Tame Galois realizations of GSp4 (Fℓ) over Q

arXiv:0910.1445v1 [math.NT] 8 Oct 2009 Tame Galois realizations of GSp4(Fℓ) over Q Sara Arias-de-Reyna and N´uria Vila Abstract In this paper we obtain realizations of the 4-dimensional general symplectic group over a prime field of characteristic ℓ > 3 as the Galois group of a tamely ramified Galois extension of Q. The strategy is to consider the Galois representation ρℓattached to the Tate module at ℓof a suitable abelian surface. We need to choose the abelian varieties carefully in order to ensure that the image of ρℓis large and simultaneously maintain a control on the ramification of the corresponding Galois extension. We obtain an explicit family of curves of genus 2 such that the Galois representation attached to the ℓ-torsion points of their Jacobian varieties provide tame Galois realizations of the desired symplectic groups. 1 Introduction A central problem in number theory is the study of the structure of the absolute Galois group GQ= Gal(Q/Q). The Inverse Galois Problem over Q, first considered by D. Hilbert, can be reformulated as a question about the finite quotients of this absolute Galois group. In spite of the many efforts made to solve the Inverse Galois Problem, it still remains open. Of course, these attempts have not been fruitless, and there are many finite groups that are known to be Galois groups over Q. For an idea of the progress made, the reader can leaf through [32] or [23]. Assume that a finite group Gcan be realized as a Galois group over Q, say G≃Gal(K1/Q), where K1/Qis a finite Galois extension. But perhaps 2010 Mathematics Subject Classification: 12F12, 11F80, 11G30, 11R32 Research supported by a FPU predoctoral grant AP-20040601 of the MEC and partially supported by MEC grant MTM2006-04895. 1 we are interested in field extensions with some ramification property, and K1/Qdoes not satisfy it. We can ask whether there exists some other finite Galois extension, K2/Q, with Galois group Gand enjoying this additional property. In this connection, several variants of the Inverse Galois Problem have been studied. This paper is concerned with the following problem, posed by Brian Birch (cf. [5], Section 2). Problem 1.1 (Tame Inverse Galois Problem).Given a finite group G, is there a tamely ramified Galois extension K/Qwith Gal(K/Q)≃G? This problem has already been dealt with for certain families of groups. It is known that solvable groups, all symmetric groups Sn, all alternating groups An, the Mathieu groups M11 and M12, and their finite central extensions are realizable as the Galois group of a tamely ramified extension (cf. [17], [28], [27]). One way to deal with the Inverse Galois Problem, and eventually with Problem 1.1, is to consider continuous Galois representations of the absolute Galois group of the rational field. Let Vbe a finite dimensional vector space over a finite field F, and consider a continuous representation of GQ (endowed with the Krull topology) in the group of automorphisms of V(with the discrete topology), say ρ:GQ→GL(V). Then Imρ≃GQ/ker ρ≃Gal(K/Q), where K/Qis a finite Galois extension. In other words, a continuous Galois representation of GQprovides a realization of Imρas a Galois group over Q. This strategy has already been used to address the Inverse Galois Problem for families of linear groups (cf. [35], [8], [9], [36], [10], [16]). In this paper we will study the Galois representations that arise through the action of the absolute Galois group GQon the ℓ-torsion points of abelian surfaces. For each prime number p, let us fix an immersion Q֒→Qp. This induces an inclusion of Galois groups Gal(Qp/Qp)⊂GQ. Inside Gal(Qp/Qp) we can consider the inertia subgroup Ip= Gal(Qp/Qp,unr) and the wild inertia 2 subgroup Ip,w= Gal(Qp/Qp,t), where Qp,unr and Qp,tdenote the maximal unramified extension and the maximal tamely ramified extension of Qp, respectively. A prime pis unramified (respectively tamely ramified) in the Galois extension K/Qif and only if ρ(Ip) = {Id}(respectively ρ(Ip,w) = {Id}). Therefore, if we want a Galois representation ρto yield a tamely ramified Galois extension K/Q, we will have to ensure that ρ(Ip,w) = {Id}for all prime numbers p. In [2], the authors apply this strategy to solve Problem 1.1 for the family of 2-dimensional linear groups GL2(Fℓ). More precisely, they consider the Galois representations attached to the ℓ-torsion points of elliptic curves, and construct explicitly elliptic curves such that this representation is surjective and moreover the image of the wild inertia group Ip,wis trivial for all primes p. In this paper we will extend these results to the case of curves of genus 2. Namely, we will prove that, for all prime number ℓ≥5 there exist (infinitely many) genus 2 curves Csuch that the Galois representation attached to the ℓ-torsion points of the Jacobian of Cprovides a tamely ramified Galois realization of GSp4(Fℓ) (cf. Theorem 5.5). We want to thank Luis Dieulefait for many enlightening conversations. 