ON THE GENUS HELD AND 3T3 APPLECATEO-fis TO FOUR PROBLEMS {N ALGEBRAIC NUMBER THEORY ‘Fhesis for the Begree of Ph. D. MSCHEGRN STATE flNEVERSiTY THOMAS RANDLE BUTTS 1973 MICHIGAN snug W am 16 30V13 N3 11H”!!! W ItIIIIIUMWIIHIHMW 3 3200627 9636 ll! mg“, L I B R A R Y Michigan State Uni-H313; :y L . .oh'alu“ '9‘“ - ' This is to certify that the thesis entitled On the Genus Field and its Applications to Four Problems in Algebraic Number Theory presented by Thomas Randle Butts has been accepted towards fulfillment of the requirements for PhoDo degree in Mathematics QLM n rug“ Major professor 0-7639 timomo av HMS & SflNS' 300K BINDERY lNC. LIBRARY amp Ins 3PM ”GPCE 1 My: likx. A I\——-—-—-\.‘ _ W;\\_/J H ABSTRACT ON THE GENUS FIELD AND ITS APPLICATIONS TO FOUR PROBLEMS IN ALGEBRAIC NUMBER THEORY BY Thomas Randle Butts The genus field of an algebraic number field K, denoted by GSF(K), is the maximal unramified extension of K of the form AK where A/Q is abelian. In this dissertation I first construct the genus field of most algebraic number fields. This construction and the theory underlying it are then applied to two recent problems: (1) (MacCluer) For which normal extensions K/Q does the multiplicative group “I”K generated by the absolute norms of all fractional ideals of K coincide with the group HHHK of absolute norms of all principal fractional ideals of K? (2) (Burgess) If f is a polynomial with rational integral coefficients, let Vf be the multiplicative group generated by the non-zero values of f for integral values of the variable. Does Vf consist Thomas Randle Butts of all rational numbers not excluded by obvious algebraic conditions? and to two classical problems: (3) Which abelian groups occur as ideal class groups of algebraic number fields? (4) Construct the Hilbert class field of an algebraic number field. A sampling of the results obtained involving these problems is: (1) HIHK = HHHK <==e K = GSF(K) = ZCF(K) where ZCF(K) is the central class field of K. (2) GSF(K) # K ==€> Vf # HIKH, meaning the answer to (2) is usually ”no". (3) Every abelian group occurs as a subgroup of the ideal class group of infinitely many a) abelian b) non- abelian and c) non-normal algebraic number fields. (4) If the exponent of the ideal class group of a quadratic number field K divides 12, the Hilbert class field of K is constructed. Many examples are given to illustrate the constructions and theorems. ON THE GENUS FIELD AND ITS APPLICATIONS TO FOUR PROBLEMS IN ALGEBRAIC NUMBER THEORY BY Thomas Randle Butts A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1973 5 3‘ .33 TABLE OF CONTENTS a INTRODUCTION . O O O O O O O O O O O O 0 CHAPTER I PRELIMINARIES . . . . . . . . 1. SOME HILBERT THEORY. . . . . . 2. HILBERT CLASS FIELDS . . . . . . 3. BETWEEN K AND HCF(K). . . . . . 4 0 AN EXAMPLE 0 O O O O O O O O O O 0 CHAPTER II GENUS FIELDS . . . . . . . . 1. MORE HILBERT THEORY. . . . . . . 2. CONSTRUCTION OF THE GENUS FIELD. CHAPTER III NORM GROUPS . . . . . . . . CHAPTER IV THE BURGESS PROBLEM. . . . . CHAPTER V IDEAL CLASS GROUPS. . . . . . CHAPTER VI CONSTRUCTION OF HILBERT CLASS FIELDS l. QUADRATIC FIELDS . . . . . . . . . 2. COMPOSITUMS OF QUADRATIC FIELDS. 4. PURE CUBIC FIELDS. . . . . . . . . 5. QUADRATIC FIELD K Q/m where 3|h. 6. QUADRATIC FIELDS K = Q(\/m) where BIBLIOGRAPHY . . . . . . . . . . . . . . ii 4|h Page 10 12 15 15 17 33 52 56 6O 6O 62 65 66 66 71 74 INTRODUCTION The value group V of a polynomial f(x) in z[x] f is defined by V = . At the 1969 AMS Number f Theory Conference, two problems concerning the value group of a polynomial were posed: Problem 1. (Stolarsky) Let f(x) = x4+x3+x2+x+l. If p a 1 mod 10. does pEVf? Problem 1a. (Burgess) If f(x)€Ez[x], does Vf contain all rational numbers not excluded by obvious algebraic conditions? If HIKH denotes the group generated by the absolute norms of all fractional ideals of an algebraic number field K and ”HKH denotes the group generated by the absolute norms of all principalfractional ideals of K, then Vf C HHKH C HIKH where K is the splitting field of f. This observation gives rise to a stronger version of Problem 1, namely, Problem 1'. If f(x) x . Hence we can choose a generator 0 of Z(T) so that I I lplk 0(x) a x mod ‘13 for all x 6 S . This unique element of Z(T) is called the Frobenius Automorphism of T over k . Q The symbol [ T J = O is called the Frobenius symbol of T over k . Remark: The Frobenius automorphisms of the prime factors of p are conjugate under G . PROOF: Note that for T E G , x E S , 0(T_1X) E (T-lx)”””k a T-1(annk) mod $ 80 that TOT-1(X) E prHk mod (Tfi) Hence [“3] = TF-‘fll - TT The conjugacy class formed by the Frobenius Symbols of the factor of p,<1fa prime unramified in K , 8 is called the Artin symbol of p and is denoted by K k . . . K—é;j . As is the common practice for abelian extensions, the Artin symbol will be thought of as element valued. 2;? HILBERT CLASS FIELDS Let K be an algebraic number field and let I denote the group of fractional ideals of K . I is free on the prime ideals of K . Let H denote the subgroup of I consisting of all principal fractional ideals of K . Then I/H = a is called the ideal class group of K and h = [I:H] = \I/H] is called the class number of K . The fact that relations existed between the ideal class group of a number field K and its abelian exten— sion fields was first observed toward the end of the nineteenth century. Hilbert defined the class field of K as that extension field of K where exactly the prime ideals in the unit class split completely. 9 He conjectured that the galois group of the class field of K with respect to K was isomorphic to the ideal class group of K . Furtwangler (1907) was the first to verify his conjecture. During the next twenty years, Artin (and Takagi) constructed general class field theory and gave another proof of Hilbert's conjecture based on the Artin symbol defined in §1. He noted that the Artin symbol associates to each prime ideal T of K an element in the galois group of every abelian extension of K in which T is unramified. Artin's formulation of Hilbert's class field is Definition D: The Hilbert class field of an algebraic number field of K , denoted by HCF(K) , is the maximal abelian unramified extension of K . Most of the properties of the Hilbert class field of K are summarized in the Artin Reciprocity Theorem: The homomorphism f defined by linearly extending the map '13 __> K HCF(:)ZK ) to all of I is surjective with kernel H . 10 Thus the galois group of HCF(K)/K is canonically isomorphic to the class group G, of K , that is Arfiiq— l is short exact where G is the galois group of HCF(K)/K . That the Artin map from I into G is surjective even on the primes of K is seen via the Cebotarev density theorem. The deep insight afforded by the Reciprocity Theorem is that the kernel is H , that is HCF(K21K K T ) = l e T E 1 mod H . When K/Q is a normal extension, HCF(K)/Q is also normal since it is unique. Thus if F denotes the galois group of HCF(K)/Q , we have the following Artin diagram denoting the galois correspondence: HCF(K) K 0 l C; T 3. BETWEEN K AND HCF(K) Of the fields between K and HCF(K) for K an algebraic number field, two are of interest in this dissertation: 11 Definition E: The genus field (Geschlecter- k6rper) of an algebraic number field K , denoted by GSF(K) , is the maximal unramified extension of K of the form AK where 'A/Q is a finite abelian extension. In fact A/Q is the maximal abelian extension of Q contained in HCF(K)/Q . Notice that the genus field is defined even for non-normal extensions of Q . Definition F: For a normal algebraic number field K/Q , the central class field (Zentralen Klassenkdrper) of K , denoted by ZCF(K) , is the maximal abelian unramified extension of K normal over Q such that the galois group of ZCF(K)/K is contained in the center of the galois group of ZCF(K)/Q If t denotes the galois group of HCF(K)/Q , then it is clear that the genus field. GSF(K) corres- ponds to 6'0 F, under the Galois correspondence. If N is the normal subgroup of F corresponding to ZCF(K) under the Galois correspondence, G/N is contained in the center of r/N . But for any I normal subgroup B of P contained in GI . 