Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 49 (2008), No. 2, 449-479.

Multiplication Modules and Homogeneous Idealization III

Majid M. Ali

Department of Mathematics and Statistics, College of Science Sultan Qaboos University, P.O. Box 36, P.C. 123, Al-Khod, Sultanate of Oman

e-mail: mali@squ.edu.om

Abstract. In our recent work we gave a treatment of certain aspects of multiplication modules, projective modules, flat modules and cancellation-like modules via idealization. The purpose of this work is to continue our study and develop the tool of idealization, particularly in the context of closed, divisible injective, and simple modules. We determine when a ring R (M), the idealization of M, is a quasi-Frobenius or a distinguished ring. We also introduce and investigate the concept of M-\frac{1}{2} (weak) cancellation ideals. MSC 2000: 13C13, 13C05, 13A15 Keywords: closed submodule, divisible module, multiplication module, quasi-Frobenius ring

0. Introduction All rings are commutative with identity and all modules are unital. Let R be a ring and M an R-module. M is a multiplication module if every submodule N of M has the form IM for some ideal I of R. Equivalently, N = [N : M] M, [18]. A submodule K of M is multiplication if and only if N ∩ K = [(N ∩ K) : K] K = [N : K] K for all submodules N of M.

Let P be a maximal ideal of R and let T_{P} (M) = {m ∈ M : (1 − p) m = 0 for some m ∈ M}. Then T_{P} (M) is a submodule of M. M is called P -torsion if T_{P} (M) = M. On the other hand M is called P -cyclic provided there exist m ∈ M and q ∈ P such that (1 − q) M ⊆ Rm. El-Bast and P. F. Smith, [19, Theorem 1.2], showed that M is multiplication if and only if M is P -torsion or P -cyclic for each maximal ideal P of R. A multiplication module M is locally

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2008 Heldermann Verlag

450 M. M. Ali: Multiplication Modules …

cyclic and the converse is true if M is finitely generated, [18, Proposition 4]. A submodule N of M is called a pure submodule of M if IN = N ∩ IM for every ideal I of R, [20]. An ideal I is a pure ideal of R if and only if I is locally either R or zero, [22].

Let N be a submodule of M and I an ideal of R. The residual submodule N by I is [N :_{M} I] = {m ∈ M : Im ⊆ N}, [24] and [25]. Obviously, [N : IM] M ⊆ [N :_{M} I]. The reverse inclusion is true if M is multiplication. If M is faithful multiplication or projective then [0 :_{M} I] = (annI) M. See [7] and [2, Lemma 5].

For a ring R and an R-module M, the idealization of M is the ring R(+)M = R (M) which is formed from the direct sum R ⊕ M by defining multiplication of elements (r, m) and (s, n) of R (M) by (r, m) (s, n) = (rs, rn + sm). The purpose of idealization is to put M inside a commutative ring A so that the structure of M as an R-module is essentially the same as that of M as an A-module, that is an ideal of A. 0(+)M is an ideal of R (M) satisfying (0(+)M)^{2} = 0. Every ideal contained in 0(+)M has the form 0(+)N for some submodule N of M and every ideal contains 0(+)M has the form I (+)M for some ideal I of R. Since R ≅ R (M) /0(+)M, I → I (+)M gives a one-to-one correspondence between the ideals of R and the ideals of R (M) containing 0(+)M. Thus prime (maximal) ideals of R (M) have the form P (+)M for some prime (maximal) ideals P of R. Homogeneous ideals of R (M) have the form I (+)N where I is an ideal of R, N a submodule of M and IM ⊆ N. These ideals play a special role in studying properties of R (M) and showing how these properties are related to those of R and M. It is shown, [17, Theorem 3.3], that a principal ideal R (M) (a, m) is homogeneous if and only if R (M) (a, m) = Ra(+)(Rm + aM) = R (M) (a, 0)+R (M) (0, m), and every ideal of R (M) is homogeneous if and only if every principal ideal of R (M) is homogeneous. A ring R (M) is called homogeneous if every ideal of R (M) is homogeneous, [3] and [4]. It is shown, [17, Corollary 3.4], that if R is an integral domain then R (M) is homogeneous if and only if M is divisible. In this case every ideal of R (M) is comparable to 0(+)M. If R is a local ring but not an integral domain, R (M) is homogeneous if and only if M = 0, [17, Theorem 3.3]. If I (+)N and J (+)K are homogeneous ideals of R (M) then

[I (+)N :_{R(M)} J (+)K] = [I : J] ∩ [N : K] (+) [N :_{M} J]

is homogeneous, [1, Lemma 1]. In particular, ann (I (+)N) = (annI ∩ annN)(+) [0 :_{M} I]. If M is faithful multiplication or projective then ann (I (+)IM) = annI (+) (annI) M.

In our recent work on idealization we investigated the idealization of multiplication modules, projective modules, flat modules, like-cancellation modules, invertible submodules, large and small submodules. We also determined when a ring R (M) is a multiplication ring, ZPI-ring, arithmetical ring, Prüfer ring, Bezout ring, (quasi) valuation ring, Marot ring, P -ring, coherent ring, finite conductor ring or generalized GCD ring, see [1]–[5].

In the first part of this paper we investigate the idealization of modules, particularly in the context of closed, divisible, injective and simple modules, continuing and extending some of our results in [1]–[5]. In the second part we show how quasi-Frobenius and distinguished properties of R (M) are related to those of R and M. In the last part we introduce and investigate the concept of M-\frac{1}{2} (weak) cancellation and M-\frac{1}{2} join principal ideals, particularly when M is a multiplication module. In the first and third part of the paper, some of our results of [3] will be improved by weakening the required conditions.

