Circular units of an abelian field ramified at three primes

Transcription

1 Crcuar unts of an abean fed ramfed at three prmes Radan Kučera a,1, Azar Saam b a Facuty of Scence, Masaryk unversty, Brno, Czech Repubc b Department of mathematcs, Amercan Unversty of the Mdde East, Kuwat Abstract Ths paper constructs a bass and gves a presentaton of Snnott s group of crcuar unts for a rea abean fed k ramfed at three prmes whose genus fed K n the narrow sense has cycc reatve Gaos group Ga(K/k). It s shown that, for ths type of feds, the quotent of the modue of a reatons satsfed by crcuar unts by the submodue generated by a norm reatons s a cycc modue generated by an Ennoa reaton. Keywords: Abean fed, crcuar (cycotomc) unts, Ennoa reaton MSC: 11R20 Introducton Snnott s group of crcuar unts of an abean fed k has appeared n the semna paper [13]. Snnott s defnton gves a group whch mproves many prevous versons of groups of cycotomc unts due to Kummer, Hasse, Leopodt, Gard and others (an overvew of these groups can be found for exampe n [10]). Snnott s group C k s used n penty of nterestng modern appcatons: an nverse mt of C k appears n Iwasawa Theory, Snnott s crcuar unts gve an exampe of an Euer system etc. In appcatons t s usefu to have a bass of C k n hand; for exampe n [5] God and Km used ther bass of the group of crcuar unts of a cycotomc fed to show that, for cycotomc feds, the groups of crcuar unts satsfy Gaos descent,.e. f K L are two cycotomc feds then C K = K C L. But such a bass has been constructed ony for a few very specfc types of abean feds, for exampe for a compostum of abean feds of prme power conductor, ether a of them beng magnary (see [9]) or a of them beng rea (see [14]), for a compostum of quadratc feds (see [8]), for a non-cycc fed whose degree s the square of an odd prme (see [7]), or for k/q beng ramfed at exacty two prmes (see [2] and [3]). Ema addresses: kucera@math.mun.cz (Radan Kučera), azarsaam@gma.com (Azar Saam) 1 The frst author was supported under projects P201/11/0276 and S of the Czech Scence Foundaton. Preprnt submtted to Journa of Number Theory October 5, 2015

2 The am of ths paper s to construct a bass of Snnott s group of crcuar unts for an abean fed k of another type, namey for a rea abean fed k ramfed at exacty three prmes and satsfyng the assumptons gven beow. It appears that n our constructon (descrbed by Theorem 19) a cruca roe s payed by an Ennoa reaton among the generators of the group of crcuar unts; ths reaton s gven by Theorem 11. In order to understand what we have n mnd by an Ennoa reaton we must frst expan the foowng detas: The group of crcuar unts of a rea abean fed k s defned as the ntersecton of the group E k of a unts and the group of crcuar numbers D k, whch s a Z[G]-modue, where G = Ga(k/Q), and t has penty of generators. In fact, there s one generator for each non-empty subset of the set of a prme numbers ramfed n k/q. When we compute the norm of such a generator gong from the subfed of k where the generator s defned to a subfed havng ess ramfed prmes, we obtan a reaton between the generators, whch we ca a norm reaton. Now, an Ennoa reaton s a vad reaton among the generators whch s not a consequence of the norm reatons. More precsey, a reatons form a Z[G]- modue contanng the Z[G]-submodue generated by a norm reatons; Ennoa reatons are just reatons outsde of ths submodue. The frst appearance of a reaton of ths knd can be found n [4] where Ennoa expans the necessty to work wth these reatons n the paper [1] of Bass. Ennoa reatons aso form a subject of recent research. For exampe n [6], these Ennoa reatons are expcty decrbed for the maxma rea subfed of a cycotomc fed ramfed exacty at three odd prmes, whch s a fed beongng to the famy of feds studed by ths paper. It s aso worth notng that n the case of the consdered fed k, the quotent of the Z[G]-modue of a reatons by the submodue generated by norm reatons s a cycc Z[G]-modue havng trva G-acton, see Theorem 21, whch we hope to be an nterestng resut on ts own. Sectons 1-3 of ths paper are based on the PhD. thess [12] of the atter author. 1. A certan cass of rea abean feds Let k be a rea abean fed ramfed exacty at three prmes p 1, p 2, p 3. Let K be the genus fed of k n the narrow sense and G = Ga(K/Q) be ts Gaos group (so we are changng the use of G n Introducton). For each = 1, 2, 3 et T be the nerta subgroup of G for p and et K be the maxma subfed of K ramfed ony at p. Hence G s the drect product G = T 1 T 2 T 3 and T 1 = Ga(K/K 2 K 3 ) = Ga(K 1 /Q), T 2 = Ga(K/K 1 K 3 ) = Ga(K 2 /Q), T 3 = Ga(K/K 1 K 2 ) = Ga(K 3 /Q), the mentoned somorphsms beng defned by restrctons. 2

