FACTORIZATION IN KRULL MONOIDS WITH INFINITE CLASS GROUP

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1 C O L L O Q U I U M M A T H E M A T I C U M VOL NO. 1 FACTORIZATION IN KRULL MONOIDS WITH INFINITE CLASS GROUP BY FLORIAN K A I N R A T H (GRAZ) Abstract. Let H be a Krull monod wth nfnte class group and such that each dvsor class of H contans a prme dvsor. We show that for each fnte set L of ntegers 2 there exsts some h H such that the followng are equvalent: () h has a representaton h = u 1... u k for some rreducble elements u, () k L. 1. Introducton and notatons. Let H be a Krull monod. For an element h of H ts set of lengths L(h) s defned as the set of all ntegers k such that there exst rreducble u 1,..., u k wth h = u 1... u k. If the class group of H s fnte, then the sets L(h) have a specal structure: L(h) = {x 1,..., x α, y 1,... y l, y 1 + d,... y l + d, y 1 + kd,... y l + kd, z 1,..., z β }, where x 1 <... < x α < y 1 <... < y l < y 1 + d < y l + kd < z 1 <... < z β and α, β, d M for some constant M dependng only on the class group of H ([1], Theorem 2.13). In ths paper we look at the sets L(h) when the class group of H s nfnte and each dvsor class of H contans a prme dvsor. Our man result states that n ths case every fnte set of ntegers 2 occurs as a set of lengths of an element n H. We apply ths result also to certan ntegral domans. Throughout ths paper the followng notatons wll be used. We let N be the set of all nonnegatve ntegers, N + = N \ {0} and N 2 the set of all ntegers 2. For a fnte set X we denote by X the number of elements of X. 2. Sets of lengths. In the followng let H be a commutatve, cancellatve monod wth unt element. By a factorzaton of an element h H we mean a representaton of the form h = u 1... u k wth rreducble u H. The nteger k s called the length of the factorzaton. Two factorzatons 1991 Mathematcs Subject Classfcaton: 11R27, 13G05. [23]

2 24 F. KAINRATH h = u 1... u k = v 1... v l are sad to be essentally the same f k = l and after some renumberng u =e v for some unt e ; they are called essentally dfferent f they are not essentally the same. We denote by L(h) = L H (h) the set of lengths of factorzatons of h and defne a functon v h = v H,h : L(h) N + by v h (k) = the number of essentally dfferent factorzatons of h havng length k. Now let H be Krull monod (see for example [1]), : H D ts dvsor theory and G = D/ (H) ts class group. We denote the canoncal map D G by d [d]. We say that every dvsor class of H contans a prme dvsor f for every g G there exsts a prme element p D wth [p] = g. Now we can state our man result. Theorem 1. Let H be a Krull monod wth nfnte class group n whch every dvsor class contans a prme dvsor. For a fnte subset L N 2 there exsts some h H such that L H (h) = L. If the class group of H s not of the form (Z/2Z) (N) Γ wth an nfnte set N and a fnte group Γ, then there s such an h satsfyng v h = v, where v s any gven functon L N +. For the proof of ths theorem we need the concept of block monods. Let G be the class group of H. We let F(G) be the free abelan monod wth bass G. The block monod B(G) over G s the submonod of F(G) defned by { } B(G) = n g g = 0. g G g ng F(G) : g G We say that a block g 1... g n B(G) s square free f the g are parwse dstnct. For an element h H defne β(h) B(G) by β(h) = [p 1 ]... [p n ] where (h) = p 1... p n s the prme factorzaton of (h) n D. Then we have L B(G) (β(h)) = L H (h) (see [1], Lemma 3.2). Moreover, t s easy to to see that v H,h = v B(G),β(h) f β(h) s square free. For the proof of Theorem 1 we also need the followng proposton whose proof wll be gven n the next secton. Proposton. Let C be a nonzero cyclc group, L N 2 a fnte set and v : L N + a functon. Then there exsts a block B n B(C k ) for some k 1 such that L(B) = L. If C Z/2Z, then there s a square free block B B(C k ) such that L = L(B) and v B = v. Proof of Theorem 1. Let H be as n Theorem 1, G ts class group and choose some fnte L N 2 and v : L N +. We show that there s a block

