Fermat's Last Theorem and Modular Curves over Real Quadratic Fields
aa r X i v : . [ m a t h . N T ] M a r Fermat’s Last Theorem and Modular Curves overReal Quadratic Fields
Philippe Michaud-RodgersMarch 3, 2021
Abstract
In this paper we study the Fermat equation x n + y n = z n overquadratic fields Q ( √ d ) for squarefree d with 26 ≤ d ≤
97. By study-ing quadratic points on the modular curves X ( N ), and working withHecke operators on spaces of Hilbert newforms, we extend work of Fre-itas and Siksek to show that for most squarefree d in this range thereare no non-trivial solutions to this equation for n ≥ Contents
There has been much recent interest in the study of the Fermat equation x n + y n = z n , n ≥ x, y, z ∈ K , a number field. By a non-trivial solution, we mean xyz = 0. The studyof the Fermat equation over number fields dates back to the work of Mailletin the late 19th century [11, p. 578].Asymptotic versions of Fermat’s Last Theorem over number fields (prov-ing there are no non-trivial solutions if all the prime factors of n are greaterthan some bound dependent on K ) have been proven over various numberfields (see [12, 16] for example). In this paper, we are concerned with tryingto prove the non-existence of non-trivial solutions for all n ≥ K . Jarvis and Meekin proved this statement over the field Q ( √
2) in [18]. This work was then extended by Freitas and Siksek [14] to thequadratic fields Q ( √ d ), with d squarefree, d = 5 ,
17 in the range 3 ≤ d ≤ n ≥
4. Kraus,in [22], proved that for K a cubic field of discriminant 148 , n ≥
4. The aim of this paper is to extendthe result of Freitas and Siksek to squarefree d in the range 26 ≤ d ≤ d in this range, issues arise surrounding irreducibility of Galoisrepresentations and the computation of Hilbert newforms. In many cases,we overcome these issues to obtain the following result. Theorem 1.1.
The equation x n + y n = z n , x, y, z ∈ K, has no non-trivial solutions for n ≥ and K = Q ( √ d ) , when d ∈ D , where D = { , , , , , , , , , , , , , , , , , , , , , , , , , , , } . We note that the case d = 79 is considered in the paper of Freitas andSiksek [14, p. 14]. We also obtain the following partial results. Theorem 1.2.
The equation x p + y p = z p , x, y, z ∈ K, has no non-trivial solutions for p ≥ , prime, and K = Q ( √ d ) , when d =26 , , , , , , , , unless p appears in the following table. d
26 34 35 37 61 94 97 p ,
103 83 23 ,
97 37 127 151 467
Email: [email protected] d = 35, all other primes in Theorem1.2 appear due to being unable to prove that the corresponding mod p Galoisrepresentation is irreducible.In some cases, extending the work of Kraus [21], we can obtain a lowerbound on the size p . Theorem 1.3.
The equation x p + y p = z p , x, y, z ∈ K, has no non-trivial solutions for ≤ p ≤ , prime, and K = Q ( √ d ) , when d = 17 , , , , , , unless d = 57 and p = 19 . Moreover, if d = 89 ,then we also have no non-trivial solutions when p ≥ and p ≡ ± . We now outline the rest of the paper. In Section 2 we overview the gen-eral proof strategy and state the key properties of the Frey elliptic curve,mainly following [14]. In Sections 3 and 4 we study the question of irre-ducibility. In Section 3 we use techniques from class field theory and variousirreducibility criteria to do this, and in Section 4 we deal with many re-maining cases by studying quadratic points on the modular curves X ( N ):using modular parametrisations, local obstructions, sieving techniques, andChabauty over number fields. Finally, in Section 5 we work with Hecke oper-ators to partially reconstruct and subsequently eliminate Hilbert newformsof large level.The Magma [4] and
Sage [30] code used to support the computations inthis paper can be found at: https://warwick.ac.uk/fac/sci/maths/people/staff/michaud/
I would like to express my sincere gratitude to my supervisors SamirSiksek and Damiano Testa for many useful discussions and their support inwriting this paper.
In this section we provide a brief overview of how one associates an ellipticcurve to a putative solution of the Fermat equation, and we state someproperties of this curve. We then state the level-lowering theorem used to(hopefully) obtain a contradiction, proving the non-existence of non-trivialsolutions. Throughout this section we mainly follow [14, pp. 4-8].We start by fixing some notation. Let K be a real quadratic field Q ( √ d )for some squarefree d with 26 ≤ d ≤
97. Let ǫ be a fundamental unit for K .Write Cl( K ) for the class group of K . We denote a prime of K by q . Write S = { q : q | } , which consists of a single prime if 2 is inert or ramifies in K , and two primes if 2 splits in K . 3n all cases we consider, Cl( K ) has order 1, 2, or 3. It follows that thequotient group Cl( K ) / Cl( K ) has size r = 1 or 2. If r = 2 then choose m an odd prime ideal such that [ m ] represents the non-trivial element ofCl( K ) / Cl( K ) . The table in the appendix displays our choice for the ideal m in each case.Although Theorem 1.1 is stated for any n ≥ x, y, z ∈ K , wecan reduce (as in the case of Fermat’s Last Theorem over Q ) to the case of n = p , prime, and x, y, z ∈ O K . We therefore study the equation a p + b p + c p = 0 , a, b, c ∈ O K , p ≥ . As discussed in [14, p. 2], it has been shown that this equation has no non-trivial solutions for primes 5 ≤ p ≤
13 (as well as for the same equation withexponent n = 4) over real quadratic fields, and so we will assume that wehave a non-trivial solution ( a, b, c ) with p ≥
17. To this solution we associatethe Frey elliptic curve E a,b,c,p : Y = X ( X − a p )( X + b p ) , and write E = E a,b,c,p . We write N for the conductor of E which is an idealin O K . Lemma 2.1 (Frey curve invariants, [14, p. 5]) . The curve E has the follow-ing invariants c = 16( a p − b p c p ) , c = − a p − b p )( a p − c p )( b p − c p ) , ∆ = 16 a p b p c p , j = c / ∆ . Lemma 2.2 (Reduction of the Frey curve, [14, pp. 5-7]) . The curve E hasadditive reduction at the primes in S and also at m when r = 2 . If splits orramifies in K , then E has potentially multiplicative reduction at the primesabove . If q / ∈ S , and q = m (when r = 2 ), then E is semistable at q and p | v q (∆ q ) , where ∆ q is the minimal discriminant at q . Even if we know that a prime q is of semistable reduction for E , we willnot know whether it is of good or multiplicative reduction. The followingresults gives a way of producing, for a fixed prime p , a prime of multiplicativereduction for E = E a,b,c,p . Lemma 2.3 (Kraus [21, p. 9]) . Let p ≥ be a prime and suppose thereexists a natural number n satisfying the following conditions: • We have q := np + 1 is a prime that splits in O K ; • We have q ∤ Res( X n − , X n +1 − . hen the primes q and q of K above q are of multiplicative reduction for E .Proof. This result is proven in the case K = Q ( √
5) in [21, pp. 9-10]. Theproof immediately generalises to Q ( √ d ).We would like to level-lower the Frey curve E . We introduce the followingnotation. Let f be a Hilbert eigenform of parallel weight 2. We write Q f forits Hecke eigenfield: the field generated by its eigenvalues under the Heckeoperators. When f is irrational (i.e. Q f = Q ), we write ̟ for a prime abovethe rational prime p . Theorem 2.4 (Level-Lowering, [14, p. 4]) . Let p ≥ and suppose ρ E,p isirreducible. Define N p = N / Y q || Np | v q (∆ q ) q . Then we can level-lower
E. That is, there exists a Hilbert newform f at level N p such that ρ E,p ∼ ρ f ,̟ , for some prime ̟ of Q f above p .Proof. As E is defined over a real quadratic field, it is modular [15]. Lemma2.2 gives the other conditions needed to level lower E , other than irreducibil-ity which is assumed in the statement of the theorem.We then need to calculate the various possibilities for the level N p . Weuse the method described in [14, pp. 4-8] to obtain a list of possibilities for N p . The table in the appendix displays the possible levels N p we obtained.At this point, there are three main issues we need to overcome.1. Proving irreducibility of ρ E,p .2. Calculating the Hilbert newforms at each level N p .3. Eliminating the Hilbert newforms at each level N p .Sections 3 and 4 are devoted to proving irreducibility. Section 5 is thenconcerned with calculating and eliminating newforms. In order to level-lower our Frey curve (Theorem 2.4), we need irreducibilityof the mod p Galois representation ρ E,p . We aim to prove the followingresult.