3 Assumpton. We assume that the Gaos group H = Ga(K/k) s cycc, so H = τ for a generator τ whch we fx through the whoe paper. Moreover, we aso assume that the group T s cycc for each = 1, 2, 3. The frst part of the assumpton s qute restrctve whe the second one s ess so: we coud have a noncycc nerta group ony for the prme 2. Lemma 1. We have K = kk 1 K 2 = kk 1 K 3 = kk 2 K 3, hence H T 1 = H T 2 = H T 3 = {1}. Proof. Ths s gven by an easy ramfcaton argument: K/K 1 K 2 s totay ramfed at prmes above p 3 and K/k s unramfed at a fnte prmes. We denote n = [K : k] = H, r = [K : kk ], and a = [(k K ) : Q] for each = 1, 2, 3. Then [kk : k] = [K : (k K )] = n r, so T = [K : Q] = na r and [K : Q] = n3 a 1a 2a 3 r 1r 2. Lemma 2. The numbers r 1, r 2, are parwse coprme, hence r 1 r 2 n. Proof. Let π : G T be the projectons, so res K/K π (γ) = res K/K γ for any γ G and we have γ = π 1 (γ)π 2 (γ)π 3 (γ). For any 1 < j 3, Lemma 1 mpes that the mappng γ π (γ)π j (γ) s njectve on H, so π (τ)π j (τ) s of order n n G. Nevertheess res K/K τ generates Ga(K /(k K )) of order n r, and so π (γ) s of order n r n G. Snce G s abean, we get (r, r j ) = 1. We denote m = n r 1r 2. For any permutaton of ndces {, j, } = {1, 2, 3} we have the foowng dagrams of feds where we mark the degrees: K K = kk K j (1) r n kk K K j k mr jr K k n k K K j ma r r j mr jr ma a jr k K Q a Q Exampe 3. To gve an exampe of a fed k satsfyng our assumptons, we consder the foowng smpest nontrva nstance, whch we ca the Prototypca Case: fx an odd prme and assume a a and r are 1, and n =, therefore m =. Ths mpes that each p 1 (mod ) or p = and K s the subfed of the p -th or the 2 -th cycotomc fed of absoute degree, respectvey. Then K = K 1 K 2 K 3 s ther compostum, hence Ga(K/Q) s eementary abean of 3

4 order 3 ; and k s a subfed of K of absoute degree 2 n genera poston,.e., dfferent from K 1 K 2, K 1 K 3, and K 2 K 3. Thus Ga(K/k) s cycc of order. Ths Prototypca Case covers an nfnte famy of abean feds but st most of the computatons beow smpfy drastcay for ths case, so we sha use t as an ustraton of a smper verson of some technca formuas we have to face. Any subgroup of ndex d of a fnte cycc group s aso cycc and any generator of the subgroup s the d-th power of a sutabe generator of the group (just because for any postve ntegers r s the natura mappng (Z/sZ) (Z/rZ) s surjectve). Hence for each = 1, 2, 3 we can fx a generator σ of T satsfyng σ a = π (τ). Hence we have We have aso the foowng dagram σ a1 1 σa2 2 σa3 3 = τ. (2) K K j (3) r K (k K K j ) mr jr ma jr k K K j K ma jr mr jr k K a Q Hence there s the exact sequence 1 Ga(K (k K K j )/K ) Ga((k K K j )/Q) Ga((k K )/Q) 1, where both maps are restrctons. Snce Ga(K (k K K j )/K ) s generated by the restrcton of σ j, we have res K/(k KK j) σ majr j = 1. (4) Smary Ga((k K )/Q) s generated by the restrcton of σ, whch gves Lemma 4. Let, j {1, 2, 3}, j. The Gaos group of (k K K j )/Q s gven by Ga((k K K j )/Q) = {(σ x σ y j ) k K K j 0 x < a, 0 y < ma j r }, where each automorphsm of k K K j determnes the coupe (x, y) unquey. 4

5 The rght hand sde dagram n (1) gves the exact sequence 1 Ga(K/K 2 K 3 ) Ga(k/Q) Ga((k K 2 K 3 )/Q) 1, where both maps are restrctons. Snce Ga(K/K 2 K 3 ) s generated by σ 1, Lemma 4 for = 3 and j = 2 mpes Lemma 5. The Gaos group of k/q s gven by Ga(k/Q) = {(σ x 1 σ y 2 σz 3) k 0 x < ma 1 r 2, 0 y < ma 2 r 1, 0 z < a 3 }, where each automorphsm of k determnes the trpe (x, y, z) unquey. 2. Crcuar numbers of k Let η = N Q(ζcond K )/k(1 ζ cond K ), where ζ r = e 2π/r for any postve nteger r and cond L s the conductor of an abean fed L. Smary for any 1 < j 3 et and for any = 1, 2, 3 et η j = N Q(ζcond K K j )/(K K j k)(1 ζ cond KK j ), η = N Q(ζcond K )/(K k)(1 ζ cond K ). Then, foowng Lett s modfcaton of Snnott s defnton descrbed by [11, Proposton 1], we ca D = 1, η, η 12, η 13, η 23, η 1, η 2, η 3 Z[G] (5) the group of crcuar numbers of k; the ntersecton of D and of the group of a unts of k s C = 1, η, η 12, η 13, η 23, η 1 σ1 1, η 1 σ2 2, η 1 σ3 3 Z[G], (6) the Snnott group of crcuar unts of k defned n [13]. The group D j and D of crcuar numbers of k K K j and of k K are defned smary,.e., D j = 1, η j, η, η j Z[G] and D = 1, η Z[G]. For any = 1, 2, 3 and any nteger h 0 et h 1 S (h) = σ u Z[G], so (1 σ )S (h) = 1 σ h, u=0 5

