Zero-divisors of semigroup modules

See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/45900208 Zero-divisors of Semigroup Modules Article in Kyungpook Mathematical Journal · February 2010 DOI: 10.5666/KMJ.2011.51.1.037 · Source: arXiv CITATIONS READS 2 30 1 author: Peyman Nasehpour University of Tehran 14 PUBLICATIONS 17 CITATIONS SEE PROFILE All content following this page was uploaded by Peyman Nasehpour on 17 April 2015. The user has requested enhancement of the downloaded file. ZERO-DIVISORS OF SEMIGROUP MODULES arXiv:1002.1869v1 [math.AC] 9 Feb 2010 PEYMAN NASEHPOUR A BSTRACT. Let M be an R-module and S a semigroup. Our goal is to discuss zerodivisors of the semigroup module M[S]. Particularly we show that if M is an R-module and S a commutative, cancellative and torsion-free monoid, then the R[S]-module M[S] has few zero-divisors of degree n if and only if the R-module M has few zero-divisors of degree n and Property (A). 1. I NTRODUCTION Let S be a commutative semigroup and M be an R-module. One can define the semigroup module M[S] as an R[S]-module constructed from the semigroup S and the Rmodule M simply similar to standard definition of semigroup rings. Obviously similar to semigroup rings, the zero-divisors of the semigroup module M[S] are interesting to investigate ([G1, p. 82] and [J]). Let g = m1 X s1 + m2 X s2 + · · · + mn X sn , where m1 , · · · , mn ∈ M and s1 , · · · , sn ∈ S. We define the content of g to be the R-submodule of M generated by the coefficients of g. Northcott gave a nice generalization of Dedekind-Mertens Lemma as follows: if S is a commutative, cancellative and torsion-free monoid and M is an R-module, then for all f ∈ R[S] and g ∈ M[S], there exists a natural number k such that c( f )k c(g) = c( f )k−1 c( f g) ([N2]). Dedekind-Mertens Lemma has different versions with various applications ([AG], [AK], [BG], [GGP], [HH], [N1], [NY] and [OR] and [R]). One of its interesting results is McCoy’s Theorem on zero-divisors ([G1, p. 96] and [M]): If M is a nonzero R-module and S is a commutative, cancellative and torsion-free monoid, then for all f ∈ R[S] and g ∈ M[S] − {0}, if f g = 0, then there exists an m ∈ M − {0} such that f .m = 0. An R-module M is said to have few zero-divisors of degree n, if ZR (M) is a finite union of n prime ideals p1 , · · · , pn of R such that pi * p j for all i 6= j. We use McCoy’s Theorem to prove that if M is an R-module and S a commutative, cancellative and torsion-free monoid. Then the R[S]-module M[S] has few zero-divisors of degree n, if and only if the R-module M has few zero-divisors of degree n and Property (A). In this paper all rings are commutative with identity and all modules are unital. Unless otherwise stated, our notation and terminology will follow as closely as possible that of Gilmer [G1]. 2. Z ERO -D IVISORS OF S EMIGROUP M ODULES Let us recall that if R is a ring and f = a0 + a1 X + · · · + an X n is a polynomial on the ring R, then content of f is defined as the R-ideal, generated by the coefficients of f , i.e. c( f ) = (a0 , a1 , · · · , an ). The content of an element of a semigroup module is a natural generalization of the content of a polynomial as follows: 1 2 PEYMAN NASEHPOUR Definition 1. Let M be an R-module and S be a commutative semigroup. Let g ∈ M[S] and put g = m1 X s1 + m2 X s2 + · · · + mn X sn , where m1 , · · · , mn ∈ M and s1 , · · · , sn ∈ S. We define the content of g to be the R-submodule of M generated by the coefficients of g, i.e. c(g) = (m1 , · · · , mn ). Theorem 2. Let S be a commutative monoid and M be a nonzero R-module. Then the following statements are equivalent: (1) S is a cancellative and torsion-free monoid. (2) For all f ∈ R[S] and g ∈ M[S], there is a natural number k such that c( f )k c(g) = c( f )k−1 c( f g). (3) (McCoy’s Property) For all f ∈ R[S] and g ∈ M[S] − {0}, if f g = 0, then there exists an m ∈ M − {0} such that f .m = 0. (4) For all f ∈ R[S], AnnM (c( f )) = 0 if and only if f ∈ / ZR[S] (M[S]). Proof. (1) → (2) has been proved in [N2]. For (2) → (3), assume that f ∈ R[S] and g ∈ M[S] − {0}, such that f g = 0. So there exists a natural number k such that c( f )k c(g) = c( f )k−1 c( f g) = (0). Take t the smallest natural number such that c( f )t c(g) = (0) and choose m a nonzero element of c( f )t−1 c(g). It is easy to check that f .m = 0. For (3) → (1), we prove that if S is not cancellative nor torsion-free then (1) cannot hold. For the moment, suppose that S is not cancellative, so there exist s,t, u ∈ S such that s + t = s + u while t 6= u. Put f = X s and g = (qX t − qX u ), where q is a nonzero element of M. Then obviously f g = 0, while f .m 6= 0 for all m ∈ M − {0}. Finally suppose that S is cancellative but not torsion-free. Let s,t ∈ S be such that s 6= t, while ns = nt for some natural n. Choose the natural number k minimal so that ks = kt. Then we have (k−i−1)s+it )(qX s − qX t ), where q is a nonzero element of M. 0 = qX ks − qX kt = (∑k−1 i=0 X Since S is cancellative, the choice of k implies that (k −i1 −1)s+i1t 6= (k −i2 −1)s+i2t (k−i−1)s+it 6= 0, and this completes the proof. for 0 ≤ i1 < i2 ≤ k − 1. Therefore ∑k−1 i=0 X (3) ↔ (4) is obvious.  Corollary 3. Let M be an R-module and S be a commutative, cancellative and torsionfree monoid. Then the following statements hold: (1) R is a domain if and only if R[S] is a domain. (2) If p is a prime ideal of R, then p[S] is a prime ideal of R[S]. (3) If p is in AssR (M), then p[S] is in AssR[S] (M[S]). Definition 4. Let M be an R-module and P be a proper R-submodule of M. P is said to be a prime submodule (primary submodule) of M, if rx ∈ P implies x ∈ P or rM ⊆ P (there exists a natural number n such that rn M ⊆ P), for each r ∈ R and x ∈ M. Corollary 5. Let M be an R-module and S be a commutative, cancellative and torsionfree monoid. Then the following statements hold: (1) (0) is a prime (primary) submodule of M if and only if (0) is a prime (primary) submodule of M[S]. (2) If P is a prime (primary) submodule of M, then P[S] is a prime (primary) submodule of M[S]. ZERO-DIVISORS OF SEMIGROUP MODULES 3 In [D], it has been defined that a ring R has few zero-divisors, if Z(R) is a finite union of prime ideals. We give the following definition and prove some interesting results about zero-divisors of semigroup modules. Modules having (very) few zero-divisors, introduced in [N1], have also some interesting homological properties [NP]. Definition 6. An R-module M has very few zero-divisors, if ZR (M) is a finite union of prime ideals in AssR (M). Remark 7. Examples of modules having very few zero-divisors. If R is a Noetherian ring and M is an R-module such that AssR (M) is finite, then obviously M has very few zero-divisors. For example AssR (M) is finite if M is a finitely generated R-module [K, p. 55]. Also if R is a Noetherian quasi-local ring and M is a balanced big Cohen-Macaulay R-module, then AssR (M) is finite [BH, Proposition 8.5.5, p. 344]. Remark 8. Let R be a ring and consider the following three conditions on R: (1) R is a Noetherian ring. (2) R has very few zero-divisors. (3) R has few zero-divisors. Then, (1) → (2) → (3) and none of the implications are reversible. Proof. For (1) → (2) use [K, p. 55]. It is obvious that (2) → (3). Suppose k is a field, A = k[X1, X2 , X3 , · · · , Xn , · · ·] and m = (X1 , X2, X3 , · · · , Xn, · · · ) and at last a = (X12, X22 , X32, · · · , Xn2, · · · ). Since A is a content k-algebra and k has very few zerodivisors, A has very few zero-divisors while it is not a Noetherian ring. Also consider the ring R = A/a. It is easy to check that R is a quasi-local ring with the only prime ideal m/a and Z(R) = m/a and finally m/a ∈ / AssR (R). Note that AssR (R) = 0/ [N1].  