Theorem 3.1.
The representation ρ E,p is irreducible for all squarefree d with ≤ d ≤ and p ≥ , unless the values d and p appear in Table 2below.
26 34 35 37 39 55 61 94 95 97 p ,
103 83 97 37 227 179 , ρ E,p to be irreducible in these cases too, but we were unableto show this.In this section, we reduce the problem to only needing to deal withfinitely many primes p . We will in fact often obtain the full irreducibilitystatement we need.As before, E = E a,b,c,p denotes our Frey curve defined over the quadraticfield K = Q ( √ d ), and we suppose for a contradiction that the mod- p Galoisrepresentation ρ E,p is reducible. Then ρ E,p ∼ (cid:16) θ ∗ θ ′ (cid:17) , where θ, θ ′ : G K → F p are the isogeny characters of the elliptic curve at p and satisfy θθ ′ = χ p , where χ p is the mod p cyclotomic character. Wecan interchange θ and θ ′ by replacing E with an isogenous curve. Thecharacters θ and θ ′ are unramified away from p and the additive primes for E . The additive primes for E are the primes of K above 2, and m when r = 2 (Lemma 2.2). We write N θ and N θ ′ for the conductors of θ and θ ′ respectively. For q an additive prime of E with q ∤ p , we have that v q ( N ) iseven, and v q ( N θ ) = v q ( N θ ′ ) = v q ( N ) /
2. So if p is coprime to N θ then fromour list of possibilities for N p we obtain a list of possibilities for N θ , sincethe additive part of N is the square root of the additive part of N θ . [14,p. 10].We now consider two cases. The first is when p is coprime to one of N θ and N θ ′ . By interchanging θ and θ ′ we may assume p is coprime to N θ . Lemma 3.2.
Suppose p is coprime to N θ . Let G be the ray class group forthe modulus N θ ∞ ∞ , where ∞ , ∞ denote the two real places of K . Let n denote the exponent of G . Then E has a point of order p defined over anumber field L of degree m , where m = n if n is even, and m = 2 n if n isodd.Proof. (See [14, p. 10]). The order of θ divides n . We may assume that θ has order n , as otherwise we can reduce to a case where the exponent of G isless than n . Suppose n is odd. As θ has order n it cuts out a field extensionof degree n over K , which we denote L , such that θ | G L = 1. This gives apoint of order p over L , and L is a number field of degree 2 n .6uppose now that n is even. Let L be the field cut out by θ . As θ hasorder n/ L is a number field of degree 2 · n/ n . Then θ | G L has order 2,and so twisting by θ gives a point of order p defined over L .This lemma then combines with the following classification of p -torsionof elliptic curve defined over number fields, obtained by studying the pointson the n th symmetric power of X ( p ). Theorem 3.3 ([9, pp. 1-2]) . let L be a number field of degree n . Let E ′ /L be an elliptic curve with a point of order p over L . We have thatif n = 2 then p ≤ if n = 3 then p ≤ if n = 4 then p ≤ if n = 5 then p ≤ if n = 6 then p ≤ or p = 37; if n = 7 then p ≤ . Moreover, if E ′ has a point P of order defined over a field L of degree ,then j ( E ′ ) = − . We obtain a ray class group with exponent 6, and consequently a pointon E defined over a number field of degree 6, in the cases d = 37 and d = 79. In each case, the elliptic curves with j -invariant − Q ( √ d ), and so do not arise from the Frey curve E .When n ≥ Theorem 3.4 (Oesterl´e, [9, p. 21]) . let L be a number field of degree n . Let E ′ /L be an elliptic curve with a point of order p over L . Then p ≤ (3 n/ +1) . In most cases we consider the exponent of the ray class groups are ≤ p right away. The cases where thereis exponent is 6 are discussed above. However, for certain cases, namelywhen d = 26 , , , , or 82, a ray class group has exponent 8 and so wecannot apply Theorem 3.3. Applying Oesterl´e’s bound gives p ≤ (3 + 1) =6724, which is rather large. Instead we use the following strategy to obtaina better bound on p . The idea revolves around the following result. Theorem 3.5 ([24, p. 2]) . Suppose q is a prime above q > of multiplicativereduction for E , and that q is inert in K . Then ρ E,p is irreducible for p = 67 and p > . We note that this theorem is stated in [24] without mentioning p = 67,but this comes from the result of Kamienny [19, Proposition 3.2] on whichTheorem 3.5 is based. 7e fix the following notation (which will also be used in Section 5). For q , a prime of O K , write n q for the Norm of q , and define A q := { a ∈ Z : | a | ≤ p Norm( q ) , Norm( q ) + 1 − a ≡ } . If q is a prime of good reduction for E , we know that a q ( E ) ∈ A . Thisfollows from the Hasse-Weil bounds and the fact that E has full two-torsionover K . We then define, for a ∈ A , P q ,a := X − aX + n q . When a = a q ( E ), this is the characteristic polynomial of Frobenius at q . Proposition 3.6.
Suppose ρ E,p is reducible with p ≥ . Define B :=Norm( ε − . Let q be a prime of O K above q with q ∤ , , , p, m , and with q inert in K . Define r q = 1 if q is a principal ideal, and r q = 2 otherwise.Define R q := lcm { Res( P q ,a ( X ) , X r q −
1) : a ∈ A } , where Res denotes the resultant of the two polynomials. Then p | ∆ K · B · R q or p ∈ P , where P = { ≤ p ≤ } ∪ { } . We can then choose a set of primes { q , · · · , q t } , and let R := gcd { q i } ,so that either p | ∆ K · B · R or p ∈ P . For the cases we considered we foundthis to give much better bounds than applying Oesterl´e’s bound for n = 8. Proof.