6 and S (0) = 0 by defnton. Let R = S (a ), so (1 σ )R = 1 σ a, and et N Z[G]. Moreover et = σav Z[G], and = vσav H = r jr 1 σ mav Z[G], and H j = r j 1 σmav {, j, } = {1, 2, 3}. Fnay, we defne Γ = N 1 2 N N 3 1. It s easy to see usng (1) and (3) that the restrcton ma r jr 1 res K/k R N H = res K/k Z[G], where as usuay s the norm operator wth respect to k/(k K j K ) and so the we-known norm reatons of crcuar unts gve σ v Smary the restrcton η RNH = η 1 Frob(p) 1 j. (7) res K/(k KK ) R N H j = res K/(k KK ) ma r j 1 s the norm operator wth respect to (k K K )/(k K ) and so we have η RNHj = η 1 Frob(p) 1. (8) Besdes the prevous reatons (7) and (8), we sha need another reaton whch we ca the Ennoa reaton; t s descrbed by Theorem 11 beow. Lemma 6. If {, j, } = {1, 2, 3} then res K/(k KK j) N H = res K/(k KK j) N j H j. Proof. The defntons and (2) gve res K/(k KK j) N H = res K/(k KK j) and the emma foows from (4). mr 1 σ v mr 1 σ av = res K/(k KK j) Lemma 7. res K/k H H j s the same for a j. For nstance res K/k H 1 H 21 = res K/k H 2 H 12 = res K/k H 3 H 13. σ ajv j Proof. Let {, j, } = {1, 2, 3}. Lemma 2 mpes the exstence of d j, d j Z such that d j r + d j r j = 1. Then (2) and σ majrr j = σ marjr = 1 gve τ mr r d j = σ ma r r d j σ mar (1 r jd j) = σ ma r r d j σ mar, so res K/k σ mar = res K/k σ ma r r d j. Snce σ ma H = H (9) 6

7 we get and r j 1 res K/k res K/k σ mar H = res K/k σ ma r r d j H = res K/k H (10) σ mar v = res K/k r j 1 σ ma r r d jv = res K/k r j 1 σ ma r v, (11) because σ ma r r j = 1 and ( r d j, r j ) = 1. Therefore r j 1 res K/k H H = res K/k H H = res K/k H H. By (2), (9), and (10) we have r 1 res K/k H H j = res K/k H The emma foows. Proposton 8. We have σ mar v σ majv j r ( 1 = res K/k H 1 + v=1 = res K/k H H = res K/k H σ mav res K/k ΓH 1 H 21 = m res K/k Proof. It s easy to see that N 1 2 N 2 1 = u=0 v=u+1 u=0 r 1 r j 1 σ ma r v σ mav σ ma v σ mar ) = resk/k H H. σ a1u 1 σ a2v 2 H 1 H 21. (v u)σ a1u 1 σ a2v 2. (12) Lemma 7 gves res K/k H 1 H 21 = res K/k H 3 H 23. By (9) and (2) we have and we get σ ma3 3 H 3 = H 3, res K/k σ a3 3 = res K/k σ a1 1 σa2 2, res K/k N 3 1 H 1 H 21 = res K/k = res K/k σ a3v 3 w=0 w=0 wσ a1w 1 H 3 H 23 wσ a1(w+v) 1 σ a2v 2 H 1 H 21. 7

8 Let [x] mean the ntegra part of a rea number x and x = x [x] mean ts fractona part. Snce σ ma1 1 H 1 = H 1 by (9), we have res K/k N 3 1 H 1 H 21 = m res K/k whch together wth (12) gves res K/k ΓH 1 H 21 = m res K/k and the proposton foows. Proposton 9. If {, j, } = {1, 2, 3} then Proof. We have u=0 u=0 u v m [u v m σ a 1u 1 σ a2v 2 H 1 H 21, ] σ a 1u 1 σ a2v 2 H 1 H 21 (1 σ aj j ) j = N j 1 (m 1)σ maj j, (13) (1 σ aj j ) jh j = (N j m)h j, (14) res K/(k KK j)(1 σ aj j ) jh j = res K/(k KK j)(n j m)h j. (15) (1 σ aj j ) j = (1 σ aj j ) vσ ajv j = vσ ajv j m (v 1)σ ajv whch gves the frst dentty; the second one foows due to (9). By (4) we see that res K/(k KK j) σ maj j H j = res K/(k KK j) H j, whch mpes the thrd dentty. Proposton 10. We have η ΓH1H21R1R2R3 η m 12, η m 13, η m 23 Z[G]. Proof. The defnton of Γ, Lemma 7, and (7) gve η ΓH1H21R1R2R3 = η 2H21R2R3(1 Frob(p1) 1 ) 23 v=1 j, η 1H12R1R3(1 Frob(p2) 1 ) 13 η 1H13R1R2(1 Frob(p3) 1 ) 12. (16) Let us defne ntegers 0 b j < mr r, 0 c j < a j, where as usuay {, j, } = {1, 2, 3}, by Frob(p 1 ) 1 = σ a2b12+c12 2 σ a3b13+c13 3, Frob(p 2 ) 1 = σ a1b21+c21 1 σ a3b23+c23 3, Frob(p 3 ) 1 = σ a1b31+c31 1 σ a2b32+c

