Abelian extensions of arbitrary fields

Math Ann 216,99—104(1975) © by Springer-Verlag 1975 Abelian Extensions of Arbitrary Fields W. Kuyk and H. W. Lenstra, Jr. 0. Introduction and Summary Let k be an Hilbertian field, i.e. a field for which Hilbert's irreducibility theorem holds (cf. [l, 5]). It is obvious that the degree of the algebraic closure k of k is infinite with respect to k. It is not obvious that the same is true for the maximal p-extension of k, p a prime number. Let A be a finite abelian group. The question whether there exists a Galoisian extension l/k with Galois group A is, classically, known to be solvable if there exists a finite group G, and a surjective homomorphism G­+A, such that the following condition is satisfied. Suppose M is a faithful k[G]-module, and let Sk(M) denote its Symmetrie algebra over k. The group G acts upon Sk(M) and on its field of quotients k(M) in a natural way. Then the condition is that the subfield k(M)G of k(M) of all G-invariants is a purely transcendental field extension of k (cf. [6, 5]). This applies in particular to the case G = A, and M is the group ring k[Ä\. In that case we denote k(M)° bykA. Let k be an arbitrary field. Recently, the second named author [4] gave necessary and sufficient conditions in order that, for given k and A, the extension kA/k is purely transcendental, äs follows. To check the pure transcendency of kA one has to look at a finite set of Dedekind domains Dq(A) = Z[_^q(A^\, where the positive integer q(A) runs through a finite subset of Z and ζ9(Α) is a primitive q(A)-ih root of unity. Then one can determine in every Dq(A) an ideal Iq(A) with the property: kA is purely transcendental over k if and only if the two following conditions are satisfied: (i) every ideal lq(A) is a principal ideal, r (ii) if 2 is the highest power of 2 dividing the exponent of A and if the characteristic of k is not equa) to 2, then the extension k(^2r}/k has cyclic Galois group. This leads to Theorem l ([4], Corollary (7.5)). Let A be a finite abelian group. Let k be any field satisjying the condition (ii) above. There exists a natural number n such that the field of invanants kA„ of the group A" = A®...@A is a purely transcendental extension of k. Aquadruple (G, φ, A, k), with φ : G-+A a surjective continuous homomorphism of (not necessarily abelian) (pro-)finite groups and k a field, is called a Galoisian extension problem. Such an extension problem is said to be solvable if for every Galoisian extension field l/k with Gal(l//c) = ,4, there exists a Galoisian extension m/k, mDl, such that Gäl(m/k)^ G and the Galois map Gal(m//c)->Gal(//k) coincides with φ. 100 W Kuyk and H W Lenstra Jr For G = Z/pmZ, A = Z/p"Z, p a pnme number, n and m positive mtegers satisfying m ^ n ^ l , we denote the natural surjective homomorphism G-+A by 0m„, if G = Zp, the additive group of p-adic mtegers, then we wnte </>00„ mstead of φηη It is clear that the problem P(m, n, k) = (Z/pmZ, φηη, Z/p" Z, k) is solvable for all m and «, if and only if the problem P(oo, n, k) = (Zp, φχη, Zjp"Z, k) is solvable for all n ä; l With these notations we prove Theorem 2. Let k be any field Ij p = char(k), then the extension problem P(oo, n, k) is solvable for all positive mtegers «g: l ///^char(k), let Ep denote the set {x | xpm = l e K for some m e Z, m Ϊ: 0} of all pm-th roots ofumty, and put K - k(Ep) Furthermore, suppose that the degree [K k] of K/k is finite Ij p3=2 then the extension problem P(oo, n, k) is solvable for all n S: l If p = 2, then let l/k be Galois with Gal(I/fc) = Z/2"Z, n> l Then k admits a Z2-extension If, on the contrary, [K k] is infinite, then there exists at least one Galois extension of k with Galois group isomorphic to Zp Corollary 1. Let k be a field, and let p be a pnme number φ 2 The followmg conditions (i) and (n) are equivalent (i) there exists a Galois extension l/k with Gal(i//c)sZ/pZ, (π) there exists a Galois extension l/k with Gal(//fc)sZ for p —2 there is equivalence between (in) there exists a Galois extension l/k with Gal(//fc)sZ/4Z, (iv) there exists a Galois extension l/k with Gal(i/k)^Z2 Puttmg Theorems l and 2 together we get Corollary 2. Let k be an Hilbertian field and let A be a finite abehan group satisfying the condition (u) above There exists a Galois extension l/k with Gal(//k) = Z χ A, where Z = /IZp is the pro-cychc group on one generator Prooj Corollary l is immediately clear from Theorem 2 For Corollary 2 one applies Corollary