Let q be a prime as defined in the proposition. Suppose q is a primeof good reduction. We then apply the result of [13, Theorem 1]: the possiblenon-constant isogeny characters are { , } and { , } , and we obtain B =Norm( ǫ − p ∤ B · ∆ K then p | Res( P q ,a q ( E ) ( X ) , X r q − a q ( E ), but as discussed above, we know it lies in A q ,and so p | R q as defined in the proposition.If instead, q is a prime of multiplicative reduction for E , then we applyTheorem 3.5 to conclude that p ∈ P .We will now consider the cases where p is neither coprime to N θ nor N θ ′ .In these cases, p must either split or ramify in K (see [20, p. 247]). Lemma 3.7.
Suppose p neither coprime to N θ nor N θ ′ and that p ramifiesin K . Suppose is not inert in K . Let n be the exponent of the ray classgroup modulo √N A ∞ ∞ , where N A is the additive part of the conductor N . Then p | n − .Proof. As 2 is not inert in K , any prime dividing 2 is a prime of potentiallymultiplicative reduction for E (Lemma 2.2). We then follow the argumentof [14, p. 10] and apply [14, Proposition 6.2], which is stated for K = Q ( √ p )but also holds for K = Q ( √ d ) with p ramifying in K .8 emma 3.8 ([14, p. 11]) . Suppose p neither coprime to N θ nor N θ ′ and that p splits K . Set n = 6 if is inert in K , and set n = 2 otherwise. Then p | Norm( ε n − if ε or − ε is totally positive, and otherwise p | Norm( ε n − . Combining the results stated in this section so far reduces the problemto dealing with a finite number of primes in each case. These are shown inTable 3 below. We have not included the primes 17 and 19 in the table,as we will see at the start of Section 4 that ρ E,p is irreducible over all realquadratic fields when p = 17 or 19. d
26 29 34 35 37 39 p ≥ P , ,
103 29 P , P , ,
97 37 P , d
53 55 59 61 71 73 p ≥
23 53 P , , , ,
127 59 89 d
74 82 89 94 95 97 p ≥
23 43 P ,
109 53 151 P , , Table 3: Irreducibility Step 1
In the previous section we saw how to go from proving irreducibility for allprimes p ≥
17 to a finite (and often empty) subset of primes p (see Table3). In this section, we use study the quadratic points on certain modularcurves X ( N ) to complete the proof of Theorem 3.1. Suppose ρ E,p is reducible for p = 17 ,
19, or p ≥
23 appearing in Table 3. As E has full two-torsion over K , it gives rise to a non-cuspidal K -point on themodular curves X ( p ), X (2 p ), and X (4 p ). It is therefore enough to showthat one of X ( p )( K ), X (2 p )( K ), or X (4 p )( K ) has no points that couldarise from E . Usually, this consists in showing that X ( np )( K ) consistssolely of cusps, but the existence of a non-cuspidal K -point can also bedealt with if we have access to the j -invariant of the point. For example,if the point corresponds to an elliptic curve without full two-torsion, thenwe know it did not arise from E (see for example the argument given afterTheorem 3.4). 9ecent works [27, 7, 5] have studied quadratic points on X ( N ) of genus2 ≤ g ≤
5, as well as the genus 6 hyperelliptic curve X (71). Here, allquadratic points are considered, rather than working over a fixed quadraticfield as we wish to do. We note that extending these results to (non-hyperelliptic) curves of genus ≥ X ( N ) has finitely many quadratic points or infinitely many quadraticpoints. A curve of genus ≥ ≥ non-exceptional and points which do not arise in this way are said to be exceptional . Theorem 4.1 (Ogg [25, p. 451], Bars [1, p. 11]) . The curve X ( N ) is hyper-elliptic of genus ≥ if and only if N ∈ { , , , , , , , , , , , , , , , , , , } . The curve X ( N ) is bielliptic with an ellip-tic quotient of positive rank if and only if N ∈ { , , , , , , , , , } . We note that the curve X (37) appears is both hyperelliptic and biel-liptic, with an elliptic quotient of positive rank. Consequently, we do notuse the terms exceptional and non-exceptional in relation to this curve.For X ( N ) non-hyperelliptic, the degree 2 elliptic quotients of X ( N ) withinfinitely many rational points are all of the form X +0 ( N ) = X ( N ) / h w N i ,where w N denotes the Atkin-Lehner involution corresponding to N . The pa-pers [5, 27] classify all quadratic points in the cases where there are finitelymany such points and 2 ≤ g ≤
5. The values N with N of the form p, p, or4 p with p ≥
17 in this list are 34 and 38. In these cases, all quadratic pointsare defined over imaginary quadratic fields [27], and so ρ E,p is reducible for p = 17 ,
19 over all real quadratic fields.When the curve X ( N ) has infinitely many points and 2 ≤ g ≤
5, or X ( N ) is hyperelliptic, the papers [5, 7] obtain a classification of its ex-ceptional quadratic points. In each case, no exceptional points could arisefrom E : they are either defined over imaginary quadratic fields or over realquadratic fields not appearing in Table 3. Lemma 4.2.
Suppose X ( N ) is hyperelliptic with N = 37 , and let P ∈ X ( N )( Q ( √ d )) be a non-exceptional quadratic point. Then P correspondsto a rational point on the quadratic twist of X ( N ) by d , which we denoteby X d ( N ) .Proof. We can take a model for the hyperelliptic curve X ( N ) of the form y = f ( x ), with the hyperelliptic involution given by ( x, y ) ( x, − y ). If P = ( u, v ) ∈ X ( N )( Q ( √ d )) is a non-exceptional quadratic point, then10 u, v ) = ( u σ , − v σ ), where σ generates Gal( Q ( √ d ) / Q ). So u ∈ Q and v = b √ d with b ∈ Q . So db = f ( u ), so ( u, b ) ∈ X d ( N )( Q ).It follows that if X d ( N )( Q ) = ∅ , then we have a contradiction. Thisis often easily checked by seeing whether or not we have points everywherelocally, we see this in Example 4.1 below. This idea is taken further in[26]: for X ( N ), possibly non-hyperelliptic, we denote by X d ( N ) the twistof X ( N ) by w N over Q ( √ d ). Rational points on X d ( N ) correspond topairs of quadratic points in X ( N )( Q ( √ d )) that map to rational points inthe quotient X +0 ( N ). We then have the following result. The full theoremis stronger than the version we present here. Theorem 4.3 (Ozman, [26, p. 2]) . Let N be squarefree and l a prime.Write M = Q ( √− N ) , and H for its Hilbert Class Field. Suppose one of thefollowing two conditions holds: • the prime is inert in Q ( √ d ) , and N is divisible by and a prime p ≡ ; or • the prime l is ramified in Q ( √ d ) , l ∤ N , and each prime of M above l is not totally split in H ;then X d ( N )( Q l ) = ∅ , so X d ( N )( Q ) = ∅ . Example 4.1.
