Euclidean Systems. Stefan G. Treatman* Communicated by A. Granville. Received November 8, 1996; revised March 6,

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1 Journal of Number Theory 73, 7791 (1998) Artcle No. NT9884 Eucldean Systems Stefan G. Treatman* Department of Mathematcs and Computer Scence, Sant Joseph's Unversty, Phladelpha, Pennsylvana E-mal: treatmanerols.com Communcated by A. Granvlle Receved November 8, 1996; revsed March 6, 1998 A Eucldean rng such as the ntegers s equpped wth span algorthm for dvson wth remander. In non-eucldean Dedeknd domans wth cyclc class group, the defnton of a Eucldean deal class generalzes the noton of a Eucldean rng. Ths generalzes the algorthm for dvson wth remander. Let K be a number feld and let S be any fnte set of prmes of K whch contans the nfnte prmes S. Then for any rng of S-ntegers of K, we defne a Eucldean system. Ths further generalzes the noton of a Eucldean rng and the algorthm for dvson wth remander. We show that under certan condtons, a rng of S-ntegers has a Eucldean system Academc Press 1. INTRODUCTION 1.1. Outlne A Eucldean rng such as the ntegers s equpped wth an algorthm for dvson wth remander. In non-eucldean Dedeknd domans wth cyclc class group, Lenstra [9] generalzed the noton of a Eucldean rng wth hs defnton of a Eucldean deal class. Now, let K be a number feld and let S be a fnte set of prmes of K whch contans the nfnte prmes S. Then for any rng of S-ntegers of K, we defne a Eucldean system, a certan set of deal classes. Ths further generalzes the noton of a Eucldean rng. In partcular, we see n Theorem 8 that the deal classes n a Eucldean system generate the deal class group of the rng of S-ntegers. Furthermore, our man result s Theorem 10 n whch we show that for suffcently large S, f we assume a generalzed Remann hypothess, then a rng of S-ntegers has a Eucldean system. Ths generalzes work of Wenberger [13] n whch the class group n trval. * Current address: 4861 Battery Lane 3, Bethesda, MD X Copyrght 1998 by Academc Press All rghts of reproducton n any form reserved.

2 78 STEFAN G. TREATMAN 1.. Notatons We wll use the followng notatons and conventons. Let K be a number feld, wth O K ts rng of ntegers and S a fnte set of prmes of K contanng S, the set of nfnte prmes. v N, Q, and Z denote the natural numbers, ratonal numbers, and ntegers, respectvely. v Ideals wll be ndcated n boldface, such as a, b, p. v R S =[x # K : ord p x0 for all prmes p S] s called the rng of S-ntegers of K. v The deals of R S are n one-to-one correspondence wth the deals of O K whch are not dvsble by any prmes of S. v I S =[> p S p n(p), such that n(p)#z and n(p)=0 for all but fntely many prmes p]. We dentfy I S wth the group of all fractonal deals of R S. v For an deal b of O K wth ord p b=0 for all p # S, I S, b = [> p S, p % b p n(p) where n(p)#z and n(p)=0 for all but fntely many prmes p]. We dentfy I S, b wth the group of fractonal deals of R S generated by prmes not dvdng the deal br S. v For an deal b of O K, K b,1 =[x # K: (x)=(a)(b), ((a), b)=((b), b) =O K,andord p (x&1)m(p), where b=> p m(p) ]. v For an deal b of O K, I b =[> p % b p n(p), wth n(p)#z and n(p)=0 for all but fntely many p]. v If p s a prme of O K,thenK p s the completon at p. v For any n, `n s a prmtve nth root of unty. v I s the group of all fractonal deals of O K whch we dentfy wth [> p prme, p fnte p n(p), wth n(p)#z and n(p)=0 for all but fntely many p]. v s the natural map of K _ nto I. v Cl RS s the deal class group for the rng R S.. HISTORY Defnton 1. A rng R s sad to be Eucldean f there s a functon,: R N such that \a, b # R, wth b{0, _q, r # R such that a=bq+r wth,(r)<,(b). We say that, s a Eucldean algorthm for R. In 1979, Lenstra [9] generalzed the noton of a Eucldean rng by defnng a Eucldean deal class. We gve a verson of that defnton here. Let R be a Dedeknd doman wth deal class group Cl R and feld of fractons K. LetE=[b : b s a fractonal deal of R, and b$r].