9 Usng (2) we get η σa = η σ a2 2 23, so η 1 Frob(p1) 1 23 = η 1 σc η σc12 23 η σc (17) 2 (1 σ c 13 3 ) 2 σ c 13 3 (1 σ a 2 (b 12 b 13 ) 2 ) Let us appy 2 H 21 R 2 R 3 and compute the factors of (17) successvey. Identty (15) and Lemma 6 gve η 2H21R2R3(1 σc 12 2 ) By (8), we get moduo η m 23 Z[G] 2 ) 23 = η 2H21S2(c12)R3(1 σa2 23 = η (N2H21 mh21)s2(c12)r3 23 = η (N3H31 mh21)s2(c12)r3 23. η 2H21R2R3(1 σc 12 2 ) 23 η N3R3H31S2(c12) 23 = η (1 σc32 2 )S 2(c 12) 2 = η (1 σ2)s2(c32)s2(c12) For the second factor of (17) we obtan smary η σc H 21R 2R 3(1 σ c 13 3 ) 23 = η σc12 23 = η σc H 21R 2S 3(c 13)(1 σ a 3 3 ) 2 2H 21R 2S 3(c 13)(1 σ a 2 2 ) 23 = η σc 12 a 2 2 (N 2 m)h 21R 2S 3(c 13) 23 η σc 12 a 2 2 N 2R 2H 21S 3(c 13) 23 = η σc 12 a (1 σ c 23 3 )S 3(c 13) = η (1 σ3)s3(c23)s3(c13) 3 moduo η m 23 Z[G]. Fnay, for the thrd factor of (17) we get η σc 12 2 σ c H 21R 2R 3(1 σ a 2 (b 12 b 13 ) 2 ) 23 = η σc12 By dentty (15) 2. 2 σ c H 21R 2R 3(1 σ a 2 2 ) b 12 b 13 +n 1 u=0 σ a 2 u η 2H21R2R3(1 σa 2 2 ) 23 = η (N2 m)h21r2r moduo η m 23 Z[G], because η N2H21R2R3 23 = N (k K2K 3)/Q(η 23 ) = 1 by Lemma 4. Puttng the three prevous factors together, we have η 2H21R2R3(1 Frob(p1) 1 ) 23 η (1 σ2)s2(c32)s2(c12) 2 η (1 σ3)s3(c23)s3(c13) 3 moduo η m 23 Z[G]. Usng symmetry 1 2 we get η 1H12R1R3(1 Frob(p2) 1 ) 13 η (1 σ1)s1(c31)s1(c21) 1 η (1 σ3)s3(c13)s3(c23) 3 moduo η m 13 Z[G]. Now usng symmetry 2 3 we obtan η 1H13R1R2(1 Frob(p3) 1 ) 12 η (1 σ1)s1(c21)s1(c31) 1 η (1 σ2)s2(c12)s2(c32) 2 moduo η m 12 Z[G]. Proposton foows by substtutng nto (16). 9

10 Theorem 11. We have the foowng Ennoa reaton u=0 v=u+1 η σa 1 u 1 σ a 2 v 2 H 1H 21R 1R 2R 3 η 12, η 13, η 23 Z[G]. (18) Proof. Compare Proposton 8 and Proposton 10 and use the fact that η, η 12, η 13, η 23 are a totay postve. Let us descrbe what Theorem 11 says for our Prototypca Case (see Exampe 3). It s easy to see that a quanttes R, H and H j smpfy to 1. Hence the somewhat ugy product (18) smpfes to η σu 1 σv 2 = η A, 0 u<v< where the coeffcents of the eement A of the group rng Z[Ga(k/Q)] can be pctured by an matrx wth zeros on and beow the dagona, and ones above the dagona. 3. Bases of the groups of crcuar unts Lemma 12. For any = 1, 2, 3, the set B conjugates of η forms a Z-bass of D. = { η σx 0 x < a } of a Proof. The Z-rank of D s [(k K ) : Q] = a, and Ga((k K )/Q) = σ k K s of order a. Lemma 13. For any 1 < j 3, et B j be the set B j = { η σx σy j,j 0 x < a, 0 y < ma j r 1, (x = 0 = y < (mr 1)a j }. Then B j B B j s a Z-bass of D j. Proof. In fact, t s enough to prove that B j B B j s a system of generators, because B j B B j (mr 1)a j + (a 1)(mr a j 1) + a + a j = ma a j r + 1, whch s equa to [(k K K j ) : Q] + 1, the Z-rank of D j. Actuay, we can show that the mssng generators of D j are (mutpcatve) Z-near combnatons of eements of ths set. Indeed, usng (8) we get ma jr 1 y=0 η σx σy j j = η σx RjNjH j j = η (1 Frob(pj) 1 )σ x 10