All rings are commutative with identity and all modules are unital. For the basic concepts used, we refer the reader to [20]–[25].

1. Some properties of homogeneous ideals

In this section we explore several properties of a homogeneous ideal of R(M) and show how these properties can be transferred to its components and conversely.

Let R be a ring and N a submodule of an R-module M. Then N is called an addition complement (resp. intersection complement), briefly adco (resp. inco), of a submodule K of M if K + N = M (resp. K ∩ N = 0) and N is minimal (resp. maximal) in K + N = M (resp. K ∩ N = 0), that is for all submodules L of M with L + K = M (resp. L ∩ K = 0) and L ⊆ N (resp. N ⊆ L) then N = L. It is shown, [23, Corollary 5.2.2], that for all submodules K and N of M, K ⊕ N = M if and only if N is adco and inco of K in M. It is also well-known, [23, Lemmas 5.2.4 and 5.2.5], that for all submodules N, K and L of M if N is adco (resp. inco) of K and K is adco (resp. inco) of L then N is adco (resp. inco) of L. A submodule N of M is large in M if for all submodules K of M, N ∩K = 0 implies K = 0. A submodule X of M is said to be closed in M if whenever X is a large submodule of a submodule U of M then X = U. It is proved that a submodule X of M is closed in M if and only if X is inco of a submodule of M, [23]. Every submodule of M is a direct summand in M if and only if every submodule of M is closed in M. For properties of adco, inco, large and closed submodules and ideals, see for example [20] and [23].

The following result gives some basic properties of the addition complement, intersection complement and closed ideals and submodules.

Proposition 1. Let R be a ring, M an R-module, I and J ideals of R and K and N submodules of M.

(1) Let M be finitely generated multiplication. If I + annM is adco of J then IM is adco of JM and the converse is true if we assume further that M is faithful.

(2) Let M be faithful multiplication. If I is inco of J then IM is inco of JM and the converse is true if we assume further that M is finitely generated.

(3) Let M be faithful multiplication. If I is closed in R then IM is closed in M and the converse is true if we assume further that M is finitely generated.

(4) Let M be finitely generated multiplication. If [N : M] is adco of [K : M] then N is adco of K and the converse is true if we assume further that M is faithful.

(5) Let M be faithful multiplication. If [N : M] is inco of [K : M] then N is inco of K and the converse is true if we assume further that M is finitely generated.

(6) Let M be faithful multiplication. If [N : M] is closed in R then N is closed in M and the converse is true if we assume further that M is finitely generated.

Proof. (1) Let M be finitely generated multiplication. Let I+annM be adco of J. Then I + annM + J = R, and hence IM + JM = M. Suppose L be a submodule of M such that L + JM = M and L ⊆ IM. It follows that M = ([L : M] + J)M, and by, [22, Theorem 76], we have that

R = [L : M] + J + annM = [L : M] + J.

Since [L : M] ⊆ [IM : M] = I + annM, we get that I + annM = [L : M] and hence IM = L. So IM is adco of JM. Conversely, assume M is finitely generated, faithful and multiplication. Let IM be adco of JM. Then (I + J) M = IM + JM = M, and hence I + J = R, [29, Corollary 1 to Theorem 9]. Assume A is an ideal of R such that A + J = R and A ⊆ I. Then AM + JM = M and AM ⊆ IM. Since IM is adco of JM, AM = IM, and hence A = I. So I is adco of J.

(2) Suppose M is faithful and multiplication. Let I be inco of J. Then I ∩ J = 0 and by, [19, Corollary 1.7], IM ∩ JM = 0. Let L be a submodule of M such that L ∩ JM = 0 and IM ⊆ L. So [L : M] ∩ J ⊆ [L : M] ∩ [JM : M] = [(L ∩ JM) : M] = [0 : M] = 0 and I ⊆ [L : M]. It follows that I = [L : M] and hence IM = L. Then IM is inco of JM. Conversely, let IM be inco of JM. Then IM ∩ JM = 0, and hence (I ∩ J) M = 0. This implies that I ∩ J ⊆ [0 : M] = 0. Let A be an ideal of R such that A ∩ J = 0 and I ⊆ A. Then AM ∩ JM = 0 and IM ⊆ AM. Hence IM = AM and hence I = A. This gives that I is inco of J.

(3) Let I be closed in R and let IM be large in a submodule L of M. Then I is large in [L : M]. For, let J ⊆ [L : M] and I ∩ J = 0. Since M is faithful multiplication, it follows by, [19, Corollary 1.7], that IM ∩JM = 0. Since JM ⊆ L and IM is large in L, JM = 0, and hence J = 0. As I is closed in R, we get that I = [L : M] and hence L = IM. So IM is closed in M. For the converse, suppose that IM is closed in M. Let I be large in J. Then IM is large in JM. For, if L ⊆ JM and IM ∩ L = 0, then

0 = [0 : M] = [(IM ∩ L) : M] = I ∩ [L : M] .

It follows that [L : M] = 0, and hence L = 0. This implies that IM = JM, and hence I = J. So I is closed in R. The proofs of (4), (5) and (6) follow by (1), (2) and (3) respectively. □

The next two results give some properties of idealization of addition complement, intersection complement and closed ideals and submodules.

Theorem 2. Let R be a ring, M an R-module and R (M) the idealization of M. Let I be an ideal of R and N a submodule of M.

(1) If 0(+)N is adco of an ideal H of R (M) then H = R(M).

(2) Let M be faithful. If I (+)M is inco of an ideal H of R (M) then H = 0.

(3) N is adco of a submodule K of M if and only 0(+)N is adco of 0(+)K in 0(+)M.