We consider the hyperelliptic curve X (41). A model forthis curve is given by y = x − x − x + 10 x + 20 x + 8 x − x − x − . let d = 35. We can check directly that X d (41) has no points over Q .Alternatively, we can apply Theorem 4.3 with l = 5. The prime l ramifiesin Q ( √ d ), and l ∤
41. Write M = Q ( √−
41) and H for its Hilbert class field,which is a degree 8 extension of M since M has class number 8. The prime l = 5 splits in O M , and the two primes of M above 5 are not totally splitin H . It follows that X d (41)( Q ) = ∅ . We deduce that for d = 35, ρ E, isirreducible. The methods described above deal with many of the cases appearing inTable 3, but not all. Here we work with modular parametrisations to provemore irreducibility results. We note that there is often a significant overlapin the choice of methods which can be used. Let E ′ be an elliptic curvedefined over Q of conductor N . Then E ′ admits a map, called the modularparametrisation of E ′ : ϕ : X ( N ) → E ′ . Q and takes cusps to torsion points of E ′ . Wewill assume E ′ is optimal so that ϕ is unique up to sign and maps the cuspat infinity on X ( N ) to the identity of E . Write m for the degree of themodular parametrisation, which we refer to as the modular degree of E ′ .We note here that the curve E ′ and the Frey curve E are not related. Theirconductors N and N are also not related. For background on the modularparametrisation map we refer the reader to [10, 32, 8].Using the map ϕ to understand quadratic points over a fixed field isbased on the following observation. Lemma 4.4.
Let E ′ be an optimal elliptic curve of conductor N and write ϕ : X ( N ) → E ′ for its modular parametrisation. Suppose E ′ ( Q ) = E ′ ( L ) for L a number field. Then X ( N )( L ) ⊆ ϕ − ( E ′ ( Q )) . We would like to compute the fields of definition of preimages of pointsunder the modular parametrisation map. For our purposes, this is onlyuseful when E ′ ( Q ) = E ′ ( K ) is finite, but the techniques we develop applyeven if this is not the case.Suppose E ′ is given by a Weierstrass equation y + a xy + a y = x + a x + a x + a , with a i ∈ Q . Then x and y are rational functions on E ′ of degrees 2 and 3respectively. We can then pull these back via ϕ to obtain rational functionson X ( N ) = Γ ( N ) \H ∗ , where q = e πiz for z ∈ C : x ( q ) = ϕ ∗ ( x ) and y ( q ) = ϕ ∗ ( y ) . Using
Sage we can compute the q expansions of x ( q ) and y ( q ). Therational functions x ( q ) and y ( q ) satisfy the equation of the elliptic curve E ′ as well as the relation dx ( q )2 y ( q ) + a x ( q ) + a = f ( q ) dqq , where f is the rational newform of level N corresponding to the isogenyclass of E ′ . The rational functions x ( q ) and y ( q ) on X ( N ) have degrees 2 m and 3 m respectively.We would like to obtain a planar model for the curve X ( N ) by findinga relation between x ( q ) (or y ( q )) and another rational function on X ( N ).We can always find such a relation. Lemma 4.5 ([8, p. 24]) . Let r ( q ) , s ( q ) ∈ Q ( X ( N )) be rational functionson X ( N ) . Then there exists an irreducible polynomial F ∈ Q [ R, S ] , whichwe call a minimal polynomial relation , such that F ( r ( q ) , s ( q )) = 0 , and deg R ( F ) ≤ deg( s ( q )) and deg S ( F ) ≤ deg( r ( q )) . Lemma 4.6 ([8, p. 28]) . Let r ( q ) , s ( q ) ∈ Q ( X ( N )) be rational functionson X ( N ) . Suppose G ∈ Q [ R, S ] satisfies G ( r ( q ) , s ( q )) = O ( q M ) for someinteger M > r ) deg( s ) . Then G ( r ( q ) , s ( q )) = 0 . Let r ( q ) , s ( q ) ∈ Q ( X ( N )) be rational functions on X ( N ) and let F ∈ Q [ R, S ] be a minimal polynomial relation for these rational functions. Thenusually F will have degree deg( s ) in R and degree deg( r ) in S. If this isthe case then F gives a planar model for the modular curve X ( N ). Ifthe degrees are less than these maxima, then we will obtain a model for aquotient of the modular curve X . For example, the equation of the ellipticcurve E ′ is a minimal polynomial relation between x ( q ) and y ( q ).We aim to find a minimal polynomial relation involving between x ( q )and another rational function on X ( N ) that will give us a planar model forour modular curve. A natural first choice is the j -function: j ( q ) = q − + 744 + 196884 q + 21493760 q + O ( q ) ∈ Q ( X ( N )) for all N. The degree of j ( q ) is given by the index of Γ ( N ) in the full modular groupSL ( Z ). As N gets large, say N ≥
50, it quickly becomes impractical tocompute a minimal polynomial relation between x ( q ) and j ( q ), and so weseek to replace j ( q ) with a rational function on X ( N ) of smaller degree.We do this using eta products.First define Dirichlet’s eta function as η ( q ) := 1 q ∞ Y n =1 (1 − q n ) . An eta product of level N (also referred to in the literature as an eta quo-tient) is then given by s ( q ) = Y d | N η ( q d ) r d = Y d | N η r d d , for some integers r d , and where we write η d for η ( q d ). Such an eta productneed not be a rational function on X ( N ), but if the integers r d satisfycertain conditions, then it is. Theorem 4.7 (Ligozat’s Criteria [23, p. 28]) . An eta product of level N , η d , is a rational function on X ( N ) if the following conditions are satisfied: . X d r d Nd ≡ . X d r d d ≡ . X d r d = 0; 4 . X d | N (cid:18) Nd (cid:19) r d ∈ Q .