11 and we see that any η σx σy j j for 0 < x < a and y = ma j r 1 s a Z-near combnaton of generators n B j B B j. Usng (2) we have η σ a j and (8) gves for any c {0, 1,..., a j 1} a 1 x=0 mr 1 z=0 η σx σzaj +c j j = a 1 x=0 mr 1 z=0 = η σ a j j j, η σx za σ c j j = η σc j RNH j = η (1 Frob(p) 1 )σ c j j. Hence any η σy j j for (mr 1)a j y ma j r 1 s a Z-near combnaton of generators n B j B B j and, f y = ma j r 1, of aready obtaned conjugates of η j. We want to construct a Z-bass B of D of Z-rank [k : Q] + 2 as foows: B = B S for a set S of some sutaby chosen conjugates of η (to be defned ater on) and B = B 1 B 2 B 3 B 12 B 13 B 23. (19) At the moment we cannot guarantee that the eements of B are neary ndependent, ths w be obtaned at the very end. But Lemma 13 gves that B { 1} generates D 12 D 13 D 23 = 1, η 12, η 13, η 23, η 1, η 2, η 3 Z[G]. At frst, assumng the ndependence of eements of B, et us compute how many eements the set S shoud contan: S = ([k : Q] + 2) B = nma 1 a 2 a (ma 1 a ma 1 a 3 r ma 2 a 3 r a 1 a 2 a 3 ) = nma 1 a 2 a 3 ma 1 a 2 ma 1 a 3 r 2 ma 2 a 3 r 1 + a 1 + a 2 + a 3 1. (20) The desred set S of sutabe chosen conjugates of η must be a subset of the set of a conjugates {η σx 1 σy 2 σz 3 0 x < ma1 r 2, 0 y < ma 2 r 1, 0 z < a 3 }, (21) see Lemma 5. We defne S to be the set of a η σx 1 σy 2 σz 3 such that the trpe (x, y, z) satsfes 0 x < ma 1 r 2, 0 y < ma 2 r 1, 0 z < a 3 but does not ft to any of the foowng sx cases: x = y = z = 0; 1 x < ma 1, 0 y < a 2, z = 0; 0 x < ma 1 r 2, y = 0, 1 z < a 3 ; ma 1 x < ma 1 (r 2 + 1), y = 0, z = 0; 11

12 x = 0, 1 y < ma 2 r 1, 0 z < a 3 ; 1 x (m 1)a 1, y = a 2, z = 0. The sx cases are vsby parwse dsjont, so the number of trpes (x, y, z) satsfyng one of the sx cases s equa to 1 + (ma 1 1)a 2 + ma 1 r 2 (a 3 1) + ma 1 (r 2 1) + (ma 2 r 1 1)a 3 + (m 1)a 1 n accord wth (20). Exampe 14. Let us have a ook at what happens for our Prototypca Case (see Exampe 3). The sx buet ponts above bo down to four dsjont forbdden regons n two-dmensona space F F {0} (z beng zero by defnton): the orgn, and three punctured nes gven n turn by x 0, y = z = 0; y 0, x = z = 0; x 0, y = 1, z = 0. Therefore the set of remanng ponts s the set {1, 2,..., 1} {2, 3,..., 1} {0} of cardnaty ( 1)( 2) = 2 3+2, whch s the vaue n equaton (20) for the Prototypca Case. Our am s to show that eements of S together wth D 12 D 13 D 23 generate a conjugates of η correspondng to the sx cases above. Therefore we need to study reatons satsfed by conjugates of η n (21) moduo D 12 D 13 D 23. Lemma 15. Let 0 y < ma 2 r 1, 0 z < a 3. If a subgroup of D contanng D 12 D 13 D 23 contans a eements of the set but one, then t contans a of them. {η σx 1 σy 2 σz 3 0 x < ma1 r 2 } Proof. The product of a eements n ths set s equa to by (7). η σy 2 σz 3 R1N1H1 = η (1 Frob(p1) 1 )σ y 2 σz 3 23 D 23 Lemma 16. Let 0 x < ma 1 r 2, 0 z < a 3. If a subgroup of D contanng D 12 D 13 D 23 contans a eements of the set {η σx+ma 1 r 2 u 1 σ y 2 σz 3 0 y < ma2 r 1, 0 u < } but one, then t contans a of them. Proof. By (11) we have r 3 1 u=0 r 3 1 η σma 1 r 2 u 1 = u=0 η σma 2 r 1 u 2, so the product of a eements n ths set s equa to by (7). η σx 1 σz 3 R2N2H2 = η (1 Frob(p2) 1 )σ x 1 σz 3 13 D 13 12