(4) N is inco of a submodule K of M if and only if 0(+)N is inco of 0(+)K in 0(+)M.

(5) N is closed in M if and only if 0(+)N is closed in 0(+)M.

(6) Let M be faithful multiplication. If I (+)IM is closed in R(M) then I is closed in R. The converse is true if we assume further that M is finitely generated.

(7) If I (+)IM is adco of an ideal J (+)JM of R (M) then I is adco of J. The converse is true if R (M) is homogeneous and M finitely generated and faithful.

(8) Let M be faithful multiplication. If I (+)IM is inco of an ideal J (+)JM of R (M) then I is inco of J. The converse is true if we assume further that R (M) is homogeneous.

Proof. (1) If 0(+)N is adco of H then 0(+)N + H = R (M). Hence H (0(+)N) = 0(+)N, and hence 0(+)N ⊆ H. So H = R (M).

(2) If I (+)M is inco of H then I (+)M ∩ H = 0, and hence 0(+)M ∩ H = 0. Since M is faithful, 0(+)M is large in R (M), [3, Proposition 14]. Hence H = 0. The proofs of (3), (4) and (5) are easy exercises by using the fact that any ideal H ⊆ 0(+)M has the form 0(+)K for some submodule K of M.

(6) Suppose I (+)IM is closed. Let J be an ideal of R such that I is large in J. Then I (+)IM is large in J (+)JM. For, let H ⊆ J (+)JM such that I (+)IM ∩ H = 0. Then I (+)IM ∩H ∩0(+)M = 0. Assume H ∩0(+)M = 0(+)K for some submodule K of M. Then 0(+)IM ∩ K = 0, and hence IM ∩ K = 0. Next, 0(+)K = H ∩ 0(+)M ⊆ J (+)JM ∩ 0(+)M = 0(+)JM, and hence K ⊆ JM. Since I is large in J, IM is large in JM and this implies that K = 0 and hence 0 = H ∩0(+)M. Since 0(+)M is large in R(M), H = 0. So I (+)IM is large in J (+)JM gives that I (+)IM = J (+)JM, and hence I = J. So I is closed in R. Conversely, let I be closed in R and suppose H is an ideal of R (M) such that I (+)IM is large in H. Then I (+)IM is large in H ∩ 0(+)M = 0(+)K for some submodule K of M. We show that IM is large in K. Let L ⊆ K with IM ∩ L = 0. Then I (+)IM ∩ 0(+)L = 0 where 0(+)L ⊆ 0(+)K. It follows that 0(+)L = 0 and hence L = 0. As I is closed in R, we get from Proposition 1 that IM is closed in M and hence IM = K. This implies that

I (+)IM ∩ 0(+)M = 0 + IM = 0(+)K = H ∩ 0(+)M.

Similarly, I (+)IM is large in H (0(+)M) = 0(+)X for some submodule X of M and hence IM is large in X. Since IM is closed in M, IM = X, and hence

(I (+)IM) (0(+)M) = 0(+)IM = 0(+)X = H (0(+)M) .

Since M is finitely generated faithful multiplication, 0(+)M is finitely generated multiplication, [1, Theorem 2] and [15, Theorem 3.1], and moreover ann (0(+)M) = 0(+)M. So by, [29, Theorem 9], we have that I (+)IM + 0(+)M = H + 0(+)M. Using the modular law which states that for all submodules K, L and N of M such that K ⊆ N then (K + L) ∩ N = K + (L ∩ N), one gets that

H = (H + 0(+)M) ∩ H = (I (+)IM + 0(+)M) ∩ H
= I (+)IM + (H ∩ 0(+)M) = I (+)IM + 0(+)IM
= I (+)IM.

Hence I (+)IM is closed in R(M).

(7) Suppose I (+)IM is adco of J (+)JM. Then (I + J)(+)(I + J) M = I (+)IM + J (+)JM = R (M) and hence I + J = R. Suppose A is an ideal of R such that A+J = R and A ⊆ I. Then A(+)AM +J (+)JM = A+J (+)(A + J) M = R (M) and A(+)AM ⊆ I (+)IM implies that A(+)AM = I (+)IM. Hence A = I and I is adco of J. Suppose now M is finitely generated faithful multiplication and R (M) is homogeneous. Let I be adco of J. Then I +J = R and hence I (+)IM +J (+)JM = R (M). Suppose A(+)K is an ideal of R (M) such that A(+)K + J (+)JM = R (M) and A(+)K ⊆ I (+)IM. Then A + J = R with A ⊆ I and K + JM = M with K ⊆ IM. Since I is adco of J, A = I. Since M is finitely generated faithful multiplication, it follows by Proposition 1 that IM is adco of JM and hence K = IM. So A(+)K = I (+)IM and this shows that I (+)IM is adco of J (+)JM.

(8) If I (+)IM is inco of J (+)JM then I (+)IM ∩ J (+)JM = 0 implies that I ∩ J = 0. Suppose B is an ideal of R such that B ∩ J = 0 and I ⊆ B. Since M is faithful multiplication, we have that B(+)BM ∩ J (+)JM = B ∩ J (+)(B ∩ J) M = 0. As I (+)IM ⊆ B(+)BM, we get that I (+)IM = B(+)BM, and hence I = B. So I is inco of J. Conversely, suppose R (M) is homogeneous. Let B(+)L be an ideal of R (M) such that B(+)L ∩ J (+)JM = 0 and I (+)IM ⊆ B(+)L. Then B ∩ J = 0 with I ⊆ B and L ∩ JM = 0 with IM ⊆ L. Since I is inco of J, I = B. By Proposition 1, IM is inco of JM and hence L = IM. This shows that B(+)L = I (+)IM and hence I (+)IM is inco of J (+)JM. □

Proposition 3. Let R be a ring, M an R-module and I (+)N and J (+)K homogeneous ideals of R (M) .