12 a/B c 2(r/B) e B :> [6.1“] where [6,F] is the group generated by all commutators c-ly-lcy where c E G and y E F Thus N = [6,F] and [6,FJ czc.fl F1 An Artin diagram illustrating these relationships when K/Q is normal is, HCF (K) 1 | ZCF (K) [(5. I‘] l GSF(K) 0 0 F' l K c i r 4. AN EXAMPLE: K = Q(\/-44fl In this example h(K) = 20 , 6' is cyclic, and so |r] = 40 . The Sylow S-subgroup C5 is normal in F , so I‘/C5 is either abelian (A) , dihedral (D4) , or quaternion (2) . Let L denote the subfield of HCF(K) with galois group I‘/C5 . If 1"/C5 = A or 2. , then every subgroup of 1"/CS is normal. Thus the inertia fields over Q of all prime 13 divisors in L of 2 or 449 , the only primes ramifying in K , coincide. Since L/K is unramified, the ramification indices of 2 and 449 are 2 in L/Q . Then the intersection of the inertia fields J(2) n J(449) is an unramified extension of degree at least 2 over Q which contradicts the Dedekind- Minkowski Theorem that there are no unramified exten- sions of Q . Hence I‘/C5 = D 4 Since C Q P and (5,8) = l , F is a semi- 5 direct product of C5 and D4 . These groups are determined by the homomorphisms e of D4 into Aut C5 = C4 . Now [ker 9] ¥ 2 since the only normal subgroup of order 2 in D4 is contained in the cyclic 2 4 -l subgroup of order 4 . D4 = , (3) ker e = . Alternative (3) is impossible since this group contains no cyclic subgroup of order 20 so two possibilities remain: (l) I‘ = C5 9 D4 and (2) I‘ = D20 . In (1), D4 4 T , so there exists an abelian subfield M/Q of HCF(K)/Q of degree 5 . But M/Q must contain a prime of ramification index 5 over Q contra- dicting the fact that HCF(K)/K is unramified. Thus 2 20 -1 we finally obtain that F = D20 = 14 . -l -1 -2 Now [F'] = 10 Since xyx y = y has order 10 so that GSF(K)/Q has degree 4 . The only normal subgroup of F properly contained in F' is c5 . but |z(1“/c5) | = 2 , so ZCF(K) = GSF(K) = Q(¢C449,i) as can be shown. We have the lattice of fields HCF(K) 1 lo egg/4449.1) = GSF(K) = ZCF(K) — D'zo = [Czo'D 20] 2 l 05/4149) = K c20 2 | Q D20 Now take K = Q(v/:ZZ9,i) . In this case h(K) = 10 and the group-field structure is the same as before. We note GSF(K) = K and since ]Z(D4)l = 2 , ZCF(K)/K has degree 2 . It can be shown that ZCF(K) = K(va) where g is a funda- mental unit of Q(v/ZZ§) , giving the lattice of fields: HCF(K) l ZCF(K) = K(\/—§I)_C5 = [ClO'DZO] GSF(K) = K = Q(V’-44 ,1) D'20 = C10 Q D20 CHAPTER II GENUS FIELDS Let K be an algebraic number field and GSF(K) be its genus field. In this chapter I give a construction of GSF(K) when K/Q is normal and determine a formula for the genus number of K, g(K) = [GSF (K) :K]. In addition, the genus field of a non-normal extension is discussed briefly. In order to understand this construction, it is necessary to extend our knowledge of the way a rational prime ramifies in a normal extension K/Q, so we develop §l More Hilbert Theory Throughout this section K/Q is a normal extension with galois group G and ring of integers S. If T is a prime ideal of K, then e denotes the ramification index of T over Q, f denotes the degree of T over Q, Z(T) denotes the decomposition group of T over Q and T(T) denotes the inertia group of T over Q. 15 16 Definition A: The subgroup Vn(T) of T(T) defined n+1 . by Vn(T) = [O 6 G]O(x) E x mod T for all x in S} is called the nth ramification group of T over Q. As is well known, the higher ramification groups of T over Q form a finite strictly decreasing normal series, UWDTW)D%W)D~-DL Definition B: The sequence of groups G 3 z D T 3 V1 3 --- D l is called the long Hilbert sequence of T over Q. As is well known, Result I: T(T)/V1(T) is cyclic with order dividing pf-l. Vi(T)/Vi+l(T) are elementary abelian p-groups where (p) = T n Q. 00 Note that e = Z)(V.:V. ). O For abelian extensions, there is the not so well known delicate result of Speiser [22]. 17 Result II: If Z(T) is abelian, then [T(T):V1(®)]\p-l- If (p,e) = l, T is said to be tamely ramified and it is clear that V1(T) = V2(T) = --- = 1. Thus in this case Results I and II become Result III: If T is tamely ramified over Q, T(T) is cyclic and ]T(T)]‘pf-1. Furthermore if Z(T) is abelian, then IT(T)]'p-l. If we localize at a prime T of K, then E(T)‘B = KT becomes the decomposition field of T and the galois group of KT/Qp 3(T) is Z(T). The long Hilbert sequence for / . T/Qp K.13 Qp is Z 3 T D‘vl D --- = l and all global results are also local results. §2 Construction of the Genus Field Recall that the genus field GSF(K) of an algebraic number field K is the maximal unramified extension of K 18 of the form AK where A/Q is a finite abelian extension and the genus number of K, g(K) = [GSF(K):K]. The genus field and consequently the genus number of K will be determined in steps when (A) K/Q is abelian, (B) K/Q is non-abelian, and then when (C) K/Q is non- normal. (A) If K/Q is abelian, GSF(K) becomes the maximal unramified extension of K which is abelian over Q. As it turns out, it suffices to consider abelian extensions of degree pa because of the following Lemma A: If Kl/Q and KZ/Q are finite and normal, then GSF(KlKZ) 2 GSF(K1)'GSF(K2). 239g: GSF(KlKZ) l /:§F(K1)GSF(K2) \ GSF(K1)K2 /GSF(K2)K1 111Kl = GSF (K1) K1K2 GSF(KZ) = A2K2 / K 1 2 \ \ / \ \/ Q 19 By tracing through this Artin diagram applying the results (1) M/N unramified = MK/NK is unramified (2) M/k, N/k abelian (unramified) =MN/k is abelian (unramified), we see that A1A2K1K2 = GSF(K1)'GSF(K2) lS unramified over K1K2 and therefore is contained in the maximal unramified extension of Kle of the form AKlK2 where A/Q is abelian. Consequently let K/Q be abelian of degree pa. We show that GSF(K) is a certain subfield of the minimal cyclotomic field containing K guaranteed by the famous Kronecker-weber Theorem: Every abelian extension of Q is contained in a cyclotomic field. Specifically I prove, Proposition: Let K/Q be abelian of degree pa with finite ramified primes {pj]:=l' p = p5. having ramification indices (e. i=1 over Q. 3 Then GSF(K) is the inertia field of p0° in s [1 Lj over K where p is any one of the j=1 °° infinite primes of K and where Lj is the subfield of Q(gp ) of degree ej over Q, j 1 3.3 S s-l, and LS is the subfield of Q(g Y+1) of degree pY over Q where eS = pY if p is odd: or either Q(g Y+l) or the maximal real 2 . _ Y . _ subfield of Q(C2(42)‘where eS — 2 , if p — 2. 20 Moreover s H e .=1 5 g(K) = 3— p05 Q where s 2 if p ramifies in n L./K a: .