Modular forms with large coefficient fields via congruences
Abstract
We use the theory of congruences between modular forms to prove the existence of newforms with square-free level having a fixed number of prime factors such that the degree of their coefficient fields is arbitrarily large. We also prove a similar result for certain almost square-free levels.
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Dieulefait et al. Research in Number Theory (2015) 1:2 DOI 10.1007/s40993-015-0003-9 RESEARCH ARTICLE Open Access Modular forms with large coefficient fields via congruences Luis Víctor Dieulefait1*, Jorge Jiménez Urroz2and Kenneth Alan Ribet3 *Correspondence: [email protected] 1Departament d’Algebra i Geometria, Universitat de Barcelona, Gran Via de les Corts Catalanes 585, 08007 Barcelona, Spain Full list of author information is available at the end of the article Abstract We use the theory of congruences between modular forms to prove the existence of newforms with square-free level having a fixed number of prime factors such that the degree of their coefficient fields is arbitrarily large. We also prove a similar result for certain almost square-free levels. 1 Background In this paper we will exploit the theory of congruences between modular forms to deduce the existence of newforms (in particular, cuspidal Hecke eigenforms) with levels of certain specific types having arbitrarily large coefficient fields. This property was already studied by Serre in [16] with other techniques. We will only consider newforms of weight 2 and trivial nebentypus. If the level is allowed to be divisible by a large power nof a fixed prime, or by the cube of a large prime p, then the coefficient fields of all newforms of thislevelwillgrowwithn(with p, respectively) due to results of Hiroshi Saito (cf. [15], Corollary 3.4; see also [1]) showing that the maximal real subfield of certain cyclotomic field whosedegree grows with n(with p, respectively) will be contained in these fields of coefficients. Thus, it is natural to deal with the question when the levels are square-free or almost-square-free, i.e., square-free except for the fact that they are divisible by a fixed power of a small prime. In the square-free case, for any given number t, we will prove that in levels which are the product of exactly tprimes there are newforms with arbitrarily large coefficient fields. We will recall results of Mazur on reducible primes for newforms of prime level that give the case of t=1. Then, a generalization of these results will allow us to deduce the case t=2. For t≥3 we follow a completely different approach, namely, we exploit congruences involving certain elliptic curves whose construction is on the one hand related to Chen’s celebrated results on (a partial answer to) Goldbach’s conjecture (cf. [2]) and on the other hand inspired by Frey curves as in the proof of Fermat’s Last Theorem (the diophantine problem that we will consider will be a sort of Fermat-Goldbach mixed problem). It is via the level lowering results in [14] that the desired congruence will be guaranteed. The precise statement of our first main result is the following: Theorem 1. Let B and t be two given positive integers. Then, there exist t different primes p1,p2, ...., ptsuch that if we call N their product, in the space of cuspforms of weight © 2015 Dieulefait et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.
Dieulefait et al. Research in Number Theory (2015) 1:2 Page 2 of 14 2, level N and trivial nebentypus there exists a newform f whose field of coefficients Qf satisfies: [Qf:Q]>B. In the almost-square-free case, we will consider levels N=2kp1...ptwhich are square-free except that they are divisible by a small power of 2. We study these particular levelsbecausenewformsoftheselevelscanberelatedtoprimevaluesofcertainbinomials already studied in the literature (for a more thorough discussion, see Remark 1 on section 5). We will prove that for any fixed t, among newforms with such levels the fields of coefficients have unbounded degree. We will use congruences with certain Q-curves constructed from solutions to the problem of finding prime values attained by the expression x4+y2/c. Again, these Q-curves will also have some features inspired by Frey curves, and the existence of the desired congruences will be a consequence of level lowering. For c=1 it is a celebrated result of Friedlander and Iwaniec that infinitely many primes are of the form x4+y2(cf. [8]). Here we give the following generalization. Theorem 2. Let B >0fixed, and the usual Von Mangold function. Then, we have uniformly in c ≤(log x)B c|a2+b4≤cx a2+b4/c=K(c)x3/4+ox3/4(1) where a,b run over positive integers, and K is completely explicit in terms of c. In Theorem 5 in section 6 we give the precise value of K(c). For our application to congruences between modular forms we only need a mild version of the particular case with fixed cof the form c=5. The following is a direct consequence of the previous theorem. Corollary 3. Let c be a positive odd integer. Then there are infinitely many primes of the form x4+y2/c if and only if c can be written as the sum of two squares. The precise statement of our second main result, the one covering the almost-squarefree level case, is the following: Theorem 4. Let B and t be two given positive integers. Then, there exist α∈{5, 8}and t different odd primes p1,p2, ...., ptsuch that if we call N the product of these t primes, in the space of cuspforms of weight 2, level 2αN and trivial nebentypus there exists a newform f with field of coefficients Qfsatisfying: [Qf:Q]>B. Letusstressthattheresultsonprimevaluesofx4+y2/c, in Theorem 2 and its corollary, are interesting in its own right, independently of the application to finding newforms with large coefficient fields. On the other hand it is also important to clarify that the proof of Theorem 2 is a generalization of Theorem 1.1 in [8], and does not contain
