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 $d = 1$.


3 Answers 3


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


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


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 $d > 1$.

  • $\begingroup$ 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? $\endgroup$
    – JSE
    Nov 14, 2009 at 4:09
  • $\begingroup$ 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... $\endgroup$ Nov 15, 2009 at 11:20
  • $\begingroup$ 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? $\endgroup$ Nov 15, 2009 at 12:18
  • $\begingroup$ @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. $\endgroup$ 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?

  • $\begingroup$ 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? $\endgroup$
    – 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.

  • $\begingroup$ 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. $\endgroup$ Nov 14, 2009 at 9:08

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