13 Lemma 17. Let 0 x < ma 1, 0 y < a 2. If a subgroup of D contanng D 12 D 13 D 23 contans a eements of the set {η σx+a 1 v+ma 1 u 1 σ y+a 2 v 2 σ z 3 0 v < mr1, 0 u < r 2, 0 z < a 3 } but one, then t contans a of them. Proof. By (2) and (11) we have mr 1 1 r 2 1 u=0 η σa 1 v+ma 1 u 1 σ a 2 v 2 = mr 1 1 r 2 1 so the product of a eements n ths set s equa to u=0 η σma 3 r 1 u a 3 v 3, by (7). η σx 1 σy 2 R3N3H3 = η (1 Frob(p3) 1 )σ x 1 σy 2 12 D 12 Lemma 18. Let 0 x (m 1)a 1. If a subgroup of D contanng D 12 D 13 D 23 contans a eements of the set {η σx+w+a 1 u+ma 1 s 1 σ y+a 2 v+ma 2 t 2 σ z 3 0 u < v < m, 0 s < r2, 0 t < r 1, but one, then t contans a of them. 0 w < a 1, 0 y < a 2, 0 z < a 3 } Proof. The product of a eements n ths set s equa to u=0 v=u+1 η σx+a 1 u 1 σ a 2 v 2 H 1H 21R 1R 2R 3, whch beongs to D 12 D 13 D 23 by Theorem 11. Theorem 19. The set B = B S (see (19) and the defnton of S on page 11) s a Z-bass of D. Proof. By (20) the cardnaty of B s equa to the Z-rank of D, so we need to show that B { 1} generates D. We know by Lemma 13 that B { 1} generates D 12 D 13 D 23. We sha use Lemmas for the subgroup A = B { 1} Z of D to show that a conjugates of η beong to A. Notce that f η σa 1 σb 2 σc 3 has appeared n the sets descrbed n these emmas then 0 a, 0 b < ma 2 r 1, 0 c < a 3 ; f a ma 1 r 2, whch can be the case ony n Lemmas 17 and 18, we can change a to ts remander upon dvson by ma 1 r 2. For any 1 y < ma 2 r 1, 1 z < a 3, Lemma 15 gves η σy 2 σz 3 A and for any a 2 < y < ma 2 r 1, Lemma 15 gves η σy 2 A. Then for any 0 x < ma1 r 2, 1 z < a 3, Lemma 16 gves η σx 1 σz 3 A. We have got that a eements η σx 1 σy 2 σz 3 of the set (21) beong to A but the foowng four cases 13

14 1 x < ma 1, 0 y < a 2, z = 0; ma 1 x < ma 1 (r 2 + 1), y = 0, z = 0; x = 0, 0 y a 2, z = 0; 1 x (m 1)a 1, y = a 2, z = 0. Consder whch eement of the set mentoned n Lemma 18 for x = (m 1)a 1 s not known to beong to A yet. Such an eement can be obtaned ony for z = 0, y = 0, v = 1, t = 0, hence u = 0 and x + w + ma 1 s < ma 1 r 2, so w = s = 0. Thus there s just one such eement and we get η σ()a 1 1 σ a 2 2 A. Ths can be repeated for x = (m 1)a 1 1,..., 1, 0 (n ths order!) to get that a eements η σx 1 σy 2 σz 3 of the set (21) beong to A but the foowng two cases 0 x < ma 1, 0 y < a 2, z = 0; ma 1 x < ma 1 (r 2 + 1), y = 0, z = 0. For any 0 x < ma 1, 1 y < a 2, Lemma 17 gves η σx 1 σy 2 eements η σx 1 σy 2 σz 3 of the set (21) beong to A but the case A. Therefore a 0 x < ma 1 (r 2 + 1), y = 0, z = 0. On one hand, suppose r 2. For any ma 1 ( 1) x < ma 1 r 2, Lemma 16 gves η σx 1. If r3 = 1 then we are fnshed. Otherwse a eements η σx 1 σy 2 σz 3 of the set (21) beong to A but the foowng two cases 0 x < ma 1 ( 1), y = 0, z = 0; ma 1 r 2 x < ma 1 (r 2 + 1), y = 0, z = 0. Let x: {0, 1,..., 1} {1, 2,..., ma 1 } {0, 1,..., ma 1 1} be the map defned by x(v, w) = ma 1 w ma 1 vr2. By Lemma 2 we have (r 2, ) = 1, and so the map x s a bjecton. By nducton wth respect to v = 0, 1,..., 1 we sha prove for each w {1, 2,..., ma 1 } that both η σx(v,w) 1 and η σx(v,w)+ma 1 r 2 1 beong to A. Snce x(0, w) = ma 1 w there s nothng to prove for v = 0. Let us assume for any 0 < v < that we aready know η σx(v 1,w) 1 A. We can use Lemma 17 for x = x(v 1, w) and y = 0. A eements of the correspondng set beong to A but η σx+ma 1 u 1 such that 0 u < r 2 and ma 1 r 2 x(v 1, w) + ma 1 u < ma 1 (r 2 + 1). As ma 1 (r 2 + 1) ma 1 r 2 < ma 1, there s at most one u satsfyng the prevous nequates. Takng u = [ r2 + (v 1)r2 ] we get 0 u r2+r3 1 < r 2 14

