# 2-adic Coefficients of Modular Hecke Eigenforms

Suppose that $N$ is prime, and consider the normalized cuspidal Hecke eigenforms of weight 2 and level $\Gamma_0(N)$.

For such an eigenform $f$, the coefficients generate (an order in) the ring of integers of a totally real field $E$. The corresponding simple abelian variety quotient $A_f$ of $J_0(N)$ has dimension $[E:\mathbb{Q}]$. Let $C(N)$ denote the maximal degree $[E:\mathbb{Q}]$ amongst all such eigenforms. It is not too hard to prove that $C(N)$ is unbounded, and not much harder to show that the $\lim \inf$ of $C(N)$ as $N\to \infty$ is infinite. (I don't want to mention the argument here because I don't think it will apply to my question below.)

Is the same result true over $\mathbb{Q}_2$? Namely, fix an embedding of $\bar{\mathbb{Q}}$ into $\bar{\mathbb{Q}}_2$. Then, for any $d$, is it true that for all sufficiently large prime $N$ there exists a cuspidal eigenform of weight 2 and level $\Gamma_0(N)$ whose coefficients generate some field $E/\mathbb{Q}_2$ with $[E:\mathbb{Q}_2] > d$?

This would be interesting to know even for $d = 1$.

A natural way to approach this question is to ask for an eigenform $f$ with coefficients in a local field $K$ with residue field degree $\ge 2$. Or, asking for slightly more, with coefficient ring $\mathcal{O}$ admitting a surjective map to a field $\mathbf{F}$ of order divisible by $4$ (this is slighly more since the rings $\mathcal{O}$ can typically be non-trivial orders in $\mathcal{O}_K$). By Serre's conjecture, this is equivalent to asking for the existence of an irreducible Galois representation

$$\rho: \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \rightarrow \mathrm{GL}_2(\mathbf{F})$$

with Serre weight and conductor $(2,p)$ (note, no oddness condition!). It's hard to construct such representations, but one natural way is to use induced ($=$ projectively dihedral) representations.

If $E$ is the corresponding quadratic extension, and $F/E$ the cyclic extension, then the Serre weight and level will be $(2,p)$ providing that $F/E$ is totally unramified and $E = \mathbf{Q}(\sqrt{-1})$, $\mathbf{Q}(\sqrt{p})$, or $\mathbf{Q}(\sqrt{-p})$. (EDIT: In a previous version of this post, I only thought about extensions that were totally unramified at $2$ for some unknown reason.) One is in good shape providing that the odd part of the class group of one of the fields $\mathbf{Q}(\sqrt{\pm p})$ is not $(\mathbf{Z}/3 \mathbf{Z})^n$. Since one can't really say much about real quadratic fields, let's think about the imaginary quadratic fields. It's a theorem, proved by (amongst other people) Lillian Pierce, and Jordan Ellenberg and Akshay Venketesh, that the 3-torsion part of the class group is a negligible part of the class group. Thus we are done if $p$ is big providing that the $2$-part of the class group is not so big. The $2$-part is trivial if $p$ is $3$ mod $4$, and has order $2$ if $p$ is $5$ mod $8$. However, it is quite possible that the class group of $\mathbf{Q}(\sqrt{-p})$ is $\mathbf{Z}/2^n \mathbf{Z}$, and very rough heuristics suggest that this might happen infinitely often.

[EDIT: LaTeX issues now mostly sorted in the sequel]

If $\mathbf{T}$ denotes the Hecke algebra tensored with $\mathbf{Z}_2$, then $\mathbf{T}$ is a free $\mathbf{Z}_2$-module of rank $n = \mathrm{dim}(S_2(\Gamma_0(p)))$. It is also a semi-local ring. For each maximal ideal $m$ of $\mathbf{T}$, we are asking that

$$\mathbf{T}_{m} \otimes \mathbf{Q}$$

is not a number of copies of the 2-adic numbers. If $m$ is the $2$-adic Eisenstein ideal, then in a paper with Matthew Emerton, I prove that