2 The Action of the Inertia Group Let A/Qbe an abelian variety of dimension n, and let us fix a prime ℓ. Consider the Galois representation ρℓ:GQ→Aut(A[ℓ]) ≃GL2n(Fℓ). attached to the ℓ-torsion points of A. This representation gives rise to a realization of Imρℓas Galois group over Q, say Imρℓ≃Gal(K/Q), where K is the number field fixed by ker ρℓ. In this section we are interested in finding conditions over the variety Athat ensure that K/Qis tamely ramified. Note that it suffices to control the image by ρℓof Ip,wjust for a finite quantity of primes p. Indeed, Ip,wis a pro-p-group, so that if pdoes not divide the cardinal of GL2n(Fℓ), the cardinal of Im(Ip,w) must be p0= 1. We will first address the problem of obtaining tame ramification at a prime p6=ℓ. Let us place ourselves in a more general setting. Let Fbe a number field, and let us consider an abelian variety A/F. Let pbe a prime ideal of the ring of integers OFof F. We begin by recalling what it means for Ato have semistable reduction at p(see [13], §A.9.4.). 3 Definition 2.1. We will say that A/F has semistable reduction at a prime p∈Spec OFif the connected component of the special fibre of the N´eron model of Aat p,A0 p, is the extension of an abelian variety by a torus. The kind of reduction of an abelian variety at a prime pis reflected in the action of the inertia group at pon the Tate module of the variety. For instance, if the variety has good reduction, the N´eron-Ogg-Shafarevich criterion ensures that the inertia group acts trivially on the Tate module. Moreover, a result of Grothendieck characterizes the case when the reduction is semistable in terms of the action of the inertia group. As a consequence, we have the following result. Theorem 2.2. Let A/Qbe an abelian variety, and let ℓ, p be two different prime numbers. Assume that Ahas semistable reduction at p. Then the image of the wild inertia group Ip,wby the Galois representation ρℓis trivial. Proof. Let us see that Ip,wacts trivially on Tℓ(A). By Proposition 3.5 of [12], we know that there exists a submodule T′⊂Tℓ(A), fixed by the action of Ip, and such that Ipacts trivially on both T′and Tℓ(A)/T′. Taking a suitable basis of Tℓ(A), Ipacts through a matrix of the shape Idr∗ 0 Ids. But the order of such a matrix is a divisor of ℓ. Since Ip,wis a pro-p-group, its elements all act as the identity. To control the image of the wild inertia group at the prime ℓis a much more subtle matter. The first author addresses this problem in [3], and obtains a condition that guarantees that the action is trivial (Theorem 3.3). Let us recall the statement here. Theorem 2.3. Let ℓbe a prime number, and let A/Qbe an abelian variety of dimension nwith good supersingular reduction at ℓ. Call Fthe formal group law attached to Aat ℓ,vℓthe ℓ-adic valuation on Qℓand Vthe group of ℓ-torsion points attached to F. Assume that there exists a positive α∈Q such that, for all non-zero (x1,...,xn)∈V, it holds that min{vℓ(xi) : 1 ≤ i≤n}=α. Then the image of the wild inertia group by the Galois representation attached to the ℓ-torsion points of Ais trivial. This result will allow us to hold in check the action of the wild inertia group at ℓ. 4 3 Image of the Representation In the previous section we considered the tameness condition. Our aim in this section is to obtain some control on the image of the representation. Let us fix a prime ℓ, let A/Qbe an abelian variety of dimension n, and let us denote by ρℓ:GQ→GL2n(Fℓ) the Galois representation attached to the ℓ-torsion points of A. Assume Ais principally polarized. Then the Weil pairing gives rise to a non-degenerated symplectic form on the group of ℓtorsion points of A,h·,·i :A[ℓ]×A[ℓ]→F∗ ℓ. Furthermore, the elements of the Galois group GQbehave well with respect to this pairing. This compels the image of the representation to be contained in the general symplectic group GSp2n(Fℓ). Now a well-known result of Serre states that, if the principally polarized abelian variety has the endomorphism ring equal to Z, and furthermore its dimension is either odd or equal to 2 or 6, then the image of the representation ρℓis the whole general symplectic group for all but finitely many primes ℓ (see Theorem 3 of 137 in [34]). We are particularly interested in the case of abelian surfaces over Q. In