A NOTE ON RINGS OF FINITE RANK

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1 A NOTE ON RINGS OF FINITE RANK PETE L. CLARK Abstract. The rank rk(r) of a rng R s the supremum of mnmal cardnaltes of generatng sets of I as I ranges over deals of R. Matsuda and Matson showed that every n Z + occurs as the rank of some rng R. Motvated by the result of Cohen and Glmer that a rng of fnte rank has Krull dmenson 0 or 1, we gve four dfferent constructons of rngs of rank n (for all n Z + ). Two constructons use one-dmensonal domans. Our thrd constructon uses Artnan rngs (dmenson zero), and our last constructon uses polynomal rngs over local Artnan rngs (dmenson one, rreducble, not a doman). 1. Introducton For a module M over a rng 1 R, let µ(m) be the mnmal cardnalty of a set of generators of M as an R-module, and let µ (M) be the supremum of µ(n) as N ranges over all R-submodules of M. We say M has fnte rank f µ (M) < ℵ 0. Ths mples that M s a Noetheran R-module. We defne the rank rk(r) as µ (R). Thus for n N, rk(r) = n means that every deal of R can be generated by n elements and some deal of R cannot be generated by fewer than n elements Motvaton and Man Results. Ths note s drectly motvated by the followng result. Theorem 1.1. a) (Matsuda [Ma84]) For all N Z +, there s a doman R N wth rank(r N ) = N. b) (Matson [Ma09]) One may take for R N a subrng of the rng of ntegers of Q(2 1 N ). Here we wll explore the class of rngs of fnte rank wth an eye to constructng further famles wth all possble fnte ranks. We begn n 2 wth the case of domans. We revew the poneerng work of Cohen and use t to deduce a local-global prncple for domans of fnte rank: Theorem 2.6. In 3 we show that gven any PID A wth fracton feld F A and any feld extenson K/F of degree N Z 2, there s a nonmaxmal A-order n K of rank N, generalzng Theorem 1.1. We also construct, for any 2 n N, a Z-order n a degree N number feld wth rank n. In 4 we consder the case of rngs whch are not domans. We gve a general dscusson of Artnan rngs (whch always have fnte rank) and show that there are Artnan rngs of rank n for any n Z +. Fnally we determne when a polynomal rng has fnte rank, show that for any local Artnan rng r, the rank of r[t] s bounded above by the length of r, and show that we have equalty when r s moreover prncpal, so that e.g. for all n Z +, Z/2 n Z[t] has rank n Prelmnares on Change of Rngs. Remark 1.2. Let ι : R S be a rng map. For an R-module M, let ι (M) be the S-module M S. Then µ(ι (M)) µ(m). If every deal of S s ι (I) for some I, we get rk(s) rk(r). Ths holds when ι s a quotent or a localzaton map. 1 Here all rngs are commutatve and wth multplcatve dentty. 1

2 2 PETE L. CLARK For an R-module M and a prme deal p of R, we put M p = M R p and µ p (M) = µ(m p ). Remark 1.2 gves µ p (M) µ(m). By way of a converse, we have: Theorem 1.3. (Forster-Swan [Fo64] [Sw67]) Let M be a fntely generated module over the Noetheran rng R. Then 1.3. Acknowledgments. µ(m) sup (µ p (M) + dm R/p). p Spec R Thanks to K. Conrad, P. Pollack and L.D. Watson for useful conversatons. 