(1) If I (+)N is adco of J (+)K then I is adco of J. If I is adco of J and N is adco of K such that R (M) is homogeneous then I (+)N is adco of J (+)K.

(2) If I (+)N is inco of J (+)K then N is inco of K. If I is inco of J and N is inco of K such that R(M) is homogeneous then I (+)N is inco of J (+)K.

(3) If I (+)N is closed in R (M) then N is closed in M.

Proof. (1) Suppose I (+)N is adco of J (+)K. Then I (+)N + J (+)K = R(M) which implies that I + J = R. Let A be an ideal of R such that A + J = R and A ⊆ I. Since AM ⊆ IM ⊆ N, A(+)N is a homogeneous ideal of R (M) which satisfies that A(+)N ⊆ I (+)N and A(+)N + J (+)K = R(M). This gives that A(+)N = I (+)N and hence A = I. So I is adco of J. Now, let R (M) be a homogeneous ring, I is adco of J and N is adco of K. Then I + J = R and N + K = M, and hence (I + J)(+)(N + K) = I (+)N +J (+)K = R (M). Suppose A(+)L be an ideal of R (M) such that A(+)L + J (+)K = R (M) and A(+)L ⊆ I (+)N. It follows that A + J = R with A ⊆ I and L + K = M with L ⊆ K. Hence A = I and L = N from which we get that A(+)L = I (+)N, and hence I (+)N is adco of J (+)K.

(2) Let I (+)N be inco of J (+)K. Then I (+)N ∩ J (+)K = 0 and hence N ∩ K = 0. Suppose L is a submodule of M such that L ∩ K = 0 and N ⊆ L. Since IM ⊆ N ⊆ L, I (+)L is a homogeneous ideal of R (M). Moreover, I (+)L ∩ J (+)K = 0 and I (+)N ⊆ I (+)L. This implies that I (+)N = I (+)L and hence N = L. Hence N is inco of K. Assume now R (M) is a homogeneous ring, I inco of J and N inco of K. Then I ∩ J = 0 and N ∩ K = 0 which shows that I (+)N ∩ J (+)K = 0. Let A(+)L be an ideal of R (M) such that A(+)L ∩ J (+)K = 0 and I (+)N ⊆ A(+)L. Hence A ∩ J = 0 with I ⊆ A and L ∩ K = 0 with L ⊆ N . Therefore I = A and N = L from which we infer that I (+)N = A(+)L and hence I (+)N is inco of J (+)K.

(3) Suppose I (+)N is closed in R (M). Then I (+)N is inco of some ideal H of R (M). It follows that I (+)N ∩ H = 0, and hence I (+)N ∩ H ∩ 0(+)M = 0. Let H ∩ 0(+)M = 0(+)L for some submodule L of M. Then 0(+)(N ∩ L) = I (+)N ∩ 0(+)L = 0, and hence N ∩ L = 0. We claim that N is inco of L. Let X be a submodule of M such that X ∩ L = 0 and N ⊆ X. Since IM ⊆ N ⊆ X, I (+)X is a homogeneous ideal of R (M) which satisfies I (+)X ∩ 0(+)L = 0 and I (+)N ⊆ I (+)X. This gives that I (+)N = I (+)X and hence N = X. Hence N is closed in M. □

An R-module M is called finitely cogenerated if for every non-empty collection of submodules N_{λ} (λ ∈ Λ) of M with ⋂_{λ∈Λ}N_{λ} = 0, there exists a finite subset Λ′ of Λ

such that ⋂_{λ∈Λ′}N_{λ} = 0. M is called uniform if the intersection of any two non-zero

submodules of M is non-zero, while M has finite uniform dimension if it does not contain an infinite direct sum of non-zero submodules, [23]. As a dual of large submodules, a submodule N of M is called small in M if for all submodules K of M, N +K = M implies K = M. It is shown, [23, Lemma 5.1.4], that a submodule N of M is not small in M if and only if there exists a maximal submodule Q of M with N ⊊ Q.

Compare the next result with [3, Propositions 17 and 18].

Proposition 4. Let R be a ring, M an R-module and I (+)N a homogeneous ideal of R (M).

(1) I (+)N is small if and only if I is a small ideal of R.

(2) Let M be finitely generated and faithful. If N is a small submodule of M then I (+)N is small.

(3) Let M be faithful multiplication. If I is a large ideal of R then I (+)N is large. The converse is true if we assume further that I (+)N is pure.

(4) Let M be faithful. If N is a finitely cogenerated (resp. uniform, has finite uniform dimension) submodule of M then I (+)N is finitely cogenerated (resp. uniform, has finite uniform dimension).

Proof. (1) Suppose I (+)N is not small. There exists a maximal ideal P (+)M of R (M) such that I (+)N ⊈ P (+)M. Hence I ⊈ P, and hence I is not small. The statement is reversible.

(2) Suppose M is finitely generated and faithful. Let N be a small submodule of M. Then [N : M] is a small ideal of R. For, let A be an ideal of R such that [N : M] + A = R then M = [N : M] M + AM ⊆ N + AM ⊆ M, so that N + AM = M. Hence AM = M. Since M is finitely generated faithful, it follows by, [22, Theorem 76], that A = R, and hence [N : M] is small. It follows by (1) that [N : M](+)N is a small ideal of R (M). Since I (+)N ⊆ [N : M](+)N, we infer that I (+)N is small. Alternatively, since [N : M] is small we get that I ⊆ [N : M] is small and by (1), I (+)N is small.