this case, the result of Serre boils down to: Theorem 3.1. Let A/Qbe an abelian surface, principally polarized, such that EndQ(A) = Z. Then, for all but finitely many primes ℓ, it holds that Imρℓ= GSp4(Fℓ). We will take the explicit results of P. Le Duff [18] as our starting point. We will sketch some of his reasoning in order to make use of it later on, and then we will dwell upon some specific points where we need to modify it. The method of Le Duff rests upon finding a certain set of elements in the image of the representation which generate the whole GSp4(Fℓ). Therefore, at the core of the method lies a theorem about generators of this group. More specifically, the main result upon which Le Duff builds his method is the following (see Theorem 2.7 of [18]): Proposition 3.2. The symplectic group Sp4(Fℓ)is generated by a transvection and an element whose characteristic polynomial is irreducible. It is easy to see that, if the image of the Galois representation contains the subgroup Sp4(Fℓ), then it necessarily contains the whole group GSp4(Fℓ), since the composition of the representation with the projection 5 GSp4(Fℓ)→GSp4(Fℓ)/Sp4(Fℓ) equals the cyclotomic character, which is surjective. Therefore, the problem boils down to finding two elements in Imρℓ, satisfying that one is a transvection and that the other has an irreducible characteristic polynomial. This is the way we approached the matter at first, but in order to produce an element with irreducible characteristic polynomial in Imρℓwe needed to make use of a conjecture of Hardy and Littlewood (namely Conjecture (F) of [14]). Following a suggestion by L. Dieulefait we have managed to devise a way to ensure a large image without resorting to this conjecture. Let us consider the following result (Theorem 2.2 of [18]). Theorem 3.3. Let Gbe a proper subgroup of Sp4(Fℓ), and assume that G contains a transvection. Then one of the following three assertions holds. (1) Gstabilizes a hyperplane and a line belonging to it. (2) Gstabilizes a totally isotropic plane. (3) The elements of Gstabilize or exchange two orthogonal supplementary non-totally isotropic planes. Remark 3.4. If Gis a subgroup of Sp4(Fℓ) which contains an element with irreducible characteristic polynomial, it cannot satisfy any of the three assertions of the theorem (see Theorem 2.7 of [18]). Therefore Proposition 3.2 is an easy consequence of this result. Our strategy will be the following: for each of the three assertions, we shall ensure the existence of an element of G= Imρℓ, contained in Sp4(Fℓ), which does not satisfy it. In this way, we will prove that Gcannot be a proper subgroup of Sp4(Fℓ). Instead of asking directly that there exists an element with irreducible characteristic polynomial, which rules out the three possibilities at once, we will require that there are elements such that the corresponding characteristic polynomial decomposes in different ways. Remark 3.5. The second assertion in the above theorem occurs when Gis contained in a maximal parabolic subgroup. In this case, it is easy to check that, choosing a suitable symplectic basis, this maximal subgroup consists of matrices of the form A∗ 0 (A−1)t, where (A−1)tdenotes the transpose of the inverse matrix of A(cf. the remark following the proof of Theorem 2.2 6 in [18]). On the other hand, if the third assertion holds, then the elements of Gleave two supplementary orthogonal non-totally isotropic planes stable, or else interchange them (this is case (3) of Proposition 2 of [15]). If an element of Sp4(Fℓ) interchanges two such planes, then it can be seen that its trace is zero. Therefore an element which belongs to this kind of maximal subgroup either has trace 0 or stabilizes two planes. Moreover, if it stabilizes two planes, it can be expressed as A∗ 0Bwith respect to some suitable basis, where Aand Bbelong to SL2(Fℓ). Let us consider an element in Sp4(Fℓ), and call its characteristic polynomial P(X). It is easy to see that this polynomial can be written as P(X) = X4+aX3+bX2+aX + 1 for some a, b ∈Fℓ. In turn, this implies that there exist α, β ∈Fℓsuch that P(X) decomposes as (X−α)(X−β)(X− 1/α)(X−1/β) over Fℓ(for P(α) = 0 implies that P(1/α) = 0). Now there are essentially two ways in which such a polynomial can break up in quadratic factors, namely P(X) =   (X−α)(X−1/α)·(X−β)(X−1/β) (X−α)(X−β)·(X−1/α)(X−1/β). The first case is labelled “unrelated” 2-dimensional constituents and the second one “related” 2-dimensional constituents in [9]. There is a nice way to discern whether the first decomposition takes place. Namely, consider the polynomial P0(X) = X2+aX+(b−2). The roots of this polynomial are