Dieulefait et al. Research in Number Theory (2015) 1:2 Page 3 of 14 any new conceptual ingredient not present in [8]. It is however the case that some of the computations done in [8] do not apply to this case in a straightforward manner and, hence, they must be done now with the required level of generality in the variable c. 2 Theorem 1 for the case of prime level: Mazur’s argument Suppose that the level Nis prime and that >3 is a prime that divides N−1. Then it is proved in [11] that the prime is Eisenstein, meaning that there is a newform fof weight 2andlevelNsuch that if we call Kfits field of coefficients there is a prime λdividing in the ring of integers of Kffor which we have ap≡1+pmod λfor all primes p.The residual mod λGalois representation attached to fis reducible. In particular, we have a2≡3modλ(1) The coefficients of the modular form fand those of any Galois conjugate fσall satisfy the bound |ap|≤2√p,inparticulara2and all its Galois conjugates have absolute value bounded above by 2√2<3. Then, a2−3 is a non-zero algebraic integer, whose norm is divisible by because of congruence (1) and with absolute value at most 3+2√2deg kf. Hence ≤3+2√2deg kf Thus, deg Kfis bigger than a fixed constant times log .Takingbig and using Dirichlet’s theorem to find an N≡1mod,wecanmakedegKfas big as we like. This proves the case of prime level (t=1) of Theorem 1. 3 A Frey curve adapted to Chen results, and the case t≥3ofTheorem1 Let be a (large) prime number, and assuming for the moment the truth of Goldbach’s conjecture let us write the even number 2+4as the sum of two prime numbers: 2+4= p+q.Sincepand qare clearly non-congruent modulo 4, we assume without loss of generality that p≡3mod4.LetFbe the semistable Frey curve associated to the triple p,q,2 +4: y2=x(x−p)x+2+4 Its conductor is 2pq, while its minimal discriminant is =2+4pq2/28=2pq2. The modularity of all semistable elliptic curves, proved by Wiles in [17], implies that there is a newform fof weight 2 and level 2pq corresponding to F. The mod Galois representation F[]ofGal(¯ Q/Q)is irreducible by results of Mazur, and unramified at 2 because the 2-adic valuation of the discriminant is divisible by (as in the original Frey curves related to solutions of Fermat’s Last Theorem). Although it comes initially from a newform fof level 2pq, by level-lowering (cf. [14]) it arises also from a newform fof level pq. The trace of the action of Frob 2 on F[]is±(1+2), because this is the well-known necessary condition for level-raising, i.e., for the existence of an -adic Galois representation with semistable ramification at 2 providing a lift of F[], and we have such a lift by construction: it is given by the Galois action on the full -adic Tate module T(F)of the curve F.Soifwecall{ap}the coefficients of fwe get a2≡±3modλ(2)
Dieulefait et al. Research in Number Theory (2015) 1:2 Page 4 of 14 for a prime λdividing in the field of coefficients of f. From this congruence we can argue as we did in the previous section using congruence (1) an conclude easily that just by choosing the prime sufficiently large we can construct newforms of weight 2 and level N=pq a product of two different primes with arbitrarily large field of coefficients. Since Goldbach’s conjecture remains open, in order to get an unconditional result we need to move to the case of three primes in the level. Using the results of Chen on Goldbach’s problem (cf. [2]) we know that for sufficiently large 2+4can be written as the sum of a prime and a “pseudo-prime”, i.e., a number that is either a prime or the product of two different primes. Then, in particular, one of the following is true for infinitely many : 2+4canbewrittenasthesumoftwoprimespand q,or2 +4can be written as the sum of a prime pand the product of two primes qr. If the first is true, as we have just seen, this will prove Theorem 1 for levels which are the product of two primes, and if the second is true a similar argument with the triple p,qr,2 +4shows that the theorem is true for levels which are the product of three primes. Thus, to finish the proof of the case t=3 of Theorem 1, it remains to show that if the result holds for t=2thenitalsoholdsfort=3. But this is just an application of raising the level (cf. [13]), because whenever we have a modular form of level pq and an irreducible mod Galois representation attached to it, the same residual representation is also realized in some newform