13e note here that the support of the divisor of an eta product s ( q ) ∈ Q ( X ( N )) is contained in the set of cusps of X ( N ). Using Sage we canfind a basis for the group of eta products of level N that are rational func-tions on X ( N ) (i.e. that satisfy Ligozat’s criteria). We can then chooseany one of these (or some combination), say s ( q ). It is natural to start bychoosing a basis element of minimal degree. We find a minimal polynomialrelation between x ( q ) and s ( q ), and substituting in x -coordinates of pointsin E ′ ( Q ) will give the s -values of the preimages of points under the mod-ular parametrisation map, and from this we can often deduce their field ofdefinition.There are two issues that can arise here. The first is that the minimalpolynomial relation may not be of maximal degree in its two variables. Thiswill occur when the map x ( q ) × s ( q ) : X ( N ) → P × P (viewing x ( q ) and s ( q ) as morphisms from X ( N ) to P ) is not injective, and so the equationwe obtain gives a planar model for a curve Y which is a finite quotient of X ( N ). It is still possible to recover some information in this case.The other issue is that the s -values we obtain may not give the field ofdefinition of the points in the preimage. For example, if a degree 2 rationaldivisor on X ( N ) has an s -value of a ∈ Q , with multiplicity 2, then thiscould be due to two rational points, or a quadratic point. By consideringthe cusps it is most likely we can conclude it is a quadratic point, but we stilldo not know its field of definition, which is what we are ultimately interestedin. We will see this in the example below.To overcome both of these problems, we can usually simply replace oureta product by a different one, and if necessary combine information frommultiple eta products as we see below. Example 4.2 (Eta Product Method for X (116)) . We consider here thecase K = Q ( √
29) and p = 29. There are no elliptic curves of rank 0 over K with conductor 29 or 58, but the elliptic curve E ′ with Cremona label‘116b1’ has Mordell-Weil group Z / Z over both Q and K . The curve isgiven by E ′ : y = x + x − x + 4 . We have E ( Q ) = E ( Q ( √ { E , R, − R } , where R has x -coordinate 0. The modular degree of E is 8.We work with the modular curve X (116) which is of genus 13 and has sixrational points: the six cusps. We denote these cusps by ∞ , , C , C , C , and C . We would like to show in this case that X (116)( Q ( √ X (116)( Q ) as this will prove that ρ E, is irreducible.We find a basis, using Sage , for the group of eta products at level 116.This basis has five elements. The first four have degree 12 as rational func-tions on X (116), and the fifth has degree 14. We start by choosing the first14asis element and find the minimal polynomial relation F ( X, S ) between x ( q ) and s ( q ). This polynomial has degree 6 in X and 8 in S , and so doesnot give a planar model for X , but rather a degree 2 quotient of X . Al-though we can still obtain information from this, we instead work with thesecond basis element s = η − · η · η − · η − · η · η − , with divisor( s ) = − ∞ ) + 5( C ) + ( C ) + ( C ) + 5( C ) − . We calculate a minimal polynomial relation F ( X, S ) for x ( q ) and s ( q ) andfind that this time it has degree 12 in X and degree 16 = 2 m in S , so wehave obtained a planar model for our curve. Some terms of this polynomialare as follows F ( X, S ) = X S + · · · + X S + · · · + 1048576 XS − S . Substituting in the value 0 = x ( R ) = x ( − R ) we find that F (0 , s ) = − S ) ( S − ( S − S + 8) ( S + 2 S + 2)( S − S + 6 S − S + 2) . The factor S corresponds to three cusps in the preimage. The factor S − S + 6 S − S + 2 corresponds to a tuple of quartic points defined over thecyclotomic extension Q ( ζ ), and the factor S + 2 S + 2 corresponds to a pairof quadratic points defined over Q ( √− S − S + 8) couldeither correspond to a pair of quadratic points defined over Q ( √ S − may correspond to a pair of rationalpoints or a quadratic point. We know | X (116)( Q ) | = 6 and so we willshortly be able to see that these rational s -values do not arise from a pairof rational points, and so ( S − corresponds to a pair of quadratic points,but we cannot say over which quadratic field they are defined. Finally, thisfactorisation does not display any poles of s appearing in the preimage of R or − R . In the factorisation of T F (0 , /T ) (which gives the 1 /s -valuesin the preimage) there is a factor T . This corresponds to a pole of s , andshows that there is a fourth cusp in preimage.In order to understand the preimage of 0 E we first define G ( Z, S ) := Z (1 /Z, S ), as setting Z = 0 will correspond to setting X = ∞ . We find G (0 , S ) = S ( S + 2 S + 2) ( S − S + 6 S − S + 2) . We note G (0 , S ) is a square since − E = 0 E . By considering this factori-sation in conjunction with T G (0 , /T ) we see that we have two cusps, a15air of quadratic points defined over Q ( √− Q ( ζ ).In order to understand more about the fields of definition of the preim-ages of R and − R we use the fifth basis element s = η · η − · η · η − · η · η − , with divisor ( s ) = − ∞ ) − C ) + 7( C ) + 7(0) . Note that this divisor is only supported on four of the six cusps. This etaproduct has degree 14. We find the minimal polynomial relation F ( X, S ).It has degree 14 in X and 16 in S , and we have F (0 , S ) =16384( S )( S + 1)( S + 29)( S − S + 29) ( S + 4 S + 29)( S + 10 S + 29)( S − S + 272 S − S + 841) . As before, we recover four cusps, and a tuple of quartic points. The quadraticfactors ( S + 4 S + 29) and ( S + 10 S + 29) both correspond to pairs ofquadratic points defined over Q ( √− s , but the other pair had s -values 2 and wecould not deduce its field of definition. We now see that this quadraticpoint is defined over Q ( √− S − S + 29) corresponds to either a pair of quadratic points defined over Q ( √− s -value, we know it must be either apair of quadratic points defined over Q ( √ Q ( √−
1) and Q ( √− Q ( √− , √
2) = Q ( ζ ).We conclude that ϕ − ( E ( Q ) = ϕ − ( Q ( √ Q ( √−
1) and threetuples of quartic points defined over Q ( ζ ), making up 24 points in total.This proves that X (116)( Q ( √ X (116)( Q ), and in fact determines X (116)( K ) for any field K with E ( Q ) = E ( K ). Studying twists of X ( N ) and modular parametrisations still does not dealwith all the cases we would like to consider. In particular we would stilllike to show that ρ E,p is irreducible for p = 37 , ,