2. Domans of Fnte Rank I.S. Cohen ntated the study of ranks of domans. We recall some of hs results. Theorem 2.1. (Cohen [Co50]) If R s a doman of fnte rank, then dm R 1. For R Noetheran and p Spec R, let k(p) be the fracton feld of R/p; put z p (R) = dm k(p) pr p /p 2 R p. Then z p (R) dm R p ; p s regular f equalty holds, otherwse sngular. Put z(r) = sup z p (R). Suppose R s a one-dmensonal Noetheran doman. Then by Krull-Akzuk [CA, Thm. 18.7] the normalzaton R s Dedeknd and Spec R Spec R s surjectve and fnte-to-one. The followng well known result dspenses wth the normal case. Proposton 2.2. We have rk(r) 2, wth equalty ff R s not prncpal. Proof. By Forster-Swan, a non-prncpal Dedeknd doman has rank 2. Or: let I be an deal of R, let 0 x I, and factor (x) as p a1 1 par r. Then R/(x) = r =1 R/pa = r =1 R p /(p R p ) a s prncpal, so I = x, y for some y I. Henceforth we suppose R s not normal. Let c = (R : R) = {x K xr R} be the conductor of R: t s the largest deal of R whch s also an deal of R. If p s regular then pr s also prme and R p RpR. We say R s nearly Dedeknd f c 0; equvalently, f R s a fntely generated R-module. In a nearly Dedeknd doman, the sngular prmes are characterzed as: the prmes p such that p + c R; the radcals of the deals n a prmary decomposton of c; or the prmes of R lyng under prmes of R whch dvde c. They are fnte n number. Theorem 2.3. (Cohen [Co50]) A nearly Dedeknd doman has fnte rank. Let (R, m) be a one-dmensonal local Noetheran doman. Then the sequence {dm R/m m /m +1 } =1 s eventually constant [Sa78, p. 40]; ts eventual value s the multplcty e(r) of R. If R s a one-dmensonal Noetheran doman and q MaxSpec R, then we put e q (R) = e(r q ) and e(r) = sup e p (R). Example 2.4. (Sally [Sa78, p. 5]) For a feld k, put R = k[[t 3, t 4 ]]. Then R s a one-dmensonal Noetheran local doman wth maxmal deal p = t 3, t 4 and R/p = k. Moreover we have p = t 3, t 3+1, t 3+2 = t 3 k[[t]] for all 2, so Thus z(r) = 2 < 3 = e(r). µ(p) = dm k p/p 2 = 2; 2, µ(p ) = dm k p /p +1 = 3.

3 A NOTE ON RINGS OF FINITE RANK 3 Proposton 2.5. Let R be a one-dmensonal local Noetheran doman. Then rk(r) = e(r). Proof. Nakayama s Lemma gves e(r) rk(r). Snce R s Cohen-Macaulay, the nequalty rk(r) e(r) follows from [Sa78, Thm. 3.1]. The concluson of Proposton 2.5 need not hold f R s not assumed to be local. Indeed, f R s a nonprncpal Dedeknd doman, then rk(r) = 2 by Proposton 2.2, whle e(r) = 1 snce R s locally prncpal. However, ths s the only problem case, as s shown by the followng result. Theorem 2.6. Let R be a one-dmensonal, non-normal Noetheran doman. Then rk(r) = e(r) = sup rk(r p ). Proof. By Proposton 2.5, for all p MaxSpec R we have e p (R) = rk(r p ) and thus By Remark 1.2 we have e(r) = sup rk(r p ). sup rk(r p ) rk(r). Snce R s not normal, there s some p MaxSpec R such that R p s not prncpal, and thus by Proposton 2.5 we have e(r) e p (R) = rk(r p ) 2. Let I be an deal of R. Applyng Forster-Swan and Proposton 2.5, we get ( ) ( ) µ(i) max sup µ(ir p ), 2 max sup e p (R), 2 e(r), so rk(r) e(r). Remark 2.7. In Theorem 2.6 one can have rk(r) = ℵ 0 [Co50, pp ] A Frst Example. 3. Nonmaxmal Orders One knows examples of local nearly Dedeknd domans R wth multplcty e(r) any gven n Z + : e.g. [Wa73]. The followng s perhaps the most famlar. Example 3.1. For a feld k and n Z +, let R n = k[[t n, t n+1,..., t 2n 1 ]] = k[[t n ]] + t n k[[t]] = k + t n k[[t]]. Then R n s local nearly Dedeknd wth maxmal deal m = t n,..., t 2n 1 = t n k[[t]]] and R n /m = k. For Z + we have m = t n,..., t (+1)n 1 = t n k[[t]], so rk(r n ) = e(r n ) = lm dm Rn/m m /m +1 = lm n = n = z(r n ).

