Arithmetic Congruence Monoids: A Survey

Transcription

1 Arthmetc Congruence Monods: A Survey Paul Bagnsk and Scott Chapman Abstract We consder multplcatve monods of the postve ntegers defned by a sngle congruence. If a and b are postve ntegers such that a b and a 2 a mod b, then such a monod (known as an arthmetc congruence monod or an ACM) can be descrbed as M a,b = (a + bn 0 ) {1}. In lectures on elementary number theory, Hlbert demonstrated to students the utlty of the proof of the Fundamental Theorem of Arthmetc for Z by consderng the arthmetc congruence monod wth a = 1 and b = 4. In M 1,4, the element 441 has a nonunque factorzaton nto rreducble elements as 9 49 = ACMs have appeared frequently n the mathematcal lterature over the last decade. Whle ther structures can be understood merely wth ratonal number theory, ther multplcatve behavor can become qute complex. We show that all ACMs fall nto one of three mutually exclusve classes: regular (relatng to a = 1), local (relatng to gcd(a,b) = p k for some ratonal prme p), and global (gcd(a,b) s not a power of a prme). In each case, we examne the behavor of varous nvarants wdely studed n the theory of nonunque factorzatons. Our prncpal tool wll be the constructon of transfer homomorphsms from the M a,b to monods wth smpler multplcatve structure. Key words: monod, arthmetc progressons, nonunque factorzaton, elastcty of factorzaton, Krull monods Early n the study of number theory, one encounters the obstacle of nonunque factorzaton, snce elements n many rngs of algebrac ntegers do not always factor unquely nto products of rreducble elements. The multplcatve structure of these number rngs has been classcally used to demonstrate the prckly ssue of nonunque factorzaton. However, we can also exhbt the phenomenon of nonunque factorzaton usng a common object from addtve number theory, arthmetc progressons. An arthmetc congruence monod (ACM) s an arthmetc Department of Mathematcs and Statstcs, Smth College, Northampton, MA 01063, USA. Department of Mathematcs and Statstcs, Sam Houston State Unversty, Huntsvlle, TX 77341, USA. 1

2 2 Paul Bagnsk and Scott Chapman progresson whch naturally possesses a multplcatve structure. Specfcally, an arthmetc congruence monod s the monod: 1 M a,b = (a + bn 0 ) {1} = {1} {a,a + b,a + 2b,...}, where a and b are postve ntegers satsfyng 0 < a b and a 2 a mod b. The congruence demanded upon a and b s both suffcent and necessary for the arthmetc progresson a + bn 0 to be closed under multplcaton. The trval values a = 1 or a = b satsfy ths congruence for any b 1, but nontrval pars, such as a = 4 and b = 6, also exst. In general, for a gven b, there are 2 r choces for a wth 0 < a b, where r s the number of dstnct prmes dvdng b (cf. Secton 4). As we shall descrbe n ths survey, ACMs exhbt both unque and nonunque factorzaton of elements, exhbtng the wdely varyng behavor one encounters n algebrac number rngs, as well as some more pathologcal behavor such as the bfurcus property (cf. Secton 2). In contrast to number rngs, however, very lttle mathematcal background s necessary to grasp the dea of an ACM and uncover many of ts factorzaton propertes. As such, ACMs can be a valuable pedagogcal tool, snce arguments can often be phrased wth elementary arthmetc and wthout resortng to norms or other mathematcal machnery to determne, say, whether an element s rreducble. Despte the low barrer to ntatng the study of ACMs, ther factorzaton theory s surprsngly complex: many questons reman open and the fner study of the factorzatons often requres more nvolved number theoretc and combnatoral arguments. We therefore present ACMs as both an alternatve and a complement n number theory to the study of nonunque factorzaton n algebrac number rngs. After a bref secton ntroducng terms and notaton from the theory of nonunque factorzatons, we proceed wth the study of factorzaton n ACMs. ACMs naturally dvde nto three classes (regular, sngular local, sngular global), each exhbtng starkly dfferent behavor due to ther connectons wth other classes of monods. In Secton 2, we consder the class of ACMs where a = b. The specal case where a = b = p r, for p a ratonal prme, s fully descrbed n Proposton 2.2. In Proposton 2.3 we generalze the argument to the case where a = b s not a power of a prme. Secton 3 consders the case of regular ACMs, whch are those wth a = 1. In Theorem 3.2, we show that these ACMs are specal n the sense that they belong to the class of Krull monods (cf. Defnton 1.3). Usng ths Krull structure, Theorem 3.4 gves a strkng overvew of the factorzaton propertes of these monods. In Secton 4, we consder the case where a 1. Such ACMs are called sngular and nclude the ACMs of Secton 2. The sngular ACMs break nto two subclasses: 1) local ACMs where gcd(a,b) = p k for p a ratonal prme, and 2) global ACMs where gcd(a,b) s not a power of a prme. After the development of some machnery applcable to all sngular ACMs, we analyze the local case n Theorem 4.9. We show n Lemma 4.15 that for each global ACM M a,b there s a constant λ such that every 1 Several authors defne M a,b to equal just the arthmetc progresson, so that t s a semgroup. We shall nclude a unty, snce t does not affect the structure but allows the factorzaton-theoretc defntons to be smpler and concde wth the lterature.

3 Arthmetc Congruence Monods: A Survey 3 nonunt of M a,b has an rreducble factorzaton of length at most λ. Whle ths behavor s more unruly than the behavor encountered n algebrac number rngs, there are stll commonaltes. For example, n Theorem 4.17, we demonstrate that n a global ACM there s a fnte bound N such that for every element, f we have two factorzatons whose lengths dffer by more than N, then there s a factorzaton of that element whose length les between the two other lengths. 1 Terms and Notaton The followng notaton draws from the theory of nonunque factorzatons of rngs and monods; see the monograph [18] of Geroldnger and Halter-Koch for comprehensve references and for undefned terms. The symbol N denotes the natural numbers {1,2,3,...} and N 0 = N {0}. The ntegers are denoted Z and for n Z, Z n denotes the quotent rng Z/nZ. The mage of x Z n Z n shall be denoted x. If gcd(k,n) = 1, then ord k (n) shall denote the order of n n the group of unts of Z k, whch we denote by Z k. We use ϕ(n) to denote the Euler totent functon of n. We open by formally defnng the objects we are about to study. Defnton 1.1 Gven a,b N wth 0 < a b and a 2 a mod b, the arthmetc congruence monod defned by a and b s, M a,b := {n N n a mod b} {1}. We note that for any a and b satsfyng Defnton 1.1, M a,b s both commutatve and cancellatve. We state the man defntons and notaton for the theory of nonunque factorzatons n terms of a general commutatve cancellatve monod. In such a general monod M, f x,y M then x dvdes y (wrtten x y) f there exsts a z M such that xz = y. We wll wrte M when we need to dstngush monods; most commonly we wll use N for the regular dvdes relaton n the natural numbers and reserve for the dvdes relaton n our arthmetc congruence monods M a,b. If M s a monod, M wll denote the unts of M. A nonunt x M s rreducble (or an atom) f whenever x = yz for some y,z M, then ether y or z s a unt. We wrte A (M) for the set of rreducbles of M. A nonunt x M s prme f whenever x yz for some y,z M, ether x y or x z. As n the theory of ntegral domans, prme elements are rreducble but not vce versa. A monod M s atomc f every nonunt x can be wrtten as a product of rreducbles of M. The Fundamental Theorem of Arthmetc states that (N, ) s atomc (and furthermore that the factorzatons nto rreducbles are unque). Snce M a,b s a submonod of the unque factorzaton monod (N, ), ts atomcty s mmedate. Indeed, f x M a,b, then x = p 1 p n, a unque product of prme numbers n N, and so x can be wrtten as a product of at most n elements of M a,b. For the rest of the artcle, all monods wll be assumed to be atomc.

