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I suspect that I'm asking (familiar?) questions from deformation theory in a different language. But I'm an illiterate in deformation theory language; if my suspicion is correct I'd be grateful for an explanation of the connection between the two languages.


Fix a prime $\ell$, an $N>0$ prime to $\ell$, and an even integer $w$ in $[0,\ell-1)$.

Definition----An element $h$ of $\mathbb{Z}/\ell [[x]]$ is in $M(w)$ if there is a modular form for $\Gamma_0 (N)$, of weight congruent to $w \mod \ell-1$, whose expansion at infinity lies in $\mathbb{Z}[[x]]$ and reduces mod $\ell$ to $h$. (It's easy to see that $M(w)$ is a $\mathbb{Z}/\ell$ vector-space).

Now for each $n$ prime to $\ell N$ there is a formal Hecke operator $T_n$, $\mathbb{Z}/\ell[[x]]\to \mathbb{Z}/\ell[[x]]$, depending on the "weight" $w$. For our fixed $w$, these $T_n$ span a Hecke algebra $HE(w)$; furthermore $HE(w)$ stabilizes $M(w)$.

Definition----An ideal of $HE(w)$ is modular if it is the annihilator of some non-zero $h$ in $M(w)$. It is maximal modular if it is maximal among such annihilators. (If $u$ is in $HE(w)$, then the annihilator of $u(h)$ contains the annihilator of $h$. So maximal modulars are prime. Also if $J=Ann(h)$ is maximal modular, then since all transforms of $h$ under $HE(w)$ lie in a finite dimensional $\mathbb{Z}/\ell$ space, $HE(w)/J$ is a domain finite over $\mathbb{Z}/\ell$. So maximal modulars are maximal ideals).

Definition----If $J$ is maximal modular, $M(w,J)$ is the set of $h$ in $M(w)$ annihilated by some power of $J$.


1)---Is it true that there are only finitely maximal $J$ in $HE(w)$? Is $M(w)$ the direct sum of the $M(w,J)$?

---Suppose now that $\ell=2$ so that $w=0$. Joel Bellaiche told me that when $N=1,3,5,7$ or $9$ and $F=x+x^9+x^25+...$, then $Ann(F)$ acts locally nilpotently on $M(w)$. So $Ann(F)$ is the only maximal modular in these cases, and one sees easily that $Ann(F)=Ann(r)$ for some $r$ "coming from a weight $4$ form for $\Gamma_0 (N)$".

2)----Suppose $\ell=2$ so $w=0$. Does every maximal modular have the form $Ann(h)$ for some $h$ coming from a form of weight 2 or 4? Is there some corresponding result for odd $\ell$?

---For a given maximal modular $J$ it seems to me that one can imitate the construction done by Nicolas and Serre in the case $\ell=2$, $N=1$, getting a "J-completed" version of $HE(w)$ acting faithfully on $M(w,J)$. In the Nicolas-Serre situation this J-completion is a two-variable power series ring over $\mathbb{Z}/2$.

3)----What is known about the $J$-completed algebra for other values of $\ell$, $N$ and $w$?

EDIT___I'm grateful to Kevin Ventullo for calling my attention to the preprint of Bellaiche and Khare for the case N=1. Though my definition of the J-completion of HE and theirs differ in some ways, I imagine that they and I are talking about the same objects, and that when N=1, their conjecture that one always gets a 2-variable power series ring and their positive results supporting this conjecture apply to my question.

For N>1, I've only looked at the case ell=2, N an odd prime. Since ell=2, w=0, and I'll omit the w in my notation, writing M,M(J) and HE. Let J in Z/2[[x]] be x+x^9+x^25+... as in question 1. When J=Ann(F) (which is the maximal ideal of HE containing all the T_p, p an odd prime other than N) it seems that my definition of the J-completion is "wrong". This is because of the presence of "oldforms" such as powers of F and powers of G=F(x^N) in M(J). One should probably replace M(J) by its quotient by the space spanned by these elements, and use the action of HE on the quotient to define the J-completion. I'll leave this aside for now and assume that J isn't Ann(F). Then N is at least 11. When N=11, there's only one other choice for J; Ann(t) where t is the reduction of the weight 2 newform for Gamma_0 (11). This J contains the elements T_3 +I, T_5 +I, and T_7.

I'll discuss this case, N=11 and J=Ann(t), for the rest of this edit. Though I have no proof, the computer convincingly suggests that the J-completion contains nilpotents, and is in fact a 2-variable power series ring over Z/2, with an element of square 0 adjoined.

Let M(odd,J) be the subspace of M(J) consisting of h in which all the exponents of x are odd. It's easy to see that the J-completion of HE doesn't change when the action on M(J) is replaced by the action on M(odd,J). Now M(odd,J) is on the face of it a very complicated HE-module. However it has an HE-stable subspace C(J) that is apparently easier to handle. (I'll define C(J) shortly). Let pr:Z/2[[x]]-->Z/2[[x]] be the map that removes from h all terms of the form x^11k. It turns out that pr is 1-1 on C(J) so that studying the action of HE on C(J) is the same as studying the action on pr(C(J)). I've made computer calculations concerning this action, and will report the results on question 135902--"Higher level analogs of Nicolas-Serre theory". What the computer suggests is the following. Let x,y and z be the images of T_3 +I, T_5 +I and T_7 in the J-completion of HE acting on pr(C(J). Then the map from the 3-variable power series ring to this J-completion is onto, and the kernel is a principal ideal generated by f^2, where f=Z+X+Y+X^2+XY+Y^2+(X^2)Y+X(Y^2)+Y^3+ higher degree stuff in X and Y.