of level pqr,aslongastheprimersatisfies the required condition for level raising (and it is well known that there are infinitely many primes r that satisfy this condition, cf. [13]). Thus, whenever we have found a newform fas in the previous argument, of level pq and satisfying (2), there are also newforms with levels of the form pqr also satisfying (2) and from this the proof of the theorem for the case t=3 follows exactly as explained above. To treat the case of more than three primes, we modify the argument above by further raising the level. Starting with the irreducible mod representation afforded by F[]of conductor either pq or pqr,andforanygivent≥4, we can find forms giving the same residual representations in levels equal to the product of tprimes by just raising the level t−2(t−3, respectively) times. For this, we have to take care not to lose spurious primes as we add on new ones. The required analysis is carried out in [4]. We conclude that Theorem 1 is true for any t≥3. Since the case of prime level was dealt with in the previous section, at this point only the case of t=2 remains unsolved (and a proof of Goldbach’s conjecture would be enough to handle it). 4 The case t=2 of Theorem 1 via a result of Ogg One way to treat Theorem 1 for the case where N=pq (without proving Goldbach’s conjecture)istoappealtotheresultsofOggin[12].Ifpand qare distinct primes, Ogg finds a degree 0 cuspidal divisor on X0(pq)whose image on J0(pq)has order equal to the numerator of the fraction (p−1)(q+1)/24. Take >3. If divides (p−1)(q+1), using exactly the same arguments applied in (cf. [11]) in the case of prime level we find an eigenform fat level pq that is Eisenstein mod , therefore giving a reducible residual mod representation. In particular, this means that the coefficient a2satisfies again congruence (1) as in section 2, for some prime λdividing in its field of coefficients Kf,andwe deduce as before that the degree of Kfis large (it grows with ). We need to ensure that fis genuinely a newform, i.e., that its eigenvalues do not arise at level por at level q.We begin as before by taking large. Then we find q≡−1modand pick pto be a random
Dieulefait et al. Research in Number Theory (2015) 1:2 Page 5 of 14 prime that is not congruent to 1 mod . Since the Eisenstein primes at prime level Nare divisors of N−1, we see that is not an Eisenstein prime at either level por level qwhile it is an Eisenstein prime at level pq,thustheformfmust be a newform of level pq.This completes the proof of the case t=2 of Theorem 1. Thus, putting together the results of the last three sections, we conclude that Theorem 1 for any positive value of t. 5 The proof of Theorem 4 To prove Theorem 4 we follow a strategy similar to the one explained in section 3, except that now we will start from a diophantine equation such that the elliptic curve corresponding to any solution is a Q-curve defined over Q(i). For the Fermat-type equation x4+y2=zpan attached Q-curve was proposed by Darmon in [3] and in the work of Ellenberg [6] it was shown using the modularity of this curve that the diophantine equation does not have non-trivial solutions for large p. We will consider instead the diophantine problem: x4+y2=5p The result that we will prove in the next sections (see Theorem 5 in section 6, specialized to the case c=5), which is a generalization of the case =0solvedby Friedlander-Iwaniec in [8], implies that for any prime exponent there exist infinitely many primes psuch that there are integer solutions A,Bto this equation. Thus, if is a given prime and A,B,psatisfy A4+B2=5p with pprime, we consider, as in the work of Darmon and Ellenberg, the elliptic curve E: y2=x3+4Ax2+2A2+iBx For simplicity, and since we have infinitely many primes psatisfying the equation, we assume p= . The following properties of this curve are known (cf. [6]): Eis 2-isogenous to its Galois conjugate, in particular it is a Q-curve. If we call Wits Weil restriction defined over Q,itisaGL 2-type abelian surface and thus it has a compatible family of 2-dimensional Galois representations of GQattached. These representations have coefficients in Q√2. This abelian surface is semistable outside 2 and it is modular. Computing the conductor of Wit follows that the modular form fattached to Whas level 2α5p,withα=5 or 8. It has weight 2 and trivial nebentypus. This newform has coefficients in Q√2and it has an inner twist. We now consider for the prime westartedwithandλ|in Q√2the residual mod λrepresentation ¯ρW,λattached to W. Assuming that >13 it is known that this representation is irreducible (cf. [6]). Since the discriminant of Eis 512 A2+iB5pwe can, as in [6], apply the Frey trick at the semistable prime 5 (observe that 5 is unramified in Q(i)/Q): locally at 5 the valuation of the discriminant is divisible by (on the other hand, this does not happen locally at the prime p). Thus, we conclude that ¯ρW,λis unramified at 5: more precisely it has conductor 2αpwith α∈{5, 8}. If we apply lowering the level (cf. [14]) we see that there is a newform fof level 2αp, weight 2 and trivial nebentypus such that some of its corresponding residual Galois representations in characteristic is ¯ρW,λ.