4 4 Paul Bagnsk and Scott Chapman If every nonunt x M has a unque factorzaton nto rreducbles of M, then M s sad to be factoral. M s factoral f and only f all ts rreducbles are prme. M s factoral f and only f t s somorphc to the free abelan monod over ts rreducbles. When M s factoral, the greatest common dvsor s well defned: gven a fnte, nonempty X M, gcd(x) s the unque (up to assocates) element g M such that g x for all x X but for each nonunt h M, there s an x X such that gh x. For x M \ M, we defne L (x) = {n : there are α 1,...,α n A (M) wth x = α 1 α n }, whch s known as the set of lengths of x. We collect these sets together as L (M) = {L (x) : x M\M }, the set of lengths of M. The rato ρ(x) = supl (x)/mnl (x) s called the elastcty of x. The elastcty of the monod M s defned by ρ(m) = sup{ρ(x) : x M \ M } (see [18, Chapter 1.4] or the survey paper [4]). If ρ(m) = 1, then M s called halffactoral. A survey of half-factoral ntegral domans and monods can be found n [12]. M s sad to be fully elastc f for every ratonal q wth 1 q < ρ(m), there exsts an x M \ M such that ρ(x) = q. Many common objects of study n factorzaton theory are fully elastc (for nstance, the rng of nteger-valued polynomals [15]), but numercal monods (cf. Defnton 2.1) are not ([14]). If there exsts an x M \ M such that ρ(m) = ρ(x), then the elastcty of M s sad to be accepted. Rngs of algebrac ntegers have accepted elastcty. Non-examples also exst (see [10] and later our Example 4.10), even ones havng ratonal elastcty that s not accepted. However, these non-examples do not abound n the lterature. Gven x M \ M, wrte ts length set n ncreasng order as L (x) = {n 1,n 2,...,n k }, where n < n +1 for 1 k 1. The delta set of x s defned by (x) = {n n 1 2 k and the delta set of M by (M) = x M\M (x) (see agan [18, Chapter 1.4]). As wth elastcty, the study of the delta sets of partcular monods has an actve hstory, and varous calculatons n specfc cases can be found n [7] and [11]. As n any feld of mathematcs, we wsh to reduce the study of complex objects nto questons about smpler objects. In the realm of factorzaton theory, the collapsng of structure s acheved usng the concept of transfer homomorphsms. Defnton 1.2 Let M and N be commutatve, cancellatve, atomc monods and σ : M N be a monod homomorphsm. The map σ s a transfer homomorphsm f σ(u) N for any u M,

5 Arthmetc Congruence Monods: A Survey 5 σ(x) / N for any x / M, (Surjectvty up to assocates) For every a N, there exsts a unt u N and an x M such that σ(x) = ua, and whenever x M and a,b N such that σ(x) = ab, there exst y,z M and unts u,v N such that x = yz, σ(y) = ua, and σ(z) = vb. Intutvely, a transfer homomorphsm from M to N ensures that N has (up to nose from unts) all the basc factorzaton theory of M. Specfcally, dvsblty relatons from M are preserved n N and L M (x) = L N (σ(x)) for all x M, so that by surjectvty up to assocates, L (M) = L (N). The cost of a transfer homomorphsm les n forgettng whch exact factors appear n factorzatons. Indeed, for factorzaton propertes not concerned solely wth length sets, ths can be a true concern. As we shall see n Secton 2, the ACM M 2,2 (namely, the even numbers along wth 1) s half-factoral but not factoral. Nonetheless, M 2,2 has a transfer homomorphsm nto the free monod (N 0,+), whch s factoral. A more surprsng example occurs n Secton 3, where the half-factoral ACM M 1,4 has a transfer homomorphsm nto the factoral monod B(Z 4 ) (defned n that secton). There are other factorzaton nvarants, such as the catenary and tame degrees, whch also requre extra care under ths caveat, but for the scope of the concepts under revew n ths survey, only the dstncton between factoralty and half-factoralty merts vglance. In all cases we shall encounter, the presence or lack of unque factorzaton can be easly verfed. Thus for our purposes, other than ths mnor excepton, one can consder transfer homomorphsms as ndcatng that M and N have the same factorzaton-theoretc propertes. As we shall see, many ACMs can be reduced to other better studed monods n much the same way as the factorzaton theory of algebrac number rngs transfers to smpler combnatoral monods over the class group (see [5] for an ntroductory exposton on algebrac number rngs from the perspectve of nonunque factorzaton theory). A smlar concept to that of a transfer homomorphsm s that of a dvsor theory. Defnton 1.3 Let M be a monod. A dvsor theory for M s a free commutatve monod F (P) and a monod homomorphsm σ : M F (P) satsfyng the followng propertes: σ(u) = 1 for any u M. σ(u) 1 for any u / M. For any nonunts x,y M, σ(x) σ(y) mples x y. For every p P, there s a fnte subset X M such that p = gcd(σ(x)). A monod M whch has a dvsor theory s sad to be a Krull monod. The generators P are sad to be the prme dvsors of M and the quotent monod F (P)/σ(M) (whch can be shown to be an abelan group) s known as the class group of M. Krull monods abound n mathematcs. For nstance, f D s a Dedeknd doman, D\{0} s a Krull monod under multplcaton. Hence, the multplcatve monod of a rng of algebrac ntegers s a Krull monod. The defnton of a Krull doman, usually gven va v-deals, can be restated more smply usng Krull monods as follows:

