0%! A PROBLEM OF SCHENZEL CONSERE‘HEG 'PRE‘NCIPAL DIVSSGRS EN ARETHMETiC PROGRESSWNS Thesis for the Degree 0? Ph. D. MfG-HEGAN STATE UHWERSITY CHARLES SEGHN PARRY 1970 LIBRARY Michigan State ‘, fl , as University This is to certify that the thesis entitled "ON A PROBLEM OF SCHINZEL CONCERNING PRINCIPAL DIVISORS IN ARITHMETIC PROGRESSIONS" presented by Charles John Parry has been accepted towards fulfillment of the requirements for Ph . D. degree inmmtj-CS (\ng [\ WMLQWM Major professor Date May 4, 1970 0-169 -=. W . ' amnmu av " one & snu5' i ' H I 300K BINDERY INC. ' I unsunv DINOERS , ABSTRACT ON A PROBLEM OF SCHINZEL CONCERNING PRINCIPAL DIVISORS IN ARITHMETIC PROGRESSIONS BY Charles J. Parry The following problem was proposed by A. Schinzel at the A. M. S. Number Theory Summer Institute held at Stony Brook in July 1969: "Let f(x) be a primitive polynomial and k an algebraicnumber field. Do there exist infinitely many integers x such that f(x) factors into principal ideals in k? (unknown even for f linear)." I have solved this problem in the affirmative when f is linear. My proof uses Frobenius and Artin symbols in certain extensions of the Hilbert class field of k. ON A PROBLEM OF SCHINZEL CONCERNING PRINCIPAL DIVISORS IN ARITHMETIC PROGRESSIONS BY Charles John Parry A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1970 ACKNOWLEDGMENTS I would like to express my thanks to my advisor Charles R. MacCluer for his helpful suggestions and for the many hours he has spent discussing this problem with me. Also I would like to thank Andrzej Schinzel for calling the problem to my attention and for the helpful comments which he has communicated to me. ii TABLE OF CONTENTS Page CHAPTER I INTRODUCTION. . . . . . . . . . . . . l 1. STATEMENT OF PROBLEM . . . . . . . . . . . l 2. SOME HILBERT THEORY. . . . . . . . . . . . 2 3. THE CEBOTAREV DENSITY THEOREM. . . . . . . 7 4. RESULT FROM CLASS FIELD THEORY . . . . . . 10 CHAPTER II PRELIMINARY RESULTS. . . . . . . . . 12 1. AN EXAMPLE . . . . . . . . . . . . . . . . 12 2. SPECIAL CASES. . . . . . . . . . . . . . . 12 CHAPTER III RESOLUTION OF THE LINEAR CASE . . . 18 REFERENCES . . . . . . . . . . . . . . . . . . . 26 iii CHAPTER I INTRODUCTION 1. STATEMENT OF PROBLEM The following problem was proposed by Andrzej Schinzel at the A. M. S. Number Theory Summer Institute held at Stony Brook, New York during July, 1969: Question I: "Let f(x) be a primitive polynomial and K an algebraic number field. Do there exist infinitely many integers x such that f(x) fac- torizes in K into principal ideals? (unknown even for f linear)." In this paper I shall prove the answer is yes when f is linear. It has been noted [1] for polynomials of higher degree that the following additional assumptions on f(x) are necessary: (i) The content of any factor of f(x) in K is principal (MacCluer). (ii) Each fixed divisor of f(x) is principal (Schinzel). In the linear case, i.e. when f(x) = mx + b, it seems reasonable to ask the slightly stronger Question II: Do there exist infinitely many primes of the form mx + b which have principal prime factors in K? Although the answer to question II can be seen to be no by an example, this question is worth examining more closely as it suggests an approach to the first question. First, however, I shall present some basic definitions and theorems of algebraic number theory and class field theory which are not readily available in the literature. 2. SOME HILBERT THEORY. Throughout this section let K be a finite galois extension of the number field k with galois group G of order n. Let R and S denote the rings of algebraic integers in k’ and K respectively. Suppose B is a prime of K. Definition A: Z($) = {Old 6 G, 0(fi) = B} is called the decomposition group of 3. Definition B: T($) = [olo(x) E x mod n for all x E S} is called the inertial group of m. The subfield I of K corresponding to T is called the inertial field of T. It is easy to verify that Z($) is a subgroup of G and that T($) is a normal subgroup of Z($). Furthermore suppose p = $ 0 k and p = (Tlfiz...