Dieulefait et al. Research in Number Theory (2015) 1:2 Page 6 of 14 Now we can conclude as in section 3: we focus on the coefficient a5of fand we want to use the fact that there is a mod λcongruence between fand a newform fcorresponding to an abelian variety which is semistable at 5 (namely, the abelian surface W). Then, the necessary condition for level-raising must be satisfied and this translates into the following congruence: a5≡±6modλ From this congruence we see, as in sections 2 and 3, that the minimal field of definition of a5, and a fortiori the field of coefficients of f,hasadegreethatgrowswith.This solves the case of almost-square-free level 2αpwith α=5 or 8, i.e., the case t=1, of Theorem 4. The case of Theorem 4 for levels of the form 2αp1·.... ·pt,witht>1fixed and the piodd, different primes, can be deduced from this by t−1 applications of raising the level as explained in section 3. Remark 1. The referee asks whether or not we can extend Theorem 4 to cover other almost-square-free levels by choosing other Frey curves or Q-curves. This seems plausible, but will require extra work because of the following reason. On one hand, several Frey curves or Q-curves have been proposed in the literature to attack diophantine equations of Fermat type, and for all of them the resulting modular forms have almost-square-free level, for example the equations x4+2y2=zpand x4+3y2=zpare solved in [5] using Frey Q-curves. On the other hand, the problem of prime values for the kind of binomials appearing on the left hand side of these equations is solved in the literature only for two cases, namely, the case we have considered and the case of irreducible cubic forms (cf. [9] and [10]). Consider for example x3+2y3. We can apply to it the standard trick of factorizing the form and construct a Frey elliptic curve, but this curve will be defined over Q3 √−2. There is no reason for this curve to be a Q-curve, so probably there is no way to relate this equation to classical modular forms. 6 Prime values of (x4+y2)/c We now introduce some notation which will be used from now on. For any given prime p, and any integer dwe denote vp(d)the highest power of pdividing d. Moreover, we will write das d=d1d2 2=d1d2 3d4 4,whered1,d3are square-free. We will consider (r)the usual Von Mangoldt function, extended as zero over non integer numbers. Then, the main result of this section is the following theorem. Theorem 5. Let B >0fixed. We have uniformly in c ≤(log x)B (a2+b4)/c≤x a2+b4/c=4π−1κG(c)(cx)3/4+ox3/4(2) where a,b run over positive integers, G is a multiplicative function, and κ=1 01−t41/2dt =1 42 /6√2π.(3)
Dieulefait et al. Research in Number Theory (2015) 1:2 Page 7 of 14 Remark 2. It is important to emphasizes here that the constant G(c), explicitly described in Lemma 8, can take the zero value, and it does precisely in those cwhich are non representable as the sum of two squares, or such that v2(c)≡3(mod 4).Fortrivial reasons there is at most one prime in the sequence a2+b4/cin these cases, since none of the elements is in fact coprime with cif cis non representable as the sum of two squares, and if v2(c)≡3(mod 4), every integer of the sequence has to be even. Hence, the proof that follows restricts to those values of csuch that this constant is non zero since, in any other case, the result is trivial. The proof of Theorem 5 relies in the verification of the hypothesis needed to apply the Asymptotic Sieve due to Friedlander and Iwaniec in [7], but now for the sequence a(c)n=0 for any (n,c)>1, and for ncoprime to cgiven by a(c)n= (a2+b2)/c=n Z(b),(4) where a,bare integers non necessarily positive, and Zis the function with value Z(m2)= 2, for any integer m= 0, Z(0)=1, and Z(b)=0 in any other case. From now on we will only consider integers ncoprime to cand, then, we have a(c)n=aold cn where aold nis the sequence related with Theorem 1 of [8]. We now include for reading convenience the hypotheses and main result of the Asymptotic Sieve. The following, with the exception of (13), is basically a copy of Section 2 in [8]. We explain the difference between (13) and (2.8) in [8] at the end of this section. Consider a sequence of real, nonnegative, numbers A=(an)n≥1,andxa positive number. We want to obtain an asymptotic formula for S(x)= p≤x aplog p, where the sum runs over prime numbers, in terms of A(x)=n≤xan.Wesuppose A(x)A√x(log x)2,(5) A(x)x1/3 n≤x a2 n1/2 .(6) As usual in sieve theory, we will assume that for any integer d>1 Ad(x)= n≤x d|n an=g(d)A(x)+rd(x), where gis a multiplicative function, and rd(x)is regarded as an error term. For the function gwe assume the following hypotheses 0≤g(p2)≤g(p)<1, (7) g(p)p−1,(8) g(p2)p−2,and (9) p≤y g(p)=log log y+e+Olog y−10, (10)