3 EUCLIDEAN SYSTEMS 79 Defnton. Let C be an deal class of Cl R wth c # C. We say a functon : E N s a Eucldean algorthm for C f \b # E and \x # bc"c, _z # x+c such that (bcz &1 )<(b). In ths stuaton, we call C a Eucldean deal class. It s routne to check that ths defnton s ndependent of the choce of representatve c # C and that bcz &1 # E. The noton of a Eucldean deal class generalzes that of a Eucldean rng. Ths s seen n the followng lemma orgnally stated by Lenstra [9]. Lemma 3. Let R be a Dedeknd doman. Then R s a Eucldean rng f and only f [R] s a Eucldean deal class. Proof. Assume R s a Eucldean rng. Then R s a prncpal deal doman so that Cl R s a trval. Let C=[R] and let c=r be the representatve. We show that [R] s a Eucldean deal class. As R s Eucldean, Samuel [1, p. 84] shows that there s a mnmal Eucldean algorthm, for R (n the classcal sense) satsfyng () \a, b # R, ab{0,,(ab),(b), wth equalty a # R _, and (),(a)=1 a # R _. Snce R s a prncpal deal doman, any fractonal deal b of R whch contans R can be wrtten as b=(1b)=[tb: t # R], for some b # R. Let E=[b: b s a fractonal deal of R, and b$r]. Defne : E N by ((1b))=,(b)&1. Ths s well-defned, for f we express b=(1b)=(1b$), then b$=ab, for some a # R _. Thus,(b$)=,(ab)=,(b). Let b=(1b) be gven. Then for all x # br"r, we must fnd z # x+r such that (brz &1 ) <(b). Now any x # br s of the form x=tb for some t # R. Because, s a Eucldean algorthm for R, we can fnd q, r # R such that t=bq+r, wth,(r)<,(b). Note that as x # b"r, r{0. Let z=(tb)&q=(t&bq) b # x+r. Then \\1 b+ Rz&1+ = \\ 1 b+\ b t&bq++ = \\ t&bq++ 1 = \\1 r++ =,(r)&1 <,(b)&1= \\1 b++.

4 80 STEFAN G. TREATMAN Now assume that [R] s a Eucldean deal class wth Eucldean algorthm and let R represent the class [R]. Defne,: R N by,(b)= ((1b))+1, f b{0 and,(0)=0. Gven a, b # R wth b{0, we seek q, r # R such that a=bq+r, wth,(r)<,(b). Assume frst that b does not dvde a. Let b=(1b) and consder x=ab # br"r. Then snce s Eucldean for [R], there s some z # ab+r, say z=(ab)&q, such that (brz &1 )<(b). Ths mples b b+\ a&bq++ \\1 < \\ 1 b++, O \\ 1 a&bq++ < \\ 1 b++, O O,(a&bq)&1<,(b)&1,,(a&bq)<,(b). Thus f b does not dvde a, we may wrte a=bq+r, wth r=a&bq and,(r)<,(b). If b dvdes a, then we may wrte a=bq for some q # R. Thus r=0 and,(r)=0<11+((1b))=,(b). K Now let R be a rng of S-ntegers of a number feld K. LetN be the usual norm on R. That s, N(x)=*R(x). If we extend N to N : K Q by N (ab)=n(a)n(b) andn (0)=0, then t s routne to check that N s a Eucldean algorthm for a rng R f and only f \x # K, _y # R such that N (x& y)<1. The noton of Eucldean deal class becomes more accessble when the functon s gven by the norm by (b)=n(b &1 ). Here N s defned on all ntegral deals a by N(a)=*Ra and extended by multplcatvty to all fractonal deals, wth N((0))=0. Then N s a Eucldean algorthm for a class C, wth c # C, f and only f \x # K, _y # c such that N((x& y))<n(c). Note that when C=[R] and c=r, then N(c)=1. Thus, N s a Eucldean algorthm for R s a Eucldean algorthm for [R]. Ths we already knew from Lemma 3. Untl recently, f one dd not assume a generalzed Remann hypothess, there were no known examples of number felds K wth full rng of ntegers R for whch N was not a Eucldean algorthm, but some other functon was. It s known that N s not a Eucldean algorthm for Z[- 14] and much has been done to fnd another functon whch makes Z[- 14] Eucldean (see [6, 10]). In 1994, Clark [] gave the frst known example for quadratc felds when he demonstrated that the full rng of ntegers n Q[- 69] s Eucldean, but not norm-eucldean. In 199, Clark [1] showed that certan totally real quartc Galos extensons of Q have rngs of ntegers whch are Eucldean, but not norm-eucldean. In 1995, Clark and R. Murty extended ths result to totally real Galos extensons of Q of