$g)e in K where $1 = T. Then since G acts transitively on the primes $I°"$g it follows that the index (G=Z) = 9 and SO Z($) has order n/g. Definition C: The sequence of groups G a Z 2 T 2 l is called the (short) Hilbert sequence of 3 over k. The importance of the Hilbert sequence is due to the following: Result I: For each prime $ of K 2(s) / T(s) is naturally isomorphic to G(s/$ I R/p). the galois group of S/T over R/p. Result II: p is totally ramified over its inertial field I($). Moreover, $1 = 3 n I is unramified over k and (Tzl) = (K:I) e, the ramification index of T over k. For proofs of these results we refer the reader to Weiss [2]. Now suppose and D is unramified over k, so T($) = l 2(a) e G(S/$ | R/p). But R/p =GF(|InHk) and S/‘B=GF(H:>H]f) where Hka is the absolute norm of p. Thus G(S/® l R/p) is cyclic and generated by the map HDHk rm—L xF——)x . Hence we can choose a generator 0 of Z($) so that 0(x) a x mod T This unique element of Z($) is called f~§ for all x 6 S. the Frobenius Automorphism of 3 over k. The symbol [591” is called the Ergbeniug Symbol,of n over k. Remark I: factors of PROOF: Note that so that Hence The Frobenius automorphisms of the prime p are all conjugate under G. forTEGoXES -1X)npnk :quka E (T ) mod T Hka (x) E x mod (T11). 5A]: ”’ -1 — T [ 3 J T The conjugacy class to which the Frobenius symbols of the factors of p belong is called the Artin Symbol of p and is denoted by C 5&3 D . If G is abelian then the Artin Symbol becomes a unique element of G. Now assume that k C L C K and let P = n m L. Remark II: If P is of degree f over k then [ KZL ] = [ ng If :3 ‘13 J PROOF: Let L I—(‘é‘k‘ ] = O H Hf Then of(x) E x p k mod n for all x E S. But MPH: = HPUL. Remark III: If L/k is galois then Herve];- PROOF: Obvious. I now consider the Artin Symbol in the case that K is a cyclotomic extension of k, i.e. K = k(§) where Q is a primitive mth root of unity. In this case all elements of the galois group G(k(g)/k) can be obtained by a substi- tution of the form g r——->ga for some a with (a,m) = l. Remark IE: Suppose oa(§) = ((a,m) = 1, then ga is in G(k(g)/k) 1215.11.15) _ = <‘ p :> — Ga e Hka — a mod m. PROOF: Note C (x) E x mod p a for all integers x of k(§). In particular HpH oa = ta 2 g k mod p. a _ b . . However Q = Q mod p implies ga(1 - gb'a) 2 0 mod p and hence l - Qb-a E 0 mod p. Now if b-a ¥ 0 mod m then m—l . m = n (1 - g3) a 0 mod p i=1 contradicting that (p,m) = 1. Thus b - a E 0 mod m. Substituting Hka for b we get “ka E a mod m. From Result II we obtain some properties of inertial fields. First, Lemma I: If k C k’ C K and T’ group of P over k’, then G’ = G(K/k’) . T’ = G n T where is the inertial and PROOF: Clear from the definition. Corollary A: (Maximal Property) If n n k’ is unramified over k then k’ C I(n). PROOF: Let I’ be the inertial field of n over k’. Then G(K/I’) = T’= T 0 G’, hence I C I’. But n O I’ is unramified over k’ and hence over k. Thus I = I’ and k’ CI’ = I. Corollary B: If a prime p of k is unramified in k’, then it is unramified in the galois closure k7 of k’. .ggggg: Note that p is unramified in each conjugate field of k’ since it has a factorization there identical to that in k. If B is any factor of p in k7 then the inertial field I($) contains k’ and all its conjugates by Corollary A. Thus I = k7: 3. THE CEBOTAREV DENSITY THEOREM In this section I state the theorem.which is the key to most results of this paper. But first, Definition: Let H be a set of prime ideals of k. The limit am) = ii? 2‘ who”: 2 mun: p6“ pék (if it exists) is called the Dedekind density of H. Result: The set of primes p in k of degree greater than 1 over Q has Dedekind density 0. PROOF: As 3 4 1+ ‘1 “'1 l 1 S s (k=Q) Z #55 = 0(1) degree ”DHk PEQ p p>l Cebotarev Densitnyheorem: If G 6 G(K/k), then the Dedekind density of all primes p of k with (5612) = flow) is IRG(O)| / (G:l). (36(0) denotes the conjugacy class of o in G and IRG(0)| denotes the order of this class). Corollary: The set of primes n of K with [.Kéh J = 0 has Dedekind density l/(G:l)- Recall the Dedekind zeta function QK(s) of a number field K is defined