6 6 Paul Bagnsk and Scott Chapman an ntegral doman s a Krull doman f and only f ts multplcatve monod s a Krull monod (see [23]). The dstnctons between a dvsor theory and a transfer homomorphsm are subtle. Frst, a dvsor theory need not be (and usually s not) surjectve. More subtly, the property that σ(x) σ(y) mples x y s not a consequence of the last property of a transfer homomorphsm. For both transfer homomorphsms and dvsor theores, f σ(x) σ(y), then σ(x) σ(y) mples σ(y) s reducble and thus so s y. Yet, unlke a dvsor theory, a transfer homomorphsm does not guarantee a choce of x as a wtness to the reducblty of y n M. 2 Multples The smplest of all ACMs are those n whch a = b (.e. M a,b = bn {1}), the set of postve multples of b along wth the element 1. If b = 1, then M a,b = N, whch has unque factorzaton, so let us assume b > 1. Our analyss shall dvde nto two cases, when b s a power of a prme and when t s not. If b = p r, where p s a prme and r 1, then all nonunts of M a,b must be dvsble n N by p r. Therefore an element x M a,b s reducble f and only f p 2r N x. Ths smple observaton leads to a complete characterzaton of M p r,pr n terms of another, well-studed monod known as a numercal monod. Defnton 2.1 Gven x 1,...,x n N, the numercal monod generated by x 1,...,x n, denoted x 1,...,x n, s the set of nonnegatve lnear combnatons of the x. In other words, x 1,...,x n = { n =1a x a N 0 } Numercal monods have deceptvely complcated combnatoral structure. For nstance, f the generators are mnmal and have no common factor, there s a least nteger not contaned n the numercal monod. Ths nteger s known as the Frobenus number, and whle formulas exst for the Frobenus number of a two- or threegenerator numercal monod, computaton of the Frobenus number n general s NP-hard. The nterested reader may consult [24], a recent reference work on numercal monods and ther occurrence n mathematcs. For our purposes, we wll only need to explore a partcularly smple and very well-understood class of numercal monods: those generated by a full nterval of ntegers, {r,r + 1,...,2r 1}. In ths case, set S r := r,r + 1,...,2r 1 = (r + N 0 ) {0}. The Frobenus number of S r s clearly r 1. Though ths monod may resemble an ACM, bear n mnd that the operaton on numercal monods s addton, whle t s multplcaton for ACMs. Proposton 2.2 Let p be prme and r 1. The map.

7 Arthmetc Congruence Monods: A Survey 7 defned by σ : M p r,p r S r = r,r + 1,...2r 1 σ(x) = v p (x), where v p (x) s the p-adc valuaton of x N, s a transfer homomorphsm. Therefore we have the followng. 1. Gven a nonunt x M p r,p r, { L (x) = l N vp (x) l 2r 1 } vp (x). r 2. The elastcty s gven by ρ(m p r,pr) 2r 1 = r and s accepted. 3. M p r,pr s not fully elastc, unless r = M p r,pr s half-factoral f and only f r = 1. However t s never factoral. 5. (M p r,p r) = {1} f r > 1 and (M p r,pr) = /0 f r = M p r,pr has no prme elements. Proof. Ths functon σ s clearly a monod homomorphsm and σ(x) = 0 f and only f x = 1. Based on the membershp crteron for M p r,pr, σ clearly maps surjectvely onto the numercal monod N = r,...,2r 1. We now demonstrate that σ s a transfer homomorphsm. Snce nether M p r,pr nor the numercal monod have any unts other than ther respectve denttes, we have only one condton left to verfy. Suppose x M p r,pr and n,m r,r + 1,...,2r 1 such that σ(x) = n + m. By the defnton of σ, p n+m N x. Snce n,m r, we fnd that p n M p r,p r and x/pn M p r,p r. Thus x = p n (x/p n ), σ(p n ) = n, σ(x/p n ) = m, and σ s a transfer homomorphsm. Snce σ s a transfer homomorphsm, for every nonunt x M p r,p r, L M p r,p r (x) = L N (σ(x)) = L N (v p (x)). Yet for any k r (n partcular k = v p (x)), the length set of k n N can easly be computed to be equal to the set on the rght hand of clam 1. Such a basc computaton appears n [11], where t s mmedately concluded that ρ(n) = 2r 1 r, so that N s half-factoral f and only f r = 1. When r = 1, clearly (N) = /0, whle (N) = {1} when r > 1 (see [2] or [11]). Because σ s a transfer homomorphsm, the values of all these nvarants are dentcal for M p r,p r and M p r,p r s half-factoral f and only f r = 1. However M p,p s never factoral, for f q s a prme dfferent than p, then (pq)(pq) = p(pq 2 ) n M p,p and all these factors are rreducble. We note that, alternately, clam 4 follows from the man proposton of [9]. It s a general fact that all fntely-generated monods have accepted elastcty [3, Thm. 7]; n ths case, the element 2r 2 r wtnesses the accepted elastcty of N and hence any premage under σ, such as p 2r2 r, wtnesses the accepted elastcty for M p r,p r. Usng the transfer homomorphsm, we obtan that M p r,pr s not fully elastc for r 2 by [14, Theorem 2.2]; M p,p s trvally fully elastc snce t s half factoral. Lastly, we show that M p r,pr has no prme elements. If x were a prme element of M p r,p r, then x would be rreducble and hence have the form ps k for some r s 2r 1 and nteger k 1 relatvely prme to p. Let m,n > 1 be ntegers relatvely prme to x. Then x p r+s nmk, but x does not dvde ether p s m or p r nk.

8 8 Paul Bagnsk and Scott Chapman Note that the prme p plays no role n the factorzaton propertes of the monods M p r,pr; all the nvarants can be characterzed solely n terms of r, the exponent. Indeed, for any two prmes p and q, M p r,p r = Mq r,qr. Ths somorphsm s just the restrcton to M p r,pr of the somorphsm σ : N N gven by σ(x) = xp v q(x) v p (x) q v p(x) v q (x), whch swaps all nstances of p and q n the prme factorzaton of x. The case M p r,p r generalzes to a correspondng prme power case n Secton 4.1, known as the class of local ACMs. Even there we shall see that the prme p plays a mnmal role n the factorzaton propertes of the ACM. In contrast, when b s not a power of a prme, factorzaton n M b,b becomes qute wld. As we shall see n Proposton 2.3, we can explot the trck from the prevous proof, whch showed that M p r,p r contans no prme elements, to show that for M b,b wth b not a prme power, any reducble element has a factorzaton of length 2. These monods are natural examples of the pathologcal class known as bfurcus monods [1]. In the prme power case, we dscovered that M b,b has a transfer homomorphsm to a translate r + N 0 of the unque factorzaton monod (N 0,+). Smlarly, n the case where b s not a power of a prme, we shall show the exstence of a transfer homomorphsm from M b,b to a translate (v 1,...,v n ) + N n 0 of the addtve, unquefactorzaton monod (N n 0,+), for an approprate n 1. Proposton 2.3 Let b be a postve nteger whch s not a prme power. In N, wrte b as p e 1 1 pe n n, where the p are dstnct prmes and the e 1 and set N = (e 1,...,e n ) + N n 0. The map defned by σ : M b,b N σ(x) = (v p1 (x),...,v pn (x)) s a transfer homomorphsm. Therefore we have the followng. 1. Gven a nonunt x M b,b, wrte t n N as b k m, wth k 1 and b N m. If x s reducble, then L (x) = {l N 2 l k}. 2. ρ(m b,b ) =. 3. M b,b s not fully elastc. 4. (M b,b ) = {1}. 5. M b,b has no prme elements. Proof. Let b = p e 1 1 pe n n and N be as n the hypotheses. Clearly the map σ s a monod homomorphsm and σ(x) = (0,...,0) f and only f x = 1. Snce every element of M b,b s dvsble by b, σ maps nto N; conversely, gven (v 1,...,v n ) N, we know that v e for all and hence x = p v 1 1 pv n n M b,b s a premage of (v 1,...,v n ) under σ. Thus σ s surjectve onto N. Snce M b,b and N have no