3 FACTORIZATION IN KRULL MONOIDS 25 B B(G) wth L(B) = L. If G s not of the form (Z/2Z) (N) Γ wth an nfnte set N and a fnte group Γ, then we wll choose B such that t s square free and satsfes v B = v. By the above consderatons ths wll prove Theorem 1. We consder three cases. Case 1: G s not a torson group. Then G contans a subgroup somorphc to Z, so we may assume G = Z. By the Proposton (wth C = Z) there s a square free block B B(Z k ) for some k such that L(B) = L and v B = v, say B = u 1... u n. Choose some homomorphsm f : Z k Z such that f(u ) 0 f u 0, I {1,..., n}, and I I f(u ) f(u j ) f j. Then t s clear that the square free block C = f(u 1 )... f(u n ) B(Z) satsfes L(C) = L and v C = v. Case 2: G s a torson group whch contans elements of arbtrarly hgh order. Choose frst a square free block B = u 1... u n B(Z) such that L(B) = L and v B = v. Ths s possble by Case 1. Defne M N by ({ } ) M = max u : I {1,..., n} { u u j :, j = 1,..., n}. I Then t s obvous that for every N > M the square free block B N = (u 1 + NZ)... (u n + NZ) B(Z/NZ) satsfes L(B N ) = L and v BN = v as well. By our hypothess on G there exsts an element of order greater than M, whch means that G contans a subgroup somorphc to Z/NZ for some N > M. Therefore the theorem s proved n ths case. Case 3: G s a torson group n whch the orders of all elements are bounded. By Theorem 6 of [4], G s a drect sum of cyclc groups G = Z/n Z I for some bounded famly of ntegers n 2. Snce by assumpton G s nfnte there s an nteger m such that G contans a subgroup somorphc to (Z/mZ) (N). If G s not of the form (Z/2Z) (N) Γ wth an nfnte set N and a fnte group Γ, we may suppose m > 2. Usng the Proposton wth C = Z/mZ, we see that the theorem s proved n ths case as well. In the followng we want to apply Theorem 1 to certan ntegral domans. Let R be a noetheran doman whose ntegral closure R s a fntely generated R-module. Denote by H R the set of all nonzero dvsors of R/R: H R = {r R \ {0} : rr R for all r R \ R}. Then H R s a dvsor closed Krull submonod of R = R \ {0} whose class

4 26 F. KAINRATH group s somorphc to the v-class group of R (cf. [2]). Therefore L HR (r) = L R (r) and v HR,r = v R,r for all r H R. Hence we get the followng theorem. Theorem 2. Let R be a noetheran doman wth fntely generated ntegral closure and nfnte v-class group. Suppose that n the monod H R every dvsor class contans a prme dvsor. Then for every fnte set L N 2 there exsts an element r R such L(r) = L. If the v-class group of R s not of the form (Z/2Z) (N) Γ wth an nfnte set N and a fnte group Γ, then there s such an r satsfyng v r = v, where v s any gven functon L N +. Remark. Examples of domans satsfyng the condton on the dvsor classes may be found n [3]. 3. Proof of the Proposton. Let C = Z/cZ, c 1, be some cyclc group. In ths secton we regard C as a rng. Let X 1,..., X n be fnte sets. We suppose that X 2 for all and that and n = 2 and X 3 for at least one or n 3 n 3 For a subset J {1,..., n} we put f C = Z/2Z. X J = j J X j f C Z/2Z, and let X = X {1,...,n} for short. The ponts x of X wll always be wrtten as x = (x 1,..., x n ). We denote by p J : X X J the projecton mappng. For a pont z X we defne X (z) = X \ {z } and X (z) J = j J X (z) j. If x X s a second pont we let J z (x) be the set of all ndces wth x z. We denote by C X the C-algebra of all functons X C. For a subset M of X we let χ M C X be ts characterstc functon. If A C X then C A s the C-submodule generated by A. We now proceed n 10 steps. In Steps 1 to 8 we construct a block n C X /V for some submodule V and calculate ts set of lengths. In Steps 9 and 10 we use ths constructon to prove the proposton. Step 1. For z X the set { J : y X(z) J, J {1,..., n}} s a bass of C X. For each the set {1} {χ y : y X (z) } s obvously a bass of C X. Now, by takng tensor products and by usng the canoncal somorphsm α : C X 1... C X n = C X, α(f 1... f n )(x 1,..., x n ) = f 1 (x 1 )... f n (x n ),