5 EUCLIDEAN SYSTEMS 81 degree 3 satsfyng certan crtera. In 1996, Clark [3] gave examples of non-galos cubc felds whose rngs of ntegers are Eucldean, but not for the norm. Wthout relyng on a Remann hypothess, R. Gupta et al. [4] showed f K s number feld and S s a set of prmes of K, wth *S suffcently large, then f the rng of S-ntegers has class number 1, t must be a Eucldean rng. The norm s not often the rght functon for Eucldean deal classes. In fact, f R s the rng of ntegers n a quadratc extenson of Q, there are only a few rngs [9, p. 13] wth class number >1 for whch N s a Eucldean algorthm for a non-prncpal class C. One such example s the rng Z[- 10], whch has class number. The man result proved by Lenstra [9] about Eucldean deal classes s the followng. Theorem 4 (Lenstra). Let R be a Dedeknd doman wth Eucldean deal class C. Then Cl R s cyclc and s generated by C. Ths generalzes the fact that f R s a Eucldean rng, then R s a prncpal deal doman. 3. EUCLIDEAN SYSTEMS In the prevous secton, we saw that through the defnton of a Eucldean deal class, we can generalze the noton of a Eucldean rng to certan Dedeknd domans wth cyclc class group. In fact, we wll see that for *S, assumng a generalzed Remann hypothess, a rng of S-ntegers wth cyclc class group has a Eucldean deal class. Thus such a rng can be gven an arthmetc structure whch generalzes that of a Eucldean rng. We now explore more generally what can be sad f the class group s any fnte Abelan group. Defnton 5. Gven a number feld K and a Dedeknd doman R whose feld of fractons s K, let [C 1,..., C k ] be dstnct classes n the deal class group of R. Letc # C be a representatve of each class wth all c parwse co-prme. Let E=[b: b$r] be the set of all fractonal deals of R whch contan R. Letc= k c =1. We say that [C 1,..., C k ] s a Eucldean system for R f there s a functon : E N such that \b # E and \x # bc"c, there s some c j for whch there exsts z # x+c j wth (bc j z &1 )<(b). Such a functon s sad to be Eucldean for [C 1,..., C k ] or a Eucldean algorthm for [C 1,..., C k ] We call [C 1,..., C k ] a mnmal Eucldean system for R f no proper subset forms a Eucldean system.

6 8 STEFAN G. TREATMAN We note that f k=1, [C 1 ] s a Eucldean system f and only f C 1 s a Eucldean deal class. It s not hard to see that the defnton s ndependent of the choce of representatve for each class C. Lemma 6. Let [C 1,..., C k ] be a Eucldean system for R wth functon. Then \a, b # E, (ab)(b), wth equalty a=r. Proof. Fx b # E. Choose a$# E such that (a$b) s mnmal. Let x # a$c"c. Then x # a$bc"c as well. Thus by defnton, there s some c j and z # x+c j wth (a$bc j z &1 )<(a$b). But as x # a$ca$c j, we see z # a$c j + c j a$c j. Thus z # a$c j so we have a$c j z &1 # E. Thus mples that ((a$c j z &1 ) b)<(a$b) contradcts the mnmalty of (a$b). The only way to avod ths s for a$c"c to be empty. Ths occurs f and only f a$=r. Thus (ab) takes ts mnmal value f and only f a=r n whch case we have equalty (ab)=(b). Ths mples that for all a{r, (ab)> (b). K Corollary 7. If [C 1,..., C k ] s a Eucldean system wth functon, then (a) assumes ts mnmal value only when a=r. Proof. Take b=r n Lemma 6. K Theorem 8. Let K be a number feld wth R a Dedeknd doman whose feld of fractons s K. Let [C 1,..., C k ] be a Eucldean system wth functon. Then [C 1,..., C k ] generates the deal class group Cl R. Proof. We frst prove the followng lemma. Lemma 9. Let [C 1,..., C k ] be a Eucldean system wth functon. If b # E"[R], then b # > k =1 C&n wth n 0 and k n =1 (b). Proof of Lemma. Let c # C and c= k c =1.Letb # E"[R]. Then bc"c s non-empty so that we can fnd x # bc"c. By defnton, there s some c j and z # x+c j such that (bc j z &1 )<(b). If bc j z &1 =R, then [b][c j ]= [R] whch means b # C &1 j and by Corollary 7, 1(b). (By Corollary 7, bc j z &1 =R wll always occur f (b) s the mnmal value taken by on E"[R].) Now assume the lemma s true for all a # E wth (a)<(b). Then f bc j z &1 {R we have by assumpton that bc j z &1 # > k =1 C&n wth k n =1 (bc j z &1 ). Therefore, b # C &1 j > k =1 C&n => k =1 C&n$ where n$ =n for { j, n$ j =n j +1. Thus, k n$ =1 =1+ k n =1 1+(bc j z &1 ) (b). K To prove the theorem, we note that Cl R s generated by [[b]: b # E]. The lemma then shows that each generator can be wrtten as [b]= > k =1 C&n whch shows that [C 1,...C k ] generate the class group. K