4 4 PETE L. CLARK 3.2. Nonmaxmal Orders I: Maxmal Rank. Let A be a PID wth fracton feld F, and let K/F be a feld extenson of degree N Z 2. An A-order n K s an A-subalgebra R of K whch s fntely generated as an A-module and such that F A R = K. We say R s an A-order of degree N. The structure theory of modules over a PID mples R = A A N. Let R be an A-order n K. Then ts normalzaton R s the ntegral closure of A n K. By Krull-Akzuk, R s a Dedeknd doman. If R s fntely generated as an A-module then t s an R- order n K, the unque maxmal order. It can happen that R s not a fntely generated A-module, but R s fntely generated f K/F s separable or A s fntely generated over a feld. Remark 3.2. Let A be a PID, not a feld, wth fracton feld F, and let K/F be a feld extenson of degree N Z 2. If the ntegral closure S of A n K s not fntely generated as an A-module, then K admts no normal A-order. But t always admts some A-order: start wth an F -bass of K, scale to get an F -bass α 1,..., α N of elements ntegral over A, and take S = A[α 1,..., α N ]. Proposton 3.3. Let N Z 2, let A be a PID, and let R be a non-normal A-order of degree N. If there s p MaxSpec R such that z p (R) = N, then rk(r) = N. Proof. Because R s free of rank N as a module over the PID A, every deal of I of R s a free R-module of rank at most N and thus µ(i) N, so rk(r) N. On the other hand rk(r) e p (R) z p (R) = N. We say an A-order n K has maxmal rank f rk(r) = N = [K : F ]. Theorem 3.4. Let A be a PID wth fracton feld F A, and let K/F be a feld extenson of degree N Z 2. Then there s an A-order R n K of maxmal rank. Proof. By Remark 3.2, there s an A-order S n K. Let x be a nonzero, nonunt n A, so x = ɛp a1 1 par r where r Z +, ɛ A, p 1,..., p r are nonassocate prme elements and a 1,..., a r Z +. Put R = R(S, x) = A + xs. We clam: () R s an A-order; () for 1 r there s a unque p Spec R wth p A = (p ); and () z p (R) = N for all. Assumng the clam, Proposton 3.3 gves rk(r) = N. We show the clam: () Certanly R s a subrng of K. If α 1 = 1, α 2,..., α N s an A-bass for S, 2 then 1, xα 2,..., xα N s an A-bass for R. So R s an A-order n K. () Fx 1 r and defne p = p A + xs. Then p s an deal of R and R/p = A/(p ) s a feld, so p s a prme deal of R contanng p. Let q be a prme deal of R such that q A = (p ). Then snce p x we have p 2 = (p A + xs) 2 = p 2 A + p xs + x 2 S = p 2 A + p xs p R q. Snce q s prme we get q p and thus (snce dm R = 1) q = p. () Snce p, xα 2,..., xα N s an A-bass for p and p 2, p xα 2,..., p xα N s an A-bass for p 2, we have p /p 2 = A (A/(p )) N. Thus for all 1 r we have z p (R) = dm R/p p /p 2 = dm A/(p)(A/(p )) N = N. Remark 3.5. a) If R s a PID wth fracton feld F R, then F admts a degree N feld extenson for all N Z 2 : for (p) MaxSpec R, we can take K = F (p 1 N ). b) The R = A + xs constructon s modelled on [Co, Thm. 3.15]. 2 We are permtted to take α1 = 1 by Hermte s Lemma [CA, Prop. 6.14].

