We will give an explicit polynomial over ℚ with 3 parameters of degree 40 as a result of the inverse Galois problem. Its Galois group over ℚ (resp. ℚ(√-3)) is isomorphic to PGSp4(3) (resp. PSp4(3)) and it is a regular PSp4(3)-polynomial over ℚ(p√−3). To construct the polynomial and prove its properties above we use some results of Siegel modular forms and permutation group theory

Kitayama, Hidetaka

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AN EXPLICIT PSp<sub>4</sub>(3)-POLYNOMIAL WITH 3 PARAMETERS OF DEGREE 40

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Math. J. Okayama Univ. 53 (2011), 155–165AN EXPLICIT PSp4(3)-POLYNOMIAL WITH 3PARAMETERS OF DEGREE 40Hidetaka KitayamaAbstract. We will give an explicit polynomial over Q with 3 param-eters of degree 40 as a result of the inverse Galois problem. Its Ga-lois group over Q (resp. Q(√−3)) is isomorphic to PGSp4(3) (resp.PSp4(3)) and it is a regular PSp4(3)-polynomial over Q(√−3). To con-struct the polynomial and prove its properties above we use some resultsof Siegel modular forms and permutation group theory.1. IntroductionIn this paper, we will construct an explicit polynomial with 3 parametersof degree 40 which has properties in Theorem 1.1 below. In this section, wewill explain its background.Definition 1.1. Let G be a finite group and K be a field. An extension L/Kis called G-extension over K if L/K is a Galois extension and the Galoisgroup Gal(L/K) is isomorphic to G. A polynomial f(X) ∈ K[X] is called aG-polynomial over K if the Galois group Gal(f(X)/K) is isomorphic to G.The inverse Galois problem asks whether there exists a G-extension overK for a given field K and a finite group G. This problem has been studiedby many mathematicians as one of the most important problems in numbertheory, especially for the case of the rational number field Q. It is stillunknown whether this problem is affirmative for every finite group, but ithas been proven affirmatively for a lot of kinds of finite groups. (See [10].) In this paper, we will consider the constructive aspects of this problem.Our problem is formulated as follows:Problem 1.1. Construct an explicit polynomial having G as the Galoisgroup for a given transitive permutation group G.Note that Problem 1.1 for groups which are not conjugate as subgroupsof the symmetric group are distinct from each other even if they are isomor-phic as abstract groups. Complete results of Problem 1.1 for all transitivepermutation groups of degree up to 15 were given in [8]. The purpose ofthis paper is to give results for transitive permutation groups of degree 40.Our main theorem is as follows. We will prove this theorem in section 5.Mathematics Subject Classification. Primary 12Y05, 11C08.Key words and phrases. inverse Galois problem, explicit polynomials, Siegel modularforms.155156 HIDETAKA KITAYAMATheorem 1.1. A polynomial F (x, y, z;X) ∈ Q(x, y, z)[X] with 3 parame-ters x, y, z of degree 40, constructed in section 4, has the following properties:(1) the Galois group of F (x, y, z;X) over Q(x, y, z) is conjugate to the prim-itive group (40, 4) in the GAP code. It is isomorphic to PGSp4(3),(2) the Galois group of F (x, y, z;X) over Q(√−3)(x, y, z) is conjugate tothe