61 (for various quadraticfields Q ( √ d )). The case p = 37 is a little different and we treat it separatelyafterwards. The modular curves X (43) and X (61) are bielliptic and non-hyperelliptic. Their quotients X +0 (43) and X +0 (61) are the elliptic curveswith Cremona labels ‘43A1’ and ‘61A1’ respectively, each of which has rank16 and trivial torsion. We employ a version of the Mordell-Weil sieve to study X d ( N )( Q ). We illustrate the sieving method for X (61), although the sametechniques will apply for other modular curves. This method also overlapswith the other methods we have seen. For a general introduction to theMordell-Weil sieve, we refer the reader to [6].The curve X (61) has genus 4. Using the ‘small modular curves’ pack-age in Magma , we obtain a smooth model for this curve, the Atkin-Lehnerinvolution w , and the j -map. We start by obtaining a model for whichthe Atkin-Lehner involution is diagonalised. We do this by finding a matrixdiagonalising w and applying the corresponding coordinate change to theequations of our curve. We obtain the following model in P : − Y − XZ + Z = T ,X − X Y − XY − X Z + XY Z + Y Z − Y Z = 0 . We see that this is the intersection of a quadric surface and a cubicsurface. We write F ( X, Y, Z ) = − Y − XZ + Z , and G ( X, Y, Z ) for thehomogeneous cubic in the second defining equation of the above model. Thecoordinate change we have applied introduces 2 as a prime of bad reductionfor this model. The Atkin-Lehner involution is now given by w : X (61) −→ X (61)( x : y : z : t :) ( x : y : z : − t ) . We see from these equation that quotienting X (61) by the group generat-edby w is given by the degree 2 map ϕ : X (61) −→ X +0 (61)( x : y : z : t ) ( x : y : z ) , with X +0 (61) = X (61) / h w i the elliptic curve defined by G ( X, Y, Z ) = 0in P .We suppose, hoping to obtain a contradiction, that our Frey curve E gives rise to a non-exceptional point P on X (61)( Q ( √ d )). We are interestedin the case d = 61, as we can prove irreducibility for the other values of d we are concerned with using Theorem 4.3, but we in fact obtained acontradiction for all d > P is a non-exceptional point, P andits Galois conjugate P σ are interchanged by the Atkin-Lehner involution;that is, w ( P ) = P σ , where σ generates Gal( Q ( √ d ) / Q ). It follows that P can be expressed as P = ( x : y : z : b √ d ) with x, y, z, b ∈ Q . As ϕ ( P ) ∈ X +0 (61)( Q ), we have that ϕ ( P ) = m · R , for some m ∈ Z , where R generates the Mordell-Weil group X +0 (61)( Q ).Choose a prime l such that l ∤ · · d . Then l is a prime of good reductionfor both X (61) and X +0 (61). Write N l for the order of R in the reductionof X +0 (61) modulo l . Write k for the residue field of K modulo a prime17bove l . This will either be F l or F l . We have the following commutativediagram, where ∼ denotes reduction modulo l : X (61) X +0 (61) e X (61) e X +0 (61) ϕ ∼ ∼ e ϕ Since ϕ ( P ) = m · R , we see that e ϕ ( e P ) = m · e R , where m ≡ m (mod N l ).So e P ∈ e ϕ − ( m · e R ). Fix m ∈ { , · · · , N l − } . Then we can explicitlycompute the set e ϕ − ( m · e R ) = { Q , Q } , where Q = ( u, v, w, s ) and Q = g w ( Q ) = ( u, v, w, − s ), with u, v, w ∈ F l and s ∈ k . We note that we mayhave Q = Q .We would like to try and argue that e P / ∈ { Q , Q } if possible, as we canthen conclude that m m (mod N l ). There are two strategies we can use.1. The point P = ( x : y : z : b √ d ) satisfies F ( x, y, z ) = db , and soreducing mod l we have F ( e x, e y, e z ) · e d − ≡ e b (mod l ) . It follows that F ( u, v, w ) · e d − is a square mod l , so if this is not thecase, then m m (mod N l ).2. The Frey curve E has full two-torsion over K . If e P is not a cusp onthe reduced modular curve e X (61), then it corresponds to an ellipticcurve with full two-torsion over k . Also, since w ( P ) corresponds tothe elliptic curve E/C where C is a cyclic subgroup of order 61, itfollows that w ( P ) also has full two-torsion over K . Suppose l > j -invariant in characteristic 3). Then if Q is not a cusp,and all elliptic curves over k with j -invariant e j ( Q ) (this consists oftwo elliptic curves when e j ( Q ) , l ), and a maximum ofsix elliptic curves otherwise) do not have full-two torsion over k , thenwe have a contradiction, and we conclude m m (mod N l ).We combine these two methods of elimination to obtain a list of possi-bilities for m (mod N l ). We then repeat this process with a list of primes { l , · · · , l r } with l i ∤ · · d for each i , and use the Chinese remain-der theorem to obtain a list of possibilities for m (mod N ), where N :=gcd( N l , · · · , N l r ). If this list of possibilities is empty then we obtain our de-sired contradiction. We choose our primes l i so that the orders N l i are smalland share many prime factors. This helps avoid a combinatorial explosiondue to the Chinese remainder theorem, and also increases the likelihood ofobtaining contradictory information.18or the curve X (61) and d = 61, using the primes 5 , , ,
13, and 17sufficed to reach a contradiction. We note the importance of using bothelimination steps in the sieve. If we do not sieve using j -invariants, then wefound that the sieve did not eliminate enough possibilities for m (mod N l )and we could not reach a contradiction. If we do not sieve using the firstmethod of elimination, then we are unable to eliminate the possibility that P reduces to a cusp modulo l for each prime in our list (i.e. that each primeis a prime of multiplicative reduction for E ), and so we will not obtain acontradiction.The sieving method for X (43) is identical. The curve is of genus 3 andwe used the following plane quartic model in P :64 X + 48 X Y + 16 X Y + 8 XY − Y +(16 X + 8 XY + 2 Y ) T + T = 0 , with the Atkin-Lehner involution given by ( x : y : t ) ( x : y : − t ), and themap ϕ to X +0 (43) ∈ P (1 , ,
2) given by ( x : y : t ) ( x : y : t ). We appliedthe sieve for d = 26 , , , , , , ,
95. In each case we obtained acontradiction using the primes 3 , , , , , , , , , , , , , In this section we consider the modular curve X (37). There are two ellipticcurves of conductor 37 over Q up to isogeny. We denote these by E and E . These have Mordell-Weil groups Z and Z / Z over Q respectively. If E ( Q ( √ d )) = E ( Q ) then we can use the modular parametrisation mapas in Section 4.3 to understand X (37)( Q ( √ d )). This is not the case for d = 35 , , , , and 82, so we seek a different argument in these cases.The curve X (37) is rather special. It is both hyperelliptic and bielliptic,with an elliptic quotient of positive rank over Q . This makes its quadraticpoints hard to classify. A description is given in [5, p. 16], but it does notseem to help us pin down quadratic points over a fixed quadratic field.The idea we use is as follows. Here, p = 37 is fixed. Let q be a primethat is inert in K , and of semistable reduction for E . We write q = q O K .We consider the reduction map ∼ q : X (37)( K ) −→ X (37( k ) . Write c ∞ , c ∈ X (37)( Q ) for the two cusps of X (37).For Q ∈ X (37)( K ) we write, following the notation of [29], B q ( Q ) := { S ∈ X (37)( K q ) : e S = e Q } for the q -unit ball around Q . Note that as q is inert in O K this is thesame as the intersection of the q -unit balls of Q with q | q . The Chabauty19riterion of Siksek [29] gives a way of testing whether or not B q ( Q ) = Q .In order to apply the criterion we need that r K ≤ g − r K isthe rank of J (37)( K ), 2 is the degree of the number field K , and g is thegenus of the curve. Here we have 2( g −