precisely α+1/α, β +1/β. Therefore, if the first factorization occurs, this polynomial is reducible and its discriminant ∆0=a2−4b+ 8 is a square in Fℓ. To determine whether the other factorization takes place is more difficult. Given an element of Sp4(Fℓ) with characteristic polynomial P(X) = X4+aX3+bX2+aX + 1, we will denote ∆0(P) = a2−4b+ 8. Theorem 3.6. Let Gbe a subgroup of Sp4(Fℓ), and assume that Gcontains a transvection. Furthermore, assume that it contains two elements whose characteristic polynomials, P1(X)and P2(X), satisfy the following: denoting by αi,1/αi, βi,1/βithe four roots of Pi(X),i= 1,2, •α1+ 1/α1, β1+ 1/β16∈ Fℓand α1+ 1/α1+β1+ 1/β16= 0. •α2+ 1/α2, β2+ 1/β2∈Fℓ,∆0(P2)6= 0 and α26∈ Fℓ. 7 Then Gequals Sp4(Fℓ). Proof. Since Gcontains a transvection, Theorem 3.3 implies that either G is the whole symplectic group or else one of the assertions (1), (2) or (3) of the theorem holds. We will see that, in fact, none of them is satisfied. If assertion (1) holds, then all the elements of Gmust leave a line invariant. But this implies that each element has one eigenvalue which belongs to Fℓ. But α1+ 1/α1and β1+ 1/β1do not belong to Fℓ. Therefore assertion (1) does not hold. Assume now that assertion (2) holds. Then Gis contained in a group which stabilizes a totally isotropic plane. Therefore, with respect to a suitable symplectic basis, it is contained in a subgroup of the shape A∗ 0 (A−1)t(see Remark 3.5). In particular, this implies that if P(X) is the characteristic polynomial of an element of G, it must factor over Fℓinto two polynomials of degree two. Call the roots of one of the factors αand β. Then the roots of the other factor are 1/α and 1/β. Let us consider the polynomial P2(X) = (x−α2)(x−β2)(x−1/α2)(x− 1/β2). Labelling the roots anew if necessary, we can assume that α2and β2 are the roots of one of the quadratic factors, as above. We can consider two cases: •β2= 1/α2or β2=α2. In this case P(X) can be factored as P(X) = (X2−AX +1)2for a certain A∈Fℓ. If we work out this expression, we obtain P(X) = X4+2AX3+(A2+2)X2+2AX+1. Writing out P(X) = X4+a2X3+b2X2+a2X+ 1 and comparing these two expressions we obtain that ∆0(P2) = 0, which contradicts our hypotheses on P2(X). •β26=α2,1/α2. In this case, the polynomials (X−α2)(X−1/α2) and (X−α2)(X−β2) are different, and both are defined over Fℓ. Therefore, one can use the Euclid algorithm to compute their greatest common divisor (X−α2). This implies that α2∈Fℓ, which contradicts our hypotheses on P2(X). Finally, assume that Assertion (3) holds. Then any element in Gsatisfies that either its trace is zero or it stabilizes two planes, which are supplementary, orthogonal and are not totally isotropic (see Remark 3.5). Consider again the element with characteristic polynomial P1(X). Since it has nonzero trace, it must stabilize two such planes. But then P1(X) should break 8 into two quadratic factors defined over Fℓ. Moreover, since the determinant of the corresponding matrix is 1 for each of the factors (cf. Remark 3.5), their independent terms must be 1. But this means that the factors have to be (X−α1)(X−1/α1) and (X−β1)(X−1/β1), and we know these polynomials are not defined over Fℓ. Therefore Assertion (3) cannot hold. 4 Explicit Construction In this section we will face the problem of constructing explicitly, for a given prime number ℓ, a genus 2 curve such that the Jacobian variety attached to it gives rise to a Galois representation yielding a finite Galois extension K/Q, tamely ramified, with Galois group GSp4(Fℓ), thus providing an affirmative answer to the Tame Inverse Galois Problem for this group. In the preceding sections we have worked out some statements that give very accurate and explicit conditions for the Galois representation attached to the ℓ-torsion points of an abelian surface to satisfy the desired properties. Our aim in this section is to replace all these conditions by others, which are more restrictive, but are simply congruences modulo powers of different primes. We shall tackle each of the conditions separately; thus the section will be split in several different subsections. In the last section we shall state a theorem possessing a very explicit flavour. 4.1 Explicit control of the ramification Assume we have a hyperelliptic curve Cof genus gdefined over a certain field kby an equation of the shape y2+Q(x)·y=P(x), where P(X) has degree 2g+ 2. The discriminant of this equation is defined as ∆ = 2−4(g+1) ·disc(4P(x) + Q(x)2). It holds that if ∆ 6= 0, the curve Cis smooth (see [20], §2). In fact, if we fix a prime p, the p-adic valuation of the discriminant is a bound of the conductor exponent of C, which in turn coincides with the conductor exponent of the Jacobian variety attached to C. Consider the genus 2 curve defined over Qby the hyperelliptic equation y2=x6+ 1. 