7 EUCLIDEAN SYSTEMS 83 Remark. It s of nterest to note that f b{r s such that (b) s the mnmal value assumed by on E"[R], then for each x # bc"c, there s exactly one c j such that there s a z # x+c j wth (bc j z &1 )<(b). Suppose there are c j1 and c j for whch ths happens, wth z j1 # x+c j1 and z j # x+c j. Then by mnmalty, we must have (bc j1 z &1 j 1 )=R=(bc j z &1 j ). Ths mples that [c j1 ]=[c j ] whch shows that j 1 = j as the C are dstnct classes. We now come to our man result. Theorem 10. Let K be a number feld and O K be ts rng of ntegers. Let S be a fnte set of prmes ncludng S and let R S be the rng of S-ntegers. Suppose the class number of R S s h and the deal class group Cl RS $ Zd 1 ZZd Z }}} Zd n Z, wth ths structure gven unquely by d 1 d }}} d n. Let [C 1, C..., C n ] be generators of Cl RS wth the order of C =d. Suppose the rank of the unt group n R S s s. Suppose further that for every squarefree nteger m and for every subset S$/S, the zeta-functon for K(`m, R 1m S$ ) satsfes the generalzed Remann hypothess. Then f smax[1, n&1], [C 1,..., C n ] s a mnmal Eucldean system for R S. Note that as [C 1,..., C n ] s a mnmal set of generators for Cl RS, Theorem 8 shows that f t s a Eucldean system, then t must be mnmal. Proof. The proof s constructve. Gven a set of generators [C 1,..., C n ], we shall wrte down a functon and show that t makes [C 1,..., C n ] nto a Eucldean system. For each class C, we choose a prme deal c of R S as a representatve and wrte c= n =1 c.lete=[b: b$r S ] be the set of all fractonal deals of R S whch contan R S. Here, we dentfy all fractonal deals of R S wth dvsors b=> p # RS p n(p), wth n(p)#z and n(p)=0 for all but fntely many prmes p. We defne : E N by (b)= : p prme p # R S ord p (b &1 ) n p, (3.1) where f p # > n =1 C m s wrtten unquely wth 1m 1 d 1 and 0m d &1 for n, then

8 84 STEFAN G. TREATMAN h+1, f p # C 1 and the projecton R _ S (R S p) _ s surjectve, h+, f p # C 1, C 1 C,..., or, C 1 C n, n p ={b b h+k, f n =1 m =k, k{1, h+1+d 1, f p # C 1 and R _ S (R S p) _ s not surjectve. For example, f Cl k $ZZZZZ4Z, thenh=16 and the values of n p are gven n the followng table. Value of n p Class of Cl K h+1=17 C 1 wth R _ S (R S p) _ h+=18 C, C 1 1C, C 1 C 3 h+3=19 h+4=0 C C 1, C C 1 3, C 1 C C 3, C 1 C, C 3 1 wth R _ % (R S Sp) _ C C 1 C 3, C 1 C, C 3 1C C, C 3 1C 3 3 h+5=1 C C 1 C, 3 C 1 C3, C 3 1C C 3 3 h+6= C C 1 C 3 3 In ths example, note that for a prme p # C 1 wth R _ (R S Sp) _ not surjectve, we have n p =h+3=h+1+d 1 as d 1 =. Also note that prncpal prmes p have n p =h+=h+d 1. More generally, a prncpal prme p wll always have n p =h+d 1.Ifn>1, the largest value assumed by any n p s always h+(d 1 +}}}+d n )&(n&1). Ths occurs when all the m are maxmal. In the above example, h+(d 1 +}}}+d n )&(n&1)=h+ (++4)&(3&1)=h+6. If n=1, the class group s cyclc and the largest value assumed by any n p s h+d 1 +1=h+1. In any case, the maxmal value of n p h+1. (3.) Havng now defned our functon : E N n (3.1), we must show that t s Eucldean for [C 1,..., C n ].LetE$=[b : b s an ntegral deal of R S ]. Defne a functon : I S Z by (a)= : p prme p # R S ord p (a) n p, where n p s defned as above. Note that f a s an ntegral deal or a fractonal deal contanng R S, we have the relatonshp: (a)= {&(a), (a &1 )=& (a &1 ), f a # E. f a # E$.