1) = 2. The rank of J (37)( K )is the rank of J (37)( Q ) plus the rank of J d (37)( Q ), where J d (37) is theJacobian of the quadratic twist of X (37) by d . For d = 35 , , , and 82, r K = 1 + 1 = 2, and so we are in a position to apply the criterion, but for d = 37, r K = 2 + 1 = 3, meaning we cannot use the same method. Wenote that it is likely that to have any chance of success, since X (37) /K isthe base change of a curve defined over Q , we also need the usual Chabautycondition of r Q ≤ g − J ( K ) which is saturatedat q ; that is, the index of the subgroup in J ( K ) is not divisible by q . Toobtain this we pull back Mordell-Weil generators of the free parts of E ( K )and E ( K ), each of which has rank 1, and map them into the Jacobian. Wecan then verify that the subgroup of J ( K ) generated by these two elementsis saturated at q for our choices of prime.We now apply the Chabauty criterion. Suppose E gives rise to a (non-cuspidal) point P ∈ X (37)( K ). We would like to show that E has goodreduction at q if possible. Suppose instead that E has multiplicative at q .Then P reduces to a cusp on X (37)( k ). If we can show that B q ( c ∞ ) = c ∞ and B q ( c ) = c , then since e P = e c or f c ∞ , we must have P = c ∞ or P = c ,a contradiction. In the cases d = 35 and d = 82 we find that 59 is inertin O K and is a prime of good reduction for E in both these cases, and weverify that B q ( c ∞ ) = c ∞ and B q ( c ) = c . For the cases d = 39 and d = 55,we obtain the same results with the primes q = 29 and q = 107 respectively.Using Lemma 2.3 we can also obtain some primes of multiplicative re-duction for E as p is fixed. In the cases d = 35 and 82 we find that theprimes above 1481 are of multiplicative reduction for E . For d = 39, theprimes above 3923 are of multiplicative reduction, and for d = 55, the primesabove 10139 are of multiplicative reduction. We now combine Mordell-Weilinformation using our primes of good and multiplicative reduction to obtaina contradiction. We have the following commutative diagram: X (37)( K ) E ( K ) = Z X (37)( k ) E ( k ) ϕ ∼ q ∼ q e ϕ We have that ϕ ( P ) ∈ E ( K ), so ϕ ( P ) = m · R for some m ∈ Z , where R generates E ( K ) = E ( Q ). We write q m for our prime of multiplicativereduction and q g for our prime of good reduction. We write N m and N g forthe orders of e R m and e R g respectively, where the subscripts m and g denotereduction modulo the corresponding prime ideals.20e find that ϕ ( c ∞ ) = ϕ ( c ) = 3 · R . So e ϕ ( f c ∞ ) = e ϕ ( e c ) = 3 · e R , andso reducing modulo q m we have ϕ ( e P m ) = 3 · e R m , meaning that m ≡ N m ).Then reducing mod q g , e P g / ∈ e ϕ − g (3 · e R g ) = { f c ∞ g , e c g } , since e P g is not acusp, because q g is a prime of good reduction for E . It follows that m N g ).For d = 35 and 82, with our choice of primes q m and q g we find that N m = 1469 = 13 ·
113 and N g = 13. So m ≡ m ≡ m d = 39, we have N m = 3933 = 3 · ·
23 and N g = 57 = 3 ·
19, and soconsidering m mod 57 leads to a contradiction. Finally, for d = 55, we have N m = 1720 and N g = 20, again leading to a contradiction.We note that using these same ideas, applying a version of Chabauty forthe symmetric square of curves [28] would give similar, and in fact stronger,results in principal, but here the rank condition required is that r Q ≤ g − X (37). Once we have obtained irreducibility of the mod p Galois representations ofour Frey curve, the next step is to apply the level-lowering theorem (Theorem2.4). By our previous work, we have a list of possible levels N p , for ourHilbert newform, f . These are displayed in the appendix. We consider eachpossibility separately, and aim to discard all isomorphisms between our Freycurve and the newforms at this level. If we can do this at all the possiblelevels then we will obtain our desired contradiction.The standard idea, as used in [14, p. 12], is as follows: compute thenewforms at the level N p and combine the local information mod q for manyprimes to obtain a contradiction, one newform at a time. For q a prime of K not dividing N p , recall from Section 3 the notation A q := { a ∈ Z : | a | ≤ p Norm( q ) , Norm( q ) + 1 − a ≡ } . If q is a prime of good reduction for E , then as discussed in Section 3, a q ( E ) ∈ A .The following lemma of Freitas and Siksek, gives the standard methodof bounding the prime p . We use the same notation as in Section 2. Lemma 5.1 ([14, p. 12]) . Suppose ρ E,p ∼ ρ f ,̟ . Let T be a set of primeideals q which do not divide N p . For each q ∈ T define the principal ideal B f , q := (cid:0) n q ( n q + 1 − a q f )( n q + 1 + a q f ) Y a ∈A ( a − a q f ) (cid:1) · O Q f . et B f := P q ∈ T B f , q , and denote by C f the norm of this ideal. Then p | C f . If C f is non-zero we obtain a bound on p . If all the prime factors of C f areless than 17, then this discards the isomorphism for all p we are concernedwith. We discuss the case C f = 0 at the end of this section.For the levels N p appearing in [14] (i.e. when 2 ≤ d ≤ q ∈ T . Bydoing this, we lose out on knowing what the Hecke eigenfield Q f is (eventhough we can usually be fairly certain what it is), and so computing thenorm of a sum of ideals in O Q f as in Lemma 5.1 is impossible. The followinglemma addresses this issue. Lemma 5.2.
Suppose ρ E,p ∼ ρ f ,̟ . Let T be a set of prime ideals q whichdo not divide N p . For each q , define the field L q := Q ( a q f ) , and define theelement b f , q := n q ( n q + 1 − a q f )( n q + 1 + a q f ) Y a ∈A ( a − a q f ) ∈ L q . Let c f , q := Norm L q / Q ( b f , q ) . Write c f := gcd { c f , q : q ∈ T } . Then p | c f .Proof. We start by noting that the Hecke eigenfield Q f of f contains L q foreach q ∈ T , and we view L q as a subfield of Q f . Following the notation ofthe previous lemma, we note that b f , q ∈ B f for all q ∈ T . The norm of anideal is the greatest common divisor of the norm of all of its elements, andso p | C f | gcd { Norm Q f / Q ( b f , q ) : q ∈ T } . As L is a subfield of Q f we have thatNorm Q f / Q ( b f , q ) = (cid:16) Norm L q / Q ( b f , q ) (cid:17) [ Q f : L q ] , so it follows that p | c f .The prime factors of C f are contained in the set of prime factors of c f ,and so this version may give worse bounds, but we found that in practice,by considering enough primes q , the two sets of prime factors coincide. Remark . We can simplify matters by working directly with the minimalpolynomial of a q ( f ) over Q , which we denote µ , to define c f , q . Denote by µ c ( m ) the constant term of µ evaluated at m . We then have c f , q = n q · µ c ( n q + 1) · µ c ( − n q − · Y a ∈A µ c ( a ) . .2 Reconstructing Hilbert Newforms Write T q for the Hecke operator on the space of newforms at level N p , andwrite χ q for its characteristic polynomial. We view T q as a matrix. We thenhave a factorisation