9 that the characteristic polynomial of φqhas the following shape: P(X) = X4+aX3+bX2+aqX +q2, for certain a, b ∈Z. More precisely, if we denote by N1(resp. N2) the number of points of Cover Fq(resp. Fq2), we can compute aand bby using the formulae (a:= N1−q−1 b:= (N2−q2−1 + a2)/2.(3) Furthermore, it can be proven (cf. Proposition 10.20 in [24]) that the polynomial obtained from P(X) by reducing its coefficients modulo ℓcoincides with the characteristic polynomial of ρℓ(Frobq). In the previous section our problem was to construct a curve such that, at a certain well chosen prime p, it satisfied some condition. Our strategy there was to choose the prime ponce and for all at the beginning; namely, we took p= 5, and we established a congruence condition modulo 52such that, whenever it is satisfied by a genus 2 curve C, we achieve our objective. The first point in which this section differs from the previous one is that now we will not choose the primes q1and q2beforehand; the primes q1and q2will actually depend on ℓ. When we face this problem, a natural question arises: Given a prime q, what conditions must a pair (a, b) satisfy in order to ensure that the polynomial P(X) = X4+aX3+bX2+qaX +q2is the characteristic polynomial of the Frobenius endomorphism at qof a genus 2 curve? In [22], the authors address the problem of determining whether, given finite field Fqand a pair of positive natural numbers (N1, N2), there exists a genus 2 curve Cwith N1points over Fqand N2points over Fq2. We shall make use of their results. Firstly, let us recall the definition of a Weil polynomial (see Definition 2.2 of [22]). Definition 4.11. Let qbe a power of a prime number. We will say that the polynomial P(X) = X4+aX3+bX2+qaX +q2∈Z[X] is a Weil polynomial if |a| ≤ 4√qand 2|a|√q−2q≤b≤a2 4+ 2q. Remark 4.12. To simplify the problem, we will always choose a= 1, so we will only have to take care to choose bsatisfying 2√q−2q≤b≤1 4+ 2q. Let us fix an odd prime number q. Collecting Theorem 2.15 and Theorem 4.3 of [22], we can state the following result: 16 Theorem 4.13. Let P(X) = X4+aX3+bX2+qaX +q2∈Z[X]be a Weil polynomial, and let ∆0=a2−4b+ 8q. Assume that ∆0is not a square in Z, q∤band a26∈ {0, q +b, 2b, 3(b−q)}. Then there exists a smooth projective curve of genus 2, defined over Fq, with N1=q+ 1 + apoints over Fqand N2= 2b−a2+q2+ 1 points over Fq2. Remark 4.14. The previous result claims the existence of a genus 2 curve, say C, defined over Fqwith N1points over Fqand N2points over Fq2. If q is odd, we know that there exists a hyperelliptic equation y2=f(x) defining C, with f(x)∈Fq[x] a polynomial of degree 6 and without multiple roots. Since there is only a finite number of such polynomials f(x)∈Fq[x], one can compute the curve Csimply by an exhaustive search, so one can say that this construction is effective. Nevertheless, there are algorithms to compute genus 2 curves with a given number of points over Fqand over Fq2. For instance, see [11]. Keeping this result in mind, the following two propositions show us how to construct suitable q1and q2. Proposition 4.15. Let ℓbe an odd prime number. Choose q1such that q1≡1 (mod ℓ). Then there exists a projective curve of genus 2,C1/Q, such that it has good reduction at q1, and the characteristic polynomial of the Frobenius endomorphism at q1,P1(X) = X4+a1X3+b1X2+q1a1X+q2 1 satisfies that ∆0(P1)is not a square in Fℓand a16≡ 0 (mod ℓ). Proof. Fix a1= 1. Since q1≡1 (mod ℓ), it follows that q1> ℓ. Therefore, if we choose any element b1∈Fℓ, there exists b1∈Z, 0 < b1< q1mapping into b1. Therefore P1(X) = X4+a1X3+b1X2+q1a1X+q2 1is a Weil polynomial. We will choose b1such that 1 −4b1+ 8q1is not a square in Fℓ(since 4 is prime to ℓ, the expression 1 −4b1+ 8q1runs through all the elements of Fℓ as b1varies, so this is clearly feasible). Now it is easy to check that the pair (a1, b1) satisfies all the conditions in Theorem 4.13, so that there exists a smooth projective curve of genus 2 defined over Fq1with a suitable number of points over Fq1and Fq2 1. Lifting this curve to Q, we obtain the curve we were seeking. Proposition 4.16. Let ℓbe an odd prime number. Choose q2such that q2≡1 (mod ℓ)and q2>3ℓ. Then there exists a projective curve of genus 2, C2/Q, such that it has good reduction at q2, and the characteristic polynomial of the Frobenius endomorphism at q2,P2(X) = X4+a2X3+b2X2+q2a2X+q2 2 17 satisfies that ∆0(P2)is a non-zero square in Fℓbut is not a square in Z, and P2(X)does not break up in