‘l j j— 5 = 1 otherwise 5 We observe that the Proposition implies that p0] H ej. j=1 As well as being a step in the proof)this fact is of independent interest, so we state it as Lemma B: If K/Q is abelian of degree n with finite ramified primes [pj): having ramification indices [ej}: , then 0 mod n. ll :21 m (D Ill 1 3 Lemma B can then be used to show equality holds in Lemma A. By putting everything together we will obtain the main Theorem 1: Let K/Q be abelian of degree m a. n = n qil with finite ramified primes {pj}:=l i=1 having ram where K = Moreover where 6 a: 21 ification indices [ej m GSF(K) = n GSF(K.) . i i=1 m H K, and [K.:Q] = i=1 1 l s H ej n6 :9 2 if 5 = 2 for I 1 otherwise Now for the proofs. PROOF (Lemma B): The inertia fields of s I. over Then 3 1 Q' (7,. l q . o l the prime divisors in K of any ramified prime in Q are conjugate. Thus if K/Q is abelian, the inertia fields of these divisors are equal, so we speak of the inertia field of p. K Let J, denote the inertia field of pl so that l e 1 . . . Tl [K.J1] — e1. Let 45 be the inertia field of p2 I J2 e2 45 and define recursively Jj as the inertia field I : of . in J. settin e! = J.:J. . Since ~ P3 3-1' 9 J [ 3 3‘1] there are no unramified extensions of Q by the I Q Dedekind-Minkowski Theorem, Js = Q so that s H e! a 0 mod n. But e{]e. for all j, so j=1 J 3 in 22 s H e. E 0 mod n completing the proof. PROOF: (Proposition) The portions of this proof which are identical with the steps in Speiser's [22] proof of the Kronecker-Weber Theorem will only be sketched. For an elementary proof of the Kronecker—Weber Theorem see Zassenhaus [26]. By Result III (Speiser), eIIpl-l. Now Q(gp ) contains 1 a unique subfield of degree el which we will denote by L 1 Let M1 be the inertia field of pl in the abelian extension KLl/Q, an extension of pth power degree. K M1 L1 d ///// P\\\\ I el Q Because pl is tamely ramified in KLl' its inertia group is a cyclic subgroup of G(KLl/Q) and of order divisible by e1. But by Galois theory, G(KLl/Q) is isomorphic to a subgroup of the external direct product G(K/Q) s G(Ll/Q) . an abelian p—group together with a cyclic p-group of order e1. Consequently 23 el = KLl/Ml that is the index 9; ramification 9f, pl in KLl/Q is still e1! Moreover since pl is totally ramified in L1 yet unramified in M1, and so [M1L1=Q] = [MleIILl=Q] = [Ml=Q]el = [Klee] that is KL = M L and so a fortiori K c MlLl° Thus Ml/Q is abelian of degree p6 with finite ramified primes p2,p3,...,ps where ej = ej(Ml/Q) for j = 2,3,...,s. Because pj is unramified in Ll/Q, it is also unramified in LlMl/Ml and KLl/K. This caistruction, then, effectively isolates the ramification of pl in the field Ll' while not disturbing the ramification of the other primes. Applying this argument s-2 more times, we obtain a sequence of fields Ll'L2'°°°'Ls-l where Lj is the subfield th of the pj roots of unity Q(gp ) of degree ej over Q. j Then K c M L ...L where M = M is abelian of 5-1 1 s-l s-l degree pY over Q where only p and possibly the infinite prime ramify. 24 As usual the cases p odd and p = 2 must be considered separately with p = 2 being the more difficult. (1) When p is odd, M turns out to be the subfield of Q(Q Y+1) of degree pY over Q. Setting M = LS, we 5 see that L = H Lj is an abelian extension of Q of degree i=1 5 H e, since Li H Lj = Q for all i,j as different primes '_ J j—l ramify totally in each Li' Furthermore L/K is unramified s }l and ej(L/Q) = ej(K/Q) for all j. As p is odd, the since [pj are the only finite primes ramifying in L/Q infinite prime is also unramified since normal extensions of odd degree are real. Thus L c:GSF(K) implying s s H e.][GSF(K):Q]. But by Lemma B, [GSF(K):Q]] H e. so j=1 3 j=1 3 GSF(K) = L and the description of GSF(K) is valid. Since IGSF(K):QI [K:Q] ' thus completing the proof of the Proposition when p is odd. [GSF(K):K] = the formula for g(K) is also clear (2) When p = 2, M = L is either Q(g ) on the S 2y+1 maximal real subfield of Q(g Y+2) if y 2_2. If y = 2, 2 then LS is Q(i), Q(VWZ), or Q(VLZ). Arguing as in (l), s s we see that L = H Lj is abelian over Q. [L:Q] = H e., and j=l j=1 L/K is unramified at all finite primes. If the infinite prime (in Q) ramifies in K, then L/K is unramified and L = GSF(K) as in (1). If however the infinite primes of K ramify in L, then GSF(K) is their inertia field in L. The description of GSF(K) and the validity of the formula 25 for g(K) are now clear in this case, so the proof of the Proposition is complete. The proof of Theorem 1 now follows trivially from the Proposition and the two lemmas. Example 1. K = Q(gm) —-i> GSF(K) = K. Since Q(gm) = n Q(g 0) where m = n pa and p is pIm p pIm the only ramified prime in Q(Q a) and p ramifies P totally. Thus GSF(Q(g a)) = Q(Q a) and the conclusion P P follows from Theorem 1. Example 2. K = Q(V/BO), 2,3,5 are finite ramified primes, p is unramified in K. m Ste 1 . L1 = Q(V/S) since 5 e 1 mod 4 so M1 = Q(V/6). Qgflm.¢%) ow%,¢%m /|\ /|\ e(./ 30) Q(\/ 6) Q(\/ 5) Q(J6) QM —2) e(\/ -3) \\\\ | //// \\\\ l /// Q a (l) (2) Ste 2 . L2 = Q(V/-3) since 3 a 3 mod 4 so M2 = Q(V/-2). Thus L = Q(v/3o. V/6, V/-3). p a: 26 ramifies in L/K, so GSF(K) = Q(V/BO, V/6) = Q(V/S, V/6) which is HCF(K) since h(K) = 2. Example 3. K = Q(\/66). 2.3.11 are the finite ramified primes, p is unramified in K. Analogous Q reasoning to Example 2 shows L = Q(v/66, V/-3, V/-ll) so GSF(K) = Q(./66. ./33) = Q(x/Z. ./33). Example 4. K = Q(\/ 231). 2.3.7.11 are the finite ramified primes. pan is unramified in K. Analogous reasoning shows L = Q(V/23l, V/-3, v/'7v V/-ll) so GSF(K) = «ah/231, \/21, ./33) = Q(\/21. (M3. ¢11) = Q(JB. ./7. ./11). 27 (B) To recapitulate. the genus field of an abelian extension K/Q with finite ramified primes [pj}i with ramification indices [ej1i is determined by constructing an abelian exten- sion .31 Lj/Q having the same ramification as K/Q at all finitg—primes and then making allowance for the infinite primes. In fact Lj is the compositum of the subfield of Q(gp ) of 3 degree e5 over Q and the subfield of Q(gp aj+1) of degree a. J p.3 over Q (with suitable modifications for p = 2) where J aj e. = e1 . , t, . = 1. J 3P3 (e3 p3) To determine the genus field of K where K/Q is not necessarily abelian, we seek to construct a maximal abelian extension A/Q such that AK/K is unramified. This extension should have the same "abelian ramification" as K/Q, an idea I will now make precise. Let pj denote any finite ramified prime of the non- abelian extension K/Q and let Tl resp. Tb denote any prime divisor of pj in K resp. A. Then ej is the ramification index of pj in the local field K . Let ej ”1 denote the ramification index of pj in A . $2 Kh Ah / \ $l\\\ ///'AT2 KBI1A 1 T2 Qp K 28 Then the "abelian ramification" e3 of pj in K is the ramification index of pj in K H A /Qp, that is the ”l ”2 ramification index of p. in the maximal abelian subfield of a. - 3 _ K /Q . Now if e1 = e? . (e? .) — 1. then e? .-l ‘31 p 3 3p] ' J'pJ J|p3 applying the local version of Result II. Let then Lj denote the compositum of the subfield of Q(Qp ) of degree e3 over j a. Q and the subfield of Q(gp aj+l) of degree pjJ over Q j (or if pj = 2, either Q(Q2a+l) or the maximal real subfield s of Q(Q2a+2) as the case may be). If L = H Lj, LK/K is j=l unramified at all finite primes and it is again a question of the ramification of the infinite primes of K in LK/K. Thus GSF(K) is the inertia field of the infinite primes of K in LK/K so GSF(K) has the form AK where A = L if K is imaginary and A is the inertia field of the infinite rational prime in L if K is real. We observe that A contains K0 and GSF(KO) where KO/Q is the maximal abelian subfield of K/Q and that A n K = KO. Thus llznm e! 900 = [GSF(K):K] = ——'—‘ = . - . 6m [KO.Q]6co [KO.Q]5m since Lj n Li = Q for all i,j since different primes ramify totally in each extension. 