5 A NOTE ON RINGS OF FINITE RANK 5 Example 3.6. Let k be a feld. For n Z 2 put A = k[[t n ]], so F = k((t n )), and let K = k((t)). Let S = k[[t]], the maxmal A-order n K. Then R(S, t n ) = k[[t n ]] + t n k[[t]] s the rng R n of Example 3.1. Example 3.7. For n Z 2 put A = Z, K = Q(2 1 n ), S = Z[2 1 n ] and x = 2. Then R = R(S, x) = Z[2 n+1 n, 2 n+2 n,..., 2 2n 1 n ] has rank n, and t s the order n K that Matson used to prove Theorem 1.1b) Nonmaxmal Orders II: Pullbacks and Locally Maxmal Orders. Let A be a DVR wth fracton feld F and resdue feld k. Let l/k be a separable feld extenson of degree d Z 2, let K/F be the correspondng degree d unramfed (hence separable) feld extenson, and let S be the ntegral closure of A n K, so S s a DVR wth maxmal deal m (say) and S/m = l. Let q : S l be the quotent map, and put R = q 1 (k). By [AM92, pp ] and the Eakn-Nagata Theorem [CA, Cor. 8.31], R s local nearly Dedeknd wth normalzaton S and an A-order n K. In loc. ct., Anderson and Mott show m = q 1 (0) = m and R/m = k. Fnally, m/m 2 = m/m 2 s a one-dmensonal l-vector space hence a d-dmensonal k = R/m-vector space, so Proposton 3.3 apples: we have rk(r) = d. Theorem 3.8. For r Z +, prme numbers p 1,..., p r and n 1,..., n r Z 2 wth max n = n, there s a degree N number feld K and a Z-order R n K wth exactly r sngular prmes p 1,..., p r, such that e p (R) = n for all. Thus rk(r) = n. Proof. Step 1: The constructon of K and R s essentally gven n [CGP15, ]; the only dfference s that n that constructon one nverts all the regular prmes to get a semlocal doman. So we wll content ourselves wth a sketch. For 1 r, let Q n p be the unramfed extenson of Q p of degree n. By weak approxmaton / Krasner s Lemma there s a number feld K of degree N and prmes P 1,..., P r such that K P = Qp n for all. The local degree [K P : Q p ] = n s assumed (only) to be at most N; when n < N ths s handled by havng other prmes of Z K lyng over p. We have r r Z K /P 1 P r = Z K /P = F n p. Let q : Z K r =1 F p n =1 =1 be the correspondng quotent map. Then we take r R = q 1 ( F p ). For 1 r, let p = P R and let q : (Z K ) P (Z K ) P /P (Z K ) P = Fp n. Then (as s shown n detal n loc. ct.; see also [CGP15, Thm. 2.6]) for all, the rng R p s the pullback q 1 (F p ), and thus s local nearly Dedeknd wth z p (R) = z(r p ) = [F n p : F p ] = n. Step 2: We have e.g. usng CRT as above that p 1,..., p r are the only sngular prmes of R, so =1 z(r) = max 1 r n = n. For 1 r, let R p denote the p -adc completon of R p. Because there s a unque prme of R lyng over p ( analytcally rreducble ), R p s tself a one-dmensonal local Noetheran doman, wth fracton feld K p, and moreover we have e(r p ) = e( R p ) and z( R p ) = z(r p ) = n. But R p s a Z p -order n K p of degree n, so Proposton 3.3 apples to gve rk(r p ) = e(r p ) = e( R p ) = rk( R p ) = z( R p ) = z(r p ) = z p (R)

6 6 PETE L. CLARK and thus rk(r) = max 1 r rk(r p ) = max 1 r z p (R) = max 1 r n = n. Remark 3.9. a) There s an analogue of Theorem 3.8 wth Z replaced by F q [t]. b) One would lke to take n = 1 n Theorem 3.8! Unfortunately at present we cannot prove that there are nfntely many number felds wth class number one: ths s perhaps the most (n)famous open problem n algebrac number theory A Trchotomy. 4. More Rngs of Fnte Rank Theorem 4.1. a) Let R = R 1 R n be a fnte drect product of rngs. Then rk(r) = max rk(r ). b) (Glmer) For a Noetheran rng R, the followng are equvalent: () R has fnte rank. () For all mnmal prmes p Spec R, the rng R/p has fnte rank. c) (Cohen-Glmer) A rng of fnte rank has dmenson zero or one. Proof. a) Every deal n R 1 R n s of the form I 1 I n for deals I of R. b) Ths s [G72, Thm. 2]. c) Apply Theorem 2.1 and part b). So f R s a rng wth rk(r) Z +, exactly one of the followng holds: R s Noetheran of dmenson zero,.e., Artnan; R s a Noetheran doman of dmenson one; R s Noetheran of dmenson one and not a doman. We treated domans n 3, and we wll dscuss Artnan rngs n 4.2. Ths leaves us wth rngs whch are one-dmensonal Noetheran and not domans. One can show that there are such rngs of all ranks qute cheaply: f R s a doman of rank n Z + then Theorem 4.1a) gves rk(r R) = n. The more nterestng case s when the localzaton of R at some mnmal prme s not a doman, so n partcular when R has a unque mnmal prme, whch s nonzero. A good example s a polynomal rng over a local Artnan rng. In 4.3 we study the ranks of polynomal rngs Artnan Rngs. Let r be an Artnan rng. We denote by l(r) the length of r as an r-module, whch s fnte. We denote by n(r) the nlpotency ndex of r: the least n Z + such that f x r s nlpotent then x n = 0. Proposton 4.2. Let r be an Artnan rng. a) We have rk(r) l(r). b) If l(r) 2 (.e., f r s nonzero and not a feld), then rk(r) l(r) 1. Proof. The result s clear when r s the zero rng or s a feld, so assume l(r) 2. If r s prncpal then rk(r) = 1 l(r) 1. If r s not prncpal, then there s an deal I wth 2 µ(i) l(i), and such an deal s necessarly proper, so l(r) l(i) + 1. The followng result shows that the bound of Proposton 4.2b) s sharp. Corollary 4.3. Let n Z 2. Then: a) For any feld k, there s an Artnan k-algebra of rank n and length n + 1. b) There s a fnte rng R of rank n and length n + 1.