primitive group (40, 3) in the GAP code. It is isomorphic to PSp4(3),(3) F (x, y, z;X) is a regular PSp4(3)-polynomial over Q(√−3).Here a regular G-polynomial is defined as follows:Definition 1.2. A polynomial f(t;X) ∈ K(t)[X] with some parameterst = (t1, · · · , tn) is called regular if it satisfies Spl(f(t;X)/K(t)) ∩K = K,where Spl(f(t;X)/K(t)) is the splitting field of f(t;X) over K(t) and K isan algebraic closure of K.We give some remarks concerning Theorem 1.1.Remark 1.1. Explicit polynomials over Q with 1 parameter for PSp4(3)and PGSp4(3) as transitive permutation groups of degree 27 are given inp.412 of [10] and they are regular PSp4(3)- and PGSp4(3)-polynomials overQ. As mentioned in Theorem 1.1 our polynomial F (x, y, z;X) is just aregular PSp4(3)-polynomial over Q(√−3). But our result is new becauseexplicit polynomials with 3 parameters for these two groups as transitivepermutation groups of degree 40 have not been known before.Remark 1.2. We will construct our polynomial by using some results ofSiegel modular forms in section 4. Note that we only have its Galois groupover C(x, y, z) at this stage and it is a difficult problem to descend fields ofdefinition to Q. In the case of SL2(Z), Shih studied this problem in [12] byusing the theory of canonical systems of models and achieved regular Galoisextensions over Q. In our case, so far, we can not improve our polynomialto have regularity over Q.Notation. The groups PSp4(3) and PGSp4(3) are defined as follows. Wedefine the symplectic group bySp4(Z) := {g ∈M4(Z)|tg · J · g = J}where J:=(02 12−12 02)and 12 is the unit matrix. The following two sub-groups of Sp4(Z) will be important.Γ0(3) := {g ∈ Sp4(Z)|g ≡(A B02 D)mod 3},Γ(3) := {g ∈ Sp4(Z)|g ≡ 14 mod 3}.AN EXPLICIT PSp4(3)-POLYNOMIAL WITH 3 PARAMETERS OF DEGREE 40 157By definition, we have Γ(3) < Γ0(3) < Sp4(Z). It is known that the index(Sp4(Z) : Γ0(3)) = 40 and (Sp4(Z) : Γ(3)) = 51840. Γ(3) is a normalsubgroup of Sp4(Z), so we put Sp4(3) := Sp4(Z)/Γ(3) and definePSp4(3) := Sp4(3)/{±14}.This is a non-abelian simple group of order 25920. We also definePGSp4(3) := {g ∈ GL(4,F3) | tg · J · g = v · J , v ∈ F×3 }/{±14}.This is a non-abelian group of order 51840 which is isomorphic to PSp4(3)oC2.2. Preliminaries from Siegel modular formsWe will consider a PSp4(3)-extension by using some results of Siegel mod-ular forms. So we review Siegel modular forms to fix notation. We denoteby H2 the Siegel upper half plane of degree 2, that is,H2 := {Z ∈M2(C)|tZ = Z, Im(Z) > 0}.The group Sp4(Z) acts on H2 bygZ = (AZ +B)(CZ +D)−1for any g =(A BC D)∈ Sp4(Z) and any Z ∈ H2. For any natural numberk and a holomorphic function F on H2, we putF |[g]k(Z) = det(CZ +D)−kF (gZ).For a finite index subgroup Γ of Sp4(Z), we denote by Ak(Γ) the space ofall Siegel modular forms of weight k of Γ, that is,Ak(Γ) = {F : a holomorphic function on H2 | F |[g]k = F for all g ∈ Γ}.We put A(Γ) = ⊕∞k=0Ak(Γ). The space A(Γ) is a graded ring. The explicitstructure of A(Sp4(Z)) and A(Γ0(3)) is known as in the following theorems.Theorem 2.1 (Igusa[6]).∞⊕k=0Ak(Sp4(Z)) = C[φ4, φ6, χ10, χ12]⊕ χ35C[φ4, φ6, χ10, χ12].