into irreducible polynomials χ q ( X ) = r Y i =1 ( e q ,i ( X ) m i . The roots of each irreducible factor e i are the eigenvalues of a Galois con-jugacy class of Hilbert newforms. Associated to each e i , we have the corre-sponding irreducible subspace V q ,i := ker( e q ,i ( T q )) , with a basis consisting of members of Galois conjugacy classes of newformswhose eigenvalues at q satisfy e q ,i . To each e q ,i , we also associate a value c q ,i := c f , q for any newform f with eigenvalue at q a root of e q ,i .Since the Hecke operators T q commute, if T q and T q are Hecke opera-tors, and V q ,i is some irreducible subspace with respect to T q , then it ispreserved by T q and we can compute the matrix of T q restricted to V q ,i .We can then compute the characteristic polynomial of this matrix and de-compose V q ,i into irreducible subspaces under T q , which we denote by V q , q ,i ,i . This subspace will have a basis of newforms whose eigenvaluesat q and q are roots of e q ,i and e q ,i respectively. Associated to thesubspace V q , q ,i ,i is the integer c q , q ,i ,i = gcd( c f , q , c f , q ), where f is anynewform in the space V q , q ,i ,i .We continue this process. If a value c q ··· q m ,i , ··· ,i m has all its primefactors ≤
13 then we can discard the associated subspace, as we know that ρ E,p ρ f ,̟ for any newform in this subspace. We aim to discard all possiblesubspaces, hence obtaining a contradiction.We carried out this process for values d for which the maximum dimen-sion of the space of newforms was < d = 33 , , , ,
89, we obtaineda value c f = 0. We consider these cases in the next subsection. Also, inthe case d = 34, we obtained values c f , c f divisible by 23, which is why 23appears in the statement of Theorem 1.2 (but not in Table 2), as we wereunable to discard this isomorphism. In some cases we can discard an isomorphism, or discard it for certain primes,even if c f = 0 or has a prime factor ≥ Lemma 5.4 (Image of Inertia [14, p. 13]) . Suppose f is a Hilbert newformwith Q f = Q , and write E ′ for the elliptic curve associated to f . Supposethat one of E and E ′ has potentially multiplicative reduction at a prime q and that the other has potentially good reduction at q . Then ρ E,p ρ f ,̟ . We applied this argument in the case d = 34 and N p = p using the primeabove 2 which is of potentially multiplicative reduction for E (Lemma 2.2),but of potentially good reduction for the elliptic curves E ′ , E ′ correspondingto the two rational newforms f , f satisfying c f = c f = 0.Note that from a rational newform f we can obtain E ′ as follows. Wecompute all newforms at level N p and using the elliptic curve search functionin Magma we obtain a (potentially incomplete) list of elliptic curves withconductor N p . We see if we can find a curve E ′ such that a q ( E ′ ) = a q f ,say for a few primes q . If the values a q ( E ′ ) do not equal a q f ′ for any othernewform f ′ , then by modularity E ′ must correspond to f . When we cannotcalculate the newform decomposition we cannot be certain of having isolatedall newforms using the method above (two rational newforms could agreeon arbitrarily many primes), and so we cannot find a corresponding ellipticcurve. Indeed we cannot even know whether a newform is rational unlesswe can compute the decomposition.We can also often deal with fixed values of p , and hence obtain a boundon p , using a method of Kraus. The following lemma is stated in [21] for K = Q ( √ Q ( √ d ). It is based on knowingcertain primes of multiplicative reduction for E (see Lemma 2.3). Lemma 5.5 ([21, p. 2]) . Let p ≥ be a prime and suppose there exists anatural number n satisfying the following conditions: • We have n < p − and n ≡ ; • We have q := np + 1 is a prime that splits in O K ; • We have q ∤ Res( X n − , X n +1 − .Then ρ E,p ρ f ,̟ for any rational newform f . We apply this lemma in the cases d = 33 , , , ,
89, as well as for d = 17, to show that ρ E,p ρ f ,̟ for all but finitely many p ≤ when c f = 0. We then remove these leftover primes by choosing n appropriately,as in [21, pp. 10-11]. We were able to do this for each prime p other than p = 19 in the case d = 57.This strategy does not help eliminate the irrational newforms f , f with23 | c f , c f in the case d = 34, as the prime 23 is appearing as a factor of( n q + 1 − a q f i )( n q + 1 + a q f i ) for each q , and so using primes of multiplicative24eduction will not help us rule it out. When working over Q , the standardstrategy at this point is to apply an argument using a Sturm bound. Al-though Sturm bounds do exist for Hilbert newforms, they are too large tobe of use computationally [3]. Other techniques are explored in [3] for elim-inating newforms, and it may be possible to adapt these methods to dealwith the newforms f , f .Finally, in the case d = 89, we apply the result of [14, p. 13] in the case d = 89 (in the same way it was applied in the case d = 17 in [14, p. 13]) toconclude that we have no solutions if p ≡ ± References [1] F. Bars. On quadratic points of classical modular curves. arXiv preprintarXiv:1308.3267 , 2013.[2] M. Bennett, V. Patel, and S. Siksek. Superelliptic equations arisingfrom sums of consecutive powers.
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Appendix
In the table below, n and n new denote the dimensions of the spaces of Hilbertcuspforms and Hilbert newforms respectively. The column denoted RCGrecords the exponents of the ray class groups appearing in Lemma 3.2. Theremaining column headings follow the notation of the paper. d S r m N p n n new RCG26 p h , √ d + 6 i p
18 2 8 m p
388 7829 p = 2 O p p
81 4530 p h , √ d i p
28 0 2 , m p
220 28 p m p p p
16 2 2 p
93 2033 p , p p p p p
34 234 p h , √ d + 1 i p
36 4 2 , , m p
292 40 p m p p h , √ d i p
28 0 2 , m p
592 120 p
160 38 m p p = 2 O p p
135 7538 p p
18 0 2 , p p h , √ d + 3 i p
36 4 2 , , p
236 56 m p
792 156 p m p m p p , p p p p h , √ d i p
36 4 2 , m p
320 36 p m p p p
20 0 2 p
127 3346 p p
26 4 2 , p p p
24 2 2 p
135 2851 p h , √ d i p
52 4 2 , m p
468 56 p
320 84 m p p = 2 O p p
189 10555 p h , √ d + 4 i p
68 12 2 , , p
412 96 m p
584 84 p m p m p p , p p p
12 4 2 p p
82 658 p h , √ d + 1 i p
50 10 4 m p
592 9059 p p
32 4 2 p
177 4761 p = 2 O p p
295 16562 p p
32 2 2 , p p , p h , √ d + 3 i p p
24 8 4 m p p
722 5466 p h , √ d i p
76 8 2 , m p
688 88 p m p p p
36 4 2 p
247 6369 p = 2 O p
10 4 2 p
330 17770 p h , √ d + 1 i p
88 16 2 , m p
840 120 m p p p
42 8 2 p
265 5873 p , p p p
16 4 174 p h , √ d + 3 i p
72 12 4 m p p = 2 O p
10 4 2 p
330 17778 p h , √ d + 6 i p
88 8 2 , m p p m p p p
156 30 6 p p h , √ d + 4 i p
168 40 8 m p p p
44 4 2 p
265 6985 p h , √ d + 5 i p
22 10 2 , m p
178 44 p
966 540 m p p p
50 8 2 , p p h , √ d i p
88 16 2 , m p
932 116 p
656 162 m p p , p p p
20 4 191 p h , √ d + 1 i p
120 20 2 , , m p p
832 206 m p p = 2 O p
14 6 2 p
330 17794 p p
68 14 2 , p p h , √ d + 2 i p
116 20 2 , , p
756 180 m p p m p m p p , p p p
25 4 125 4 1