15 because r 2 > 1. Moreover x(v 1, w) + ma 1 u [ = ma 1 w ma 1 (v 1)r2 + ma 1 r r2 3 + (v 1)r2 ] ( = ma 1 r 2 w + ma 1 1 r 2 (v 1)r2 + [ r 2 + (v 1)r2 ]) ( = ma 1 r 2 w + ma 1 1 r 2 + (v 1)r2 ) ( = ma 1 r 2 w + ma 1 1 vr 2 ) = ma 1 r 2 + x(v, w). Therefore η σx(v,w)+ma 1 r 2 1 A and Lemma 16 for x = x(v, w) and z = 0 gves η σx(v,w) 1 A, too. On the other hand, f r 2 < then we can proceed the same way just nterchangng the use of Lemma 16 and Lemma 17 ony. Coroary 20. The set B 1 B 2 B 3 B 12 B 13 B 23 S, where B = { η 1 σx 0 < x < a }, s a Z-bass of the Snnott group of crcuar unts of k, see (6). Proof. We have C = E D, where E s the group of a unts of k, see [11, Proposton 1]. It s we known that η 12, η 13, η 23, η E whe η / E, but η 1 σx Theorem 19. E and η a p 1 E. Usng that, the coroary easy foows from 4. The group of a reatons The totay postve part D + of the group of crcuar numbers D s generated, as an abean group, by a conjugates of η, η j, η, see (5). We can consder the Z[G]-modue X = Z[Ga(k/Q)] Z[Ga(k K K j /Q)] Z[Ga(k K /Q)], 1 <j where G acts on each summand va restrcton. We denote x, x j, x the eements havng a coordnates zeros but ony 1 at the correspondng poston. Hence X = x, x 12, x 13, x 23, x 1, x 2, x 3 Z[G]. Snce η k, η j k K K j, and η k K, we have a surjectve near map ϕ: X D + of Z[G]-modues defned by ϕ(x) = η, ϕ(x j ) = η j, ϕ(x ) = η. To descrbe a presentaton of D + we need to understand the kerne of ϕ, whch s a Z[G]-submodue of X. By (7) and (8) we see that R N H x (1 Frob(p ) 1 )x j ker ϕ, (22) R N H j x (1 Frob(p ) 1 )x ker ϕ, (23) 15

16 for each {, j, } = {1, 2, 3}. But we have aso Ennoa reaton descrbed by Theorem 11 statng that there are α j Z[Ga(k K K j /Q)] such that ( u=0 v=u+1 σ a1u 1 σ a2v 2 H 1 H 21 R 1 R 2 R 3 )x (α 12 x 12 + α 13 x 13 + α 23 x 23 ) ker ϕ. (24) Theorem 21. The Z[G]-modue ker ϕ s generated by reatons (22), (23), and (24). Let M ker ϕ be the Z[G]-submodue generated by reatons (22) and (23) and et e ker ϕ be the Ennoa reaton descrbed by (24). Then the quotent (ker ϕ)/m s a cycc group of order m wth trva acton of G generated by the eement e + M. Proof. For each eement of B (see Theorem 19) we fx a premage n X wth respect to ϕ; et Y mean the set of these fxed premages. Theorem 19 gve that the eements of Y are Z-neary ndependent and that X = ker ϕ + Y Z. Snce the ony reatons we have used to prove Theorem 19 are (22), (23), and (24), we get the frst statement of the theorem. Hence the Z[G]-modue (ker ϕ)/m s generated by e + M. The way we have proved Theorem 11 shows that me M. To show that the order of the mage of e n (ker ϕ)/m s equa to m we construct, f m > 1, a homomorphsm ψ: X Z of abean groups as foows (warnng: ths s not a Z[G]-modue homomorphsm) ψ(σ u 1 x) = 1 for a u = 0, 1,..., ma 1 r 2 1, ψ(σ u 1 σ a2 2 x) = 1 for a u = 0, 1,..., ma 1r 2 1, and ψ(τx) = 0 f τ has a dfferent restrcton to Ga(k/Q) than σ1 u Moreover for any τ G we defne or σ u 1 σ a2 2. ψ(τx 12 ) = ψ(τx 13 ) = ψ(τx 23 ) = ψ(τx 1 ) = ψ(τx 2 ) = ψ(τx 3 ) = 0. Lemma 5 mpes that ψ: X Z s we-defned. For any τ G we have ψ(r 1 N 1 H 1 τx) { ma 1 r 2, 0, ma 1 r 2 }, ψ(r 2 N 2 H 2 τx) = ψ(r 3 N 3 H 3 τx) = 0, but ψ(e) = a 1 r 2. Therefore ψ gves a homomorphsm of abean groups X/M Z/(ma 1 r 2 )Z sendng e + M to an eement of order m. To prove the theorem we need to show that (1 σ )e M for each = 1, 2, 3. For any β, γ Z[G] et β γ mean res K/k β = res K/k γ. Let so α = u=0 v=u+1 σ a1u 1 σ a2v 2 H 1 H 21 R 1 R 2 R 3 Z[G], e = αx (α 12 x 12 + α 13 x 13 + α 23 x 23 ). (25) 16