9 EUCLIDEAN SYSTEMS 85 Let a # E$ (soa &1 # E). To show s Eucldean, for all x # a &1 c"c, we need to fnd some c j and z # x+c j such that (a &1 c j z &1 )<(a &1 ), or equvalently, (ac &1 j z)< (a). Snce s a homomorphsm from I S (Z, +), we see that s Eucldean f \a # E$ and \x # a &1 c"c, _c j and z # x+c j such that,.e.,.e., (a)+ (c &1 j )+ ((z))< (a), ((z))<& (c &1 j ), ((z))< (c j ). Ths formulaton s now ndependent of a. Note that K= a # E$ a &1 c. Thus n provng that s Eucldean, t suffces to show that \x # K"c, there s some c j and z # x+c j wth ((z))< (c j ). (3.3) To prove (3.3), we begn wth x # K"c. For any 1n, consder the fractonal deal (x)c of R S and wrte (x)c =a b, wth a and b unquely wrtten as co-prme ntegral deals of R S.LetF be the S-ray class feld for the modulus b so that I b H $Gal(F K), (3.4) where H s the subgroup (K b,1)}(p# S) of I b. (Note that I b H $ I S, b ($(Kb,1)), where f 6 s the projecton of I b onto I S, b,then$=? b.) As (a, b )=R S, we have a # I S, b and thus under the Artn recprocty map [7, p. 197], a corresponds to some { # Gal(F K). In fact, there are nfntely many ntegral deals a$ # I S, b such that (a$, F K)={. For any such a$, t follows that a$ #a n I S, b ($(Kb,1)). That s, a$ =(#) a for some # # K b,1. We may wrte #=1+t wth ord q (t)n(q), where b => q n(q). (3.5) Let z=x#=x(1+t)=x+xt. We now show that xt # c so that z # x+c. Snce a (#)=a$ s an ntegral deal n R S, we have that for all a # a, a# # R S. Ths mples that a+at # R S whch n turn shows that at # R S. Ths shows that a (t) s an ntegral deal. We consder (xt)=(a c b )(t)= (c b ) a (t). By (3.5), b dvdes the ntegral deal a (t) sothat(xt)=c br

10 86 STEFAN G. TREATMAN for some ntegral deal br. Ths mples that xt # c and that z # x+c. So for any x # K"c, we have found z # x+c such that ((z))= ((x#))= _ a c b = (c )+ (a$ )& (b ). (#) &= \ c a$ b + If we can choose a$ so that (a$ )< (b ), then the above shows that ((z))< (c ), satsfyng (3.3). We therefore complete the proof of the theorem by showng that for at least one, 1n, we can fnd some a$ # I S, b wth a$ #a n I S, b $(Kb,1) and (a$ )< (b ). If any b s not prme then we are done. By defnton of n p,fb s not prme, then (b )h+. Now the Chebotarev densty theorem [7, p. 169] guarantees that there are nfntely many prmes p#a n I S, b $(Kb,1). Choose any such p. Then by (3.), (p)h+1. Hence we may take a$ =p so that (a$ )h+1<h+ (b ) as requred. Henceforth, when wrtng (x)c =a b, we may assume that all b are prmes of R S. Lemma 11. Let [C 1,..., C n ] be as above wth prme c # C and c= n c =1. Let x # K"c be gven and for each, wrte (x)c =a b wth all b prmes of R S. Assume b 1 {c 1. Then for all, b =b 1. Wrte b 1 # C m 1 1 Cm, wth ths unquely defned by 1m n 1d 1, 0m d &1 for n. Then for all, a # C m 1 1 Cm }}}C m &1. n Proof. Snce (x)c 1 =a 1 b 1, t s clear that a 1 # C m 1 &1 1 C m n. Now for any {1, (x) c = (x) c 1 c 1 c = a 1 b 1 c 1 c = a b. As b s prme, t must be that (a 1 b 1 )(c 1 c ) s not n lowest terms. But a 1 and b 1 are co-prme and by assumpton, b 1 {c 1. Snce all the c are dstnct prmes, we must have that c dvdes a 1. Hence, a b =a~ 1 c 1 b 1, where a~ 1 = a 1 c. It now follows that b =b 1 and that a # C m 1 1 Cm }}}C m &1. K n We now complete the proof by consderng the three possbltes for the prme b 1.Letb 1 # C m 1 1 Cm. n Case 1. b 1 # C 1. If the projecton R _ (R S Sb 1 ) _ s not surjectve, then by defnton of n p, (b 1 )=h+1+d 1. Clearly n ths case, a 1 s prncpal. In the comments precedng (3.), we saw that any prncpal prme p has (p)=h+d 1. Snce Cl RS s a quotent of the S-ray class group, any prme p#a 1 n I S, b 1 $(Kb1,1) wll also be equvalent to a 1 n Cl RS. Thus we take a$ 1 =p