φ4, φ6, χ10, χ12 are algebraically independent over C.Theorem 2.2 (Ibukiyama[5], Aoki and Ibukiyama[1]). We putB := C[α1, β3, γ4, δ3] , C := C[α21, β23 , γ4, δ23]then∞⊕k=0Ak(Γ0(3)) = B(even) ⊕ Cα1χ14 ⊕ Cβ3χ14 ⊕ Cδ3χ14 ⊕ Cα1β3δ3χ14.158 HIDETAKA KITAYAMAα1, β3, γ4, δ3 are algebraically independent over C.3. Preliminaries from permutation group theoryIn this section, we review permutation group theory. A subgroup G ofSn, the symmetric group of degree n, is called transitive if arbitrary twoelements in {1, · · · , n} can be permuted each other by G-action. It is wellknown that the Galois group of an irreducible separable polynomial of degreen is a transitive subgroup of Sn. GAP[2] has data bases of classification oftransitive subgroups of Sn for n up to 30. These are based on Hulpke[4]. Byusing the data, we see the following fact.Lemma 3.1. The least degree of which PSp4(3) can be realized as a tran-sitive subgroup of the symmetric group is 27.Next we consider primitive groups, which are of special type in transitivegroups.Definition 3.1. Let G be a transitive subgroup of Sn. G is called primi-tive if there are no partitions of {1, · · · , n} which satisfy the following twoconditions.(1) {1, · · · , n} = ∪ri=1Bi, r ≥ 2,]Bi ≥ 2 (i = 1, . . . , r) and Bi ∩Bj = ∅ (i 6= j).(2) G induces a transitive action on {B1, . . . , Br}.GAP[2] has data bases of classification of primitive subgroups of Sn for nup to 2499. These are based on Colva M. Roney-Dougal [11]. Among these,we will use the following.Lemma 3.2. Primitive groups of degree 40 are one of the following 8 groupsup to conjugacy.PSp4(3)a, PSp4(3)b, PGSp4(3)a, PGSp4(3)b,PSL4(3), PGL4(3), A40, S40.(The symbols “a” and “b” mean that one group is isomorphic but is notconjugate to the other.)4. Construction of a PSp4(3)-polynomialIn this section, we will consider PSp4(3)-extension by using Theorem 2.1and 2.2 and construct a polynomial which has PSp4(3) as its Galois groupover the rational function field over C of dimension 3.For a finite index subgroup Γ of Sp4(Z), we denote by K(Γ) the modularfunction field of Γ, that is, the field which consists of meromorphic functionson H2 which are invariant with respect to the action of Γ. It is known thatK(Γ) is generated by fractions of modular forms for Γ of the same weight.AN EXPLICIT PSp4(3)-POLYNOMIAL WITH 3 PARAMETERS OF DEGREE 40 159(cf. corollary (i) of p.131 of [7]). We consider a sequence of the modularfunction fields of Sp4(Z), Γ0(3) and Γ(3):K(Sp4(Z)) ⊂ K(Γ0(3)) ⊂ K(Γ(3)).It is known that these three fields are purely transcendental over C of di-mension 3. (cf. [6],[5],[3]). Note that K(Γ(3))/K(Sp4(Z)) is a PSp4(3)-extension and K(Γ0(3))/K(Sp4(Z)) is a non-Galois extension of degree 40.We will compute a polynomial which defines the extensionK(Γ0(3))/K(Sp4(Z)). Then the splitting field of the polynomial overK(Sp4(Z)) isK(Γ(3)) because the Galois group Gal(K(Γ(3))/K(Sp4(Z))) 'PSp4(3) is simple. Thus the polynomial is a PSp4(3)-polynomial over Cwith 3 parameters of degree 40. (We will consider the Galois group over Qin section 5.) To carry out this computation, we need the following threesteps.1. We will compute transcendental basis over C of K(Sp4(Z)) andK(Γ0(3)).2. We will find a primitive element of the extension K(Γ0(3))/K(Sp4(Z)).3. We will compute the irreducible polynomial of the