# GENERALIZED CYCLOTOMIC PERIODS

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1 proceedings of the american mathematical society Volume 81, Number 2, February 1981 GENERALIZED CYCLOTOMIC PERIODS RONALD J. EVANS Abstract. Let n and q be relatively prime integers with n > 1, and set N equal to twice the product of the distinct prime factors of n. Let t(n) denote the order of q (mod n). Write ij = 22-<T ' aj2 > wnefe?. = exp(2m/n). Uav" 1 for all v, then ij is Kummer's cyclotomic period, and if av «expq.mv/t(n)) for each t>, then tj is a type of Lagrange resolvent. For certain classes of n,e QXi^f)» necessary and sufficient conditions for the vanishing of r are given, and the degree of t\ over Q is determined. 1. Introduction and notation. Let n and q be fixed relatively prime integers with n > 1, and set N equal to twice the product of the distinct prime factors of n. Fix an integer s prime to n and set K = Kn = Q(fJt Ç*), where Ç = exp(2m'/«). For any integer j prime to «, define Oj G Gal(Q(í )/Q(fJ)) by aßn) = ft. Let i(/j) denote the order of q (mod n). Fix ac G Kn (0 < u < t(n)) with not all a vanishing, and define the generalized cyclotomic period r\ by /(#!)-i o-o If all av equal 1, then tj is the cyclotomic period Zf* first studied for general n by Kummer [2], but studied for prime n over half a century earlier by Gauss. If ac = exp(2triv/t(n)) for each v, where n is a prime power, then rj is a type of Lagrange resolvent studied in Weber's book [5, 19]. Fuchs [1] proved in essence the following facts about cyclotomic periods. (1) If t(n) = pt(n/p) for some prime p dividing n, then Z?* = 0; and, conversely, (2) if Zf* = 0, then t(n) = pt(n/p) for some prime/» dividing n; and (3) if Z f ^. then 2fÍ nas degree <b(n)/t(n) over Q, where <f> is Eider's function. (Earlier, Kummer [2, p. 5] had stated (3) without proof.) We prove here some analogues of Fuchs' results, for certain generalized periods r\ = ~Zav\$f in place of Zf?. In the process, we obtain simple new proofs of (1), (2), and (3). In 1977, Kurt Mahler wanted to know when the period Zf 2 vanishes, in order to glean information about the behavior of the function Z^Lo-z2" near the unit circle (see [4]). Seeking to answer his query, D. H. and E. Lehmer were led to rediscover (1) and (2) (see [3], but note that the formulation given there is not quite correct). Received by the editors February 14, 1980 and, in revised form, April 3, Presented at the Western Number Theory Conference, Asilomar, December 20, AMS (MOS) subject classifications (1970). Primary 10G05. Key words and phrases. Cyclotomic periods American Mathematical Society /81/ /\$02.SO

2 208 R. J. EVANS From here on, write n = p"m, where/» is prime,p} m, and a > Generalization of (1). Theorem 1. Assume that av = aqv(a0)for each v, and suppose that t(n) = pt(n/p) for some p dividing n. Then ij = 0. Proof. Let w = t(n/p). We have '(»)-' w-\ [p-\ \ u=0 u-0 I, Jt-O J Now, qw = 1 + jn/p for some./' prime to/». Since c""" = 1 (mod n), we see that/» divides n/p, so qwx = 1 + xjn/p (mod n) for 0 < x </>. Thus, since a0 G Kn, oq~,(a0) = ö0. Therefore since the inner sum vanishes. w-\ i p-\ \ v= 2 v «< «2 i,v -o, u=0 [ x-0 J Q.E.D. Theorem 2. Assume that av = ev for each v, where e is a t(ri)th root of unity. Suppose that t(n) = pw for some prime p dividing n, where w = t(n/p). Write ew = Sp and qw = 1 + jn/p. Suppose further that -k/j (mod/») is not congruent to a power of q (mod/») (this holds, for example, if ew = 1). Then tj = 0. Proof. The argument is essentially the same as that for Theorem 1. Q.E.D. Theorem 2 proves one direction of Weber's theorem [5, 19] while Lemma 5 (below) proves the other direction Note that if a0 = e = 1, then Theorems 1 and 2 reduce to (1). 3. Generalization of (2).* Lemma 3. We have t(n) =pt(n/p) if and only if either t(ri) = t(mp) or pa = 2" > 8, t(n) = 2t(m), 2} t(m), and q = 3 (mod 4). (4) Proof. If t(n) = t(mp) or (4) holds, clearly t(n) J=pt(n/p). Conversely, suppose that t(ri) =pt(n/p) and t(n) = t(mp). We must prove (4). Note \hax pa\\(qkmp)-\) for some A with 1 < A < a. Suppose that /» is odd. Then pa+b\\(qp',(nv)-l) for each B > 0. Thus t(mpa+b) = pbt(mp) for each B > 0. Therefore, t(n) = pt(n/p), a contradiction. Thus /» = 2. We can now obtain a contradiction exactly as before, unless A = 1 and t(n) = 2f(2w). These last equalities are easily seen to imply (4). Q.E.D. Lemma 4. Suppose that p is the largest prime factor of n, and t(p") = pt(p"~l). Then t(n) = pt(n/p). Proof. If n = pa, the result is obvious, so we may assume that/» is odd. Assume that t(n) = t(mp). Then t(p")\t(mp), so by hypothesis, p\t(mp). Thus p\<b(mp), a contradiction. Therefore t(n) = t(mp). Since also /> is odd, the result follows from Lemma 3. Q.E.D. Lemma 5. Suppose that n = pa and ij = 0. Then t(ri) = pt(n/p).