9 Arthmetc Congruence Monods: A Survey 9 unts besdes the dentty, to show σ s a transfer homomorphsm, we take an arbtrary x = p v 1 1 pv n n m M b,b, where gcd(m,b) = 1. Then σ(x) = (v 1,...,v n ). Suppose (v 1,...,v n ) = (w 1,...,w n ) + (u 1,...,u n ), where w,u e for all. Then y = p w 1 1 pw n n m and z = p u 1 1 pu n n are premages of (w 1,...,w n ) and (u 1,...,u n ), respectvely, n M b,b and x = yz. So σ s ndeed a transfer homomorphsm. All the remanng clams may be computed n N, usng σ, however we shall argue them drectly n M b,b. Let p be a prme of N that dvdes b and set v = v p (b). Gven a nonunt x M b,b, wth x = b k m for some m not dvsble by b, we can factor x as b b (bm), a product of k rreducbles. No factorzatons of x n M b,b may be longer, snce each rreducble must be dvsble by b n N. If x s reducble, then b 2 N x, so we wrte ( ) x x = bp v (bp vp(x) 2v ). p(x) 2v Both factors on the rght are n M b,b snce they are dvsble n N by b. However they are rreducbles; the factor on the rght s clear because b s not a power of p, whle the factor on the left has a p-adc valuaton of v, so t can be dvsble by b exactly once. Therefore we have shown every reducble element of M b,b can be wrtten as a product of two rreducbles. Now, f x s reducble, then for any 2 < l < k, we can factor x as b b y 1 y 2, where y 1 y 2 s a factorzaton of b k l+2 m as a product of two rreducbles (note b k l+2 m M b,b s reducble). Thus we have produced a factorzaton of x of length l and the length set of x has the prescrbed form. From ths explct descrpton of the length set, the values of the elastcty and delta set of M b,b are mmedate. It s also clear that M b,b s not fully elastc snce ρ(x) = k/2 for some k 2 for any nonunt x M b,b. If x M b,b s a nonunt, then x (p 1 x)(p 2 x) but x p 1 x and x p 2 x snce b s not a power of a prme. So M b,b has no prme elements. Despte beng submonods of the unque factorzaton monod (N, ), these M b,b exhbt factorzaton propertes very unlke the unqueness enjoyed by N. Yet far from beng unusual submonods of N, these M b,b possess a partcularly smple form. They can even be expressed as fnte ntersectons of the more well-behaved M p r,p r, as stated n the proposton below, whose proof s clear. In partcular, by Proposton 2.2, for b a squarefree composte number, the badly behaved monod M b,b can be expressed as a fnte ntersecton of half-factoral monods M p,p. Proposton 2.4 If b = p e 1 1 pe n n N, where all the p are dstnct and prme and e 1 for all 1 n, then M b,b = n =1 M e p,p e. Proposton 2.4 wll parallel Proposton 4.14 below, just as the local/global dchotomy of Secton 4 parallels the dchotomy n ths secton between b a power of a prme and b not a power of a prme.

10 10 Paul Bagnsk and Scott Chapman 3 Regular Arthmetc Congruence Monods The case a = 1 bears attenton not only mathematcally, but hstorcally. The ACMs M 1,b are sometmes called Hlbert monods n honor of Davd Hlbert s early use of M 1,4 to demonstrate the occurrence of nonunque factorzaton n natural numbertheoretc settngs (see [8]). Mathematcally, they mert dstncton from other ACMs as they are the ACMs whch fall nto the mportant class of Krull monods, a generalzaton of Dedeknd domans. Our prmary goal for ths secton wll be to demonstrate the Krull property for these monods. We begn wth an elementary, yet key, observaton. Lemma 3.1 Let x,y,z N such that x = yz. If x,y M 1,b then z M 1,b. Proof. We have 1 x = yz z mod b, so z M 1,b. In more compact notaton, for any x,y M 1,b, f x N y, then we already have x y. In factorzaton theory parlance, we say that M 1,b s saturated n N. Note that the saturaton of M 1,b s n stark contrast to M b,b where, for nstance, 2 N 6 but 2 6 n M 2,2. Note that f x M 1,b and p s prme n N wth p N x, then gcd(p,b) = 1. Indeed, snce x 1 mod b, p must be nvertble modulo b. Therefore, p belongs to the set of prmes P = {p N p s prme and gcd(p,b) = 1}. Let F (P) be the free commutatve monod generated by P, whch we dentfy wth ts somorphc copy n (N, ). Under ths dentfcaton, M 1,b wll be a submonod of F (P). Ths observaton allows us to prove the followng theorem, frst shown by Halter-Koch [21]. Theorem 3.2 Let P = {p N p s prme and gcd(p,b) = 1}. The free monod F (P) (N, ) and the homomorphsm ι : M 1,b F (P) form a dvsor theory for M 1,b. Thus M 1,b s Krull. Proof. Snce ι s njectve, for any u M 1,b, ι(u) = 1 f and only f u = 1. Furthermore, f for some nonunts x,y M 1,b we have ι(x) ι(y), then x y n N, so Lemma 3.1 gves us that x y n M 1,b. Lastly, we must show that for every p P there s a fnte subset X M such that p = gcd(ι(x)). Let such a p be gven. By Drchlet s Theorem, we may choose two dstnct prmes q 1,q 2 N dstnct from p such that q 1 q 2 p 1 mod b (here we use gcd(p,b) = 1). Therefore q 1,q 2 P; pq 1, pq 2 M 1,b ; and p = gcd(pq 1, pq 2 ) = gcd(ι(pq 1 ),ι(pq 2 )). In fact, M 1,b beng Krull s an nstance of a general theorem that all saturated submonods of factoral monods are Krull [18, Prop (3)]. All other ACMs besdes N are not Krull. For ACMs that are multples, ths s due to ther correspondence wth numercal monods and translates of (N n,+), both of whch are not Krull; for general sngular ACMs ths shall be proven n Theorem 4.8. A thorough reference for the theory of Krull monods can be found n [18]; a gentle ntroducton to ths theory can be found n [5], where most of the facts below nvolvng block monods are developed n detal for algebrac number rngs, a partcularly well-behaved class of Krull monods. We shall summarze these facts from