to be the series cK 1 gK(s) = g<1 - l/HPHi) where the product is over all prime ideals of K. Now for any number field K, §K(s) can be shown by analytic continuation to have a simple pole at s = 1. Now —s - log gK(s) = log H <1 - HPHK ) P = - Z 109 (l - HPHIZS) P ‘ ' -s \‘ “ -'s =2HNK+LJ2WMJ P P j=2 _ 2 HPHK + 0(1). P Since Qk(s) has a simple pole at s = l we have lim (s-l) §k(s) = c for some c > O. sal Hence for s > 1, log QK(s) = - log (5—1) + 0(1) + and so as s 4 l , log §K(s) ~:-1og (3—1). This gives the important Result: For any two number fields K and L, as s 4 1+, 1NW§~2¢NN:~-Mgedy PEK PEL Now I prove 10 Lemma II: A finite extension K of the number field k is galois over k if and only if almost every prime P of k that has one linear factor in K splits completely in K. .PRQQF: If K/k is galois, then the condition follows easily from Kummer's Theorem. Conversely assume the con— dition holds. Let U be the set of primes of k which splits completely in K. Since a prime splits completely in K if and only if it splits completely in K, the galois closure of K, it follows easily that em) = 1/= -1 which obtains if and only if P 3 $3, $13 (mod 40). Thus film for instance, no prime of the form p = 40x + 3 has principal divisors in Q(/I6). 2. SPECIAL CASES Now question II is worthy of closer examination as it suggests an approach to the first question and is of some 13 interest in itself. Specifically I shall prove: Theorem I: Let k be a number field galois over Q, CF(k) the class field of k, and C a primitive m-th root of unity. If CF(k) n k(§) = k and if k n Q(§) = Q, then for each (a,m) = 1 there are infinitely many primes p E a(mod m) ’4” which split principally and completely in k. (I will say a rational prime p splits principally in k if each prime factor of p in k is principal in k.) PROOF: We have the following Artin diagram CFLkHQ) /////////// k(C)/// \ A prime p of k with Artin Symbol (pklélZE-j = Ua’ where Ua(§) = Qa, has absolute norm HnHk E a mod m. Thus if in addition p is linear over Q then Hka = p E a mod m. It now only remains to produce infinitely many such principal l4 primes p, i.e., with Artin symbol <_§EJ§lZ£-) = 1. But by hypothesis the galois group G(CF(k)(§)/k) a G(CF(k)/k)xG(k(§)/k). Thus by the Cebotarev density theorem l/h m(m) of the primes \ of k have C CF(k;(£)/k ) = l x Ga, where h is the class number of k. But this means (lg-Ellis) = 0a and (W) = 1. Since almost all primes of k are linear over Q, we need only consider such primes p of k. But this means Hka = p E a mod m. Also p principal and k/Q galois implies p splits principally in k. Thus at least l/h-m(m)-(k:Q) of the rational primes p split principally and completely in k and satisfy p a a mod m. Corollary I: Let k be a number field (not neces- sarily galois over Q) and A be the discriminant of k. Suppose (m,A) = 1, then there are infinitely many primes p E a (mod m) which split principally and completely in k. 2399:: Since (A,m) = 1, every prime divisor of m is unramified in k and hence unramified in the galois closure k of k. However, the primes which ramify in Q(§) are exactly the divisors of m and so Q(§) n k = Q. From this it follows that [CF(E) n i<§> E] = [(CF(E) n E(g)) n Q(€)=Q]- 15 Now because (A,m) = 1, no prime can ramify in the exten- sion (CF(k) fl k(§) fl Q(Q))/Q and so this extension is of degree 1, hence CF(k) fl k(§) = k. We can thus apply Theorem I to get infinitely many primes p E a mod m which split principally and completely in k and hence also split principally and completely in k. Remark: It is worth noting that there are always infinitely many positive rational primes p E 1 mod m (for any m) which split principally and completely in any number field k. PROOF: By the Cebotarev density theorem the set of primes which split completely in CF(k)(g) has positive density. Also under certain hypothesis question II is true for all modulii m. I now prove Theorem_II: Suppose the number field k is galois over the rational numbers Q and has class number h. Let n = (k:Q) and take m > 1 and a to be any integers with (a,m) = 1. If (n,h) = 