29 Thus we have constructed the genus field of any normal extension K/Q. We summarize this construction in the following theorem which, of course, contains Theorem 1 as a special case. Theorem 2: Let K/Q be a normal extension with finite ramified primes [pj}:_l. Let e5 denote the rami- fication index of pj in the maximal abelian subfield C1 . of the local field K /Q and let e[ = e?p. J Bj P 3 J 3 where (e3,pj)= 1. Then GSF(K) is the inertia field of the infinite primes ofI(in LK/K where s L = H L. with L. being the compositum of the j=1 3 J subfield of Q(Cp )/Q of degree e3 and the subfield j a. of Q(gqu+1)/Q of degree pj J (or if pj = 2 3 either Q(g2a+l) or the maximal real subfield of Q(C2d+2))o Moreover e' II :10) l—" g(K) = , [KO.Q]6co 30 where KO/Q is maximal abelian subfield of K/Q 2 if the infinite primes of K ramify LK/K and 5 = m 1 otherwise Example 5: Let K be the Kummer extension K = Q(Q/a, cn), n > 2, a ¥‘: 1 is square-free and odd. The primes divisors of lcm(a,n) are the finite ramified primes of K. Suppose pl.p2.....pm divide n and pm+1....,pS . . a . _ diVide TETHY° Then Lj c:Q(gn) for j — l,2,...,m. For Pm+1.....ps. Lj is the subfield of Q(gp )/Q of degree (n,pj-1), since the maximal abelian subfield of the local . n . t field Qp(\/a, gn) over Qp is Qp(./a) where t = (n,pj-l) s and p. is totally ramified in Q (E/a). Then L = H L. so 3 p j=m+l j GSF(K) = K(9m+1....,98) where ej is a primitive element for Lj/Q, j = m+1,...,s. Since K0 = Q(gn), s m s H e! H e! o n e5 3 '=1 j=l j=m+1 K = - = 1 . , .-1 = , -1 . 9( ) JERRY—I ¢Kn) j=£+l(n P ) g (n P ) p a,n (C) The genus field, unlike the central class field, is defined for non-normal extensions. In this case we Obtain Theorem 3: If K/Q is a non-normal algebraic number field and K/Q is its normal closure, then GSF(K) is the maximal unramified extension of K contained in AK where GSF(K) = AK. 31 PROOF: A prime unramified in K is unramified in K, so the same primes ramify in both K and K. Thus GSF(K) 9 AK and the conclusion is immediate. I have not investigated conditions for equality of GSF(K) and AK except for the following Example 6: K = Q(nV/a), (a,n) = l, a ¥ i l is square- free and odd GSF(K) = AK /' ‘ /.K )3 1, of an are the finite ramified The prime divisors [p. primes of K. Let [pj): denote divisors of a and )3 . denote divisors of n. j t+l {p _ _ t _ From Example 5. GSF(K) = AK = ( n Lj)K where Lj/Q i=1 is the subfield of Q(gp )/Q of degree (n,pj-l). Now 3 _ _ _ pj,j = l,2,...,t, is unramified in K/K. Since AK/K is also unramified, prime divisors of pj in K are unramified AK/K. For ,k=t+l,...,s, is unramified in A and Pk pk 32 hence unramified in AK/K. Therefore AK/K is unramified and by Theorem 3 GSF(K) = AK = K(el,...,9t) where ej is a primitive element for Lj. To my knowledge the connection between the genus field and the Kronecker-Weber Theorem has not been noted in literature. Furuta [8] has computed a formula for a general genus number g(K/k), where k is any algebraic number field, using class field theory and idele class groups. Special cases of the genus number formula have been proved in similar fashion by Yokoi [24] and Iyanga-Tamagawa [l6]. Hasse [12] and Leopoldt [19] have discussed genus fields using character theory. Frahlich [4] has computed the genus number using rational congruence groups. It appears that he is responsible for the definition of genus field used here. For cyclic extensions of prime degree, Herz [13] has constructed the genus field using a different technique. Historically the primary interest has been in genus fields of quadratic fields looked at in terms of quadratic forms. CHAPTER III NORM GROUPS 1. Recall that for K/k a finite separable extension, k a number field, the (relative)pp£m NK/k(T) of a prime ideal T in K is defined to be the ideal pf in k where p = T O k is the prime ideal of k lying below T and f is the degree of T over k . This map is extended to IK , the group of fractional ideals of K , by multiplicavity. NK/Q(T) then is a principal ideal in Z generated by its least positive integral element, f P , where p is the rational prime lying below T in . f . Q . The integer P is called the absolute norm of the prime ideal T and, in deference to the analysts, will be denoted by HTHK , or simply “T” where only one field is being discussed. An alternative characterization of HTHK is the order of the residue class ring S/T where S is the ring of integers of K . Since the absolute norm inherits the multiplicative property, fl 4 ”mHK is a group homomorphism from I K into the multiplicative group of positive rationals. 33 34 The image group HIKH is therefore generated by the absolute norms of (integral) prime ideals of K Similarly if HK denotes the subgroup of principal fractional ideals, then the homomorphic image HHKH is generated by the absolute norms of the principal integral ideals of K . In this chapter, I will investigate necessary and sufficient conditions for HIKH = HHKH where K/Q is normal. 2. Let K be a finite galois extension of Q with galois group G of order n and ideal class group O of order h . We first check that HIKH and HHKH uniquely determine K by proving the Proposition: Assume K/Q and L/Q are normal. Then (1) if HIKH ”IL” , then K = L and L . (2) if IIHKII HHLH . then K This Proposition is a consequence of: Bauer's Theorem (1916) [ 1]: Let K/k be normal and L/k be finite. Let SK denote the set of all prime ideals of k which split completely in K Then SK c SL , if and only if L c K . 35 Proof: Since a prime splitting completely in L also splits completely in L , the galois closure of L , we may assume that L/k is also normal. Since a prime ideal splitting completely in two extensions of a field k also .///KL\\\\ K L .\\\\ ,//// splits completely in their compositum k SKL = SK fl SL SO that SKL = SK For a normal extension M/k , the Dirichlet density of 1 1 l M [M:k] ' thus [K:k] = [KL:k] or KL = K implying Note that Bauer's Theorem is true under the weaker hypothesis that the Dirichlet density of SK - SL is zero. . . . f Proof of PropOSition: (l) HIKH is generated by all p where f is the degree of each prime divisor of p in K Therefore p E HIKH if and only if ;p splits completely in K . Thus if HIKH = “IL“ then SK = SL . so K = L follows by letting k = Q in Bauer's Theorem. (2) If K T L , then by Bauer's Theorem there exists rational prime p such that p E S and p $ SL . For T , K a prime divisor of p in K , there exists an integral ideal m in K with (T,fl) = 1 such that 36 f 1 mod HK . Thus HToflHK = pHWHK E HHKH . But p , T'3 f > 1 , is the minimal power of p which could occur as a factor of an integer in HHLH since L/Q is normal. Thus K D L . Exchanging the roles of K and L yields the proposition. To prove “I” = ”H“ then, it suffices to prove pf E ”H” for every rational prime p . This observation leads trivially to one class of fields for which “I“ = “H” , namely Theorem I: If (h,n) = l , then “I“ = ”H“ whether or not K/Q is normal. (h denotes the class number of K .) Proof: Let p denote an arbitrary positive rational prime and T a prime divisor of p in K of degree f . Since (h,n) = l , there exist positive integers x and y such hx hx fhx f that hx - ny = 1. Then “2—E5'6 H and H-ELE;M = nfy = P (p) (p) p completing the proof. This proof, however, sheds no light on the general problem. ”I“ = “H” means that for every rational prime p : there exists in K a prime divisor T of p , a principal ideal (B) , and an ideal u , such that 37 T = (B)u where HUHK = l and hence HTHK = “(B)H}(= Pf Such ideals u will be called unitary ideals, that is Definition: An ideal u in a finite extension K/Q for which HuHK = 1 is called a unitary ideal Since the absolute norm is multiplicative, the unitary ideals form a subgroup of the group of fractional ideals IK which will be denoted by UK . The subgroup UK/HK of the class group CK w111 be denoted by uK . subscripts will be omitted when the meaning is clear. The problem then can be reformulated as: (1) For which finite galois extensions K/Q does IK = HKUK ? or (2) For which finite galois extensions K/Q does CK = UK ? I will consider formulation (2) since it is