I've now found a proof that M(odd, J)=C(J). So the results of the last paragraph apply equally to the J-completion of HE acting on M(J). So if one believes the computer the situation is quite different from that of level 1.

C and M_0 are defined in my question 138495--"Are these two subspaces of Z/2[[x]] the same?" (When N=11 they are; this is essential to my arguments). I define M_0 to be the subspace of M(odd) consisting of elements whose trace from Z/2(F,G) to Z/2(G) is zero. (Here F is as above and G=F(x^11); M is just the integral closure of Z/2[G] in Z/2(F,G)). C=M_0 is, like M and M(odd), stable under HE. Also it admits a direct sum decomposition with J locally nilpotent on the first summand, which I call C(J), and Ann(F) locally nilpotent on the second.

To show that M(odd, J)=C(J) is to show that every element of M(odd,J) has trace 0. The argument proceeds in several steps. (The first 2 generalize to arbitrary N).

1)__Suppose m is odd and T_11(F^m) is sum ((c_i)(F^i)). Then the trace of F^m is sum((c_i)(G^i)).

To see this let K be an algebraic closure of Z/2. For alpha in K with alpha^11=1, let F_alpha be F((alpha*(x^(1/11)). Then G and the F_alpha are the conjugates of G over K(F). So the trace of G^m from Z/2(F,G) to Z/2(F) is the sum of the mth power of G and the mth powers of the F_alpha. This sum is easily seen to be T_11(F^m)=sum((c_i)(F^i)). We now use the fact that the irreducible equation between F and G is symmetric in F and G to derive 1).

2)__If m is odd, trace(T_3(F^m))=T_3(trace(F^m))

For 1) shows that the left hand side is the result of replacing F by G in the element T_11(T_3(F^m)) of Z/2[F], while the right hand side is similarly obtained from T_3(T_11(F^m)).

3)__If h is in M(odd), then trace(T_3(h))=T_3(trace(h))

Since M_0 is stable under HE (this uses the fact that M_0=C), it is stable under T_3, and the result holds for h in M_0. In view of 2) it suffices to show that M_0 and the F^m with m odd span M(odd). In view of 1) it's enough to show that T_11 maps the space spanned by the F^m with m odd ONTO itself. But this is immediate from the level 1 Nicolas-Serre theory.

The proof that each element h of M(odd,J) has trace 0 is now easy. (T_3 +I)^q annihilates h for some q, and we may take q to be a large power of 2. then h=((T_3)^q)(h)). So by 3), trace(h)=((T_3)^q)(trace h). But T_3 is locally nilpotent on the space spanned by the powers of G giving the result.

It would be good to give a proof that what the computer makes clear is true is in fact the case. But M(odd,J) is so much more complicated than the space spanned by the odd powers of F, that I'm not optimistic about an approach to this in the style of Nicolas and Serre.


In level 1, when ell-1 divides 12, it isn't hard to show that each maximal modular J in HE(w) has residue-class-field equal to Z/ell. When ell is 2 or 3 (so that w=0), it's known that there's only one J, and that the image of T_p in Z/ell is p+1.

I've made computer calculations with ell = 5,7 and 13; they suggest the following striking results (which perhaps follow from the proof of Serre's modular forms conjecture?)

(a)--- When ell is 5, for each w in 2Z/4Z there are two J. When w=0, the image of T_p in HE(0)/J is (p^3)+1 for one of these, (p^2)+p for the other. When w=2, the image of T_p in HE(2)/J is p+1 for one J, (p^3)+(p^2) for the other.

(b)---When ell is 7, for each w in 2Z/6Z there are three J. When w=0, the images of T_p are (p^5)+1, (p^4)+p, and (p^3)+(p^2). When w=2, the images are p+1, (p^4)+(p^3) and (p^5)+(p^2). When w=4, they are (p^3)+1, (p^2)+p and (p^5)+(p^4).

(c)---When ell is 13, for each w in 2Z/12Z there are eight J. For six of these J there are a and b in Z/12Z with a+b=w-1, such that the image of T_p in HE(w)/J is (p^a)+(p^b). But I don't yet understand the other two J. (I guess what I'm saying is that these six J correspond to reducible Galois representations, and I'd like a representation -theoretic description of the other two J).

EDIT--I've found a lot of what I'm looking for in Bas Edixhoven's paper, "The weight in Serre's conjecture on modular forms", Invent. Math. 109, 563-594 (1992). In his discussion of his Theorem 3.4 he writes "The fact that eigenforms of weight at most p+1 give, up to twist, all systems of eigenvalues has been known to be true for a long time .... The first published proof ... for p>5, is due to Ash and Stevens".

This answers my question on the finiteness of the number of maximal modular ideals in any level. And in level 1, it shows that my last edit indeed gives all the maximal modulars when ell (that is to say p in the notation of the last paragraph) is 5,7 or 13. (I also found an arXiv paper by Craig Citro and Alexandru Ghitza, "Enumerating Galois representations in Sage" , which lists, in level 1, the number of these ideals for ell up to 211).

share|improve this question
This paper people.brandeis.edu/~jbellaic/preprint/Heckealgebra4.pdf studies Question 3 for $N=1$ and $\ell > 3$. They mention in the second paragraph on the top of page 3 that the answer to the first part of Question 1 is yes. –  Kevin Ventullo Jul 22 '13 at 1:20

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