$$\mathbf{T}_{m}/2$$

has rank $2^{e-1}-1$, where the $2$-part of the class group of $\mathbf{Q}(\sqrt{-p})$ (we may assume that $p$ is $1$ modulo $8$) has order $2^{e}$. If $p$ is $9$ modulo $16$ then $\mathbf{T}$ localized at $m$ is also a domain, and so we are done in this case if $2^{e-1} -1 \ge d$. Even if $p$ is $1$ modulo $16$, this shows that the bulk of $\mathbf{T}/2$ must come from non-Eisenstein primes $m$, since $2^e \le h \ll p^{1/2 + \epsilon} \ll p/12 \simeq n$. So, in the case $d = 1$, one may assume that the odd part of the class group is $3$-torsion, and then hope that for the corresponding $m$, one can prove that the corresponding rings $\mathbf{T}_{m}/2$ can't be too large.

It turns out that the relevant question becomes: Given a $\overline{\rho}$ corresponding to an $S_3$ extension, and given a finite flat deformation of $\overline{\rho}$ to $\mathrm{SL}_2(A)$ for a local ring $A$ killed by $2$, can one bound the rank of $A$ as an $\mathbf{F}_2$-module? For example, can one prove that $A$ has rank $p^{1 - \epsilon}$? The example of Eisenstein $m$ suggests that one can not neccesarily do better than $p^{1/2 + \epsilon}$.

Here is an answer that shows works for d = 1, and works for general d assuming GRH.

Step I. Let $C$ be the class group of $\mathbf{Q}(\sqrt{-p})$. The group $C$ decomposes as $C_{odd} \oplus C_{even}$.

Step II. Suppose that $C_{odd}$ is not annihilated by $2^{2d} - 1$. Then there exists a dihedral representation induced from a character of $C_{odd}$ which is finite flat at $2$, ordinary at $p$, and has which has image in

$$\mathrm{SL}_2(\mathbf{F})$$

for a field $\mathbf{F}$ of degree $>d$. Thus we are done unless $C_{odd} = C_{odd}[2^{2d} - 1]$.

Step III. If $d = 1$, then then the $2^{2d} - 1 = 3$-torsion of $C_{odd}$ has small order, namely, of order at most $$p^{1/2 - \delta}$$ for some explicit $\delta > 0$. For general $d$, the $2^{2d} -1$-torsion has order at most $p^{\epsilon}$, assuming GRH.

Step IV: If $p$ is $-1$ modulo $4$ or $5$ modulo $8$, then $C_{even}$ has order $\le 2$. This implies by Steps II and III that $C$ has order $\le p^{1/2 - \delta}$, which contradicts the estimate $h \gg p^{1/2 - \epsilon}$.

Step V: We may assume that $p$ is $1$ modulo $8$. Let $|C_{even}| = 2m$. Using the estimate $h \gg p^{1/2 - \epsilon}$, we deduce that $m \gg p^{\delta - \epsilon}$ for some explicit $\delta > 0$.

Step VI: Let $\mathbf{T}$ denote the localization of the Hecke algebra at the Eisenstein prime of residual characteristic $2$. It is a consequence of one of the main theorems of FC-ME that one has $\mathbf{T} = \mathbf{Z}_2[x]/f(x)$, where:

$$f(x) \equiv x^{m - 1} \ \mathrm{mod} \ 2,$$

$$\mathbf{Z}_2/f(0) \mathbf{Z}_2 \simeq \mathbf{Z}_2/n \mathbf{Z}_2,$$

and $n$ is the numerator of $(p-1)/12$. (Compare the discussion in Mazur's Eisenstein ideal paper, section 19, page 140.)

Step VII: If the coefficients of all forms of level $\Gamma_0(p)$ define extensions of degree $\le d$, then the normalized $2$-adic valuation of every root of $f(x)$ is at least $1/d$. We deduce that

$$\frac{m - 1}{d} \le v_2(n),$$

Since $v_2(n) \le \log_2(n) < \log_2(p)$, we deduce that

$$m \le d \log_2(p).$$

This contradicts the previous estimate $m \gg p^{\delta - \epsilon}$ for sufficiently large $p$.