a problem involving finite, rather than infinite, groups. Now U is an infinite abelian group generated by unitary ideals of the form T-10(T) where T is a prime divisor of p in K . Thus, Since prime ideals are equidistributed among all ideal classes, u is a finite abelian group generated by unitary ideal classes of the 38 form c-lo(c) where c ranges over 0 and G ranges over G . The converse of Theorem 1 is true for quadratic fields K , for primes splitting completely in K factor as (p) = T T so that Tan a (T—l)2 mod H . Thus for 12 12 1 -1 -l . . . every c 6 6.. c 0(c) = (c )2 implying u is generated by the squares of elements of 6.. But A2 = A only in abelian groups A of odd order, hence u = c. only when h is odd. Sadly, however, the converse is false as evidenced in the following interesting 3 —— . . . Example 1. K = Q(~/ll, w) , w a primitive cube root of unity. If k/Q is a non-normal cubic extension and k is the normal closure of k , then h(k) is either h2(k) or h2(k)/3 (cf. [14]). Since h(Q({/Il)) = 2 , we have h(K) = 4 . Let 0 denote an automorphism in G(K/Q) = S3 of order 3 . Then as C. is C4 (cyclic group of order 4 ) or V4 (Klein four group), 0 either fixes every element in c. or permutes the three non-identity elements. The latter alternative insures that u = c. as the map c 4 c-lo(c) is an isomorphism. The impossibility of the former is 39 guaranteed by Lemma B: Let K/k be galois of degree m , (p,m) = l . Then the p-class group of k coincides with the p-class group of K iff G(K/k) fixes the p—class group of K elementwise. For a proof of Lemma B, cf. [25]. In this example let K = Q(i/Il,w) , k = Q(w) , p:= 2 and hence |G(K/k)] = 3 . Since h(Q(w)) = 1 , the 2-class group of k cannot coincide with that of K . To reiterate we have an example of a galois field K where (h,n) = 2 , yet ”I” = ”H“ . Remark: It is worth noting that if K/Q is a normal extension of degree n and (hK,n) = 1 , then at least one prime divisor of n must divide [Aut C] . Proof: If not, then every 0 E G(K/Q) must fix all elements in c. so that u = 1 contradicting Theorem 1. This means, for example, that there are no normal extensions K/Q of degree p with hK = q if p X (q-l) . 40 Examination of class number tables shows that normal extensions where (h,n) = 1 occur far less frequently than those where (h,n) # 1. This seems attributable to the limited number of ways G(K/Q) can be embedded in Aut G for (h,n) = l to be true. 3. The notation remains in effect for the remainder of this chapter: K/Q is a finite normal extension of degree n with galois group G, ideal class group C, class number h, unitary ideal group u, genus field GSF(K), Hilbert class field HCF(K), and central class field ZCF(K). We first show how our problem of when HIKH = HHKH fits into the general setting of Chapter 1. Recall the Artin diagram HCF(K) — 1 I I ZCF(K) —- [0.1“] GSl!‘(K) —-—— 0'0 1" I I K ————- C I I Q --—- P Lemma C: u = [a,P]. Thus in the galois correspondence, the unitary group and the central class field correspond! 41 PROOF: By the Artin Reciprocity Theorem, C. is canonically isomorphic to G(HCF(K)/K) under the map c F—> (EQEéELAK) where T is any prime ideal in c since (flgEfileK> = <§Q§é§L15> ¢=é m a 8 mod H for any integral ideals fl and 8. Since F/C E G, for any 0 E G we have 0 = yc for some Y E F, Y cut back to K is O. But any prime ideal T in K so that C-lO(c) c-1(yc) h_> C-1 _ , - -1 _ _ C lYKHCFgQéK>Y l = C WY 1 completing the proof. Thus the following statements are equivalent: NH” (2) UH = I (1) IIIII (3) u = c (4) 6:001“ =[o.1‘] (5) K = GSF(K) = ZCF(K). 42 4. Sufficient Conditions In this section II give some sufficient conditions for u = Cw By Theorem I, (h,n) = l is always a sufficient condition.~ When F is a semi-direct product of G and a, we show F' n G): u and hence, in this situation, the condition GSF(K) = K is also sufficient. Lemma D: If F is a semi-direct product of G and c. then I" no: u. PRQQE: It suffices to show F' n c): u. If (O,C) denotes an arbitrary element in T. a semi-direct product of G and Cu then multiplication is defined by (O,c)(T.d) = (OToT(C)d) where T represents both an element of G and its image in Aut c. Any element in P' n c. then has the form x = (O,c)(T,d)(O-l,O-l(c-l))(T-l,T-l(d-1)) where O and T commute. Thus (ow.w(c)d>(o'l.o‘l)(T‘l.w‘l)(T‘1.T‘1 K = H Kj where either a. (l) K. : Q(g a.) for any prime power p.3 3 Pj J J or (2) Kj is a real field of degree 2a over Q which has exactly two ramified primes ql, q2 such that the subfield of the cyclotomic field of degree e(q2) over Q where only q2 ramifies is imaginary and Ki n Kj = Q for all i,j. PROOF: Contemplation 45 Examples of fields of type (2) are Q(V/pq), p = 2 or p a 3 mod 4, q a 3 mod 4 and Q(V/p, V/q), p = 2 or p a 1 mod 4, q a 3 mod 4, where p > 0, q > O in both cases. Example 3. Q(Qm) = H Q(g ) where m = H pa. pIm p” pIm Example 4. Quadratic fields K/Q for which GSF(K) = K are (l) K=Q(\/p). palmod4 (2) K = Q(v/pq), p a q a 3 mod 4. p > 0. q > O (3) K = Q(v/ZP), p a 3 mod 4, p > O. The construction in (2), for instance, is L1 = Q(v/-p) since p E 3 mod 4, so M1 = Q(v/-q) Q(V/qu v/TP) / ] \ Q(x/ Pq) Q(J -q) Q(\/ 19) \I/ Thus L = Q(V’pq, v/—p) but the infinite primes of K ramify in L/K, and their inertia field is Q(\/pq). Since u = c.=€> (h,n) = 1 for quadratic fields, we remark that the fields of (l), (2), and (3) are precisely the quadratic fields with odd class number. 46 It is clear, then, how to construct abelian fields K/Q for which GSF(K) = K. However given an arbitrary abelian field L/Q, much computation may be required to determine whether or not GSF(L) = L. The non-abelian case seems even more intractible. One obvious criterion which follows from the construction of genus fields for non-abelian fields (Theorem 2, Chapter 2) is Theorem V: Suppose K/Q is non-abelian and KO/Q is the maximal abelian subfield of K/Q. Then GSF K = K 4:; GSF K ) = K and e! = e.(K /Q) ( ) ( O O 3 J O for all ramified primes [pj}:_l of K where e! is the ramification index of pj in the maximal abelian subfield of K /Q . ’“j pj The application of this criterion can, again, lead to extensive computation. 9 9 Example 5: K1 — Q( V/s. g9). K2 — Q( V/7. g9). [K1:Q] = [K2:Q] = 54. Now KO = Q(gg) in both cases and GSF(KO) = KO. By the formula for g(K) of Example 5, Chapter 2, we see g(Kl) = (9,4) = 1 and g(KZ) = (9,6) = 3. Thus GSF(Kl) = Kl while 'i ) since e' = 3 and c7 7 Q(g7 +‘i ) is the subfield of Q(g7) of degree 3 over Q. 7 GSF(KZ) = K2(g7 + 47 Salvaging what we can, we state the sometimes useful Corollary: Suppose K/Q is non-abelian and KO/Q is its maximal abelian subfield. Then GSF(K) = K 4:? GSF(KO) = K0 and every prime ramifying in K either ramifies totally or remains prime in K . O PROOF: In both cases, for any ramified prime p, Kfi/Qp = K/Q so e; = ep(KO/o). Example 6. K Q(Pv’a, gp), p odd prime, a square-free and odd. with (a,p) = l. GSF(K) = K é=§ every prime factor of a is a primitive root modulo p. The case a = 11, p = 3 is Example 2. It appears difficult to determine when GSF(K) = ZCF(K) for an arbitrary normal extension K/Q. But we do note that [ZCF(K):GSF(K)] is divisible by only primes dividing n, for Lemma E: Let c be any ideal class in c. If (|c|, n) = 1, then c 6 u. PROOF: Let T be any prime divisor of p of degree 1 over Q in c. Since (]c], n) = 1, there exist positive integers |c‘x x and y such that Ic]x - ny = -1. Then T e 2L———9 mod H and (pi, 48 fl - O lflL—-_‘H = p]c]x+l ny = 1. So c E u completing the proof. Y . (p) We now turn to p—extensions and Show that (h,p) = l is a necessary as well as sufficient condition. Theorem VI: If K/Q has degree pa, then U=G<—_—> (hop) =1 2399:: Only (=€fl need be proved. Suppose p]h. Let HCFp(K) denote the p-class field of K, that is the field corresponding to the Sylow p-subgroup of G. HCFP(K)/Q is then normal and we let Pp = G(HCFp(K)/Q). Thus we have the Artin diagram HCF K 1 . p( ) K = GSF(K) = ZCF(K) -——— c.