1, then there are infinitely many rational primes p with p E a mod m which factor into principal ideals in k. 16 23992: Let CF(k) be the Hilbert class field of k and let G and H denote the galois groups G(CF(k)/Q) and G(CF(k)/k) respectively. Then H has order h and is a normal subgroup of G. Also (GzH) = (k:Q) = n. Since (n,h) = l, the Schur—Zassenhaus Lemma [3] applies to give a subgroup A of G for which G is semi-direct product of A and H. Let L be the subfield of CF(k) with galois group G(CF(k)/L) = A. Note that CF(k) = kL and that k n L = Q. I now show that if a prime B of CF(k) has its Frobenius automorphism [.QElngQ 1 in A, then p = B O Q splits into principal prime ideals in k. We need only note that the restriction map 0! Mlk gives an isomorphism of G(CF(k)/L) and G(k/Q). Also [4119“: 15:21:: Thus if [.inngQ 1 is in A then p = B fltQ gains the same degree in both R and CF(k). Since k/Q is normal, p splits into principal prime ideals in k. Next we note that L D Q(Q) = Q where Q is an m-th root of unity. Suppose some rational prime q has ramifi- cation index e’ in L n Q(§). Then e’ divides (LzQ) = h. On the other hand e’ must divide the ramification index e of q in CF(k). But e divides n so e’ also divides n. Hence e’ = l and L 0 Q(Q) = Q. Therefore the substitution determined by Oa(C) = £3 (a,m) =1 17 is in G(L(§)/L). By the Cebotarev density theorem, the set of primes P of L with Artin Symbol _ — o P a has positive density. Since almost all primes of L are of degree 1 over Q, we need only consider such primes P. Now (gig-ML): Ga and P linear over Q implies p = HPHL E a mod m. Now let m be a divisor of P in CF(k). Since P is linear over Q ‘we have that [.QfilngQ is in A and as was shown above, p = B Q Q must split into principal prime ideals in k. This gives the desired result. 18 CHAPTER III RESOLUTION OF THE LINEAR CASE As we have just seen, there are infinitely many primes p E a mod m that split principally in k provided the modulus m contains no primes that ramify in k. On the other hand we have seen that there are no primes p e 3 mod 40 that split principally in (QQ/IO), a field in which both 2 and 5 ramify. We shall soon see that the non-existence of such primes is not solely because of the ramification of the factors 2 or 5 of m = 40, but because m = 40 has at least two distinct prime faCtors, both of which are ramified. For Theorem III: Let k/Q be galois, z be prime, (a,L) = l, and (m’,L) = 1. Then for any n 2 1 there are infinitely many positive rational primes p which split principally in k with p E a mod Ln and p E 1 mod m’. Once that we have proved Theorem III we have an immedi- ate solution to Question I for k/Q galois. That is: 19 Theorem IV: If k/Q is galois and (a,m) = l, then there are infinitely many rational integers x E a mod m all of whose prime factors split principally in k. Later I will show that the assumption of normality on k/Q can be deleted. But now I prove Theorem III via two lemmas. Lemma IV: Let M/L and N/L be finite extensions of the number field L. Suppose M/L and MN/L are galois and M n N = L. Let n be a prime of MN such that the degree of TN = B n N over L equals 1. Let p = T O M. Then the order of ['fléé J is precisely the order of [ Mglfl ]. PROOF: We first note that we have an isomorphism between the galois groups G(MN/N) and G(M/L) and that the isomorphism is given by the restriction map OF———)OIM . Let [ flflég ] = G. Since the degree of 3N over L is 1, it follows that [ MgAN ] = [ Mfléé ] = o and so 0 E G(MN/N). Thus the order of 0 equals the order of GI . But from M the definition of the Frobenius symbol O|M=iM€Ll' 20 Lemma V: Let k/Q be a finite galois extension and L a rational prime. Let Q be a prime divisor of z in the class field CF(k) of k with inertial field I = I(8) over Q. Finally let p be a prime of CF(k) unramified over Q. If the degree of the prime TI 2 n O I is 1 over Q, (or even over k D I), then the prime p=$nk is principal in k. Moreover the rational prime p = P 0 Q splits principally in k. PROOF: We have the following diagram CF(k) E i I Q Recall that CF(k)/Q is