(3) Since M is faithful multiplication and I is large, we infer that IM is a large submodule of M, [6, Proposition 12]. Since IM ⊆ N, N is a large submodule of M. Since M is faithful, it follows by, [3, Proposition 14], that I (+)N is large. Conversely, suppose I (+)N is a large and pure ideal of R (M). Let J be an ideal of R such that I ∩ J = 0. Then IJ = 0, and hence J ⊆ annI from which it follows that J (+)(annI) M is a homogeneous ideal of R (M). Next

J (+) (annI) M ∩ I (+)N = J ∩ I (+) (annI) M ∩ N
⊆ 0(+) [0 :_{M} I] ∩ N.

Using the fact that for any pure ideal A of R (hence A is locally either R or zero), A ∩ annA = 0 is true locally and hence globally, one gets that

0 = I (+)N ∩ ann (I (+)N) = I (+)N ∩ (annI ∩ annN) (+) [0 :_{M} I]
= (I ∩ annI ∩ annN) (+)N ∩ [0 :_{M} I] .

This gives that J (+)(annI) M ∩ I (+)N = 0, and hence J (+)(annI) M = 0. So J = 0 and hence I is a large ideal of R.

(4) Suppose M is faithful. Let H_{λ} (λ ∈ Λ) be a non-empty collection of ideals of R (M) that are contained in I (+)N such that ⋂_{λ∈Λ}H_{λ} = 0. Then ⋂_{λ∈Λ}(H_{λ} ∩ 0(+)M) =

( ⋂_{λ∈Λ}H_{λ}) ∩ 0(+)M = 0. Assume H_{λ} ∩ 0(+)M = 0(+)K_{λ} for some submodules K_{λ}

of M. Since H_{λ} ⊆ I (+)N, 0(+)K_{λ} ⊆ 0(+)N and hence K_{λ} ⊆ N. As N is finitely cogenerated, there exists a finite subset Λ′ of Λ such that ⋂_{λ∈Λ′}K_{λ} = 0. It follows

that

0 = 0 + ⋂_{λ∈Λ′}K_{λ} = ⋂_{λ∈Λ′}0(+)K_{λ} = ( ⋂_{λ∈Λ′}H_{λ}) ∩ 0(+)M.

Since M is faithful, 0(+)M is a large ideal of R (M) by [3, Proposition 14] and this shows that ⋂_{λ∈Λ′}H_{λ} = 0. Hence I (+)N is finitely cogenerated. The uniform

case is now obvious. Finally, if I (+)N contains a direct sum of subideals H_{λ} then 0(+)N = I (+)N ∩ 0(+)M contains a direct sum of subideals H_{λ} ∩ 0(+)M. Assume H_{λ} ∩ 0(+)M = 0(+)K_{λ} for some submodules K_{λ} of M. It follows that ∑_{λ∈Λ}K_{λ} is a

direct sum and contained in N. Since N has finite uniform dimension, all but a finite number of K_{λ} is zero. If K_{λ} = 0, then H_{λ} ∩ 0(+)M = 0(+)K_{λ} = 0. Since M is faithful, 0(+)M is large in R(M) and hence H_{λ} = 0. Hence I (+)N has finite uniform dimension. □

An R-module M is called divisible if for every regular element r ∈ R, M = rM. And M is called injective if it is a direct summand of any R-module B that is a submodule of, that is M is injective if for R-modules B with M ⊆ B, B = M ⊕ K for some submodule K of B, [23]. A ℤ-module M is injective if and only if M is divisible. More generally, if R is an integral domain then a torsion-free R-module M is injective if and only if M is divisible, [23]. In this paper we say that a submodule N of M is weak injective if it is a direct summand of any submodule of M that is contained in. In particular, if N is weak injective then it is a direct summand of M. Let I be an ideal of R and M a faithful multiplication R-module. If I is weak injective then IM is weak injective. For if N is a submodule of M with IM ⊆ N, then I ⊆ [N : M] and hence [N : M] = I ⊕ A for some ideal A of R, hence N = IM ⊕ AM. The converse is true if one assumes further that M is finitely generated. Let I ⊆ J for some ideal J of R. Then IM ⊆ JM, and hence JM = IM ⊕ K = IM ⊕ [K : M] M. This implies that J = I ⊕ [K : M]. So for a submodule N of a faithful multiplication module M, if [N : M] is a weak injective ideal of R then N is weak injective and the converse is true if we suppose further that M is finitely generated.

The next two results show how the divisibility and weak injectivity of a homogeneous ideal of R (M) can be transferred to its components and conversely.

Proposition 5. Let R be a ring, M an R-module, I an ideal of R and N a submodule of M.

(1) If N is divisible then 0(+)N is a divisible ideal of R (M) and the converse is true if M is torsion-free.

(2) If M is torsion-free and I (+)IM a divisible ideal of R (M) then I is divisible. The converse is true if we assume further that M is divisible.

(3) Let IM ⊆ N. If M is torsion-free and I (+)N a divisible ideal of R (M) then I is divisible and N divisible. The converse is true if we assume further that M is divisible.

Proof. (1) Suppose N is divisible. Let (r, m) be a regular element of R (M). Then r is a regular element of R and hence N = rN. This gives that (r, m) (0(+)N) = 0(+)rN = 0(+)N, and hence 0(+)N is divisible. Assume M is torsion-free. Suppose 0(+)N is divisible and let r ∈ R be regular. Then (r_{,}0) is regular in R (M), [1, Lemma 6], and hence 0(+)N = (r_{,}0) (0(+)N) = 0(+)rN. So N = rN, and hence N is divisible.