= [c.rp] = 6.0 T; Q Pp But if c.= [c.r]. the descending central series of Pp must break off at a. contradicting the nilpotence of pp. Thus th and the theorem is proved. Frohlich [ 7] has determined those abelian extensions K/Q of degree pa which have (h,p) = 1. Though Frohlich's 49 theorem is expressed in a way I cannot completely interpret, the gist of the theorem seems to be: If K/Q is abelian of degree pa, p odd, then (h,p) = 1 if and only if (1) K has exactly one ramified prime. (2) K = K1 K2 where each Ki has exactly one ramified prime one of which remains prime in the other extension. (3) K = KleK3 where each Ki has exactly one ramified prime and ( ?) Every extension K/Q with four or more ramified primes has (h.P) > 1- Conditions (1) and (2) follow from Lemma F: Suppose K/k is normal with exactly one ramified prime. If p]h(K), then p|h(k). 2399:: Let T be a prime divisor of the ramified K HCFP(K) prime of k in HCFp(K) and let T(T) \ i be its inertia group. Then since G(HCFp(K)/k) is a p-group, T(T) is contained in a maximal normal subgroup N of G(HCFp(K)/k) of index p. It is easy to see that the inertia groups of the other prime divisors of the ramified prime are also contained 50 in N. Let F be the intermediate field of HCFp(K)/k corresponding to N. Then F is an abelian, unramified extension of degree p over k, and the theorem follows. In (1), if p]h(K), then p|h(Q) = 1, a contradiction. In (2), suppose q1 ramifies in K, and remains prime in K2. Then Kl/KZ has exactly one ramified prime, so if p|h(Kl), then p|h(K2) contradicting (1). For (3) I have been unable to construct an example. I suspect the condition is vacuous since if K = K1K2K3 then K contains a subfield L such that more than one prime ramifies in L and K/L is not cyclic. Thus g + l for all primes in L contradicting the necessary condition that 9 = l for the ramified primes in order for the method of (l) and (2) to apply. Since the direct product of nilpotent groups is nilpotent, one attempts to extend Theorem VI to the compositum of O'.. l p-extensions. However if K = HKi and [Ki : Q] = pi , then HCF(K) D nHCF(Ki) where equality seldom obtains. Thus a general necessary and sufficient condition for an arbitrary extension K/Q seems hopeless. Summarizing those fields for which a necessary and sufficient condition does exist, we note for 51 p-extensions: u c.<:=%> (h,p) = l cyclic extensions: u G <=> GSF(K) extensions where r is semi-direct product of G and Ca u G €==€> GSF(K) CHAPTER IV THE BURGESS PROBLEM In this chapter we examine the problem, now almost forgotten, which prompted the investigation of the genus field and the central class field. Suppose r(x) isa polynomial with rational integral. coefficients. The value group of r(x), Vr' is the multiplicative group generated by the non-zero values of r(x) as x ranges over the integers. There are many unsolved problems concerning value groups of polynomials. Two of these which were posed at the 1969 AMS Number Theory Institute at Stony Brook, New'York are: Problem 1: (Kenneth Stolarsky) If r(x) = x4+x3+x2+x+l does p 6 Vr if p E 1 mod 10? Prgplem 2: (D. A. Burgess) For any polynomial r(x) with rational integral coefficients, does Vr consist of all rational numbers not excluded by obvious algebraic conditions? We show that the answer to Problem 2, is a mild-to-emphatic "no" depending on one's definition of "obvious". We then 52 53 indicate a more reasonable problem of which Problem 1 is a special case. For simplicity, let r(x) E Z[x] be armnxu: irreducible polynomial over Q and let K denote the splitting field of r(x). Then K = Q(e) where e is a primitive element n for K and r(x) = iHl(x-Oi(e)) where 01 = l, 02, ..., on are the elements of the galois group G(K/Q). For any rational n integer a, r(a) = H (a—Oi(e)) is within a Sign the absolute i=1 norm of the principal ideal (a-e). Thus Vr is a subgroup of HHK“ and HIKH. Since HIKH is generated by the rational integers ipf where f is the degree of any prime divisor of p in K over Q, it is clear that Vr t Q because not every prime splits completely in K. Suppose we ask the more plausible question: Does p 6 Vr if p splits completely in K? We see that this, too, is clearly impossible unless HIKH = ”HRH or CK = uK, which as we saw in Chapter 3 occurs very infrequently. Hence we modify Problem 2 and pose the more reasonable Problgm 3: Suppose r(x) is a monic irreducible polynomial with rational integral coefficients and splitting field K. If GSF(K) = ZCF(K) = K, does Vr contain all primes p splitting completely in K or, stronger, does Vr = HIKH? 54 Since Q(QS) is the splitting field for r(x) = x4 + x3 + x2 + x + l and primes p splitting completely in Q(QS) are precisely those p a 1 mod 10, we see that Problem 1 is indeed a special case of Problem 3. As a first case we consider quadratic polynomials r(x) = x2 - m so that K = Q(v/m). Vr can contain all primes splitting completely in K only if h(K) is odd. For some of those fields, the following ad hoc technique can be used; though it cannot be generalized to fields with degree greater than 2. Example: r(x) = x2 - 21 so K = Q(v/Zl) h(K) = 1 so GSF(K) = ZCF(K) = K. Only 3 and 7 2 2 ramify in K and -3 =';§—:—;l' and 7 = 13—2—gl . 5 - 21 5 — 21 5 and 17 split completely in K and -5 = 42 - 21, -17 = 22 - 21. 2, 11, 13, and 19 remain prime in K. Thus for every prime p, Ip] < 21 splitting completely in K, either i p belongs to Vr' Let p1,p2,...,pn,... denote the primes which split completely (or ramify) in K. Then p1 = 3, p2 = 5, p3 = 7, p4 = 17, etc. Suppose p1,...,pn_l belong the Vr for n 2_5. Then Since pn splits completely, the congruence x2 e 21 mod pn p -1 has a solution xO with Ixo] g n2 . Thus 55 p -l (I; )2 - 21 Zapn for some positive integer a or 2 - —1 Dn 21>n __i a 4p p '2 n n or P 75-8- But every prime divisor of a splits completely or ramifies in K and, by the induction hypothesis, xi-Zl belongs to V . Thus p = -——-—‘ also belongs to V r n a r completing the proof. A similar technique is valid for polynomials of the form r(x) = x2 + ax + b. We remark that Vr contains all primes splitting completely in the splitting fields of r(x) = x2 + l and r(x) = x2 + x + l which indicates the origin of Problem 1. Numerical evidence supports the conjecture that all primes splitting completely in a quadratic field K of odd class number belong to Vr where r(x) is any quadratic polynomial whose splitting field is K. In fact I conjecture that for r(x) = x2 - m, each prime p splitting completely 2 in K satisfies tp = xz-m for some integers x and y. y-m CHAPTER V IDEAL CLASS GROUPS A classical problem of algebraic number theory is the determination of all abelian groups which occur as ideal class groups of algebraic number fields. While not attempting to solve this general problem, I can show that every abelian group occurs as a subgroup of infinitely many abelian, non- abelian, and non-normal algebraic number fields, by showing every abelian group 4' is isomorphic to G(GSF(K)/K) for infinitely many number fields K. This result contains recent ones of Madan [20], [21] and Ishida [15]. I begin by proving Lemma: For every finite abelian p-group 9 of exponent pe, p prime, there exist infinitely many abelian number fields L/Q of degree pe whose ideal class group contains a subgroup isomorphic to 9. n o PROOF: Let 9 = H Pi be the decomposition of 9 into a i=1 ei product of cyclic