5 FACTORIZATION IN KRULL MONOIDS 27 we prove our clam (note also that = 1 f y s the unque element of X (z) ). Step 2. Defne submodules V, W z (z X) of C X by Then we have V = C : y X, = 1,..., n, W z = C J : J 2, y X(z) J. (1) C X = V W z for all z X. Note that V s generated by {1} { all z X. Therefore the asserton follows from Step 1. : y X(z), = 1,..., n} for Step 3. Let z, x X wth x z and M X, z M. Then there exst w k W z (k = 1, 2, 3) and Y X (z) ( = 1,..., n) such that: (2) χ z = 1 + w 1, (3) (4) =1 y X (z) { } 0 f Jz (x) 2 χ x = + w 2, (x ) f J z (x) = {} χ M = + w 3. =1 y Y Let w X. Then we have n χ w = (w ) = =1 = (w ) J z (w) J z (w) (w ) J z (w) J z (w) (1 y X (z) (z ) Expandng the last product for w = x and z yelds (2) and (3). Formula (4) s an mmedate consequence of (3). Step 4. The cosets χ x + V C X /V (x X) are parwse dstnct. Let x, z X be such that x z and suppose χ z χ x V. By (1) (3), we obtan (5) χ z χ x = 1 n =1 y X (z) χ M, where M = f J z (x) 2 and M = p (x ) f J z (x) = {}. Assume frst C Z/2Z. Choose w X such that w x and J z (w) = 2. Ths s possble ).

6 28 F. KAINRATH by our assumpton on n and the X. Evaluatng both sdes of (5) at w we get 0 on the left sde and or 2 on the rght sde. Ths contradcton proves our asserton n the case C Z/2Z. Assume now C = Z/2Z. Snce n 3 there s some w X such that w x, J z (w) = 2 and, n addton, J z (w) f J z (x) = {}. Agan evaluatng both sdes of (5) at w gves a contradcton. Step 5. Suppose C Z/2Z and let M be a subset of X such that χ M V. Then M = p (Y ) for some and some Y X. If M = X there s nothng to do. So assume z X \ M. By (4) there exst subsets Y X (z) such that χ M =. =1 y Y Takng squares we get χ M = χ 2 M = =1 + 2 y Y <j y Y Y j {,j}. Now usng Step 1 we nfer Y for at most one, whch mples the asserton. Step 6. Assume C = Z/2Z. Let M X and suppose χ M V. For any z X \ M there exst Y X (z) ( = 1,..., n) such that M = {x X : { : x Y } s odd}. By (4) there are Y X (z) such that χ M = =1 y Y = n =1 (Y ). Now the clam follows from the equaton = 0 n Z/2Z. Step 7. Let z X and Y, Y X(z) ( = 1,..., n). Set M Y = {x X : { : x Y } s odd} and defne M Y n the same manner. Suppose we have M Y M Y. Then there exsts an ndex such that M Y = p (Y ) and M Y = p (Y ),.e. Y j = Y j = for j. Let y Y j for some j. Then (z 1,..., z j, y, z j+1,..., z n ) M Y M Y, whch mples y Y j. So we conclude Y j Y j for all j. Snce M Y M Y there s some such that Y Y. Suppose now Y j for some j. Choose y j Y j and y Y \ Y. Then (z 1,..., y,..., y j,..., z n ) M Y \ M Y. Ths contradcton proves Y j = for j. Smlary, suppose y j Y j. Choose y Y. Snce Y j = we obtan (z 1,..., y,..., y j,..., z n) M Y \ M Y.