11 Arthmetc Congruence Monods: A Survey 11 the theory of Krull monods wthout further ctaton untl we return to dealng wth regular ACMs agan specfcally. As mentoned n the dscusson of dvsor theores and the Krull property n Secton 1, Krull monods have a noton of class group. Namely, for a Krull monod M wth dvsor theory σ : M F (P), the quotent monod F (P)/σ(M) forms an abelan group G generated (as a monod) by a subset S equal to the mage of P under ths quotent. The Krull monod M can then be related to a combnatoral structure bult out of G and S, known as the block monod. Defnton 3.3 Let G be an abelan multplcatve group and S G be nonempty. Let F (S) be the free commutatve monod generated by S, wth elements wrtten as [s 1 ] e 1 [s n ] e n. Let E be the dentty of F (S). Gven A = [s 1 ] e 1 [s n ] e n F (S), the length of A, denoted A, s n =1 e. There s a natural evaluaton map θ : F (S) G gven by θ([s 1 ] e 1 [s n ] e n ) = s e 1 1 se n n. The block monod of G over S s the monod defned as: B(G,S) = {[s 1 ] e1 [s n ] e n F (S) θ([s 1 ] e1 [s n ] e n ) = 1}. In other words, the block monod B(G, S) corresponds to all unordered sequences over S, such that the product of these terms n G yelds the dentty. The set B(G,S) s clearly a submonod of F (S) and hence s atomc snce F (G,S) s. Moreover, t s a saturated submonod of F (S). Block monods have been studed extensvely n the lterature and numerous factorzaton-theoretc propertes of them are known (see [5] and [18]). For example, the elastcty of B(G,S) relates to an mportant value known as the Davenport constant, D(G,S), of G wth respect to S. The Davenport constant s defned as the maxmal length of an rreducble of B(G,S). When G s an nfnte group and S = S 1, for example, then D(G,S) =, but for fnte groups t s easly shown that D(G,S) G. For any group G and subset S G, t s known that ρ(b(g,s)) max{1,d(g,s)/2}, wth equalty n many cases, such as when S = G. If M s Krull, there s a transfer homomorphsm φ : M B(G,S), where G = F (P)/σ(M) s the class group of M and S = {pσ(m) p P} s the subset of classes whch contan elements of P. For each m M, we can consder the correspondng element σ(m) = [p 1 ] e 1 [p n ] e n of σ(m) F (G). Usng ths expresson, we now are able to defne φ compactly as: φ(m) = [s 1 ] e1 [s n ] e n, where for each, s s the mage of p n G. In the next theorem, we explctly determne the class group G and subset S for a regular ACM M 1,b. We shall then use the natural transfer homomorphsm to the block monod B(G,S) to conclude many factorzaton-theoretc propertes of M 1,b. The dentty of the class group G s easy to surmse based on our prevous knowledge. Let x M 1,b be gven and factor x n N as p e 1 1 pe n n. We know all the p P. Snce x 1 mod b, each p s necessarly an nvertble element of Z b,.e. p Z b. Conversely, f we have prmes p P, then p Z b for all. If, fur-

12 12 Paul Bagnsk and Scott Chapman thermore, p e 1 1 pe n n = 1, then x = p e 1 1 pe n n belongs to M 1,b. Thus our class group s G = Z b, as we shall now verfy. Theorem 3.4 The class group G of M 1,b s Z b and the correspondng subset S also equals Z b. The map φ : M 1,b B(Z b,z b ) defned by φ(x) = φ(p e 1 1 pe n n ) = [p 1 ] e1 [p n ] e n, s a transfer homomorphsm. Thus, we have the followng. 1. M 1,b s half-factoral f and only f b = 2,3,4,6. 2. M 1,b s factoral f and only f b = The elastcty s ρ(m 1,b ) = D(B(Z b )) The elastcty of M 1,b s accepted. 5. M 1,b has full elastcty. 6. (M 1,b ) = {1,...,c} for some c N. 7. M 1,b contans nfntely many prme elements. Proof. We use the notaton of Theorem 3.2. As shown n that theorem, F (P) and the map ι : M 1,b F (P) form a dvsor theory for M 1,b. Here F (P) s beng dentfed wth ts somorphc copy n (N, ). Set σ = ι. Let n,n F (P) be gven such that n n mod b. By Drchlet s Theorem, we may choose a prme q N dstnct from n and n (f they were prme) such that q n 1 mod b. Then nq and n q are both elements of M 1,b, so σ(n) and σ(n ) are equvalent modulo σ(m 1,b ). Conversely, f n,n F (P) and n n mod b, then for no m F (P) do we have both nm M 1,b and n m M 1,b. Therefore G s somorphc to the mage of F (P) n (Z b, ). By Drchlet s Theorem, for every ω Z b, there s a p P such that p = ω, so S Z b. Yet we observed pror to ths theorem that a product p 1 p k of prmes n N s an element of M 1,b f and only f gcd(p,b) = 1 for all and p 1 p k 1 mod b. In other words, G s contaned n Z b and thus G = S = Z b. Therefore we have our transfer homomorphsm φ : M 1,b B(Z b,z b ), and all factorzaton propertes relatng to lengths are dentcal for M 1,b and B(Z b,z b ). It s well-known (see [18] or [5], for example) that f G s a fnte abelan group, then B(G,G) s half-factoral f and only f G = 1 or 2. For other fnte abelan groups, ρ(b(g,g)) = D(G)/2, where D(G) s the Davenport constant of G. In both cases, the elastcty s accepted. Thus M 1,b s half-factoral f and only f Z b 2, whch s the case precsely when b = 2,3,4, or 6. The case M 1,2 s the odd numbers, whch s factoral by the Fundamental Theorem of Arthmetc. For all other b, one can easly construct a product wth dstnct factorzatons. For example: 4 25 = n M 1,3 ; 9 49 = n M 1,4 ; and = n M 1,6. Geroldnger and Yuan [20, Theorem 1] recently determned that f G s a fnte abelan group, then (B(G,G)) s the set of consecutve ntegers {1,2,...,c} up to some nteger c 1. For many fnte groups G, they gve an explct value of c n terms of another factorzaton nvarant of B(G, G) known as the catenary degree. Namely, for those groups c s 2 less than the catenary degree.

13 Arthmetc Congruence Monods: A Survey 13 Lastly, we demonstrate that M 1,b contans nfntely many prme elements. By Drchlet s Theorem, there are nfntely many prmes p N whch are equvalent to 1 modulo b. These are all prme by Lemma 3.1. From the exstence of prme elements and accepted elastcty, we conclude M 1,b s fully elastc by [6, Corollary 2.2]. Example 3.5 Notce that the use of Drchlet s Theorem n the frst part of the proof of Theorem 3.4 ndcates that each dvsor class of the the class group Z b of M 1,b contans countably many prme dvsors. Hence, by part 2 above, M 1,3, M 1,4 and M 1,6 are half-factoral Krull monods wth class groups somorphc to Z 2 and equal dstrbutons of prme dvsors n each dvsor class. Snce all three of these monods are reduced, by [18, Theorem 2.5.4] t follows that M 1,3 = M1,4 = M1,6. Hence, ACMs wth dfferent defnng modul can stll be somorphc. Example 3.6 Consder the specal case mpled by Theorem 3.4 when b = p where p s a ratonal prme. Then Z p = Z p 1 s a cyclc group, and snce D(B(Z p 1 )) = p 1, t follows that ρ(m 1,p ) = p 1 2. It s easy here to construct a factorzaton where the elastcty s attaned. Let x be a prmtve root modulo p. By Drchlet s Theorem, choose dstnct prmes p 1 and p 2 so that n Z p we have x = p 1 and x 1 = p 2. Then p 1 p 2, p p 1 1 and p p 1 2 are all atoms of M 1,p. Moreover z = (p 1 p 2 ) p 1 = p p 1 1 p p 1 2 yelds ρ(z) = p 1 2. That the class group s cyclc yelds even further results. For nstance, by [13, Proposton 5.3.3], (B(Z n )) = {1,2,...,n 2} for all n N. Takng n = p 1 we have (M 1,p ) = {1,2,..., p 3}. 4 Sngular Arthmetc Congruence Monods If a 1, the ACM s known as sngular. We have already encountered one example of a sngular ACM n Secton 2, namely a = b. However, for a gven modulus b, there are generally many sngular ACMs. Proposton 4.1 Let b > 1 be gven and factor b = p e 1 1 pe r r as a product of prmes n N. There are 2 r choces of a wth 0 < a b such that M a,b s an ACM. In partcular, 2 r 1 of them wll be sngular. Proof. Suppose M a,b s an ACM for ths fxed b. For each 1 r, we have a 2 a mod p e, so pe a(a 1). Snce p s prme, we conclude p e a or pe a 1. Thus, each a for whch M a,b s an ACM satsfes the system of lnear congruences: a c 1 mod p e a c r mod p e r r, where for each, c s ether 1 or p e. Conversely, by the Chnese Remander Theorem, for all choces of c 1,...,c r, where each c s ether 1 or p e, we obtan an a wth