linear factors over Fℓ. Proof. As in the proof of Proposition 4.15, we will fix a2= 1. Note that, since q2>3ℓ, for each element b2∈Fℓthere exist three values of b2∈Z,0< b2< q2 such that b2maps into b2, which can be taken as b2,ℓ+b2, 2ℓ+b2. Let us choose an element z∈Fℓsuch that z2−16q2is not a square in Fℓ. Such an element exists: if we take any square x2∈Fℓ, and add −16q2as many times as we wish, we can obtain any element in Fℓthat we like. In particular, if we consider the sequence x2, x2−16q2, x2−2·16q2, x2−3·16q2,..., a point will come when we obtain a non-square element. The previous element shall be our z2. If z2= 1, we will take z= 1. Now let us choose b2< q2such that 1−4b2+ 8q2is congruent to (z+ 1)2modulo ℓ. This is possible for the same reason as in the proof of Proposition 4.15. Moreover, at the beginning of the proof we noted that there are, in fact, three possible choices for b2which are strictly smaller than q2. It is not difficult to check that for the three of them 1−4b2+8q2cannot be a square in Z. We have set this claim aside in Lemma 4.17. Therefore, we can choose b2such that 1 −4b2+ 8q2is not a square in Z, and furthermore b2is not divisible by q2. If we choose b2in this way, it is easy to check that the conditions of Theorem 4.13 hold. Therefore, there exists a smooth projective genus 2 curve over Fq2such that the characteristic polynomial of the Frobenius endomorphism of q2is P2(X) = X4+a2X3+b2X2+q2a2X+q2 2. Now we ascertain that the thesis of our Proposition holds. It is clear that ∆0(P2)≡(z+ 1)2(mod ℓ) is a non-zero square in Fℓ. It remains to show that P2(X) does not split into linear factors. Call α2, q2/α2, β2, q2/β2the roots of P2(X). The fact that ∆0(P2) is a square tells us that the polynomials (X−α2)(X−q2/α2) and (X−β2)(X−q2/β2) are defined over Fℓ. We will achieve our objective if we see that one of these polynomials is irreducible over Fℓ. Since both α2+q2/α2 and β2+q2/β2are roots of P0(X) = X2+a2X+b2−(b2−2q2), they are given by the expressions −a2±√a2 2−4(b2−2q2) 2. Interchanging α2and β2if necessary, we can assume that α2+q2/α2=−a2+√a2 2−4(b2−2q2) 2. Therefore the polynomial (X−α2)(X−q2/α2) can be written as X2−−a2+√a2 2−4(b2−2q2) 2X+q2, and its discriminant is ∆ = −a2+pa2 2−4(b2−2q2) 2!2 −4q2. 18 Let us compute this quantity modulo ℓ. Since a2= 1, ∆0= 1−4b2+8q2≡ (z+ 1)2, we obtain that ∆ ≡z−16q2 4(mod ℓ), which is not a square in Fℓ because of the choice of z. This proves that P2(X) does not decompose in linear factors over Fℓ. Lemma 4.17. Let Abe a natural number, and let ℓbe a prime. Then the three numbers A,A−4ℓ,A−8ℓcannot be squares in Z. Proof. Assume that there exist x, y, z positive integers such that A=x2, A−4ℓ=y2and A−8ℓ=z2. From the first two equations we obtain that 4ℓ=x2−y2= (x+y)(x−y). Therefore ℓ=x+y 2·x−y 2. Since ℓis a prime number, it follows that x−y= 2, and moreover ℓ=(y+2)+y 2·(y+2)−y 2=y+ 1. The same reasoning applied to the last two equations yields that y−z= 2, and we can write ℓ=y+z 2·y−z 2=(z+2)+z 2·(z+2)−z 2=z+ 1. This is clearly a contradiction. 5 Main Result In this section we will state the main result concerning tame Galois realizations of GSp4(Fℓ). Our starting point is the following straightforward statement: Let Cbe a smooth projective curve of genus 2, defined over Q and such that, if we denote by Jthe Jacobian variety attached to Cand by ρℓthe Galois representation attached to the ℓ-torsion points of J, the following conditions are satisfied: •The Galois extension obtained by adjoining to Qthe coordinates of the ℓ-torsion points of Jis tamely ramified. •The image of ρℓcoincides with the general symplectic group GSp4(Fℓ). Then ρℓprovides a tamely ramified Galois realization of GSp4(Fℓ). Throughout this paper, we have sought to remodel these conditions in order to make them look like congruences. We have succeeded to a great extent. Replacing these conditions with those (more restrictive but simpler) obtained in the previous sections, we obtain the following result: 19 Theorem 5.1. Let Cbe a genus 2curve defined by a hyperelliptic equation y2=f(x), where f(x) = f6x6+f5x5+f4x4+f3x3+f2x2+f1x+f0∈Z[x]is a polynomial of degree 6without multiple factors. Let ℓ≥7be a prime number, and let P be the set of prime numbers that divide the order of GSp4(Fℓ). Let a∈Fℓbe such that the elliptic curve defined by y2=x3+ (1 − a)/ax2+(1−a)/ax+1 is supersingular. Furthermore, let q1, q2≡1 (mod ℓ) be different prime numbers with q2>3ℓ. Let C1/Qbe a genus 2curve such that it has good reduction at q1, and the characteristic polynomial of the Frobenius endomorphism at q1,P1(X) = X4+a1X3+b1X2+q1a1X+q2 1 satisfies that ∆0(P1)is not a square