Theorem 9. Let the R-module M have very few zero-divisors. If S is a commutative, cancellative, torsion-free monoid then the R[S]-module M[S] has very few zero-divisors too. Proof. Let ZR (M) = p1 ∪ p2 ∪ · · · ∪ pn , where pi ∈ AssR (M) for all 1 ≤ i ≤ n. We will show that ZR[S] (M[S]) = p1 [S] ∪ p2 [S] ∪ · · · ∪ pn [S]. Let f ∈ ZR[S] (M[S]), so there exists an m ∈ M − {0} such that f .m = 0 and so c( f ).m = (0). Therefore c( f ) ⊆ ZR (M) and this means that c( f ) ⊆ p1 ∪ p2 ∪ · · · ∪ pn and according to Prime Avoidance Theorem, we have c( f ) ⊆ pi , for some 1 ≤ i ≤ n and therefore f ∈ pi [S]. Now let f ∈ p1 [S] ∪ p2 [S] ∪ · · · ∪ pn [S], so there exists an i such that f ∈ pi [S], so c( f ) ⊆ pi and c( f ) has a nonzero annihilator in M and this means that f is a zero-divisor of M[S]. Note that by Corollary 3, pi [S] ∈ AssR[S] (M[S]) for all 1 ≤ i ≤ n.  In [HK], it has been defined that a ring R has Property (A), if each finitely generated ideal I ⊆ Z(R) has a nonzero annihilator. We give the following definition: Definition 10. An R-module M has Property (A), if each finitely generated ideal I ⊆ ZR (M) has a nonzero annihilator in M. Remark 11. If the R-module M has very few zero-divisors, then M has Property (A). Theorem 12. Let S be a commutative, cancellative and torsion-free monoid and M be an R-module. The following statements are equivalent: 4 PEYMAN NASEHPOUR (1) The R-module M has Property (A). (2) For all f ∈ R[S], f is M[S]-regular if and only if c( f ) is M-regular. Proof. (1) → (2): Let the R-module M has Property (A). If f ∈ R[S] is M[S]-regular, then f .m 6= 0 for all nonzero m ∈ M and so c( f ).m 6= (0) for all nonzero m ∈ M and according to the definition of Property (A), c( f ) 6⊆ ZR (M). This means that c( f ) is M-regular. Now let c( f ) be M-regular, so c( f ) 6⊆ ZR (M) and this means that c( f ).m 6= (0) for all nonzero m ∈ M and hence f .m 6= 0 for all nonzero m ∈ M. Since S is a commutative, cancellative and torsion-free monoid, f is not a zero-divisor of M[S], i.e. f is M[S]-regular. (2) → (1): Let I be a finitely generated ideal of R such that I ⊆ ZR (M). Then there exists an f ∈ R[S] such that c( f ) = I. But c( f ) is not M-regular, therefore according to our assumption, f is not M[S]-regular. Therefore there exists a nonzero m ∈ M such that f .m = 0 and this means that I.m = (0), i.e. I has a nonzero annihilator in M.  Let, for the moment, M be an R-module such that the set ZR (M) of zero-divisors of M is a finite union of prime ideals. One can consider ZR (M) = ∪ni=1 pi such that pi * ∪nj=1∧ j6=i p j for all 1 ≤ i ≤ n. Obviously we have pi * p j for all i 6= j. Also, it is easy to check that, if ZR (M) = ∪ni=1 pi and ZR (M) = ∪m k=1 qk such that pi * p j for all i 6= j and qk * ql for all k 6= l, then m = n and {p1 , · · · , pn } = {q1 , · · · , qn }, i.e. these prime ideals are uniquely determined (Use Prime Avoidance Theorem). This is the base for the following definition: Definition 13. An R-module M is said to have few zero-divisors of degree n, if ZR (M) is a finite union of n prime ideals p1 , · · · , pn of R such that pi * p j for all i 6= j. Theorem 14. Let M be an R-module and S a commutative, cancellative and torsion-free monoid. Then the R[S]-module M[S] has few zero-divisors of degree n, if and only if the R-module M has few zero-divisors of degree n and Property (A). Proof. (←): By considering the R-module M having Property (A), similar to the proof of Theorem 9, we have if ZR (M) = ∪ni=1 pi , then ZR[S] (M[S]) = ∪ni=1 pi [S]. Also it is obvious that pi [S] ⊆ p j [S] if