14 14 Paul Bagnsk and Scott Chapman 0 < a b such that a 2 a mod b,.e., M a,b s an ACM. There are 2 r such choces for the c and only one of them (all c = 1) produces the regular ACM M 1,b. The structure of a sngular ACM depends heavly upon the factors n common between a and b. Let d = gcd(a,b) and f = b/d. It s easy to see from the congruence a 2 a mod b (or the prevous proof) that d = 1 f and only f a = 1, so a sngular ACM must always have d > 1. The arthmetc congruence monods of the form M b,b studed n Secton 2 fall nto the class of sngular ACMs. In the ntal exposton for ths secton, we shall allow ACMs of multples as possbltes, even though we have already studed many of ther propertes n Secton 2. One of the goals of ths ntal exposton s to show that sngular ACMs are not Krull, a fact whch we dd not yet demonstrate for ACMs of multples. Sngular ACMs are dvded nto two subclasses dependng on d: a local ACM has d a power of prme, whle a global ACM has d dvsble by at least two dstnct prmes. In the later subsectons on local and global sngular ACMs, we shall exclude the possblty that a = b. Lemma 4.2 [7, Thm. 2.1] Let a,b N wth 0 < a b and a 2 a mod b. Suppose d = gcd(a,b) > 1 and set f = b/d. Then gcd(a, f ) = gcd(d, f ) = 1 and M a,b = M d,d M 1, f. Conversely, gven d, f N wth d > 1, f 1 and gcd(d, f ) = 1, then there exsts a unque a N such that 0 < a < d f and gcd(a,d f ) = d, so that M a,d f s a sngular ACM satsfyng the above ntersecton. Proof. As n the proof of Proposton 4.1, let b = p e 1 1 pe r r. For each 1 r, we have a c mod p e, where c = 1 or c = p e. In ether case, we have M a,b M c,p e and by the Chnese Remander Theorem, M a,b = r =1 M c,p e. Let I = {1 r c = 1} and J = {1,...,r}\I. For each I, c = 1 and so p N a. Hence p e N f. Conversely, for each J, c = p e. Snce a c mod p e, we have p e Na and thus p e Nd. Together, we fnd f = I p e and d = J p e and that gcd(a, f ) = gcd(d, f ) = 1. By the Chnese Remander Theorem, J M c,p e = M e p,p e = M d,d and M c,p e = M e 1,p = M 1, f J I I Thus M a,b = M d,d M 1, f, as desred. The converse clam follows mmedately from the proof of Proposton 4.1 by takng b = d f = p e 1 pe r r and havng a p e mod p e f p N d and a 1 mod p e f p N f. The unqueness of a s guaranteed by the Chnese Remander Theorem. We already saw n Lemma 3.1 that regular ACMs are saturated n N. Usng the above characterzaton of sngular ACMs as ntersectons of a regular ACM wth an

15 Arthmetc Congruence Monods: A Survey 15 ACM of multples, we are able to demonstrate a weak form of saturaton for sngular ACMs. Lemma 4.3 [7, Corollary 2.2] Let a,b,d = gcd(a,b), and f have the usual assumptons. Let x,y M a,b wth y N x. Then: 1. x/y M a,b f and only f d N x/y, and 2. f x A (M a,b ), then y A (M a,b ). Proof. For the proof of (1), by Lemma 4.2, x,y M 1, f so by Lemma 3.1 x/y M 1, f. By Lemma 4.2 agan, x/y M a,b f and only f x/y M d,d, whch s equvalent to d N x/y. Clam (2) s mmedate from (1). Proposton 4.4 Let a,b,d = gcd(a,b), and f = b/d have the usual assumptons for a sngular ACM. Then M a,b has no prme elements. Proof. Let q be a prme number n N wth gcd(q,d) = 1 and q 1 mod f. By Lemma 4.2, for any x M a,b wth x > 1, we must have xq M a,b and xq 2 M a,b. Now let such an x be gven. Snce q / M a,b, we cannot have x xq n M a,b. Yet x(xq 2 ) = (xq)(xq), so x cannot be prme n M a,b. Although M a,b has no prmes, t has nfntely many rreducbles as shown by the followng smple argument from [7]. Proposton 4.5 Let a,b,d = gcd(a,b), and f = b/d have the usual assumptons. If x M a,b s reducble, then x + b s rreducble. Proof. If x = yz n M a,b, then by Lemma 4.2 d 2 N x. If x + b s also rreducble, then d 2 N x + b, so d 2 N b. But gcd(d,b/d) = gcd(d, f ) = 1 by Lemma 4.2, a contradcton. In the sngular ACM M p,2p, we can easly show that x s rreducble f and only f p 2 x, so n ths monod the rreducbles are perodc wth perod p. Open Queston 4.6 Determne the dstrbuton of the rreducbles n M a,b. Are they (eventually) perodc? Our remanng goal for ths ntroductory exposton on sngular ACMs s to show they are not Krull. Ths fact was orgnally proven by Halter-Koch [21, Thm. 1] and our proof shall follow smlar lnes. Snce gcd(d, f ) = 1, the nteger γ = ord f (d) 1 exsts. Thus d γ 1 mod f and d γ M a,b by Lemma 4.2. Ths element d γ shall be central to our analyss of M a,b due to the followng proposton. Proposton 4.7 Let γ = ord f (d). For all nonunts x 1,...,x γ+1 M a,b, we have d γ x 1 x γ+1. Proof. Let nonunts x 1,...,x γ+1 M a,b be gven. Wrte each x as d k m, where d N m for all and let K = γ+1 =1 k. Snce k 1 for all 1 r by Lemma 4.2, K γ 1 and so d N d K γ m 1 m γ+1. Because x 1 x γ+1 = (d γ )(d K γ m 1 m γ+1 ), by Lemma 4.3 we conclude d γ x 1 x γ+1.