7 A NOTE ON RINGS OF FINITE RANK 7 Proof. Let R be a non-normal doman of fnte rank n wth a maxmal deal p such that dm R/p p/p 2 = n. Then R/p 2 s Artnan, of rank n and length n + 1. Takng R as n Example 3.6 gves part a), and takng R as n Example 3.7 gves part b) Polynomal Rngs of Fnte Rank. Next we explore polynomal rngs of fnte rank. For any nonzero rng r, a polynomal rng over r n at least two ndetermnates has Krull dmenson at least 2 and thus has nfnte rank. So we are reduced to the case of r[t]. Theorem 4.4. For a rng r, the followng are equvalent: () The polynomal rng r[t] has fnte rank. () The rng r s Artnan. Proof. () = (): Suppose r[t] has fnte rank. Then r[t] s Noetheran, hence so s r. Thus dm r[t] = 1 + dm r. By Theorem 4.1c) we have dm r = 0, so r s Artnan. () = (): If r s Artnan, there are local Artnan rngs r 1,..., r r such that r = r 1 r r, and then r[t] = r 1 [t] r r [t]. So Theorem 4.1a) reduces us to showng: a polynomal rng r[t] over a local Artnan rng (r, m) has fnte rank. The rng r[t] s Noetheran, wth unque mnmal prme mr[t]. Snce r[t]/mr[t] = (R/m)[t] s a PID, the rng r[t] has fnte rank by Theorem 4.1b). Theorem 4.5. Let r be an Artnan local rng of length l, and let R = r[t]. Then rk R l. Proof. Let p be the unque prme deal of r, let k = r/p. Then P = p[t] s the unque mnmal prme of R. By Theorem 4.4, R has fnte rank. The man nput was Theorem 4.1b), and we wll get the upper bound rk R l by followng Glmer s proof. For 1 l, let R = R/P = r[t]/p r[t] = r/p [t]. We wll show nductvely that rk R s at most the length l(r/p ) of r/p. Base Case ( = 1): The rng R 1 = r/p[t] = k[t] s a PID, thus rk R 1 = 1 = l(k). Inductve Step: Suppose 1 < l and rk R l(r/p ). Consder the exact sequence of R-modules 0 P /P +1 R/P +1 R/P 0. For any surjectve homomorphsm of rngs R S, the rank µ (S) as an R-module s equal to ts rank rk S as a rng. Thus our nductve hypothess gves µ (R/P ) l(r/p ). Moreover the rank of P /P +1 as an R-module s ts rank as an R/P = k[t]-module. By [G72, Prop. 2] we have By [G72, Prop. 1] we have µ (P /P +1 ) rk(k[t])µ R (P /P +1 ) = µ r (p /p +1 ) = l(p /p +1 ). rk R +1 = µ (R/P +1 ) µ (P /P +1 ) + µ (R/P ) l(p /p +1 ) + l(r/p ) = l(r/p +1 ), completng the nducton. Snce the nlpotency ndex of r s at most l, we have rk R = rk r[t] = rk r/p l [t] = rk R l l(r/p l ) = l(r) = l. Snce an Artnan rng s a fnte product of local Artnan rngs, Theorem 4.5 mples: for any Artnan rng r, the rank of r[t] s bounded above by the maxmum length of the local Artnan factors of r. Corollary 4.6. Let r be a nonzero prncpal Artnan rng, whch we may decompose as r = r =1 r, wth each r a local Artnan prncpal rng. For 1 r, let n be the length of r (whch concdes wth ts nlpotency ndex), and let n = max 1 r n. Then rk r[t] = n.

