 (Dimension 1). If dim A = 1 then Y/ℓY has dimension one and we have

0 → Hom(X/ℓX, μ_{ℓ}) → A[ℓ] → Φ_{A}[ℓ] → 0.

This forces A[ℓ] to be finite, so ℓ | numer (\frac{p−1}{12}).

5 Further computations (informal)

Barry’s next email:

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!)

Condition(p; q): ∏_{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.

1I will do this soon.

I used the following Magma program to list all tough pairs p, q, with p < 2113 such that condition Condition(p; q) is satisfied:

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;

There are 52 tough primes p ≤ 2113, in the sense that Condition(p, q) is satisfied for some odd q. In the notation (p, 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:

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, 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?

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 I +q where I is the Eisenstein ideal, which is generated by 1 + ℓ − T_{ℓ}, for all ℓ ≠ p, and 1 − T_{p}.

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  <p, [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?
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

References

[1] J. Cannon, The magma computational algebra system,

   http://www.maths.usyd.edu.au:8000/u/magma/.

[2] J. W. S. Cassels and E. V. Flynn, Prolegomena to a middlebrow arithmetic of curves of genus 2, Cambridge University Press, Cambridge, 1996.

[3] J. E. Cremona, Algorithms for modular elliptic curves, second ed., Cambridge University Press, Cambridge, 1997.

[4] J. I. Manin, Parabolic points and zeta functions of modular curves, Izv. Akad. Nauk SSSR Ser. Mat. 36 (1972), 19–66.

[5] B. Mazur, Modular curves and the Eisenstein ideal, Inst. Hautes Études Sci. Publ. Math. (1977), no. 47, 33–186 (1978).

[6] J.-F. Mestre and J. Oesterlé, Courbes de Weil semi-stables de discriminant une puissance m-ième, J. Reine Angew. Math. 400 (1989), 173–184.

[7] K. A. Ribet, On modular representations of Gal(\overline{Q}/Q) arising from modular forms, Invent. Math. 100 (1990), no. 2, 431–476.

[8] W. A. Stein, Component groups of optimal quotients of Jacobians, preprint (1999).