3 GENERALIZED CYCLOTOMIC PERIODS 209 Proof. Assume that t(n) ^pt(n/p). Then by Lemma 3, either t(n) = t(p) or 8 «, t(n) = 2, q = 3 (mod 4). In the latter event, 0 = tj = a + axç, so»"' G Kn, which is impossible because 2\\(q 1). Thus t(n) t(p). Since tj = 0, /(/i) > 1. Thus/» > 2. Let rv denote the least positive residue of qv (mod/»). Then '(/>)- c-0 and the expressions in parentheses are in Kn. Since the elements * (0 < u < *(/»)) are distinct elements of a basis for K (Çn) over Kn, it follows that all av vanish, a contradiction. Q.E.D. Lemma 6. Write z = t(p"), Q = qz, and let T(m) denote the order of Q (mod m). Then where m+av) = Z2Z<>AW, (5) u-0 T(m)-\ ô = S ^.jr. (6) x-0 Proof. Since T(m) = /(«)/z and 0 = fp., 0x,u - m+pmq"(.azx + u)- (7) '(«)-! z-\ T(m)~\ and the result follows. Q.E.D. ü-0 u-0 x-0 Theorem 7. Suppose that tj = 0. 7>W t(n) = pt(n/p) for some p dividing n. Proof. The notation of Lemma 6 will be used here. We induct on the number of distinct prime factors of n. The induction starts by Lemma 5, so it can be assumed that n is not a prime power. Let p be the largest prime factor of n. First suppose that 8U ^ 0 for some u. By (6) and (7), aq.(8u) G Äp.imj. Since the left side of (5) vanishes, we may apply Lemma 5 to the generalized period on the right side of (5) to conclude that /(/»") = pt(pa~x)- Thus t(n) = pt(n/p) by Lemma 4. Finally, suppose that 8U = 0 for each u. Fix u such that 9XU =/= 0 some for x (this is possible by (7)). All of the 9XU are in Am^,., so by (6) and the induction hypothesis, T(m) = rt(m/r) for some prime r dividing m. Multiplying by t(p"), we find that t(n) = rt(n/r). Q.E.D. Remark. If n = mp" with a > 1 and L = Q( ), then tj = 0 if and only if tj G L. Assume for the purpose of contradiction that 0 # tj G L. By (5), Çp. is a zero of a polynomial fix) G L[x] with /(0) ^ 0 but with all of fs nonconstant monomials possessing the form bxr, where b G L, p} r. Since g(x) = xp Çp, is the minimal polynomial of p«over L, g(x)h(x) = f(x) for some h(x) G L[x\. Since h(0) " 0, this implies that fix) has a monomial of the form bxr with p\r, a contradiction.