11 EUCLIDEAN SYSTEMS 87 for any p#a 1 n I S, b 1 $(Kb1,1) Then p s prncpal and (a$ 1 )= (p)= h+d 1 <h+d 1 +1= (b 1 ). If the projecton R _ (R S Sb 1 ) _ s surjectve, then by defnton, (b 1 )=h+1. Agan a 1 =(a 1 ) s prncpal wth a 1 #(R S b 1 ) _. It follows that (a 1 )#R S n I S, b 1 $(Kb1,1). Hence, we may choose a$ 1 =R S so that (a$ 1 )= (R S )=0<h+1= (b 1 ). Case. b 1 C 1, (b 1 )>h+. We know by defnton that snce (b 1 )>h+ we have n m =1 >. Ths leaves three possbltes: () m 1 >, () for at least one # [,..., n] we have m, or () for some 1, # [,..., n], m 1 1 and m 1. In the case that m 1 >, we consder (x)c 1 =a 1 b 1. It follows that a 1 # C m 1 &1 1 C m n. By defnton of n p, any prme p # C m 1 &1 1 C m n wll have (p)=h+((m 1 &1)+m +}}}+m n )= (b 1 )&1. So take a$ 1 =p where p s any prme of R S and such that p#a 1 n I S, b 1 $(Kb1,1). (The Chebotarev densty theorem mples there are nfntely many such p). Then p#a 1 n Cl RS too and we have found a$ 1 such that (a$ 1 )= (p)= (b 1 )&1< (b 1 ) as requred. In (), we may assume that for some {1, m. Then consder (x)c =a b. By Lemma 11, we have b =b 1 and a # C m 1 1 Cm }}}C m &1. n Note that by defnton of n p, any prme p n the same class as a wll have (p)=h+(m 1 +m +}}}+(m &1)+ } } } +m n )= (b )&1. By Chebotarev, we can fnd a prme p # R S wth p#a n I S, b 1 $(Kb1,1). Then p#a n Cl RS as well. We may take a$ =p so that (a$ )= (p)= (b 1 )&1< (b ), as requred. In (), we have 1 and such that m 1 1 and m 1. As n (), we consder (x)c 1 =a 1 b 1. Agan by Lemma 11, we have b 1 =b 1 and a 1 # C m 1 1 Cm }}}C m 1 &1 1. So by defnton of n n p, any prme p n the same class as a 1 wll have (p)=h+(m 1 +m +}}}+(m 1 &1)+ } } } + m +}}}+m n )= (b )&1. So by Chebotarev, we can fnd a prme p # R S such that p#a 1 n I S, b 1 $(Kb1,1). Then p#a 1 n Cl RS as well. We may take a$ 1 =p so that (a$ 1 )= (p)= (b 1 )&1< (b 1 ), as requred. Case 3. b 1 C 1, (b 1 )=h+. By defnton, b 1 # C 1 C j for some j # [1,,..., n]. To mmc the above arguments, we need to fnd a class of Cl RS n whch prmes p wll have (p)=h+1. Ths can only happen f p # C 1 and R _ (R S Sp) _ s surjectve. Hence we consder (x)c j =a j b j. By Lemma 11, b j =b 1 and a j # C 1. Let {=(a j, F 1 K). We seek a prme p # R S such that p#a j n I S, b 1 $(Kb1,1) wth R _ (R S Sp) _. Ths s equvalent to fndng p such that (p, F 1 K) ={ and R _ S (R S p) _, where F 1 s the S-ray class feld for the modulus b 1. To fnd such p, we apply a theorem of Lenstra [8, (4.8) p. 08]. We consder the specal case of ths theorem n whch F 1 s the S-ray class feld for the modulus b 1, C=[{], W=R _ S, and k=1. We assume that for every