8 8 PETE L. CLARK Proof. We easly reduce to the case n whch r s local wth maxmal deal p = (π). Theorem 4.5 gves rk r[t] n. Let m = π, t, so m s a maxmal deal of r[t] and R/m = r/πr = k, say. We clam that the deal m n 1 = π n 1, π n 2 t,..., πt n 2, t n 1 requres n generators. Indeed, for a 1,..., a n k, we have that a 1 π n 1 + a 2 π n 2 t a n t n 1 cannot be expressed as an r-lnear combnaton of terms π t j wth + j n unless a 1 =... = a n = 0, so dm k m n 1 /m n = n and µ(m n 1 ) = n. Remark 4.7. a) Corollary 4.6 mples: f A s a PID, π R a prme element and n Z +, then rk A/π n A[t] = n. In fact ths s equvalent: every local prncpal Artnan rng s a quotent of a PID [Hu68, Cor. 11]. P. Pollack has shown me a thoroughly elementary proof that rk A/π n A[t] n. b) In partcular, for a prme number p and n Z + we have rk Z/p n Z[t] = n. The case p = n = 2 appears n Matson s thess [Ma08, p. 44, Example ]. Her proof uses that Z/4Z has a unque nonzero zero-dvsor. The only other rng wth ths property s Z/2Z[t]/(t 2 ), so ths argument s rather specalzed. Matson s result was our motvaton for fndng Theorem Work of Matsuda After ths paper was frst wrtten we learned of relevant of R. Matsuda [Ma77] [Ma79], [Ma84]. We end wth a bref dscusson of ts pertnence to the current work Matsuda s work on polynomal rngs. A rng R has the n-generator property f for all deals I of R, f µ(i) < ℵ 0 then µ(i) n. Thus a rng wth the n-generator property has rank at most n ff t s Noetheran. In [Ma77] and [Ma84], Matsuda studed these propertes n polynomal rngs. Our Theorem 4.4 follows from [Ma84, Thm. 15]. The other results of 4 are complementary to Matsuda s work rather than overlappng. For nstance: Theorem 5.1. (Matsuda [Ma84, Prop. 18]) Suppose r s a rng that s not a feld. 3 If r[t] has the n-generator property, then r has the (n 1)-generator property. It follows that f r s Artnan and not a feld, then rk(r) rk(r[t]) 1. Corollary 4.6 shows that the gap between the rank of r and r[t] can be arbtrarly large Matsuda s work on semgroup rngs. A numercal semgroup s a nonempty subset of the postve ntegers Z + that s closed under addton. A numercal semgroup S s prmtve f there are elements n 1,..., n k S such that gcd(n 1,..., n k ) = 1. For any numercal semgroup S, let d be the greatest common dvsor of the elements of S. Then S := { n d n S} s a prmtve numercal semgroup and S d S s a semgroup somorphsm. A famous result of Frobenus (see e.g. [RA]) asserts that for a prmtve numercal semgroup S, the set Z + \ S s fnte, and thus there s a unque element c S, the conductor of S, such that c S, c 1 / S and for all n Z +, we have c + n S. Let S be a prmtve numercal semgroup wth least element n 1, and put S 1 = nz +. If n 1 = 1 then S = S 1. Otherwse S 1 s not prmtve and thus cannot equal S, and we take n 2 to be the least element of S \ S 1. We contnue n ths manner: f the semgroup S k generated by n 1,..., n k s not all of S, let n k+1 be the least element of S \ S k. Every element of S les n some S k, and there s some k 0 Z + such that gcd(n 1,..., n k0 ) = 1, so Frobenus s theorem mples that S \ S k0 3 The hypothess that r not be a feld does not explctly appear n Matsuda s work, but t must be ntended.