Dieulefait et al. Research in Number Theory (2015) 1:2 Page 8 of 14 for every yand some edepending only on g. For the error term we will assume 3 d≤DL2|rd(t)|≤A(x)L−2, (11) uniformly in t≤x,forsomeDin the range x2/3<D<x. (12) The superscript 3 in (11) restrict the sum to cube free moduli and L=(log x)224 .Wealso require Ad(x)d−1τ(d)8A(x)log xuniformly in d≤x1/3, (13) and finally an estimate in bilinear forms like m| N<n≤2N mn≤x (n,m)=1 β(n)μ(mn)amn|≤A(x)L−4, (14) where the coefficients are given by β(n)=β(n,K)= k|n,k≤K μ(k), (15) for any Kin the range 1≤K≤xD−1, (16) Nverifying −1√D<N<δ −1√x(17) for some ≥δ≥2, and is the product of all primes p<Pfor some Pwhich can be chosen conveniently in the range 2≤P≤1/235 log log x. (18) In this conditions we have Proposition 6. Let Abe a sequence verifying the above hypotheses. Then, S(x)=HA(x)1+Olog δ log (19) where H is the positive constant given by the convergent product H= p (1−g(p)) 1−1 p−1 , (20) and the implied constant depends only on the function g. Normally δis a large power of log xand a small power of x. Remark 3. It is important to note that (13) is not the original assumption (1.6) in [7], but a slightly weaker. However, as the authors mention in that paper, (1.6) is only required to reduce the hypotheses m τ5(m)| N<n≤2N mn≤x (n,m)=1 β(n)μ(mn)amn|≤A(x)(log x)−3, (21)
Dieulefait et al. Research in Number Theory (2015) 1:2 Page 9 of 14 and d≤D μ2(d)τ5(d)|rd(t)|≤A(x)(log x)−3, (22) to (11) and (14). We just have to follow the reasoning in Section 2, p. 1047 of [7] to see that this reduction is also possible with our hypothesis (13). 7 Proof of Theorem 5 To prove Theorem 5 we will use Proposition 6 for the sequence given in (4). Hence, we have to check that the sequence verifies hypotheses (5) through (14). Given an integer d≥1, we denote Ad(x;c)=n≤x,n≡0(mod d)a(c)n. The first thing that needs to be done is to find a good approximation of Ad(x;c)in terms of a multiplicative function. Now, Ad(x;c)=0for(d,c)>1andfor(d,c)=1wehave Ad(x;c)= k|c μ(k)Aold ckd(cx), (23) and we know by [8] that Aold d(x)=g(d)Aold(x)+rold d(x), where the functions g,rold satisfy conditions (5) through (14). Note that for any integer d the definition of Aold d(x)is implicit in (23) for c=1, and observe that g,rold are precisely the functions g,rappearing in [8]. Hence, to approximate Ad(x;c)we are tempted to use the approximation of Aold(x)given in Lemma 3.4 of [8]. However, this lemma only works for cube-free integers dwhich do not cover completely our case, since cwill be any number c≤(log x)B. Because of this, an extra technical difficulty appears. Indeed, any cube-free integer can be factorized as d=d1d2 2with d1,d2square-free and coprime, conditions that are highly convenient when dealing with multiplicative functions. When the integers are not cube-free as it is our case, we need to control the contribution that comes from common divisors (c.f. Lemma 8), and we have to keep track of it throughout the argument. Hence, our next objective is to generalize Lemma 3.4 of [8] to any integer d.Asin[8], we start approximating Ad(x;c)by Md(x;c)= k|c μ(k)1 ckd 0<(a2+b2)≤cx Z(b)ρ(b;ckd), for any dcoprime to c,whereρ(b,d)denotes the number of solutions α(mod d)to the congruence α2+b2≡0(mod d),andMd(x;c)=0 otherwise. The following is a trivial consequence of Lemma 3.1 of [8]. Lemma 7. Let B >0.Foranyc≤(log x)Bwe have d≤D|Ad(x;c)−Md(x;c)|D1/4x9/16+ε for any D ≥1and ε>0and the implied constant depending only on ε. Now, we need to find out the main term of Md(x;c), as we mentioned, by generalizing Lemma 3.4 of [8].