subgroups with 'Pi‘ = p , ei‘g e. By 56 57 Dirichlet's Theorem on the infinitude of primes in an arithmetic progression, there exist infinitely many primes qi, qi # q. 3 e. satisfying qi e 1 mod p 1, eO = e, i = O,l,2,...,n. Let Ki/Q denote the unique cyclic subfield of Q(gq ) of degree e i ' n p 1, i=O,l,...,n. Then K = H Ki is a field for which i=0 GSF(K) = K. We Show how to determine a subfield L/Q of degree pe in which all the qi, i=O,l,...,n ramify with e . e(qi) = p 1. Then GSF(L) = K and G(GSF(L)/L) e 0. Let now G(Ki/Q) = <0i>. i=o.1. Let Mi denote the cyclic -1 subfield of K0 Kl/Q with G(Ml/Q) = (0001 >. Then it follows easily that deg Ml/Q = pe and ///¥OK1 K M]\K KOKl = MlK = MlKl . O l////1 Q Now q0 is unramified in Kl/Q and hence in K1M1/QM1 = KOKl/Ml. Similarly ql is unramified in MlKo/M1 = KOKl/Ml' thus KOKl/Ml is unramified. Applying this construction to M1 and K2, we obtain a field Mz/Q of degree pe where KOKle/M2 is unramified. Continuing in this manner, we Obtain a sequence of fields M3, M4, ...,Mn such that deg Mj/Q = pe and KOKl...Kj/Mj is unramified for j=3,...,n. 58 Then L = n is the desired field for which deg L/Q = pe n GSF(L) = n Ki = K and hence G(GSF(L)/L) e 9 completing i=1 the proof. From the Lemma we now obtain Theorem 1: For every finite abelian group d’ of order a and exponent m, there exist infinitely many abelian number fields of degree m whose ideal class group contains a subgroup isomorphic to d. 9i be the decomposition n o PROOF: Let m = n pil and a’= ““" = 1 1 i ll :35 of 4’ into direct product of its Sylow p-subgroups. For each 9i, we obtain, by the Lemma, infinitely many abelian fields Li/Q of degree p:i whose ideal class group has 0i as a subgroup. Then, as (deg Li/Q, deg LjflQ) = l for all i,j, it follows that for any set of fields L1,L2,...,Ln so obtained, n L = H Li is an abelian field of degree m over Q whose i=1 ideal class group contains a subgroup isomorphic to 0. thereby completing the proof. The non-abelian and non-normal cases can be proved by simply reconsidering two examples from Chapter 2. Specifically, we have, 59 Theorem 2: For every finite abelian group d’ of order a and exponent m, there exist infinitely many non-abelian number fields of degree m m (m) and non-normal number fields of degree m whose ideal class group contains a subgroup isomorphic to d. e PROOF: Let [prs} denote the invariants of d. Dirichlet's Theorem again insures that there are infinitely many primes e I I S qrs satisfying (qr ,m) — l and qrs = 1 mod pr , es+l es 1 . qrsfi mod pr for every pr For each set {quI so determined, let K = Q(my/ H q ) and then K = K(g ) Clearly r's rs m K/Q is non-normal and K/Q is non-abelian of degrees m and m T (m) respectively. Then as (m,qu-l) = prs, it follows from Examples 5 and 6 of Chapter 2 that G(GSF (K) /K) e- G(GSF(K)/K) e- a completing the proof. CHAPTER VI CONSTRUCTION OF HILBERT CLASS FIELDS In one of his typical understatements Serge Lang [18] remarks, "It becomes a problem to exhibit the Hilbert class field explicitly".I will examine the tip of this iceberg in this chapter. The algebraic number fields K for which HCF(K) is most easily determined are those where HCF(K) = GSF(K). After considering several classes of such fields, I conclude by examining the simplest class of fields for which GSF(K) ¥ HCF(K). Specifically I give a method to construct an unramified extension of a quadratic number field of degree 3 or 4. Thus if the exponent of the ideal class group of a quadratic number field divides 12, its Hilbert class field can be constructed. §l Quadratic Fields: Q(v/m) From the examples of genus fields of quadratic number fields computed earlier, the general method is apparent. Thus the known cases of quadratic fields K for which GSF(K) = HCF(K) are merely listed in tabular form. 60 61 Table - Hilbert Class Fields for Certain Quadratic Fields Q(V/m) Q(./-pqr../p*.v€l*o\/r*) 3 21./p../-q../-r../-s1 0(3) Q(x/Pox/qH/rIs/‘S) 0(1) Q<./p../q../-r../-s) 0(1) —2pqr p.q.r¥2 _pqrs pal mod 4,q§rE-SE3 mOd 4 -pqu peqsral mod 4,sa3‘mod 4 m Conditions h HCF(Qv/m=GSF(Q(v/m)) fiiineggg —p pal mod 4 2 Q(V/+p,i) 3 -2p pal mod 4 2 Q(v/p,V/-2) 2 -2p pe-3 mod 4 2 06/ -p.\/ 2) 2 -pq p53 mod 4. gal mod 4 2 QM -p,\/q) ll Pq pal mod 4 2 Q(v/PQV/Q) 73 pqr pal mod 4, r,qél mod 4 2 Q(\/ pn/ qr) 3O 2pq qu53 mod 4 2 Q(v/PQov/Z) 15 —pq paqs3 mod 4 4 o(./ -p../ -q.i) ‘ 7 -pq psqsl mod 4 4 Q(\/ pM/q, i) l -2pq p.q¥2 4 QM ‘Zqu\/P*o\/Q*) 7 -pqr pqrE3 mod 4 4 22M -pqrn/p*n/q*) 3 pqr p.q.r#2 4 o(\/ Pa\/q0\/r) 11 2pqr pel mod 4, qarEB mod 4 4 Q(v/p,v/qr,v/2) 3 2pqr peqel mod 4. re3 mod 4 4 Q(\/ p,\/q,\/2r) l pqrs peqal mod 4, r5553 mod 4 4 Q(V/r,s,v/p,v/q) O -pqr pal mod 4. qEr—='3 mod 4 8 Q(\/ p.¢pq.¢r.i) 6 8 8 8 8 -pqu peqel mod 4, r5353 mod 4 p,q,r,s represent distinct primes; 2 is possible unless in- 1 mod 4 dicated otherwise. p if p P* = -p if p _ 3 mod 4. The numbers in parentheses in the last column indicate the number of known quadratic fields with ‘m] < 500 satisfying the given conditions. 62 I do not know whether there exist any real quadratic fields K with h(K) = 2t. t 2_3, and GSF(K) = HCF(K). The problem is unsolved for arbitrary t. Chowla [ 3] proved in 1934 that there are only a finite number of imaginary quadratic fields K where HCF(K) = GSF(K). An old conjecture is that there are 65 such fields which are, in addition to those indicated in Table l: K Q(VLm), h(K) = 4: m = 555. 595. 715. 795. 1435 K Q(v/-m), h(K) = 8: m = 1155. 1365, 1995. 3003. 3315. Selfridge showed that these are the only such fields for m < 2-3-5-7~11-l3-44,838. For a complete account of this problem see Grosswald [ll]. §2 Compositums of Quadratic Fields Let K , K 1 2,...,Km be quadratic extensions of Q. Suppose these fields are indgpendent, that is the degree of m K = H K. is 2m over Q. Then the galois group of K/Q . 1 i=1 is an elementary abelian 2-group and there are t = 2m-l different quadratic subfields of K denoted by K ,K ,...,K . l 2 t Let hi and ti denote the class number and unit group of K. Then it is known (cf.[l7]) that: 1 t t H ='—;[E: Hei] Hhi 2 i=1 i=1 63 m(2m-l-1) if K is real (m-l)(2m-2-l) + 2m—l—l if K is imaginary where v = If GSF(Ki) = HCF(Ki) for all i and GSF(K) = HCF(K) then of course, the Hilbert class field of K is determined. When K is imaginary, this occurs only a few times. For example if K is imaginary biquadratic, a necessary condition that GSF(K) = HCF(K) is that exactly two primes ramify in K as I shall now show. K = Q(V/-ml, V/-m2) has three quadratic subfields Kl = Q(v/-ml), K2 = Q(v/-m2), and K3 = Q(V/m1m2)° The ramified primes of K are the prime divisors of mlmz, say pl,p2,...,pk, and each pi ramifies in two of the three quadratic subfields. If GSF(Ki) = HCF(Ki), then r.-l 2 J hj = (5 where rj is the number of primes ramifying in Ki and 5a is as defined in the genus-number formula. It 3 3 can be shown that [E: u 6.] = H 5 so that . i , w 1=1 i=1 3 r -l r -l r -l I H 5 )2 1 2 2 2 3 4 i=1 ° rl+r2+r3- 2k-4 H 2-—— 3 = 2 = 2 . However 2( H 6”) i=1 g(K) = Zk-z. so if GSF(K) = HCF(K). then 22k"4 = 2k-2 implying k = 2. 64 Example 1: K = Q(v’-2,V/—6). Here Kl Q(V/-2), =Q(\/-6), K3=Q(\/3) and hl=h3=l, h2=2, H=2. GSF(K) = HCF(K) = Q(\/ 2,\/3,i) I =-@(\/ 2 //K ‘ \m/ GSF(KZ) =@(\/- -.3/2) =9(./— -2) K3 =Q(./3) K2 =9(./- -6) \I/ Since there are only 9 imaginary quadratic fields with h = 1 and either 47 or 48 with h = 2, the number of compositums K of imaginary quadratic fields with GSF(K) = HCF(K) can be completely determined. More examples of compositums K of real quadratic fields for which GSF(K) = HCF(K) exist. I have not attempted to completely Solve this problem, though I suspect only the cases H = 2 and H = 4 are possible. Two examples will illustrate the situation. Example 2: K = Q(v/BOA/35). Here Kl = Q(\/30), K2 = Q((/35). K3 = Q(\/42) and 65 GSF(K) = HCF(K) = Q(\/2. ./3. ./5. \/7) I GSF(K1)=Q((/5.