element of Step2over K(Sp4(Z)). This is a polynomial we want.Step 1. We will compute transcendental basis over C of K(Sp4(Z)) andK(Γ0(3)) by using Theorem 2.1 and 2.2. By the structure of⊕∞k=0Ak(Sp4(Z)), we see that the former (resp. latter) part of it consistsof even (resp. odd) weight functions. We also see that every odd weightfunction is a product of χ35 and a even weight function. So we see thatK(Sp4(Z)) consists of fractions of functions of the same weight belongingto C[φ4, φ6, χ10, χ12]. Hence K(Sp4(Z)) is generated by all functions of theformφ4aφ6bχ10cχ12d, (a, b, c, d ∈ Z, 2a+ 3b+ 5c+ 6d = 0).We can determine transcendental basis of K(Sp4(Z)) over C by computingZ-basis of the free Z module of rank 3,V := {x ∈ Z4|(2 3 5 6)x = 0}because φ4, φ6, χ10, χ12 are algebraically independent over C.We putA :=1 0 0 00 1 0 02 3 6 −1−2 −3 −5 1 .160 HIDETAKA KITAYAMAThen we haveV = {x ∈ Z4|(0 0 0 1)A−1x = 0}= ZA1000⊕ ZA0100⊕ ZA0010= Z102−2⊕ Z013−3⊕ Z006−5 .It follows thatK(Sp4(Z)) = C(φ4χ102χ122,φ6χ103χ123,χ106χ125).Next we will determine transcendental basis ofK(Γ0(3)). By the structureof A(Γ0(3)), we see that the first part of it consists of even weight functionsand the other parts odd weight. We also see that every odd weight functionis a product of χ14 and α1, β3, γ4, δ3. Thus we can see K(Γ0(3)) is generatedby all functions of the formα1aβ3bγ4cδ3d, (a, b, c, d ∈ Z, a+ 3b+ 4c+ 3d = 0).We can determine transcendental basis by the same way as above and getK(Γ0(3)) = C(β3α31,γ4α41,δ3α31).For simplicity, we puta :=β3α13, b :=γ4α14, c :=δ3α13,x :=φ4χ102χ122, y :=φ6χ103χ123, z :=χ106χ125,then we haveK(Sp4(Z)) = C(x, y, z) , K(Γ0(3)) = C(a, b, c).Step 2.We will find a primitive element of the extension K(Γ0(3))/K(Sp4(Z)).We have the following relations ([1],p.259).AN EXPLICIT PSp4(3)-POLYNOMIAL WITH 3 PARAMETERS OF DEGREE 40 161φ4 = 8α1β3 + 41α41 − 162γ4 + 5α1δ3φ6 = 277α61 − 2187α21γ4 + 80α31β3 +118δ23 + β23 +72β3δ3 +912α31δ3χ10 = γ4(8α31 + 2β3 − δ3)2/6144χ12 = (−124416α81γ4 − 192α31β23δ3 − 768α61β3δ3 + 16α31δ33 + 256α31β33+4096α91β3 + 153α61β23 − 1024α91δ3 + 16β43 − δ43 − 16β33δ3 + 4β3δ33+4096α121 + 7776α21β3δ3γ4 + 10077696α41γ24 − 62208α51β3γ4+31104α51δ3γ4 + 2519424α1β3γ24 − 1944α21δ3γ4 − 7776α21β23γ4−68024448γ34 − 1259712α1δ3γ24)/3981312By definition of a, b and c, it is easy to see thatφ4/α41 = 8a− 162b+ 5c+ 41φ6/α61 = 80a+912c− 2187b+ a2 + 72ac+118c2 + 277χ10/α101 = b(8 + 2a− c)2/6144χ12/α121 = (16a4 + (−16c+ 256)a3 + (−7776b− 192c+ 1536)a2+(2519424b2 + (7776c− 62208)b+ (4c3 − 768c+ 4096))a+(−68024448b3 + (−1259712c+ 10077696)b2+(−1944c2 + 31104c− 124416)b+ (−c4 + 16c3 − 1024c))/3981312.We putθ :=χ10α12χ12.We will prove that θ is a primitive element of the extension K(Γ0(3))/K(Sp4(Z)). We haveK(Sp4(Z))(θ) = C(x, y, z, θ)= C(x/θ2, y/θ3, z/θ5, z/θ6)= C(φ4/α41, φ6/α61, χ10/α101 , χ12/α121 )= C(s, t, u, v),wheres := φ4/α41 − 41, t := φ6/α61 − 277u := 6144χ10/α101 , v := 3981312χ12/α121 .162 HIDETAKA KITAYAMANote thats = 8a− 162b+ 5ct = 80a+912c− 2187b+ a2 + 72ac+118c2u = b(8 + 2a− c)2v = 16a4 + (−16c+ 256)a3 + (−7776b− 192c+ 1536)a2+(2519424b2 + (7776c− 62208)b+ (4c3 − 768c+ 4096))a+(−68024448b3 + (−1259712c+ 10077696)b2+(−1944c2 + 31104c− 124416)b+ (−c4 + 16c3 − 1024c)).We have