Remark: For $d = 1$, it should be easy to use the unconditional results to give an explicit lower bound on possible $p$.

To summarize, for any $d$, and sufficiently large $p$, there exists either:

(i) A modular form $f$ of level $\Gamma_0(p)$ with coefficients in an extension of $\mathbf{Q}_2$ which contains an unramified extension of degree $> d$, and whose residual representation is dihedral,

(ii) A modular form $f$ of level $\Gamma_0(p)$ which coefficients in an extension of $\mathbf{Q}_2$ which contains a ramified extension of degree $> d$, and whose residual representation is Eisenstein.

Extra: For any $p$, the coefficients of an $f$ of level sufficiently divisible by $p$ contains the totally real subfield of the $p^n$th roots of unity. For any $p$, and sufficiently large $n$, this contains a large extension of the $2$-adic numbers.

Reason why this argument still sucks: Doesn't work at all for primes $> 2$, and uses GRH for $d > 1$.

• 1. Those Q-curves over Q(i) have bad reduction at 2 and p, not just p, if it matters to you. 2. I agree that it doesn't seem there'll be enough Q-curves. I wonder about RM abelian surfaces more generally. Is there a generally accepted heuristic, akin to "N^{5/6} elliptic curves of conductor at most N," for the number of RM abelian surfaces (say, with RM by a fixed real quadratic field K) of conductor at most N?
– JSE
Nov 14, 2009 at 4:09
• For inspiration in the case $p=1$ mod $16$ one could take some 3-digit prime of this form, compute, and see why there are big dim spaces over $Q_2$. Are the associated mod 2 reps big (in the sense that they're giving reps to something bigger than $GL(2,Z/2Z)$? Or are the mod 2 reps small but the deformations are big? Did you try any computations FC? I looped over $p\leq 3700$ and found that the largest prime with all local pieces of size at most 5 was 257, by the way. So the conjecture that the local $d$ is growing looks plausible... Nov 15, 2009 at 11:20
• p=257 is an interesting case, right? The class groups are Z/3 and Z/16, which are bad. Where are the reps coming from in that case? Nov 15, 2009 at 12:18
• @FC: This is a great answer. But I think it's weird to refer to people whose work you are citing only by their initials, and weirder to insist on anonymity and also refer to your own work. Jan 14, 2010 at 7:05

Not an answer, but a remark: absent any reason to think otherwise, you might think of the etale Q_p-algebra generated by the Hecke operators on S_2(Gamma_0(N),Q_p) as a "random" such algebra of the given dimension; a reasonable probability distribution of these is described by Serre, then revisited by Bhargava and (perhaps most usefully for you) in Kedlaya's paper on mass formulas:

It ought to be easy to compute the expected number of copies of Q_p in a random etale Q_p-algebra in this sense, maybe somewhat complicated to compute the probability that there's no Q_p-factor. I'll bet the former has a simple answer and I would hesitantly guess the latter goes to 0. In which case I suppose I'd even more hesitantly lean towards a negative answer to your question.

Are there good reasons on the modular forms side to lean towards a positive answer?

• No, I the misparsing was all mine. I agree -- on heuristic grounds one wouldn't expect to find that the Q_2-algebra was (Q_2)^dimension. I agree that the question in the weight aspect seems different, at least when the weight has something to do with 2. On your view, if the weight is increasing in a sequence of integers that (let's say) equidistributes in some 2-adic region, instead of approaching 0 like yours, would you expect the Q_2-algebra to look "random" again?
– JSE
Oct 25, 2009 at 4:50

One silly idea: maybe one can just write down infinitely many $\mathbb{Q}$-curves such that for each such curve C the prime 2 does not split completely in the field of definition E of C. I doubt this would work in general, but for d = 1 (so try constructing C which are defined over a quadratic field) this might be managable.

• I would imagine the hard part with this approach would be to get the conductor prime. Another idea would be to write down lots of A_5 extensions of Q and consider the mod 2 Galois reps and go from there, but again you're not in general going to be able to find an A_5 extension ramified only at 2 and a given large prime p. Nov 14, 2009 at 9:08