4 210 R. J. EVANS 4. Generalization of (3). Theorem 8. Suppose that av = oq (a0) for all v. If tj = 0, then tj has degree <b(n)/t(n) over Qi ). Proof. It suffices to show that if 0 = tj = oc(tj), then c = power of q (mod n). We will prove this by induction on the number of distinct prime factors of n. Suppose that 0 9* tj = o-c(tj). By Theorem 1, t(n) =pt(n/p), for each prime /» dividing n. Let /» now denote the largest prime factor of n. By Lemma 4, t(p") ^pt(pa~l), so by Lemma 3 (with/»" in place of n), either or t(p') = /(/») (8) = p" = 2" > 8, f(2a) = 2, 4 = 3 (mod 4). (9) Suppose first that (9) holds. Then aq\$ + axtf = oc(a0)tf + oc(ax)\$cnq, so {a0 - Ms;-1} = rrwaik*'-1' - «>} oo) We may assume without loss of generality that c = 1 (mod 4), otherwise c can be replaced by cq. Thus, the braced expressions in (10) are in Kn, and since f*-1 G KH by (9), it follows that the left side of (10) vanishes. Thus and by repeated applications ck) = ûoi '"c. (il) of ac to (11), we obtain M«o) = ^n1_c' (* > 0). (12) If c = 1, the result follows, so assume that c = 1. Then 2* (c 1) for some B > 2. Assume for the purpose of contradiction that B < a. For each A > 0, 2^ (c^-l). (13) Let A = a 1 B. Then by (13) and (12) with e = 2A, we obtain a0 = -a, which contradicts the fact that tj ^ 0. Thus B > a, which yields the desired result. Now suppose that (8) holds. Say /» = 2. Then the equality tj = ac(tj) becomes a0í = oc(a0)c. Thus (11) holds and c = 1 (mod 4), so that the desired result follows as above. It remains to consider the case/» > 2. By (8), the right side of (5) is where R= 2 v(w> «-0 7-(m)-l 8= 2 MW> *o = «W-K)- x-0 Since 0 = tj = ac(tj), it follows from (5) that 0 = R = oc(r) (and in particular, 8 ^ 0). For 0 < u < t(p), write qu=psu + ru, cq"=ps'u + ri (0 < ru, r'u </»). (14) Clearly the ru are distinct and the r'u are distinct.

5 We have '(/>)-! GENERALIZED CYCLOTOMIC PERIODS 211 '(/>)-! 2 (oa8)sfi->)s;* - r - oe(r) - 2 («vwtf-otf u-0 u-0 Since the elements ÇpK,..., Çp.~ ' are linearly independent over Q(? ), there exists a fixed value of u such that and Thus, by (14) and (15), and K. - r0-1 (15) 5 = «/j-, = ac,.(s)0-,. - c?" - 1 -/>y (16) ^(8) = «?/-'. (17) where y = -s'u. If rf = 1, the result follows, so assume that d = 1. Repeated applications of od to (17) yield o>(«) = «JK-»/«*-» (e > 0). (18) We have/»b (</ - 1) for some B > 1. Thus, by (16), pb'l\\y- (19) Assume for the purpose of contradiction that B < a. For each A > 0, pa+b\\(dpa - 1). (20) Let,4 = a B. By (20) and (18) with e = jr4, j»«-1^^ (19),pa\(dpA - 1), which contradicts (20). Thus B > a, so - l)/( - 1), so by d = 1 (mod/»"). (21) This completes the proof if n is a prime power, i.e., if m = 1. Thus assume that m> 1. By (W),/»"-1!* k> by (T7), od(8) = 5. (22) Since 8 equals the generalized period Z^ro_10g*( >o)ím * with 0 9* >0 G ATOT^,«and Oç* G Gal(Q(fmspa)/Q(fíp«)), it follows from (21), (22), and the induction hypothesis that d =Qh (mod m) for some h. Since Q s 1 (mod/»a), (21) yields d = Qh (mod/»"). Thus d = Qh (mod w) and c = power of? (mod n). Q.E.D. Theorem 8 states that if L is a field with Q(fOT) dld Q(,) such that Q( ) is cyclic over L, then the degree of Tr(a0f ) over Q(,) is either 0 or \L: Q(,), where Tr denotes the trace map from Q(Çsn) to L. In other words, if T^a^f,,) is nonzero, then it has the maximum possible degree over Q(ÇS) that any element of L can have. A modification of the proof of Theorem 8 shows that the hypothesis that Q(ÍOT)/L is cyclic may be dropped. The author is very grateful to the Lehmers for their encouragement and helpful comments.

6 212 R. J. EVANS References 1. L. Fuchs, Ueber die Perioden, welche aus den Wurzeln der Gleichung u" = 1 gebildet sind, wenn n eine zusammengesetzte Zahl ist, J. Reine Angew. Math. 61 (1863), E. Kummer, Theorie der idealen Primfaktoren der complexen Zahlen, welche aus den Wurzeln der Gleichung u" = 1 gebildet sind, wenn n eine zusammengesetzte Zahl ist, Math. Abh. Kon. Akad. Wiss. Berlin (1856), 1-47; Collected Papers, vol. 1, Springer-Verlag, Berlin and New York, 1975, pp D. H. Lehmer and E. Lehmer, Notices Amer. Math. Soc. 25 (1978), K. Mahler, On a special function, J. Number Theory 12 (1980), H. Weber, Lehrbuch der Algebra, 3rd ed., vol. 2, Chelsea, New York, Department of Mathematics, University of Caufornia-San Diego, La Jolla, California 92093

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