9 A NOTE ON RINGS OF FINITE RANK 9 s fnte. Thus S = S k for some k, and then {n 1,..., n k } s the unque mnmal generatng set. For a feld k and a numercal semgroup S, the semgroup rng k[s] s the subrng of k[t] consstng of polynomals s S {0} a st s.e., the monomal exponents must le n S {0}. Theorem 5.2. (Matsuda [Ma79, Prop. 5.6]) Let k be a feld, and let S be a prmtve numercal semgroup wth least element n 1. Then rk(k[s]) = n 1. For n 1 Z +, there are nfntely many prmtve numercal semgroups wth least element n 1 : e.g. n 2 can be prescrbed to be any nteger greater than n 1 and not dvsble by t. Thus we attrbute Theorem 1.1a) to Matsuda. Perhaps the smplest example s S n1 := {n Z n n 1 }. The mnmal generatng set of S n1 s {n 1, n 1 + 1,..., 2n 1 1}, so k[s n1 ] = k[t n1,..., t 2n1 1 ]. (The completon of k[s n1 ] at the maxmal deal t n1,..., t 2n1 1 s the rng R n1 of Examples 3.1 and 3.6, and one can easly deduce that rk R n1 = n 1 from rk k[s n1 ] = n 1.) Ths suggests a connecton between Theorem 5.2 and the nonmaxmal orders of 3.2. Indeed we can get a quck proof of Theorem 5.2 usng these deas: The rng k[t] s a free k[t n1 ]-module of dmenson n 1. The rng k[t n1 ] = k[t] s a PID, so t follows by the usual PID structure theory that k[s] s a free k[t n1 ]-module of rank r n 1. Thus any deal I of k[s] has rank at most r n 1 as a k[t n1 ]-module. It follows that rk(k[s]) n 1. Let c be the conductor of S, and let J be the deal t c, t c+1,..., t c+n1 1 of k[s]. We clam that rk(j) n 1, whch wll complete the proof of Theorem 5.2. Let m be the deal t n n S of k[s]. It s enough to show that dm k J/mJ n 1, whch amounts to showng: for α 1,..., α n1 k, α 1 t c + α 2 t c α n1 t c+n1 1 mj = α 1 =... = α n1 = 0. But every element of mj \ {0} has degree at least c + n 1, so ths s clear. References [AM92] D.D. Anderson and J.L. Mott, Cohen-Kaplansky domans: ntegral domans wth a fnte number of rreducble elements. J. Algebra 148 (1992), [CA] P.L. Clark, Commutatve Algebra. [CGP15] P.L. Clark, S. Gosav and P. Pollack, The number of atoms n an atomc doman. uga.edu/~pete/cgp15.pdf [Co] K. Conrad, The conductor deal. pdf [Co50] I.S. Cohen, Commutatve rngs wth restrcted mnmum condton. Duke Math. J. 17 (1950), [Fo64] O. Forster, Über de Anzahl der Erzeugenden enes Ideals n enem Noetherschen Rng. Math. Z. 84 (1964), [G72] R. Glmer, On commutatve rngs of fnte rank. Duke Math. J. 39 (1972), [Hu68] T.W. Hungerford, On the structure of prncpal deal rngs. Pacfc J. Math. 25 (1968), [Ma77] R. Matsuda, Torson-free abelan group rngs. III. Bull. Fac. Sc. Ibarak Unv. Ser. A No. 9 (1977), [Ma79] R. Matsuda, Torson-free abelan semgroup rngs. V. Bull. Fac. Sc. Ibarak Unv. Ser. A No. 11 (1979), [Ma84] R. Matsuda, n-generator property of a polynomal rng. Bull. Fac. Sc. Ibarak Unv. Ser. A No. 16 (1984), [Ma08] A. Matson, Results Regardng Fnte Generaton, Near Fnte Generaton and the Catenary Degree. PhD thess, North Dakota State Unversty, [Ma09] A. Matson, Rngs of fnte rank and fntely generated deals. J. Commut. Algebra 1 (2009), [RA] J.L. Ramírez Alfonsín, The Dophantne Frobenus problem. Oxford Lecture Seres n Mathematcs and ts Applcatons, 30. Oxford Unversty Press, Oxford, [Sa78] J.D. Sally, Numbers of generators of deals n local rngs. Marcel Dekker, Inc., New York-Basel, [Sw67] R.G. Swan, The number of generators of a module. Math. Z. 102 (1967), [Wa73] K. Watanabe, Some examples of one dmensonal Gorensten domans. Nagoya Math. J. 49 (1973),