12 88 STEFAN G. TREATMAN subset S$/S and for every square-free nteger m that the zeta-functon for K(`m, R 1m S$ ) satsfes the generalzed Remann hypothess. Then the theorem says that the set of prmes p, for whch (p, F 1 K)={ and R _ (R S Sp) _ s surjectve, s nfnte f and only f there s no prme l for whch there s a feld L l =K(`1, R _1l S ) such that K/L l F 1 and { # Gal(F 1 L l ). If we can show no such feld L l exsts wth K/L l F 1 and { # Gal(F 1 L l ), then we can always fnd a sutable prme p # R S wth (p)=h+1<h+= (b 1 ). In ths event, we take a$ j =p and then (a$ j )< (b) as requred. Thus we complete the proof of Theorem 10 f we can show there s no such L l as above. Henceforth, we suppose such an L l exsts as above, and derve a contradcton. Lemma 1. Wth the current defntons, f there s an L l wth K/ L l F 1 and { # Gal(F 1 L l ), then `l # K. Proof. By class feld theory, F 1 K s an Abelan (Galos) extenson. Hence any ntermedate feld must be Abelan over K as well. In partcular, let u be any unt of R _ S whch s not an lth power, for nstance, any fundamental unt. Let K$=K(u 1l )L l,so[k$:k]=l. Then K$ s Abelan over K and must be the splttng feld of x l &u over K. Ths mples that K$=K(`l, u 1l ). Clearly we have KK(`l)K$. But note that [K(`l):K]l&1 and dvdes [K$:K]=l so that [K(`l) :K] must be 1. Therefore `l # K. K Next, we note that snce Cl RS s a quotent of I S, b 1 $(K b1,1), the S-Hlbert class feld H of K s a subfeld of F 1. We consder the feld H & L l. We have { # Gal(F 1 K) such that { fxes L l. But recall that {=(a j, F 1 K) and a j # C 1. Ths means that { H =_ 1, where _ 1 corresponds to C 1 under the somorphsm Cl RS $Gal(HK). Therefore, _ 1 generates a subgroup of Gal(HK) of order d 1. It follows that f H$ denotes the fxed feld of _ 1, then Gal(H$K)$Zd ZZd 3 Z }}}Zd n Z. (3.6) Snce { fxes L l, t must be that L l & HH$. Further, by Galos theory, we know that [L: H & L l ] dvdes [F 1 : H]. (3.7) To determne [F 1 : H], we examne the exact sequence 0 (R S b 1 ) _?(R _ S ) I S, b 1 $(K b1,1) Cl RS 0, where?: R _ (R S Sb 1 ) _ s the natural projecton. To see why ths s exact, we recognze that the S-ray class group I S, b 1 $(Kb1,1) projects naturally onto Cl RS =I S $(K _ ). We determne that the kernel of ths projecton conssts

13 EUCLIDEAN SYSTEMS 89 of those fractonal deals of I S, b 1 $(K b1,1) whch are prncpal. These are of the form (ab) for some a, b #(R S b 1 ) _. Now, f b # R S represents a multplcatve nverse of b n (R S b 1 ) _, then we see that \ a b+ #(ab ) n I S, b 1 $(K b1,1). Thus every prncpal fractonal deal of I S, b 1 can be represented n I S, b 1 $(K b1,1) by an ntegral deal (ab ) where ab #(R S b 1 ) _. But we must consder that for any unt u # R _, and for any x #(R S Sb 1 ) _, we have (x)=(xu). Thus the kernel s gven by ((R S b 1 ) _?(R _ S )) and the above sequence s exact. Thus *(I S, b 1 $(Kb1,1))=*Cl RS } *((R S b 1 ) _?(R _ )). S Ths yelds [F 1 : K]=h } *((R S b 1 ) _?(R _ S )), by (3.4). As [H : K]=h, we have determned that [F 1 H]=*((R S b 1 ) _?(R _ S )). (3.8) To determne [L l : H & L l ], we note that by Lemma 1, L l s a Kummer extenson [11, p. 15] of K. In fact, f the [u 1,..., u s ] form a system of fundamental unts of R _, then L S l=k(u 1l 1,..., u1l s, `1l l r ), where r1 s maxmal such that `lr # K. Hence, [L l : K]=l s+1 and Gal(L l K)$ZlZ }}}ZlZ, where there are s+1 copes. Because Gal(H & L l K) s a quotent of Gal(L l K), we have Gal(H & L l K)$ZlZ }}} ZlZ, wth the number of copes equal to some ts+1. But Gal(H & L l K) must also be a quotent of Gal(H$K), so by (3.6), tn&1. By assumpton, s max[1, n&1] so that s+1max[, n]. It follows that t<s+1 and [L l : H & L l ]=l s+1&t. (3.9) As a result, there s some unt u # [u 1,..., u s, `lr] such that K(u 1l )/3 H & L l.letk$=k(u 1l ). Because K$/3 H and H s the maxmal unramfed Abelan extenson of K, K$K s ramfed at some prme l of R S. In fact, snce the mnmal polynomal for u 1l over K s f(x)=x l &u, we have Dsc( f(x))=dsc(x l &u) =\Nm K$K l(u 1l ) l&1 =\l l u l&1. So (Dsc( f(x)))=(l) l whch shows that Dsc(K$K) dvdes (l) l. Thus, l must be a prme of R S lyng over l. On the other hand, K$/F 1 and the only prmes of R S whch ramfy n F 1 are those dvdng the modulus b 1.