to show that a, b and c belong to C(s, t, u, v) in order to showC(s, t, u, v) = C(a, b, c). We definef := 8a− 162b+ 5c− sg := 80a+912c− 2187b+ a2 + 72ac+118c2 − th := b(8 + 2a− c)2 − uj := 16a4 + (−16c+ 256)a3 + (−7776b− 192c+ 1536)a2+(2519424b2 + (7776c− 62208)b+ (4c3 − 768c+ 4096))a+(−68024448b3 + (−1259712c+ 10077696)b2+(−1944c2 + 31104c− 124416)b+ (−c4 + 16c3 − 1024c))− v.Here we assume a, b, c, s, t, u and v are seven independent variables. Wecompute a Gro¨bner basis of the ideal < f, g, h, j > relative to the lexico-graphic order with a > c > v > b > u > t > s. Then we get a Gro¨bner basiswhich consists of 14 polynomials. Among them, one finds two polynomialsG1, G2 without a, c from which a linear equation of b over Z[s, t, u, v] can beobtained. This implies b ∈ C(s, t, u, v). Also one finds another polynomialwhich is linear in c over Z[u, v, s, t, b]. Thus we see that c ∈ C(u, v, s, t, b).Finally, by definition of s: s = 8a − 162b + 5c, we have a ∈ C(s, t, u, v). Itfollows thatK(Sp4(Z))(θ) = K(Γ0(3)).Thus Step2 has been completed.Step 3. We will compute the irreducible polynomial of the element θ ofStep2 over K(Sp4(Z)). We can get a polynomial relation of s, t, u and v byeliminating b from G1 = 0 and G2 = 0. Then we replace s, t, u and v inthis relation bys = x/θ2 − 41 , t = y/θ3 − 277u = 6144z/θ5 , v = 3981312z/θ6 − 4096AN EXPLICIT PSp4(3)-POLYNOMIAL WITH 3 PARAMETERS OF DEGREE 40 163and multiply it by θ40. Then we get a polynomial of θ over Q[x, y, z] whichhas degree 40. This is the irreducible polynomial of θ over K(Sp4(Z)) =C(x, y, z) and its Galois group over C(x, y, z) is isomorphic to PSp4(3). Weomit this polynomial here because it takes up too much space. We give anexplicit form of F (x, y, z;X) in [9].5. Proof of Theorem 1.1The polynomial F (x, y, z;X) obtained in the previous section has PSp4(3)as Galois group over C(x, y, z). Actually, the coefficients of F (x, y, z;X) arein Q[x, y, z]. What can be said about its Galois group over an intermediatefield of C/Q ? By answering this question, we can prove Theorem 1.1. Forsimplicity, we denote Q(x, y, z) and C(x, y, z) by K and K ′ respectively. Wealso denote by L and L′ the splitting field of F (x, y, z;X) over K and K ′respectively.Lemma 5.1. PSp4(3) is primitive as a transitive group of degree 40.Proof. Suppose G ' PSp4(3) is not primitive. Then G has a partition of{1, · · · , 40} satisfying the condition of Definition 3.1. The group H := {g ∈G | gBi = Bi , ∀i = 1, . . . , r} is a normal subgroup of G. Because PSp4(3)is simple, H is 1 or G. If H = G, then it contradicts transitivity of G. IfH = 1, then G ' PSp4(3) as a permutation group of {B1, . . . , Br}. Thisimplies r | 40 and it contradicts Lemma 3.1. By definition of the symbols L′ and K ′, Gal(L′/K ′) is a transitive groupof degree 40. So Gal(L′/K ′) is primitive by Lemma 5.1. BecauseGal(L/(L ∩K ′)) ' Gal(L′/K ′)and Gal(L/K) has this group as a subgroup, we see that Gal(L/K) is alsoa primitive group of degree 40. Thus Gal(L/K) is conjugate to one of thefollowing 8 groups by Lemma 3.2.PSp4(3)a , PSp4(3)b , PGSp4(3)a , PGSp4(3)b,PSL4(3) , PGL4(3) , A40 , S40.We have to determine Gal(L/K) out of them. Because the intermediate fieldL∩K ′ is a Galois extension