in Fℓand a16≡ 0 (mod ℓ). Let C2/Qbe a genus 2curve such that it has good reduction at q2, and the characteristic polynomial of the Frobenius endomorphism at q2,P2(X) = X4+a2X3+ b2X2+q2a2X+q2 2satisfies that ∆0(P2)is a non-zero square in Fℓbut is not a square in Z, and P2(X)does not break up in linear factors over Fℓ. Consider hyperelliptic equations y2=c6x6+c5x5+c4x4+c3x3+c2x2+c1x+c0 and y2=d6x6+d5x5+d4x4+d3x3+d2x2+d1x+d0defining C1and C2. Assume that the following conditions hold: •The following congruences mod 24hold: (f0≡f6≡1 (mod 16) f1≡f5≡0 (mod 16) (f2≡f4≡4 (mod 16) f3≡2 (mod 16). •The following congruences mod 3hold: (f0≡f6≡1 (mod 3) f1≡f5≡0 (mod 3) (f2≡f4≡1 (mod 3) f3≡0 (mod 3). •The following congruences mod 52hold: (f6≡f5≡f3≡f1≡f0≡1 mod 25 f4≡f2≡0 mod 25. •The following congruences mod ℓ4hold:      f6≡f0(mod ℓ4) f5≡f1(mod ℓ4) f4≡f2(mod ℓ4). 20 Furthermore, (f6≡1 (mod ℓ) f5≡0 (mod ℓ)(f4≡(1 −a)/a (mod ℓ) f3≡0 (mod ℓ). •The following congruences mod q1hold: fi≡ci(mod q1), i = 0,1,...,6. •The following congruences mod q2hold: fi≡di(mod q2), i = 0,1,...,6. •For all p∈ P different from 2,3,5,q1,q2and ℓ, the following congruences hold: (f0≡f6≡1 (mod p) f1≡f5≡0 (mod p)(f2≡f4≡0 (mod p) f3≡0 (mod p). Then the Galois representation attached to the ℓ-torsion points of the Jacobian of Cprovides a tamely ramified Galois realization of GSp4(Fℓ). Proof. The existence of a∈Fℓfollows from Theorem 1-(b) of [6] (cf. Corollary 3.6 of [2]), and the existence of the genus two curves C1and C2is proved in Propositions 4.15 and 4.16. The result is a direct consequence of the considerations made in the previous sections. A quick look at this theorem shows that, for each prime number ℓ≥7, there exists a genus 2 curve Csatisfying all the hypotheses, simply because of the Chinese Remainder Theorem. Therefore, we may write the following corollary: Corollary 5.2. For each prime number ℓ≥7, there exists a Galois extension over Qwhich is tamely ramified and has Galois group GSp4(Fℓ). Remark 5.3. As we remarked at the beginning of Section 4.2, we excluded the prime ℓ= 5 just in order to write a neat statement for all ℓ6= 5. Let us now tackle the case ℓ= 5. Consider the hyperelliptic curve Cdefined over Qby the equation y2=x6+ 391300x4+ 1170x3+ 1300x2+ 1.(4) 21 To simplify the notation, call f6= 1, f5= 0, f4= 391300, f3= 1170, f2= 1300, f1= 0, f0= 1. We can compute the reduction data of this particular curve by means of the algorithm of Liu (which is implemented in SAGE). We obtain that this curve has good reduction outside (possibly) the primes 2, 27792683 and 195476205803858674906021. In any case, at the last two primes the conductor exponent is 1. Therefore the curve has stable reduction at all odd primes. The algorithm of Liu does not compute the conductor at the prime 2. Nevertheless, in this case it is easy to check that the coefficients fiof the equation defining the curve satisfy that (f6≡f0≡1 (mod 16) f5≡f1≡0 (mod 16) (f4≡f2≡4 (mod 16) f3≡2 (mod 16). Therefore, Proposition 4.2 ensures that the reduction of Cat 2 is also stable. As a conclusion, we can say that the Galois representation attached to the 5-torsion points of the Jacobian surface of Cis tamely ramified outside the prime ℓ= 5. On the other hand, note that the equation defining Cis congruent modulo 54with the supersingular symmetric equation y2=x6+ 1300x4+ 1170x3+ 1300x2+ 1. This guarantees that the wild inertia group at 5 acts trivially on the group of 5-torsion points. Let us denote by ρℓ:GQ→GSp4(F5) the Galois representation which arises from the action of the Galois group on the points of 5-torsion of the Jacobian variety attached to C. The computation of the reduction data of Cat the prime p= 27792683 shows that the stable reduction at pis of type (II). Therefore, this prime satisfies the hypothesis of Proposition 4.6. Furthermore the order of the group of connected components of the special fibre of the N´eron model at p is 1, so it is not divisible by 5. This ensures the existence of a transvection in the group Imρℓ⊂GSp4(Fℓ). In order to prove that the image of the Galois representation is GSp4(Fℓ), we will make use of Proposition 3.2. For instance, let us consider the prime q= 19. The number of points of Cover F19 is 22, and the number of points of Cover F192is 410. Therefore, Equations (3) allow us to compute the characteristic polynomial of the Frobenius endomorphism at q, namely P(X) = X4+ 2X3+ 26X2+ 38X+ 361. It is not difficult to ascertain that this polynomial is irreducible over F5. Therefore, we conclude that the image of ρ5is GSp4(F5). Thus this group can also 22 be realized as the Galois group over Qof a tamely ramified