(2) Suppose M is torsion-free and I (+)IM is divisible. Let r ∈ R be regular then (r_{,}0) ∈ R (M) is regular. It follows that I (+)IM = (r_{,}0) (I (+)IM) = rI (+)rIM, and hence I = rI which means that I is divisible. Conversely, assume M is torsion-free and divisible and I is divisible. Let (r, m) ∈ R (M) be regular. It follows by [17, Theorem 3.9] that R (M) (r_{,}m) = R (M) (r, 0) and r is a regular element of R. This implies that I (+)IM = rI (+)rIM = (r_{,}0) (I (+)IM) = (r_{,}m) (I (+)IM). Hence I (+)IM is divisible.

(3) Since IM ⊆ N, I (+)N is a homogeneous ideal of R (M). Suppose M is torsionfree and I (+)N divisible. Let r ∈ R be regular then (r_{,}0) ∈ R (M) is regular.

Hence I (+)N = (r_{,}0) (I (+)N) = rI (+)rN, and hence I = rI and N = rN. So I is a divisible ideal of R and N a divisible submodule of M. Conversely, assume M is torsion-free and divisible. Let (r_{,}m) ∈ R (M) be regular then r is regular and by [17, Theorem 3.9], we have that

R (M) (r_{,}m) (I (+)N) = (Rr(+)M) (I (+)N) = rI (+)(rN + IM)
= rI (+)(rN + rIM) = rI (+)rN = I (+)N.

Hence I (+)N is divisible. □

Proposition 6. Let R be a ring, M an R-module, I an ideal of R and N a submodule of M.

(1) If 0(+)N is a weak injective ideal of R (M) then N = 0.

(2) Let IM ⊆ N. If I (+)N is a weak injective ideal of R(M) then I is weak injective. Assuming further that M is multiplication then N is weak injective.

(3) I is weak injective if and only if I (+)IM is a weak injective ideal of R (M).

Proof. (1) If 0(+)N is weak injective then it is a direct summand of R(M) and hence N = 0, [4, Proposition 1].

(2) Suppose I (+)N is weak injective. Then I (+)N is a direct summand of R(M) and hence R (M) = I (+)N ⊕H for some ideal H of R (M). Let H +0(+)M = A(+)M for some ideal A of R. Then R (M) = I (+)N +A(+)M and hence R = I +A. Let J be an ideal of R such that I ⊆ J. Since I + A is multiplication, we get from, [14, Corollary 1.2] and [29, Proposition 4], that J = J ∩ (I + A) = (J ∩ I) + (J ∩ A) = I + (J ∩ A). Since R (M) = I (+)N + H is multiplication, it follows again by, [14, Corollary 1.2] and [29, Propostion 4], that

0(+)M = (I (+)N ∩ H) + 0(+)M = (I (+)N + 0(+)M) ∩ (H + 0(+)M)
= I (+)M ∩ A(+)M = (I ∩ A) (+)M.

So I ∩ A = 0 and hence I ∩ (J ∩ A) = 0. This implies that J = I ⊕ (J ∩ A). Hence I is a weak injective ideal of R. Suppose now M is multiplication. Since R (M) = I (+)N + H is a multiplication ideal of R(M), we infer that

0(+)M = (I (+)N + H) ∩ 0(+)M
= (I (+)N ∩ 0(+)M) + (H ∩ 0(+)M) = 0(+)N + (H ∩ 0(+)M).

Assume H ∩ 0(+)M = 0(+)L for some submodule L of M. Then M = N + L. Let K be a submodule of M such that N ⊆ K. Hence K = K ∩ (N + L). As M = N + L is multiplication, it follows by [14, Corollary 1.2] that K = (K ∩ N) + (K ∩ L) = N + (K ∩ L). On the other hand, we have I (+)N ∩ H = 0 and hence I (+)N ∩ H ∩ 0(+)M = 0. So I (+)N ∩ 0(+)L = 0 and hence N ∩ L = 0 which implies that N ∩ (K ∩ L) = 0. This finally implies that K = N ⊕ (K ∩ L) and this shows that N is a weak injective submodule of M.

(3) Suppose I is weak injective. Then R = I ⊕ J for some ideal J of R. The fact R = I + J implies that R (M) = I (+)IM + J (+)JM. Let H be an ideal of R(M) such that I (+)IM ⊆ H. As I (+)IM + J (+)JM is a multiplication ideal of R (M), we get from, [14, Corollary 1.2], that

H = H ∩ (I (+)IM + J (+)JM)
= (H ∩ I (+)IM) + (H ∩ J (+)JM) = (I (+)IM) ∩ (H ∩ J (+)JM) .

Since I ∩ J = 0, IJ = 0 and hence

IM ∩ JM = I(IM ∩ JM) + J (IM ∩ JM) ⊆ IJM = 0.

So I (+)IM ∩ J (+)JM = I ∩ J (+)IM ∩ JM = 0. This implies that I (+)IM ∩ (H ∩ J (+)JM) = 0, and hence H = I (+)IM ⊕ (H ∩ J (+)JM). This shows that I (+)IM is a weak injective ideal of R (M). The converse follows by part (2). □

An R-module M is said to be simple if M ≠ 0 and for every submodule N of M either N = 0 or N = M, [23]. Obviously, simple modules are multiplication. The next result gives some properties of simple ideals and modules.

Proposition 7. Let R be a ring and M an R-module. Let I be an ideal of R and N a submodule of M.

(1) If M is finitely generated faithful multiplication then I is simple if and only IM is a simple submodule of M.

(2) If [N : M] is a simple ideal of R and M faithful multiplication then N is simple. The converse is true if we assume further that M is finitely generated.

(3) N is simple if and only if 0(+)N is a simple ideal of R (M).