7 FACTORIZATION IN KRULL MONOIDS 29 Step 8. For any subset M of X defne B M = (χ x + V ) F(C X /V ). x M Then B M s a block f and only f χ M V, n partcular B = B X B(C X /V ). We have L(B) = { X 1,..., X n }, v B ( X ) = {j : X j = X } f C Z/2Z and L(B) = {2, X 1,..., X n } f C = Z/2Z. By Steps 5 7 the blocks B p, = 1,..., n, y X, are rreducble. Therefore B has the followng factorzatons: (6) B =, = 1,..., n. y X B p Suppose frst C Z/2Z. We have to show that the factorzatons (6) are the only ones for B. By Step 5 the rreducble dvsors of B are gven by the B p wth y X and = 1,..., n. Now B s square free and two sets p, p j (y ) wth j have nonempty ntersecton. Hence the asserton follows. Assume now C = Z/2Z. Let z X and choose subsets Y X (z) such that Y for at least two ndces. Set M Y = {x X : { : x Y } s odd}. By Steps 6 and 7 the blocks B MY and B X\MY are rreducble. Hence we obtan 2 L(B). Suppose now that B = B 1... B k s some factorzaton dfferent from all the ones n (6). We have to show that k = 2. Snce B s square free there s some partton X = M 1... M k, M s M t = for s t, such that B s = B Ms for all s. Snce the sets p, p j (y ) for j have nonempty ntersecton, there exsts some s, say s = 1, such that M 1 and therefore also X \ M 1 are not of the form p (Y ) (for any = 1,..., n, Y X ). Hence by Steps 6 and 7 agan, B X\M1 s rreducble, and we get B 2 = B X\M1 and k = 2. Step 9. Suppose that C Z/2Z and let L N 2 be a fnte subset and v : L N + a functon. We assume frst that (7) (L, v) ({m}, m 1), ({2}, 2 2) for all m 2. Set n = l L v(l) and choose fnte sets X 1,..., X n such that for l L exactly v(l) of them have cardnalty l. Then by our assumpton (7) we have n 3, or n = 2 and X 3 for at least one. Then the block B B(C X /V ) constructed n Step 8 satsfes L(B) = L and v B = v. Note also that by Steps 1 and 2, C X /V s free. To fnsh the proof of the proposton n the case C Z/2Z we need to check the two remanng cases

8 30 F. KAINRATH (a) L = {m}, v(m) = 1 (m 2), and (b) L = {2}, v(2) = 2. In case (a) one may for example take 1 B = B(Cm ) For (b) we can choose B = = B(C 3 ) Step 10. Assume now C = Z/2Z and let L N 2 be a fnte set. Defne m = mn L. Suppose that, for some k 1, we have constructed a block B B(C k ) wth L(B) = L m + 2. Then obvously L(0 m 2 B) = L. We may therefore assume that 2 L. In ths case, choose fnte sets X 1,..., X n wth n 3 and L = { X 1,..., X n }. Then the block B constructed n Step 8 satsfes L(B) = L. REFERENCES [1] S. Chapman and A. Geroldnger, Krull domans and monods, ther sets of lengths, and assocated combnatoral problems, n: Factorzaton n Integral Domans, D. D. Anderson (ed.), Lecture Notes n Pure and Appl. Math. 189, Marcel Dekker, 1997, [2] F. K a n r a t h, A dvsor theoretc approach towards the arthmetc of noetheran domans, Arch. Math., to appear. [3], The dstrbuton of prme dvsors n fntely generated domans, preprnt. [4] I. Kaplansky, Infnte Abelan Groups, thrd prntng, The Unversty of Mchgan Press, Insttut für Mathematk Karl-Franzes-Unverstät Henrchstraße 36 A-8010 Graz, Austra E-mal: floran.kanrath@kfungraz.ac.at Receved 26 February 1998