14 90 STEFAN G. TREATMAN However, b 1 s prme and we conclude that b 1 =l and thus R S b 1 has characterstc l. From (3.8), we see now that [F 1 : H] dvdes l f &1, for some f 1. But we have already establshed n (3.9) that [L l : H & L l ]=l k, for some k1. Ths contradcts (3.7). Thus there can be no such L l wth K/L l F 1 and { # Gal(F 1 L l ). Ths completes the proof of the theorem. K Concluson. Recall that f a rng R s Eucldean wth a multplcatve functon whch s extended to : K Q by (ab)=(a)(b) and (0)=0, then the dvson algorthm can be stated as, \x # K, _y # R such that (x& y)<1. A Eucldean system generalzes ths n the followng way. Because s a homomorphsm, we have that \x # K"c, there s some c such that _y # c such that ((x& y))< (c ). Remark. In the specal case of Theorem 10 n whch n=1, d 1 1, R S has a cyclc class group of order h=d 1. The theorem then says that [C 1 ] s a mnmal Eucldean system whch s equvalent to sayng that C 1 s a Eucldean deal class. Ths proves a theorem orgnally stated by Lenstra [9, p. 17]. The algorthm s then gven by h+1, f p # C 1 and R _ (R S Sp) _ s surjectve, h+, f p # C, 1 n p ={b b h, f p # C h 1,.e., p s prncpal, h+1, f p # C 1, R _ (R S Sp) _ s not surjectve. Ths s not the mnmal possble algorthm. Remark. If we take the specal case of Theorem 10 n whch n=1 and d 1 =1, then Cl RS s trval and thus Cl RS =[[R S ]]. The theorem then says that f R S s a PID and the number of unts s nfnte, then assumng a generalzed Remann hypothess, [[R S ]] s a Eucldean system. Equvalently, [R S ] s a Eucldean deal class and thus R S s a Eucldean rng, by Lemma 3. In ths case the algorthm s gven by n p = {, f the projecton R_ S (R Sp) _ s surjectve. 3, otherwse. Ths s precsely the algorthm gven by Wenberger [13] n hs proof that for any rng R S wth nfntely many unts and S=S, assumng a generalzed Remann hypothess, R S s a PID R S s Eucldean.

15 EUCLIDEAN SYSTEMS 91 REFERENCES 1. D. A. Clark, ``The Eucldean Algorthm for Galos Extensons of the Ratonal Numbers,'' Ph.D. Thess, McGll Unversty, Montreal, D. A. Clark, A quadratc feld whch s Eucldean but not norm-eucldean, Manuscrpta Math. 83 (1994), D. A. Clark, Non-Galos cubc felds whch are Eucldean but not norm-eucldean, Math. Comp. 65 (1996), R. Gupta, M. R. Murty, and V. K. Murty, The Eucldean algorthm for S-ntegers, Canad. Math. Soc. Conf. Proc. 7 (1987), A. Fro hlch and M. J. Taylor, ``Algebrac Number Theory,'' Cambrdge Unv. Press, Cambrdge, F. Lemmermeyer, The Eucldean algorthm n algebrac number felds, Exposton. Math. 13, No. 5, S. Lang, ``Algebrac Number Theory,'' nd ed., Sprnger-Verlag, New YorkBerln, H. W. Lenstra, Jr., On Artn's conjecture and Eucld's algorthm n global felds, Invent. Math. 4 (1977), H. W. Lenstra, Jr., Eucldean deal classes, Journe es Arth. Lumny Aste rsque 61 (1979), M. Nagata, Some questons on Z[- 14], n ``Algebrac Geometry and Its Applcatons,'' pp. 3733, West Lafayette, IN, J. Neukrch, ``Class Feld Theory,'' Sprnger-Verlag, New YorkBerln, P. Samuel, About Eucldean rngs, J. Algebra 19 (1971), P. J. Wenberger, On Eucldean rngs of algebrac ntegers, Proc. Sympos. Pure Math. 4 (1973), 3133.