l, takmg into account that for <5 = Z/4Z the field k(M)G is purely transcendental over k, whence the existence of a k-extension with Galois group Z2 The existence of a Z^-extension of k, p φ 2, follows from Theorem l and Corollary l The factor A does not give any difficulty, because k bemg Hilbertian, there exists for every m an extension l of k with Gal(I/fc)^yim (Theorem l, applymg Galois theory) Remark / Note that the Hilbertian field Q admits only one Zp-extension for every p, and infmitely many (hnearly disjomt) extensions with group A (wellknown), where A is an arbitrary finite abehan group However, the pair (Q, A) does not generally saüsfy condition (n) Remark 2 Corollary 2 substanüates a claim made in [2] (p 401) and [3] (p 113) statmg that for Hilbertian k, the maximal p-extension k(p) has infinite degree over k Mr Jarden drew attention to the incompleteness of the proof in [?] l.Proofof Theorem 2 Preserving the notations of the previous paragraph and Theorem 2, let = char(k) It is well-known that the extension problem P(n+l,n,k) is solvable Abelian Extensions of Arbitrary Fields 101 for all «S: l, e.g. using Witt vectors or by induction. This means however, that the extension problem P(m, n, k) is solvable for all m ^ n g l . Next, let p=j=char(k). First we consider the case when [K : k] is infinite. It is clear from infinite Galois theory, that Gal(K/fc) is a closed subgroup of Z*. The latter group is of the form Z*^Z/(p- l)Z®Zp if p=j=2, while Z|sZ/2Z®Z2 if p = 2. In both cases Gal(K/k) = F®Zp, where F is a finite group; Galois theory finishes this case. We are left with the case when [X : k] < oo. Again, there are two possibilities, viz. X = k and K =t= k. First, if K = k, then let l /k be an extension with Gal(i/fc) s Z/qZ, and 4=p m We have 1 = /< (f/ί) for some α e fc* α <£ k*". The field L= (J„g 1 fc(j7a) is a Galois extension of k = K with Galois group Zp, satisfying our desire. Let, alternatively, K^k, Gal(K/k) = π. The group π is cyclic of order dividing p — l if p Φ 2, and of order 2 if p = 2. Let now K(p) denote the maximal abelian Galois p-extension of K. The fact that K(p) is a Galois extension of k gives the existence of an exact sequence of groups where A„ = Gal(K(p)/K) and G = Gal(K(p)/fc). The fact that over K the extension problem P(oo, m, K) is solvable, translates in terms of group theory äs follows : Lemma 1. For every continuous surjective group homomorphism α : Ap— >Z/qZ, q = pm,m^l, there exists a continuous surjective homomorphism J0\Ap-*Z.p such that the diagram is commutative; here φ denotes the natural homomorphism with kernel pm · Zp. Now the proof goes äs follows. We are given an extension l/k with Gal(///c)s Z/p"Z, where w ä l if p=)=2 and n^2 if p = 2. We wish to construct an extension M/k with Gal(M//<)^Zp. We have Gal(/ · K/K) s Z/qZ, where q=­2n­1 if p = 2, Kd, and q = p" otherwise; so q>l in all cases. The natural surjective map is denoted by a, and we let /0, φ be äs in Lemma l. We are going to change /0 in such a way that the kernel of the new map AV­*ZV defines a Z^-extension of K which is Galois and abelian over k. Then the construction of M will be immediate. In order to carry out this prograrnme we need to know how the Statement "L is Galois and abelian over k" [for an intermediate field KcLcK(p)~] translates in terms of group theory. The group π acts on Ap via ατ = τ*ατ* - 1 where aeAp, τ ε π , and τ* e G a preimage o f t . Putting A'p­{a'\a€ Ap, ie/}, where / is the augmentation ideal 102 W Kuyk and H W Lenstra, Jr of Z [π], the cychcity of π entails A'p = [G, G], the commutator group of G This follows by direct venfication, takmg into account that α 1 " 1 =τ*ατ*~1α~1 The next lemma follows immediately from this consideration Lemma 2. Let L be an intermediate field KcLcK(p) The following conditions are equwalent (i) L/k is Galois with abelian Calais group, (u) the subgroup G&\(K(p)/L) of Ap zs invariant m G with abelian faclor group, (in) the natural map ψ Ap^G'dl(L/K) has the property A',cKer(y) It follows, in particular, that ApCKer(a.) We defme s Ap~+Ap by s(a) =as where S = £ t e n τ e Ζ[π]. Note that A'pCKer(s), smce / S is the zero ideal of Ζ[π] Proposition 1. Azurne p Φ 2, and /ei the notation be äs above The diagram Z[qZ ^ > Z/qZ ­^^ Z/qZ where the map \π\ denotes the (contmuous) automorphism "multiphcation by Ιπ|" 1 on Zpi and Z/qZ, is commutatwe Moreover, the surjective map* Jf l ΗπΓ f i . Γ _ ϊ · ϊ - / / - \ \ \ J \J is such that / Prooj The commutativity of the diagram is easily venfied by a straight-forward calculation, for the surjectivity of |π| and \n\ ~J one has to note that (|π|, p) = l The mclusion ^CKer^) follows from ^4pCKer(s) Fmally, the diagram teils us that the image of /, is a closed subgroup of Zp mappmg onto Z/qZ, so the procychc structure of Zp imphes that /L is surjective This proves Proposition l Theorem 2 is now easily settled for p 4= 2 Let /, α be äs before, let /j be äs m Proposition l, and let LcK(p) be the invariant field of Ker^) Thcn IcL, Gal(L/K)^Zp, and L/k is Galois and abelian by Lemma 2 Further, Galois theory gives us an exact sequence of abelian groups 0-»Zp->Gal(Z,/fc)-»7i->0 The sequence sphts by (|π|,ρ)=1, so L-M K where Gal(M/k)^Zp Fmally, l CM again follows from (|π|, p)= l We conclude that M is the reqmred extension of k and that the problem P(oo, n, k) is solvable for all «^ l Proposition 2. Assume p = 2, and let the notation be äs before The diagram •Z.2 Z/qZ — > Z/qZ is commutatwe, but the homomorphism f± — f0° s not surjective One has A'2 Im(/1) = 2Z2, and, if /2 = 2/i, then f2 is a contmuous surjectwe homomorphism satisjymg A\ CKer(/ 2 ) Abehan Extensions of Arbitiary Fields 103 Proof The commutativity of the diagram and the mclusion A'2 go äs before Further, the diagram implies that lm(fi) is a closed subgroup of Z2 mappmg onto 2Z/qZ If q>2 this implies Im(/1) = 2Z2 by the procychc structure of Z 2 In the case q — 2 we arnve at the same conclusion by an explicit computation q = 2 implies Gal(i/k) S Z/4Z and X c / , let σ* e Gal(K(p)/k) be such that σ = σ*\Κ generates π, then σ*\1 generates Gal(l/k) so the element τ = (σ*)2 of .42 is not the identity when restncted to /, this means α(τ)φΟεΖ/2Ζ so /0(τ) e Z2\2Z2 , also τ" = τ so /1(τ) = / 0 (τ 2 )ε2Ζ 2 \4Ζ 2 , therefore 2Z2Clm(/1), and since the opposite mclusion follows from the diagram we conclude Im(/1) = 2Z2, äs required The assertions about /2 follow immediately This concludes the proof of Proposition 2 To fmish the proof of Theorem 2, let /, α be äs before and let /2 be äs m Proposition 2 Then the invariant field LCK(2) of Ker(/2) has Galois group SiZ2 over K, and L is Galois and abehan over k Theie is an exact sequence 0-»Z2-»Gal(L/fe)-»ir->O If this extension sphts then Gal(L/k) ^ Z 2 ® π, and if it does not spht then Gal(L/k) s Z2 In both cases there exists an extension M of k with Galois group isomorphic to Z2 This concludes the proof of Theorem 2 Remark A closer look at the construcüon reveals that m the case p = 2 the field M can be chosen such that the intersection Mr\l has degree 2""1 or 2" over k 2. Supplementary Remarks It is not true that any field k, admittmg a field extension / with Gal(f/k) = F4 (cf Theorem 2) admits a Z/4Z-extension (and, by consequence, a Z2-extension) The field of all totally-real algebraic numbers, for instance, admits K4-extensions and no Z/4Z-extensions The followmg is an example of a field admittmg for an arbitraiy cardmal number m an extension with Galois group (Z/2Z)'", and no Z/4Z-extension Let / be a set with \l\ = m and let F = Q ( { t t \ i e I } ) be a purely tianscendental extension of β with transcenciental degree m Choose for every ι e I an ordermg < , of F, m such a manner that t, <, 0 and 0 <, t}, for / Φ ι Let Ä„ FCR.CF, be a real-closed field the ordermg of which is an extension of <, Then k= P) le/ -R, has the required property one sees easily that Gal(k//c) is topologically generated by elements of order 2 It is also possible to give a proof of Theorem 2 (p Φ 2) more directly by usmg Kummei theory However, this method does not seem to be readily extendible to the case p = 2 References Hubert, D Über die Irreduzibihtat ganzer rationaler Funktionen mit ganzzahligen Koeffizienten J Reine Angew Math 110,104—129(1892) Kuyk,W Genenc approach to the Galois embeddmg and extension problem J Algebra 9 (4), 393-407(1968) Kuyk, W Extensions de corps hilbertiens J Algebra 14 (t), 112—124 (1970) 104 W Kuyk and H W Lenstra, Jr 4 Lenstra, Jr , H W Rational functions invariant under a flnite Abehan group Invent Math 25,299—325(1974) 5 Lang, S Diophantme geometry Interscience Tracts 11 New York-London 1962 6 Noether,E Gleichungen mit vorgeschriebener Gruppe Math Arm 78, 22j—229(1918) W Kuyk Department of Mathematics Antwerp University B-2020 Antwerp, Belgmm H W Lenstra, Jr Mathematisch Instituut Umversiteit van Amsterdam Roetersstraat 15 Amsterdam The Netherlands (Received October 30, 1974)