(4) If I (+)IM is a simple ideal of R (M) then I is simple. The converse is true if M is finitely generated faithful multiplication and every ideal contained in I (+)IM is faithful.

(5) Let IM ⊆ N. If I (+)N is a simple homogeneous ideal of R (M) and I ≠ 0 then I is simple. Assuming further that M is faithful then N is simple.

Proof. (1) Since I is simple and M faithful, IM ≠ 0. Suppose K is a submodule of M such that K ⊆ IM. Then [K : M] ⊆ [IM : M] = I. So [K : M] = 0, and hence K = [K : M] M = 0, or [K : M] = I. The latter case shows that K = IM. Hence IM is simple. Conversely, suppose IM is simple. Then I ≠ 0. Let J ⊆ I be an ideal of R. Then JM ⊆ IM. Hence JM = 0, and hence J = 0 or JM = IM. Since M is finitely generated faithful multiplication, it follows by [29, Corollary to Theorem 9] that J = I. So I is simple.

(2) Follows by (1).

(3) Suppose N is simple. Then 0(+)N ≠ 0. Every ideal contained in 0(+)N has the form 0(+)K for some submodule K of N. Hence K = 0 or K = N from which we get that 0(+)K = 0 or 0(+)K = 0(+)N. The statement is reversible.

(4) Suppose I (+)IM is simple. Then I ≠ 0. Let J ⊆ I be an ideal of R. Then J (+)JM ⊆ I (+)IM, and hence J (+)JM = 0 or J (+)JM = I (+)IM. So J = 0 or J = I, and hence I is simple. For the converse, let I be simple. Then I (+)IM ≠ 0. Let H ⊆ I (+)IM be an ideal of R (M). Since H is faithful by assumption, H ≠ 0. We show that H = I (+)IM. We have H ∩ 0(+)M ⊆ I (+)IM ∩ 0(+)M = 0(+)IM. Since M is faithful, 0(+)M is a large ideal of R (M) and hence H ∩ 0(+)M ≠ 0. By (1) and (3), 0(+)IM is a simple ideal of R (M). Hence H ∩ 0(+)M = 0(+)IM. Next, H + 0(+)M ⊆ I (+)IM + 0(+)M = I (+)M. Assume H + 0(+)M = J (+)M for some ideal J of R. Then J ⊆ I. Since J (+)JM ⊆ I (+)IM is faithful, we infer that J is faithful, [1, Theorem 9]. Hence J ≠ 0 and therefore J = I. So H + 0(+)M = I (+) M = I (+)IM + 0(+)M. Using the modular law one gets that

H = (H + 0(+)M) ∩ H = (I (+)IM + 0(+)M) ∩ H
= I (+)IM + (H ∩ 0(+)M) = I (+)IM + 0(+)IM
= I (+)IM.

Hence I (+)IM is a simple ideal of R (M).

(5) Suppose I (+)N is simple. Then I (+)N ≠ 0. Since I (+)IM ⊆ I (+)N and I ≠ 0 (and hence I (+)IM ≠ 0), we get that I (+)IM = I (+)N. The fact that I is simple follows by (4). Suppose M is faithful. Since 0(+)N ⊆ I (+)N, we have either 0(+)N = 0 (hence N = 0) or I (+)N = 0(+)N. The case that N = 0 implies that IM = 0 and hence I = 0, a contradiction. Hence I (+)N = 0(+)N and the result follows by (3). □

In the following result we show how purity, multiplicativity or flatness of a homogeneous ideal of R (M) can be transferred to its components. Compare with [2, Theorems 9(3) and 7(2)].

Proposition 8. Let R be a ring, M an R-module and I (+)N a homogeneous ideal of R (M).

(1) If I (+)N is pure then I is a pure ideal of R and N a pure submodule of M.

(2) Let every principal ideal contained in I (+)N be homogeneous. If I is a finitely generated multiplication ideal of R and N a finitely generated multiplication submodule of M then I (+)N is finitely generated multiplication.

(3) If I (+)N is multiplication then N is a multiplication submodule of M if and only if IM is.

(4) If I (+)N is flat then I is a flat ideal of R. Assuming further that M is flat then N is a flat submodule of M.

Proof. (1) Let J be an ideal of R. Since I (+)N is pure, we get that

JI (+)JN = (J (+)JM) I (+)N) = J (+))JM ∩ I (+)N
= J ∩ I (+)JM ∩ N.

Hence JI = J ∩ I and JN = JM ∩ N which shows that I is a pure ideal of R and N a pure submodule of M. Alternatively, let P be a prime ideal of R then P (+)M is a prime ideal of R (M). Hence I_{P} (+)N_{P} ≅ (I (+)N)_{P (+)M} = 0_{P (+)M} or R (M)_{P (+)M}. It follows that I_{P} = 0_{P} or I_{P} = R_{P} and hence I is a pure ideal of R and N_{P} = 0_{P} or N_{P} = M_{P}. The latter case shows that for any ideal J of R, the equality JN = JM ∩ N is true locally and hence globally. So N is a pure submodule of M.

(2) I (+)N is finitely generated, [1, Theorem 9], and hence it is enough to prove the result locally, [18, Proposition 4]. So we may assume R (M) is a local ring and hence R is local. It follows that I = Ra and N = Rn for some a ∈ I and n ∈ N. Hence I (+)N = Ra(+)Rn = Ra(+)(Rn + aM) = R (M) (a, n), [17, Theorem 3.4]. So I (+)N is multiplication.