and only if pi ⊆ p j for all 1 ≤ i, j ≤ n. These two imply that the R[S]-module M[S] has few zero-divisors of degree n. (→): Note that ZR (M) ⊆ ZR[S] (M[S]). It is easy to check that if ZR[S] (M[S]) = ∪ni=1 Qi , where Qi are prime ideals of R[S] for all 1 ≤ i ≤ n, then ZR (M) = ∪ni=1 (Qi ∩ R). Now we prove that the R-module M has Property (A). Let I ⊆ ZR (M) be a finite ideal of R. Choose f ∈ R[S] such that I = c( f ). So c( f ) ⊆ ZR (M) and obviously f ∈ ZR[S] (M[S]) and according to McCoy’s property, there exists a nonzero m ∈ M such that f .m = 0. This means that I.m = 0 and I has a nonzero annihilator in M. Consider that by a similar discussion in (←), the R-module M has few zero-divisors obviously not less than degree n and this completes the proof.  An R-module M is said to be primal, if ZR (M) is an ideal of R [D]. It is easy to check that if ZR (M) is an ideal of R, then it is a prime ideal and therefore the R-module M is primal if and only if M has few zero-divisors of degree one. Corollary 15. Let M be an R-module and S a commutative, cancellative and torsion-free monoid. Then the R[S]-module M[S] is primal, if and only if the R-module M is primal and has Property (A). ZERO-DIVISORS OF SEMIGROUP MODULES 5 3. ACKNOWLEDGMENT The author wishes to thank Prof. Winfried Bruns for his useful advice and Dr. Neil Epstein for his editorial comments. R EFERENCES [AG] J. T. Arnold and R. Gilmer, On the content of polynomials, Proc. Amer. Math. Soc. 40 (1970), 556– 562. [AK] D. D. Anderson and B. G. Kang, Content formulas for polynomials and power series and complete integral closure, J. Algebra, 181 (1996), 82–94. [BG] W. Bruns and A. Guerrieri, The Dedekind-Mertens formula and determinantal rings, Proc. Amer. Math. Soc. 127 (1999), no. 3, 657–663. [BH] W. Bruns and J. Herzog, Cohen-Macaulay Rings, revised edn., Cambridge, 1998. [D] J. Dauns, Primal modules, Comm. Algebra 25 (1997), no. 8, 2409–2435. [G1] R. Gilmer, Multiplicative Ideal Theory, Marcel Dekker, New York, 1972. [G2] R. Gilmer, Commutative Semigroup Rings, The University of Chicago Press, 1984. [GGP] R. Gilmer, A. Grams and T. Parker, Zero divisors in power series rings, J. Reine Angew. Math. 278 (1975), 145–164. [HH] W. Heinzer and C. Huneke, The Dedekind-Mertens Lemma and the content of polynomials, Proc. Amer. Math. Soc. 126 (1998), 1305–1309. [H] J. A. Huckaba, Commutative Rings with Zero Divisors, Marcel Dekker, 1988. [HK] J. A. Huckaba and J. M. Keller, Annihilation of ideals in commutative rings, Pac. J. Math. 83 (1979), 375–379. [J] B. D. Janeway, Zero divisors in commutative semigroup rings, Comm. Algebra 12 (1984), 1877–1887. [K] I. Kaplansky, Commutative Rings, Allyn and Bacon, Boston, 1970. [M] N. H. McCoy, Remarks on divisors of zero, Amer. Math. Monthly, 49 (1942), 286–295. [N1] P. Nasehpour, Content algebras over commutative rings with zero divisors, arXiv:0807.1835v3, preprint. [N2] D. G. Northcott, A generalization of a theorem on the content of polynomials, Proc. Cambridge Phil. Soc. 55 (1959), 282–288. [NP] P. Nasehpour and Sh. Payrovi, Modules having few zero-divisors, to appear in Comm. Algebra. [NY] P. Nasehpour and S. Yassemi, M-cancellation Ideals, Kyungpook Math. J., 40 (2000), 259–263. [OR] J. Ohm and D. E. Rush, Content modules and algebras, Math. Scand. 31 (1972), 49–68. [R] D. E. Rush, Content algebras, Canad. Math. Bull. Vol. 21 (3) (1978), 329–334. [T] H. Tsang, Gauss’ lemma, dissertation, University of Chicago, Chicago, 1965. [ZS] O. Zariski and P. Samuel, Commutative Algebra, Vol. I, Van Nostrand, New York, 1958. U NIVERSIT ÄT O SNABR ÜCK , FB M ATHEMATIK /I NFORMATIK , 49069 O SNABR ÜCK , G ERMANY E-mail address: Peyman.Nasehpour@mathematik.uni-osnabrueck.de E-mail address: Nasehpour@gmail.com View publication stats