17 Proposton 8 gves mα ΓH 1 H 21 R 1 R 2 R 3 = (N 1 2 N N 3 1 ) H 1 H 21 R 1 R 2 R 3. Snce (1 σ 1 )R 1 N 1 H 1 = 0 and by (14), we have Lemma 7 and (9) gve and so, by (2), (1 σ 1 )R 1 1 H 1 = (N 1 m)h 1 (26) (1 σ 1 )mα (N 3 N 2 )(N 1 m)h 1 H 21 R 2 R 3. σ ma1 1 H 1 H 21 = H 1 H 21 H 2 H 12 = σ ma2 2 H 2 H 12 σ ma2 2 H 1 H 21, N 1 N 3 H 1 H 21 = H 1 H 21 Hence H 1 H 21 u=0 u=0 therefore usng Lemma 7 agan σ a1u 1 σ a3v 3 H 1 H 21 u=0 σ a1(u v) 1 σ a2v 2 σ a1u 1 σ a2v 2 = N 1 N 2 H 1 H 21. (27) (1 σ 1 )mα m(n 2 N 3 )H 1 H 21 R 2 R 3, (1 σ 1 )α R 2 N 2 H 2 H 12 R 3 R 3 N 3 H 3 H 23 R 2 (28) because m s a non-zero-dvsor n Z[Ga(k/Q)]. Smary (1 σ 2 )mα (1 σ 2 )(N N 3 1 )H 1 H 21 R 1 R 2 R 3. We use Lemma 7, (2), and (26) to get and by (26) (1 σ 2 )N 3 1 H 1 H 21 R 2 (1 σ a2 2 )N 3 1 H 3 H 23 (1 σ a1 1 σ a3 3 )N 3 1 H 3 H 23 (1 σ a1 1 )N 3 1 H 1 H 21 = σ a1 1 N 3 (N 1 m)h 1 H 21 (1 σ 2 )N 1 2 H 1 H 21 R 2 (1 σ a2 2 )N 1 2 H 2 H 12 = (N 2 m)n 1 H 2 H 12. Hence by (27) we get (1 σ 2 )α σ a1 1 R 3 N 3 H 3 H 23 R 1 R 1 N 1 H 1 H 21 R 3. (29) 17

18 Fnay, permutng the ndces n the above computaton, (1 σ 3 )mα (1 σ 3 )(N 1 2 N 2 1 )H 1 H 21 R 1 R 2 R 3 thus ( σ a2 2 N 1 (N 2 m)h 1 H 21 + σ a1 1 N 2 (N 1 m)h 1 H 21 ) R1 R 2, (1 σ 3 )α σ a2 2 R 1 N 1 H 1 H 21 R 2 σ a1 1 R 2 N 2 H 2 H 12 R 1. (30) Therefore (25), (28), and (22) gve an eement f M such that (1 σ 1 )e + f = H 12 R 3 (1 Frob(p 2 ) 1 )x 13 H 23 R 2 (1 Frob(p 3 ) 1 )x 12 (1 σ 1 )(α 12 x 12 + α 13 x 13 + α 23 x 23 ). Snce (23) are the ony reatons we have used to prove Lemma 13, we have (1 σ 1 )e + f M + Y Z and ϕ((1 σ 1 )e + f) = 0 gves (1 σ 1 )e + f M, so (1 σ 1 )e M. Smary (29) and (30) gve (1 σ 2 )e, (1 σ 3 )e M and the theorem foows. Theorem 21 for our Prototypca Case (see Exampes 3 and 14) mpes that the norm reatons never suffce to generate a reatons, as we have m = > 1. Acknowedgment The authors thank the anonymous referee for the very usefu remarks and the advce to ustrate the resuts of ths paper on the Prototypca Case. References [1] H. Bass, Generators and reatons for cycotomc unts, Nagoya Math. J. 27 (1966), [2] K. Dohmae, On bases of groups of crcuar unts of some magnary abean number feds, J. Number Theory, 61(1996), [3] K. Dohmae, A note on Snnott s ndex formua, Acta Arth. 82(1997), [4] V. Ennoa, On reatons between cycotomc unts, J. Number Theory 4 (1972), [5] R. God, J. Km, Bases for cycotomc unts, Composto Math. 71 (1989), [6] J. M. Km, J. Ryu, A note on Ennoa reaton, Tawanese J. Math. 18 (2014), DOI: /tjm

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