over K, Gal(L/(L∩K ′)) is a normal subgroupof Gal(L/K). So we can exclude groups not containing PSp4(3) as theirnormal subgroups. Thus Gal(L/K) is conjugate to one ofPSp4(3)a , PSp4(3)b , PGSp4(3)a , PGSp4(3)b.By consulting the data base of GAP[2], we can see that PGSp4(3)b is theonly group not contained in A40 and the other three groups are containedin A40. So the discriminant of F (x, y, z;X) determines whether Gal(L/K)164 HIDETAKA KITAYAMAis conjugate to PGSp4(3)b or not. We denote by d(F (x, y, z;X)) the dis-criminant of F (x, y, z;X). We can writed(F (x, y, z;X)) = f(x, y, z)g(x, y, z)2with some f(x, y, z) and g(x, y, z) ∈ Q[x, y, z]. We can assume that f(x, y, z)is square free. On the other hand,d(F (x, y, z;X)) ∈ C[x, y, z]2because Gal(L′/K ′) ' PSp4(3), which is contained in A40. Thus f(x, y, z)must be constant andd(F (x, y, z;X)) = r · g(x, y, z)2for a square free integer r. Whether Gal(L/K) is contained in A40 or notdepends on r. By computation, it turns out that r · g(1, 1, 1)2 = −3 · n2 forsome n ∈ Z. The left hand is the specialization of d(F (x, y, z;X)) and theother hand is the discriminant of the specialized polynomial F (1, 1, 1;X).Thus we get r = −3 and d(F (x, y, z;X)) = −3 · g(x, y, z)2. It followsthat if k contains (resp. does not contain)√−3 then the Galois group ofF (x, y, z;X) over k(x, y, z) is isomorphic to PSp4(3) (resp.PGSp4(3)) foran intermediate field k of C/Q . Thus we can see that F (x, y, z;X) is aregular PSp4(3)-polynomial over Q(√−3).AcknowledgementThe author would like to thank to Professor Tomoyoshi Ibukiyama ofOsaka University for giving him this problem and various advices.References[1] H.Aoki and T.Ibukiyama, Simple graded rings of Siegel modular forms, differentialoperators and Borcherds products, Int. J. Math. 16 (2005), 249-279.[2] GAP, http://www.gap-system.org/.[3] G.van der Geer, Note on abelian schemes of level three, Math. Ann. 278 (1987), 401-408.[4] A.Hulpke, Constructing Transitive Permutation Groups, J. Symbolic. Computation.39 (2005), 1-30.[5] T.Ibukiyama, On Siegel modular varieties of level 3, Int. J. Math. 2 (1991) 17-35.[6] J.Igusa, On Siegel modular forms of genus two, Amer. J. Math. 84 (1962), 175-200; II, ibid. 86 (1964), 392-412.[7] H.Klingen, Introductory lectures on Siegel modular forms, Cambridge UniversityPress, 1990.[8] J.Klu¨ners and G.Malle, Explicit Galois realization of transitive groups of degree up to15, J.Symbolic.Comput. 30 (2000), 675-716.[9] H.Kitayama, Appendix in arXiv:0910.1755v1[math.NT][10] G.Malle, B.H.Matzat, Inverse Galois Theory, Springer, 1999.AN EXPLICIT PSp4(3)-POLYNOMIAL WITH 3 PARAMETERS OF DEGREE 40 165[11] Colva M. Roney-Dougal, The primitive permutation groups of degree less than 2500,J. Alg. 292 (2005), 154-183.[12] K.Shih, On the construction of Galois extensions of function fields and number fields,Math.Ann. 207 (1974), 99-120.Department of Mathematics, Graduate School of Science, Osaka University,Machikaneyama 1-1, Toyonaka, Osaka, 560-0043, Japane-mail address: h-kitayama@cr.math.sci.osaka-u.ac.jp(Received July 27, 2009 )(Revised October 12, 2009 )

AN EXPLICIT PSp<sub>4</sub>(3)-POLYNOMIAL WITH 3 PARAMETERS OF DEGREE 40

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