galois. Note that an is the inertial field Q and, since CF(k)/k is unramified, [CF(k):I] = [k:(kflI)]. Since k/(kflI) is normal it follows that By Lemma IV it follows that the order equals the order of [ EAL§Q1)] equals f, of Bflk over CF(k) = kI. of [going] say. Now 21 since the degree of BI over k D I is 1, the degree of 3 over Q is f. But the degree of p over k D I is also f so p must gain degree 1 in the extension CF(k)/k. Thus n is principal in k and since k is normal, p must split principally in k. PROOF OF THEOREM III: We let g be a primitive Ln—th root of unity, C’ a primitive m’-th root of unity. We have CF(k) I( g’) / up), fi’ I””””” ”””’/ \\\\ //// ~////’/’Q(C) ////////// Q ’4‘, where I is as in Lemma v. Now I(Q’) fl Q(Q) = Q Since L is totally ramified in Q(§) yet has an unramified prime factor in I(§’). Hence G(Q(C)/Q) 2' G(I(§.§’)/I(§’)). Thus the substitution Oa(§) = Qa is an automorphism of I(§,Q’)/I(§’). By the Cebotarev density theorem, the set of primes p of I(§’) with Artin Symbol ('Iigifi’3/I(Q’lj> = Ca has positive density. Since almost all primes of I(Q’) are of degree 1 over Q, we need only consider such linear primes. However, if p is such a prime then p = Hp“ E a mod Ln 22 and p a 1 mod m’. Let pl = p n I, then the degree of pI over Q is 1. So by Lemma V , p must split principally in k which proves Theorem III. I will now show that the assumption of normality on k/Q can be deleted. Lemma VI: Let k be an arbitrary number field and k be the galois closure of k. Suppose I is a rational prime and 9 is a prime factor of L in CF(k). Take I = 1(8) to be the inertial field of 9 over Q and T = T(B) the inertial group. Then T n G(CF(k)/CF(k)) = T n G(CF(k)/k) 2399:: Let I’ and I” be the inertial fields of 9 over k and CF(k) respectively. Since CF(k)/k is unramified, it follows that CF(k) C I’, and so I’ = I . However, G(CF(E)/I'> T n G(CF(k)/k) and G(CF(R)/I”) T n G(CF(k)/CF(k)) With the same notation we now have Lemma VII: If ‘13 is any prime of CF (k) such that [ EELgLZQ J E T then p = T O k is principal in k. 23 PROOF: We have the following diagram CF (1'2) / l \. .\\ / k I CF(k) 10—3 - '1 Say [ EElgllg J = O and that the degree of p over Q is fl then f [W120 1 eenrr. Hence f ’ \ o 1 e G C CF(k)/CF(k) ) n T by Lemma VI. Thus p = T O k gains degree 1 in CF(k)/k. r _ Corollary II: If L.Q§i%1£Q J 6 T then p = 3 O Q splits principally in k. 3399:: In the preceding proof we can replace k by any of its conjugate fields G(k) and CF(k) by CF(O(k)) and get that p0 = m 0 G(k) is principal. Say p0 = 0(a). Then 0-1(50) = a is principal in k. But O—1(®) lies above 0-1(p0) and since the galois group acts transitively on the primes of CF(k) dividing p, it follows that all prime factors of p are principal in k. 24 And so finally we have Theorem V: If k is an arbitrary number field and (a,m) = 1, then there are infinitely many rational integers x a a mod m all of whose prime factors split principally in k. PROOF: Using the result of the preceding corollary we can now retrace the proof of Theorem III and the desired result follows. It is now possible to slightly strengthen Corollary I of the previous chapter. Specifically I shall prove Theorem VI: Let k be a number field with discri- minant A. If m is a positive integer with (m,A) = in where z is prime, then for each a with (a,m) 1 there are infinitely many primes p a a mod m which split principally in k. PROOF: Let Q be a prime factor of z in CF(k) and I = 1(9) to be the inertial field of Q. If f is \J m-th root of unity then Q(C) n I = Q. the substitution 25 , a Oa' €t————)€ is in G(I(§)/I). Now the set of linear primes P of with has positive density. But p = HPHI E a mod m and by Corollary II, p splits principally in k. [l] [2] [3] 26 REFERENCES MacCluer, C. R., Non-principal Divisors among the Values of Polynomials: Acta. Arith. Weiss, Edwin, "Algebraic Number Theory". McGraw-Hill, New York, 1963, 172-182. Rotman, Joseph J., "Theory of Groups”. Allyn and Bacon, Boston, 1965, 144-145. i {I'lmilHIM/I3}!!!(lltljfllflltimlfljul'l’ ‘ -—.—-—.—_~._- A ._—__—_._._.-._‘_-_ _._-—__._ l __