16 16 Paul Bagnsk and Scott Chapman Theorem 4.8 Let a,b N wth 0 < a b such that M a,b s a sngular ACM. Then M a,b s not Krull. Proof. Let d = gcd(a,b) > 1 and let γ = ord f (d). Then d γ M a,b. Suppose that M a,b s Krull. By defnton, there s a free commutatve monod F (P) and a monod homomorphsm σ : M a,b F (P) yeldng a dvsor theory. Consder X = σ(d γ ) and wrte X = P e 1 1 Pe n n. By Proposton 4.7, d γ x γ+1 for any nonunt x M a,b and thus X must dvde σ(x) γ+1. Consder a. By Lemma 4.2, we know a 1 mod f and we can wrte a = d k m, for some k,m N wth d N m. Note that k γ snce d γ M a,b and a s the least element of M a,b greater than 1. Snce gcd(a, f ) = 1, by Drchlet s Theorem we may pck a prme q N such that q m 1 mod f and gcd(d,q) = 1. For all v 1, (d γ k qa) vγ+1 = (d γ qm) vγ+1 = (d γ ) vγ (d γ q vγ+1 m vγ+1 ). By the choce of q and γ, we have d γ qm M 1, f and d γ q vγ+1 m vγ+1 M 1, f. Snce d N d γ q vγ+1 m vγ+1, we conclude d γ q vγ+1 m vγ+1 M a,b by Lemma 4.2. Smlarly d γ k qa M a,b, so n M a,b we fnd (d γ ) vγ (d γ k qa) vγ+1 for all v 1. Let B = σ(d γ k qa) = σ(d γ qm). Snce X B γ+1 and we are workng n a free commutatve monod, we can wrte B as B = P g 1 1 Pg n n Q h 1 1 Qh t t, where g 1 for all 1 n and P Q j for any 1 n and 1 j t. Because σ s a monod homomorphsm and (d γ ) vγ (d γ k qa) vγ+1 for all v 1, we fnd X vγ B vγ+1 for all v 1. In other words, for all 1 n and all v 1, vγe (vγ + 1)g. Snce v 1 arbtrary and e,g N, we conclude e g for all. Thus X B. But σ s a dvsor theory, so n M a,b we fnd d γ d γ qm. Therefore qm M a,b, but ths s a contradcton by Lemma 4.2 snce d N qm. So no such dvsor theory exsts and M a,b s not Krull. Ths concludes the general statements we can make about sngular ACMs. For the remanng two subsectons, assume a < b. 4.1 Local Arthmetc Congruence Monods In ths secton, we shall assume d = p α, for p prme and α 1. Snce gcd(d, f ) = 1, p has fnte order ord f (p) modulo f. Choose a least β α such that p β 1 mod f ; by Lemma 4.2, p β s the least power of p n M a,b. In the prevous secton we had chosen a least γ 1 such that d γ 1 mod f ; snce d = p α, we conclude γ = ord f (p α ) and γ N ord f (p). Furthermore, β s a multple of ord f (p), so γ N β but

17 Arthmetc Congruence Monods: A Survey 17 they need not be equal. The nvarants α and β shall prove to be pvotal for many of the factorzaton propertes of M a,b. Theorem 4.9 Let M a,b be a sngular ACM wth d = gcd(a,b) = p α for some p N prme and α 1. Let β α be mnmal such that d β M a,b. 1. [9, Theorem 2.4 (1)] The elastcty of M a,b s ρ(m a,b ) = α + β 1. α 2. [9, Theorem 2.4 (4)] ρ(m a,b ) < 2 f and only f a = p α. 3. [9, Theorem 2.4 (3)] M a,b s half factoral f and only f a = p. 4. [7, Theorem 3.1] The Delta set of M a,b s /0 f α = β = 1 (M a,b ) = {1} f α = β > 1 N [1, β α ) f β > α. Proof. (1). If x M a,b and v p (x) α + β, then x = (p β )(x/p β ). By Lemma 4.2, x, p β M 1, f and so by Lemma 3.1, x/p β M 1, f as well. Furthermore, p α N x/p β, so by Lemma 4.2, x/p β M a,b. Thus x s reducble. Hence all rreducbles x of M a,b have v p (x) α + β 1. On the other hand, all rreducbles x of have v p (x) α by Lemma 4.2. Consequently, f y M a,b wth y > 1, then any factorzaton of y nto rreducbles nvolves at most v p (y)/α rreducbles and at least v p (y)/(α + β 1) rreducbles. Thus for all nonunt y M a,b, ρ(y) v p (y) α v p (y) α+β 1 = α + β 1. α To show that ths fracton equals ρ(m a,b ), we must fnd elements y whose elastctes approach ths value. By Drchlet s Theorem, there exst prmes q and r dstnct from p such that q p β α+1 mod f and r p β α mod f. By Lemma 4.2, both p α+β 1 q and p α r are elements of M a,b. Snce p α N x for all nonunts x M a,b and p β s the least power of p n M a,b, both p α+β 1 q and p α r are rreducble. Note that q β r β 1 mod f snce p β 1 mod f, so for smlar reasons as above, both p α q βk r and p α r βk+1 are rreducbles of M a,b for any k 0. Therefore we have for all k 1: (p α+β 1 q) kαβ (p α r βk(α+β 1)+1 ) = (p α r) kβ(α+β 1) (p α q βkα r). The factorzaton on the left has kαβ +1 rreducbles, whle the one on the rght has kβ(α + β 1) + 1 rreducbles, so the elastcty of M a,b s bounded below by: ρ((p α+β 1 q) kαβ (p α r kβ(α+β 1)+1 )) kβ(α + β 1) + 1, kαβ + 1

18 18 Paul Bagnsk and Scott Chapman whch goes to (α + β 1)/α as k goes to nfnty. Part (3) clearly follows from (1) as the followng statements are equvalent: 1. M a,b s half factoral, 2. ρ(m a,b ) = 1, 3. β = 1, 4. p M a,b, and 5. a = p by mnmalty of a. For part (2), observe that f β > α, then ρ(m a,b ) (α + (α + 1) 1)/α = 2. So ρ(m a,b ) < 2 mples β = α, so that d = p α M a,b. By mnmalty of a, we conclude a = d = p α. Conversely, f a = d = p α, then α = β and ρ(m a,b ) < 2. The proof of part (4) s nvolved; the nterested reader should consult [7]. To ths pont, we have seen that whenever an ACM has fnte (and ratonal) elastcty, ts elastcty s accepted. Ths s no longer the case for local ACMs, as the followng example testfes. Example 4.10 Consder M 4,6. Then p = 2, α = 1, and β = 2 snce clearly 4 M 4,6. By Theorem 4.9, ρ(m 4,6 ) = 2. We also know d = p and f = 3, so by Lemma 4.2, M 4,6 = M 2,2 M 1,3. Consequently, we may characterze the rreducbles of M 4,6. If 4m M 4,6, then m 1 mod 3. If m = m 1 m 2 where m 1,m 2 2 mod 3, then 4m = (2m 1 )(2m 2 ) n M 4,6. Thus 4m s rreducble f (and only f) m s a product of prmes, all equvalent to 1 modulo 3. Call rreducbles of ths form type A rreducbles. We are left wth consderng x M 4,6 of the form x = 2m, where m s odd and, necessarly, m 2 mod 3. Any such x s rreducble snce t s not dvsble (n N) by 4. Call rreducbles of ths form type B rreducbles. Suppose x M 4,6 wth ρ(x) = 2. Fx a longest factorzaton of x and suppose t has s type A rreducbles and t type B rreducbles. Then v 2 (x) = 2s + t. Fx a shortest factorzaton of x and suppose t has u type A rreducbles and v type B rreducbles. Then v 2 (x) = 2u + v. But 2 = ρ(x) = (s +t)/(u + v), so s +t = 2(u + v) = v 2 (x) + v = 2s +t + v. Thus v+s = 0 and so v = s = 0. The elastcty tells us 2u = t,.e. x may be wrtten as a product of u many rreducbles of type A, and also as a product of 2u rreducbles of type B. As the former product, x = 4 u m, where m s a product of prmes, all equvalent to 1 modulo 3. As the latter product, x = 2 2u m 1 m 2u, where each m s odd and m 2 mod 3. But ths s absurd, snce each m would consst of a product of odd prmes all equvalent to 1 modulo 3. Thus there s no x M 4,6 wth ρ(x) = 2. Note that M 4,6 falls nto the thrd class of local ACMs wth respect to ther Delta sets. Here α = 1 and β = 2 so (M 4,6 ) = {1,2}. Although there exst examples of local ACMs wthout accepted elastcty, Banster, Chaka, Chapman, and Meyerson have demonstrated a large class of local ACMs whch do have accepted elastcty.

