International Journal of Algebra, Vol. 2, 2008, no. 20, 953 - 956
Minimal Prime Submodules Atul Gaur and Alok Kumar Maloo Department of Mathematics and Statistics Indian Institute of Technology, Kanpur 208016, India
[email protected],
[email protected] Abstract In this note we give an example of a minimal prime submodule N of a module M over a ring R such that (N :R M ) is a not a minimal prime ideal of R thereby answering a question of P. F. Smith.
Mathematics Subject Classification: 13C12, 13C99 Keywords: Prime submodules, Minimal prime submodules
1
Introduction
Throughout this article, rings are assumed to be commutative with unity and modules are assumed to be unitary. Let R denote a ring and let M be an R-module. There is no guarantee that M admits prime submodules (for example Q/Z as a Z-module). However, if M does admit prime submodules then M also admits minimal prime submodules (see, Remark 2.2). In, [2, pp. 5257], the following question is attributed to P. F. Smith: If M is faithful and N is a minimal prime submodule of M , is (N :R M ) then a minimal prime ideal of R? In this note we give an example to show that the answer to this question, in general, is in negative.
2
The results
Henceforth, we shall assume that R denotes a ring. We first recall a definition. The first author was supported by SRF of CSIR, India.
954
A. Gaur and A. Kumar Maloo
Definition 2.1 Let M be an R-module. A proper submodule N of M is said to be a prime submodule of M if (N :R M ) is a prime ideal of R and M/N is a torsion-free module over the integral domain R/(N :R M ). If N is a prime submodule of M then N is also referred as a P -prime submodule of M , where P = (N :R M ). Remark 2.2 Let M be an R-module and let F = {N | N is a prime submodule of M }. If F 6= ∅, then by Zorn’s Lemma, F has minimal elements. A minimal element of F is called a minimal prime submodule of M . Notation. For an R-module M , where R is an integral domain, let T (M ) denote the torsion submodule of M . We now prove some lemmas. These lemmas establish the existence of prime submodules for finitely generated nontrivial modules. Lemma 2.3 Let M be an R-module and let S be a multiplicative set of R. Let P be a prime ideal of R such that P ∩ S = ∅. Let X = {N | N is a P - prime submodule of M } and Y = {L | L is a S −1 P - prime submodule of S −1 M }. Then there is a bijective correspondence θ from X to Y which takes N ∈ X to S −1 N ∈ Y . Proof. Let N ∈ X. Then we need to show that S −1 N ∈ Y . Let x N 0 = {x ∈ M | ∈ S −1 N }. 1 We show that N = N 0 . Clearly, N ⊆ N 0 . Let x ∈ N 0 . Then there exists some s ∈ S such that sx ∈ N . As M/N is torsion-free over R/P and s 6∈ P , we have x ∈ N . In particular, S −1 N is a proper submodule of S −1 M . Furthermore, as M/N is torsion-free over R/P , S −1 M/S −1 N is a torsion-free S −1 R/S −1 P module. Thus S −1 N is a S −1 P -prime submodule of S −1 M . Note that as N = N 0 , θ is one-one. x ∈ L}. One trivially checks that Now let L ∈ Y . Put N = {x ∈ M | 1 L = S −1 N and that N is a P -prime submodule of M . 2 Lemma 2.4 Let M be an R-module. Then we have the following: (a) Let m be a maximal ideal of R such that M 6= mM . Then mM is an m-prime submodule of M .
Minimal prime submodules
955
(b) Let P be a prime ideal of R such that MP 6= P MP . Then M admits a P -prime submodule. Proof. Clearly, m ⊆ (mM :R M ) 6= R. Therefore, m = (mM :R M ). Furthermore, M/mM is torsion-free over the field R/m. This proves part (a). We now prove part (b). By part (a), P MP is a P RP -prime submodule of x MP . Now, by Lemma 2.3, N = {x ∈ M | ∈ P MP } is a P -prime submodule 1 of M . 2 Corollary 2.5 Let M be a finitely generated faithful R-module and let P be a prime ideal of R. Then we have the following: (a) M admits a P -prime submodule. (b) If P is maximal then P M is a P -prime submodule of M . Proof. As M is finitely generated and faithful, MP 6= 0. Furthermore, by Nakayama Lemma, MP 6= P MP . Now part (a) follows by Lemma 2.4 and part (b) follows by Lemma 2.4 and the fact that M 6= P M . 2 Lemma 2.6 Let R be an integral domain and let M be a finitely generated faithful R-module. Then T (M ) is a (0)-prime submodule of M . Furthermore, T (M ) is contained in every (0)-prime submodule of M . Proof. As M is finitely generated and faithful, M 6= T (M ). As M/T (M ) is a torsion-free R-module, it follows that T (M ) is a (0)-prime submodule of M . Now, let N be a (0)-prime submodule of M and let x ∈ T (M ). Then there exists some a ∈ R \ {0} such that ax = 0 ∈ N . As M/N is torsion-free over R, we have x ∈ N . Thus T (M ) ⊆ N . 2 We now give an example of a minimal prime submodule N of a faithful R-module M such that (N :R M ) is not a minimal prime ideal of R. Example 2.7 Let k be a field and let R = k[[x, y]], where x, y are indeterminates. Put P = (x, y). Then (P y :R P ) = (y). Put R0 = R/(y), P 0 = P/(y) and M = P/P y. Clearly, M is a finitely generated faithful R0 -module. Henceforth, treat M as an R0 -module. Now, by Corollary 2.5, P 0 M = P 2 /P y is a P 0 -prime submodule of M . Note that T (M ) = R0 z, where z = y + P y. As y 6∈ P 2 , it follows that T (M ) 6⊆ P 0 M . We claim that P 0 M is a minimal prime submodule of M . Let N be a prime submodule of M such that N ⊆ P 0 M . Then N is either (0)-prime or P 0 -prime. If first is the case then by Lemma 2.6, T (M ) ⊆ N ⊆ P 0 M , a contradiction. Therefore, N is P 0 -prime. Clearly, P 0 M ⊆ N . Thus P 0 M = N . Therefore, P 0 M is a minimal prime submodule of M and P 0 = (P 0 M :R0 M ) is not a minimal prime ideal of R0 .
956
A. Gaur and A. Kumar Maloo
Remark 2.8 If M is a faithful multiplication R-module then by [1, Corollary 2.11], for every minimal prime submodule N of M , the ideal (N :R M ) is a minimal prime ideal of R.
References [1] Z. A. El-Bast and P. F. Smith, Multiplication modules, Communications in Algebra 16(4)(1988), 755-779. [2] Y. Tiras and M. Alkan, Prime modules and submodules, Communications in Algebra 31(11)(2003), 5253-5261. Received: September 24, 2008