extension. The previous remark allows us to state Corollary 5.2 without excluding the prime ℓ= 5: Corollary 5.4. For each prime number ℓ≥5, there exists a Galois extension over Qwhich is tamely ramified and has Galois group GSp4(Fℓ). As a matter of fact, we have proven not just the existence of a tamely ramified Galois realization of GSp4(Fℓ), but of infinitely many of them. Theorem 5.5. For each prime number ℓ≥5, there exist infinitely many tamely ramified Galois extensions over Qwith Galois group GSp4(Fℓ). Proof. If ℓ≥7, this is clear from the statement of Theorem 5.1. If ℓ= 5, note that the cardinal of GSp4(F5) equals 37440000 = 29·32·54·13. Therefore, each curve Cgiven by a hyperelliptic equation congruent to (4) modulo a suitable power of the primes 2,3,5,13,27792683 will provide a tamely ramified Galois realization of GSp4(F5). Remark 5.6. The symplectic group GSp4(F2) is isomorphic to the symmetric group S6, thus it is already known to be realizable as the Galois group of a tamely ramified extension of Q(cf. Proposition 1 in [29]). If ℓ= 3, we hope that a tame Galois realization of GSp4(F3) can also be obtained by using different techniques. For instance, one could make use of the isomorphism Sp4(F3)≃U(4,2) to obtain a regular realization of this group (cf. [23]), and then apply the techniques in [27]. 6 Example In this section we will present an example to illustrate how Theorem 5.1 allows us to compute explicitly genus 2 curves providing tame Galois realizations of the group GSp4(Fℓ). In fact, Theorem 5.1 can easily be turned into an algorithm to compute these genus 2 curves. Even though we have not explicitly formulated it in this way, we hope this example will clarify how the algorithm works. Firstly, note the simple well-known result: 23 Lemma 6.1. Let qbe a prime number. Then card(GSp2n(Fq)) = (q−1)q(2n)2 4 n Y j=1 (q2j−1). Let us take ℓ= 7. We will compute all the elements that appear in the statement of Theorem 5.1. •First of all, let us compute the set Pof prime numbers which divide the order of GSp4(Fℓ). Since ℓ= 7 and n= 2, Lemma 6.1 yields that card(GSp4(Fℓ)) = 1659571200 = 210 ·33·52·74. Therefore, the set of primes that divide the order of GSp4(Fℓ) is P={2,3,5,7}. •Next, we need to compute the element a∈F7such that the equation y2=x3+ (1 −a)/ax2+ (1 −a)/ax + 1 defines a supersingular elliptic curve. In order to do this, we compute the Deuring polynomial H7(x) = 3 X j=0 3 j2 ·xj=x3+ 2x2+ 2x+ 1 = (x+ 1)(x+ 3)(x+ 5). Therefore, x2−x+ 5 = (x+ 1)(x+ 5) divides the Deuring polynomial, so we can take a= 5 in F7. •The following elements that emerge are the primes q1and q2. We must choose two different prime numbers, q1and q2, which are congruent to 1 modulo 7 and such that q2>3·7 = 21. We may take q1= 29, q2= 43. Now we must find the genus 2 curves C1and C2. –Choosing the curve C1: firstly, fix a1= 1. According to Proposition 4.15, we need to find b1such that 1 −4·b1+ 8 ·29 is not a square modulo ℓ. For instance, we can take b1= 1. Now that we have a pair (a1, b1), we seek a genus 2 curve over F29 with N1= 29+1+a1= 31 points over F29 and N2= 292+1+2b1−a2 1= 843 points over F292. We know that such a curve exists, and if we scrutinize the set of all hyperelliptic curves defined over F29 (which is a finite set), we can obtain the curve given by the hyperelliptic equation y2=x6+x5+ 17x+ 5. 24 –Choosing the curve C2: again, fix a1= 1. According to Proposition 4.16, the first step is to find an element z∈F7such that z2−16 ·43 is not a square in F7. For instance, we can consider z= 1. Next, we must find b2such that 1−4·b2+8·43 is congruent to (z+ 1)2= 4 modulo 7. For instance, take b2= 3. Note that 1−4·3+8 ·43 = 333 is not a square in Z. We have a pair (a2, b2). We must seek a genus 2 curve over F43 with N1= 43+1+a2= 45 points over F43 and N2= 432+ 1 + 2b2−a2 2= 1855 points over F432. Such a curve exists, and again an exhaustive search can provide it. For instance, we have taken the curve defined by the hyperelliptic equation y2=x6+x5+ 3x2+ 13x+ 21. Let us now go through Theorem 5.1 replacing the elements that appear there by the ones we have chosen above: Proposition 6.2. Let Cbe a genus 2curve defined by a hyperelliptic equation y2=f(x), where f(x) = f6x6+f5x5+f4x4+f3x3+f2x2+f1x+f0∈Z[x]is a polynomial of degree 6 without multiple factors. Assume that the following conditions hold: •The following congruences mod 24hold: (f0≡f6≡1 (mod 16) f1≡f5≡0 (mod 16) (f2≡f4≡4 (mod 16) f3≡2 (mod 16). •The following congruences mod 3hold: (f0≡f6≡1 (mod 3) f1≡f5≡0 (mod 3) (f2≡f4≡1 (mod 3) f3≡0 (mod 3). •The following congruences mod 52hold: (f6≡f5≡f3≡f1≡f0≡1 mod 25 f4≡f2≡0 mod 25. 25