\/6) GSF(KZ) =.0\/2 \/2/l) Q(x/30, \/35) GSF(K/3) =Q(\/5. (/7) I Kl = Q(/30) = (IN/J42) Q Example 3: K = Q(v/Zov/15ov/21)- K3 = {ah/35) The seven quadratic subfields are Q(v/Z). Q(v/lS), Ink/21). o<¢30). o(./35). egg/42). rah/70). H = 4 and a diagram like that of Example 2 shows HCF(K) = GSF(K) = 0(./2.¢3../5../7). If K/Q is a cyclic extension of degree p with h(K) = p, then HCF(K)/Q is abelian since all groups of order p2 are abelian. Thus HCF(K) = GSF(K) and the Hilbert class field of K is determined. There are only eight cyclic cubic fields of class number 3 with discriminant A < 20,000, two each with 66 Example 4: K = Q(e), e3 - 219 - 35 = o. 2 1 [\(K) = 63 and HCF(K) = GSF(K) = K(g7 + ?)where 7 th C is a primitive 7 root of unity. 7 §4 Pure Cubic Fields: K = Q(3V/a). In Example 6 of Chapter 2, the genus field of the pure . n . field K = Q( v/a) (n,a) = l, a # i l is square-free and odd was determined. In that case g(K) = H (n,p-l) so for pIa n = 3, HCF(K) = GSF(K) if h(K) = 3t where t is the number of primes p e 1 mod 3 dividing a. Known examples (with small discriminants) are: Example 5: K = Q(3\/a), h(K) = 3, a = 7, 13, 19, 21. 35. 37. HCF(K) = GSF(K) = K(9) where e is a primitive element for the subfield of Q(Qp)/Q of degree 3 where p]a and p a 1 mod 3. Example 6: K = Q(3v/9l), h(K) 9, GSF(K) = HCF(K) = K(61,92) (6i determined as in Example 5). §5 Quadratic fields K = Q(v/m) where 3|h. The genus field is the "easy part" of the Hilbert class field of an algebraic number field K. To complete the construction of HCF(K) it is necessary to construct abelian unramified extensions of K. In general, this is very difficult 67 so I will focus on two cases: constructing unramified extensions of degree 3 and 4 of quadratic fields. Let K = Q(V/m) with 3]h(K). There exists, then, a field L such that L/K is unramified of degree 3. Suppose, in addition, that L/Q is normal (which occurs if 3Hh for example). Since the galois group of L/Q is S L is the splitting 3. field for a cubic polynomial f(x) = x3 - ax - b whose dis- criminant A = mkz. We seek to determine a and b so that K(e)/K is unramified where 63 - a9 — b = 0. Now (*) A = 4a3 - 27b2 = mk2 9tl if 31m Set a = 3t, b = St, k = 3tl if 3|m Then (*) becomes (**) 4t - s2 = .m, 2 Suppose first that m < 0, set m = -m. Then t is a norm from Q(\/ 3m) (or Q \/ g!) . SO taking t = i l. we can determine 3 and t by finding the fundamental unit 3 of Q(x/3m) (or OMB-B). that is e = S - “23m 1 {m s- 3 l . . ). Now K(e)/K can ramify only at primes (or 2 dividing 3 in K. Case 1: t = 1, 31m, (the most frequent case). Since 3Xm, K(e)/K is unramified 4:; 3 is unramified in Q(e)/Q. But in this case 68 f(x) = x3 - 3x - s and A = 27(4-52) so 3 is unramified in Q(e), 93 - 3e - s = O 4:? s a i 2 mod 27. Example 7: K = Q(V/-23), h = 3. 25 - 30/69 GSF (K) = K and e = 2 So HCF(K) = Q(V/’23o e) where e3 - 38 - 25 = 0. Example 8: K = Q(V/-38), h = 6. 2050 - l9gg/ll4 GSF(K) = Q(V/l9, V/-2) and c 2 So HCF(K) = Q(V/l9, v/‘Zv e) where e3 - 363- 2050 = Case 2: 't = i 1, 3|nh K(e)/K will be unramified 4:; (3) = pfpz where pl and 92 are prime ideals in Q(e) of degree 1 over In this case f(x) = x3 i 3x i S. To check the rami- fication of 3 in Q(e), 93 i 39 i s = 0, we apply Newton's polygon (see Weiss [23]). (}.w) Newton's polygon for ' (3,5) (2 1) x3 i 3x i s. I Q. B] Newton's polygon (3) = pfpz in Q(e) if s a 0 mod 9. 69 Example 9: k = {QM-231). h = 12. 9+ 77 GSF(K) = Q(\/-3, \/-7. \/—11) and c = 2 so HCF(K) = Q(\/-3, \/-7. \/-11. e) where 3 8 - 36- 9 = 0. Case 3: t = -1. Then 3]m for in any quadratic field Q(V/d) if d is divisible by a prime p a 3 mod 4, then N(e) = +1. Again K(e)/K is unramified <=? (3) = pipz in Q(e). Now f(x) = x3 + 3x + S so applying Newton's polygon directly for s i 0 mod 9 is futile. However f(x+l) = x3 + 3x2 + 6x + (s+4) and f(x+2) = x3 + 6x2 + 15x + (s+l4) so if s+4 sOmod9 or s+l4 es+5 '=‘0mod9, (3) =p21p2 in 0(6). 93 + 39 + s = 0 so that K(e)/K is unramified. Example lg) K = Q(‘/ -87), h = 6 GSF(K) = Q(\/29,\/—3) and e = 5+229 so HCF(K) = 0(./29../-3.e) where ‘ e3 + 39 + 5 = 0 Summarizing these cases is Egopogition: Let K = Q(v/-m) be an imaginary quadratic field with class number h. For Kl = Q(v/3m), let 6 denote the fundamental unit of K1, t the norm of e, and s the trace of S. Then 3]h if 7O (1) t = l s e i 2 mod 27 (2) t = i l s e 0 mod 9 (3) t = -l s a i 4 mod 9. K(e)/K is unramified of degree 3 where 93-3te-st=0. Unfortunately all cases are not covered by the Proposition. Example 11: K = Q(\/687), h = 6 15 + .1229 GSF(K) = Q(¢-3.¢229) and e = 3 Unhappily x3 + 3x + 15 is Eisenstein, so 3 ramifies totally in Q(9)/Q where Q3 + 39 + 15 = 0. If, however, we can find s,t so that s - 0 mod 9, t E 0 mod 3, Newton's polygon can then be applied to x3 - 3tx - st as in Example 7. Since 27 is the first odd multiple of 9 greater than 15, we consider 6 + 6. Now 272 - 229 5 where 93 - 3-1256 - 27:125 = 0 so that ”6 + a” = = 125. S0 (3) = 1.2102 in gum/0 HCF(K) = Q(./-3, \/229. e). A similar analysis can be applied to any imaginary quadratic field not satisfying the conditions of the Proposition. For real quadratic fields, an analogous strategy can be employed. Example 12.. K = 05/ 79). h = 3. For real quadratic fields, (**) becomes m _ sz+3ml2 3 Since 3179, we mimic Case (1) of the Proposition by seeking integers s and t such that 2 2 t = 8 +237 1 and S2 4 4t mod 27. One solution is s = 2, t = 2134 so that HCF(K) = Q(V/79, e) where 93 - 3:21349 - 2-2134 = O. This method appears capable of generalization to the construction of an unramified extension of degree p over some quadratic fields by considering f(x) = xp - ax - b P with discriminant A = (-l)2)[(p-1)P’1ap-pphp'l]. This idea will not be pursued here. §6 Quadratic fields K = Q(V/m) where 4|h. Let K = Q(\/m) with 4]h(K) and GSF(K) #HCF(K). There exists then a field L such that L/K is unramified of degree 4. Suppose, in addition, that L/Q is normal (which occurs if 4Hh for example). Since the galois group G of L/Q is non-abelian of order 8, ‘6'] = 2. So g(K) 2,2 and by the discussion in Chapter 2 there exists a subfield of GSF(K)/Q of the form M = Q(V/a,v/b) where a = -l or a e 1 mod 4 and b = 2 or is a positive prime p a 1 mod 4. Thus h(Q(v/b)) is odd and hence ”IHQ(v/b) = ”HHQ(V/b) by 72 Theorem 1 of Chapter 3. By examining the various cases it can be shown that a belongs to HI”Q(V/b) and thus the equation a = x2 - by2 has a solution where x and y are rational. Consequently M(\/a)/M where a = x + be is unramified since it is clearly unramified at all prime divisors of p in M and since a = -l or a a 1 mod 4, it is also unramified at prime divisors of 2. Hence L = K(v/b, v/a) is an unramified extension of K = Q(V/m) of degree 4. Example 13: K = Q(¢-142) h = 4. a = -71' b 2,a=l+6\/2 Q(\/ -71, ¢ 2. (/ 1+6z7i') So HCF(K) Example 14: K = Q(v/l45), h = 4. a= 5. b= 29. a: 11+2/29 or a: 29. 10: 5. a: 7+2¢5. So HCF(K) = Q((/ 5. ./29. ¢ll+2./"2"‘9) = Q(\/ 5. ¢29. 7+2./'5') Example 15: (am-65). h = 8. Here Q(V/5,i) is a subfield of GSF(K) = Q(V/5,~/13,i) Thus a=-1,b=5, 0.=./5. so HCF(K) = 06/5. ./13. i. ./2 f5). fixamplggl6: K = Q(v/-89), h = 12. GSF(K) = Q(\/ 89,i) so extensions of degree 3 and 4 must be determined. For 4, a = -l b = 89 so a 73 is the fundamental unit of Q(V/89), 1000 + 106 1/ 89 a = 2 = 500 + 53/89 2 2 Now 3189, so 1 = S 2:7 l . So 5 is determined by the fundamental unit of Q(V/267). 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