19 Arthmetc Congruence Monods: A Survey 19 Theorem 4.11 [10, Theorem 1] Let M a,b be a sngular, local ACM wth d = gcd(a,b) = p α for some prme p N and some α 1. Set f = b/d and choose β α mnmal such that p β M a,b. Let ω be the least resdue of α modulo ord f (p). Suppose a d and p s a prmtve root modulo f, so that ord f (p) = ϕ( f ). Then M a,b has accepted elastcty f and only f 1. ϕ( f ) > 5, and 2. ω 1 + ϕ( f ) 2. Example 4.12 Suppose M a,b satsfes the hypotheses of Theorem 4.11 and assume M a,b has accepted elastcty. Snce p s a prmtve root modulo f, ord f (p) = ϕ( f ) > 5 by condton 1. Condton 2 then forces α 4. Hence, for α = 1,2 or 3, the elastcty of an M a,b satsfyng the hypotheses of Theorem 4.11 s not accepted. Examples of accepted and non-accepted elastcty of monods of ths type can easly be constructed, even n the case where α = 4. For nstance, t s easy to show that M 2 4 9, does not satsfy condton 2 above snce ω = α = 4 < 1 + ϕ(11)/2. Hence M 2 4 9, does not have accepted elastcty. The ACM M 2 4 4,2 4 9 does satsfy condtons 1 and 2 and hence has accepted elastcty. In ths case, α = 4 and β = ϕ(9) = 6 yelds that ρ(m 2 4 4,2 4 9 ) = 4 9 whch s realzed by the rreducble factorzaton [(2 4 )(17)(5)] 9 = [(2 9 )(17 9 )][(2 9 )(5 3 )] 3. Whle not much s known about sngular ACMs whch are fully elastc, we do have one partal result. Theorem 4.13 [9, Corollary 3.3] Let M a,b be a local sngular ACM wth d = p α. Suppose ρ(m a,b ) < 2. Then M a,b s fully elastc f and only f α = ord f (p). Proof. By Theorem 4.9.2, ρ(m a,b ) < 2 f and only f a = d = p α. By Lemma 4.2, snce a = p α M a,b, we must have p α 1 mod f so ord f (p) dvdes α. Conversely, f ord f (p) dvdes α, then p α M a,b by Lemma 4.2. Snce d = p α dvdes every element of M a,b, we fnd that a = d = p α by the mnmalty of a. Thus we have establshed that we are n the exact stuaton proscrbed by Corollary 3.3 of [9], from whch we conclude that M a,b s fully elastc f and only f α = ord f (p). 4.2 Global Arthmetc Congruence Monods We begn by observng that every global ACM can be wrtten n terms of local ACMs. Proposton 4.14 [7, Secton 4] A global ACM M a,b s a fnte, unque ntersecton of local ACMs. Specfcally, f d = gcd(a,b) = p α 1 1 pα n n and f = b/d, then M a,b = n =1 where a s the least resde of a modulo f p α. M a, f p α

20 20 Paul Bagnsk and Scott Chapman Proof. The proof follows easly usng the Chnese Remander Theorem. Though ths observaton may appear to be valuable upon frst glance, t s actually of lttle utlty for questons of nonunque factorzaton. As we saw n Secton 2 wth M b,b for b not a power of a prme, the ntersecton of a fnte number of wellbehaved ACMs can stll result n an ACM wth poorly behaved factorzaton theory. Smlarly here, we may fnd M a,b to be an ntersecton of local ACMs all of whch have fnte elastcty, yet the ntersecton M a,b wll never have fnte elastcty due to the followng lemma. Lemma 4.15 [7, Theorem 4.2][9, Theorem 2.3] Let M a,b be a global ACM. There exsts λ 3 such that for all nonunts x M a,b, mnl (x) < λ. In partcular, f d = p α 1 1 pα n n, then { ord f (p ) λ mn γ n} Proof. Ths result was proven n both [7] and [9], wth [7] gvng a constructve method for obtanng small (occasonally sharp) values of λ. Our proof dffers from both of these proofs n the nterest of smplcty. By Lemma 4.2, M a,b = M d,d M 1, f where gcd(d, f ) = 1, and so there exsts a least γ 1 such that d γ 1 mod f. Wrte d = p α 1 1 pα n n, where n 2 and α 1 for all n. We shall prove the exstence of γ and ts bound usng p 1 ; the proof for other 1 n s analogous. We may choose a mnmal 1 v γ such that there exsts u 0 wth d v p u 1 M a,b. For ths v, choose the mnmal u 0; then d v p u 1 s rreducble n M a,b. Snce p ord f (p 1 ) 1 1 mod f, we must have 0 u < ord f (p 1 ). Now let x M a,b and wrte x = d k m, where d N m. Clearly, maxl (x) k snce every rreducble factor of x must be dvsble by d n N by Lemma 4.2. Thus, f k γ + ord f (p 1 )/α 1, we are done. Let us assume k > γ + ord f (p 1 )/α 1, so that k γ ord f (p 1 )/α 1 > v u/α 1. Wrte m = p s 1 l, where p 1 N l and s 0. Thus v p1 (x) = kα 1 + s. Snce k > v u/α 1, the nteger (k v 1)α1 + s u z = ord f (p 1 ) s nonnegatve. We have the followng equaton, x = d k p s 1 l = (dv p z ord f (p 1 )+u 1 )(p (k v)α 1+s z ord f (p 1 ) u 1 p (k v)α 2 2 p (k v)α n n l). Set y = p (k v)α 1+s z ord f (p 1 ) u 1 p (k v)α 2 2 p (k v)α n n l. Then x = (d v p z ord f (p 1 )+u 1 )y. By the choce of u and v, d v p z ord f (p 1 )+u 1 M a,b. In fact, by the mnmalty of v, ths element s rreducble n M a,b. By the choce of z, we have α α 1 (k v)α 1 + s z ord f (p 1 ) u < α 1 + ord f (p 1 ). (1)