(3) Suppose I (+)N is multiplication. We first show that I is multiplication. Let J ⊆ I be an ideal of R. Then JM ⊆ IM ⊆ N and hence J (+)N ⊆ I (+)N is a homogeneous ideal of R (M). It follows that J (+)N = H (I (+)N) for some ideal H of R(M), and hence

J (+)N = J (+)N + 0(+)IM = H (I (+)N) + 0(+)IM
= (H + 0(+)M) (I (+)N) .

Assume H + 0(+)M = A(+)M for some ideal A of R. Then J (+)N = AI (+)(AN + IM). So J = AI and hence I is multiplication. Assume now that IM is multiplication. Let P be a maximal ideal of R such that N_{P} ≠ 0_{P}. Then P (+)M is a maximal ideal of R (M) and (I (+)N)_{P (+)M} ≠ 0_{P (+)M}. Hence I (+)N is P (+)M-principal and by, [1, Theorem 9], we have that I is P -principal. From, [1, Proposition 10], we get that I = I [IM : N] and hence I_{P} = I_{P} [IM : N]_{P}. Since I_{P} is principal, we infer from, [22, Theorem 76], that R_{P} = [IM : N]_{P} + ann (I_{P}). As I is multiplication and P -principal, we get that I_{P} ≠ 0_{P}, that is, ann (I_{P}) ≠ R_{P} and hence R_{P} = [IM : N]_{P}. There exists p ∈ P such that (1 − p) ∈ [IM : N] and hence (1 − p) N ⊆ IM. The fact that N is multiplication follows by, [29, Corollary to Lemma 1]. Conversely, assume that N is multiplication. Let P be a maximal ideal of R such that (IM)_{P} ≠ 0_{P}. Hence I_{P} ≠ 0_{P} and as we have seen above we have p ∈ P such that (1 − p) N ⊆ IM. The fact that IM is multiplication follows by [13, Corollary 2] and [14, Proposition 4.1].

(4) Suppose I (+)N is a flat ideal of R(M). The fact that I is a flat ideal of R follows by [2, Theorem 7(2)]. Since I ≅ I (+)N/0(+)N is a flat R-module, we infer from, [25, Example 1, p. 54] and [20, Corollary 11.21], that 0(+)N is a pure Rsubmodule of I (+)N. Consider R(M) as a direct sum of R and M and since M is a flat R-module, one gets that R(M) is a flat R-module. The map f : R → R(M) defined by f(a) = (a, 0) is a homomorphism of rings and hence R(M) is a flat R-Algebra, [25, pp. 45–46]. Now, I (+)N is flat over R(M) and R(M) is a flat R-Algebra, so by the transitivity of flatness one obtains that I (+)N is flat over R, see for example [25, p. 46]. Using the fact that pure submodules of flat modules

are flat, [20], we infer that 0(+)N is a flat module over R. Let ∑_{i=1}^{k} r_{i}n_{i} = 0, where

r_{i} ∈ R and n_{i} ∈ N. Then ∑_{i=1}^{k} r_{i} (0, n_{i}) = (0, 0). Since 0(+)N is flat over R, it follows

by, [25, Theorem 7.6], that there exist (0, h_{j}) ∈ 0(+)N and a_{ij} ∈ R, 1 ≤ j ≤ ℓ, such that

0 = ∑_{i=1}^{k}r_{i}a_{ij}, (0, n_{i}) = ∑_{j=1}^{ℓ}a_{ij} (0, h_{j}) .

This implies that 0 = ∑_{i=1}^{k}r_{i}a_{ij} for all j and n_{i} = ∑_{j=1}^{ℓ}a_{ij}h_{j} for all i and this shows

that N is a flat R-module, [25, Theorem 7.6]. □

Let R be a ring and K, N submodules of an R-module M. Then K divides N, denoted K|N if N = IK for some ideal I of R. If K|N then N ⊆ K and the converse is true if M is multiplication. The common divisor by every common divisor of K and N (if it does exist) is denoted by GCD (K, N). Similarly, we define LCM (K, N) as a submodule of M which is a common multiple of K and N which divides every common multiple of K and N (if such exists). The existence and arithmetic properties of these in the case of finitely generated faithful multiplication modules (ideals), finitely generated projective modules (ideals) and invertible submodules (ideals) are discussed in [9]–[12]. The next theorem gives some relationships between the gcd and lcm of a homogeneous ideal and those ones of its components.

Theorem 9. Let R be a ring and M an R-module.

(1) For all submodules K and N of M, if M is multiplication and gcd (0(+)K, 0(+)N) exists then GCD(K, N) exists. The converse is true if M is a faithful prime module and in both cases

gcd (0(+)K, 0(+)N) = 0(+)GCD (K, N) .

(2) For all submodules K and N of M, lcm (0(+)K, 0(+)N) exists if and only if LCM (K, N) exists and in this case

lcm(0(+)K, 0(+)N) = 0(+)LCM (K, N) .

(3) Let M be divisible and torsion-free. For all invertible ideals I and J of R, gcd (I, J) exists if and only if gcd (I (+)IM, J (+)JM) and in this case

gcd (I (+)IM, J (+)JM) = gcd(I, J)(+) gcd (I, J) M.

(4) Let M be divisible and multiplication. For all homogeneous ideals I (+)N and J (+)K of R(M), if I and J are finitely generated faithful multiplication ideals of R and lcm (I, J) exists then lcm (I (+)N, J (+)K) exists. The converse is true if we assume further that M is faithful, and in both cases

lcm (I (+)N, J (+)K) = lcm (I, J) (+)LCM (N, K) .

Proof. (1) Suppose M is multiplication and let gcd (0(+)K, 0(+)N) = H. Since K = IM for some ideal I of R, 0(+)K = 0(+)IM = (I (+)M) (0(+)M) and hence 0(+)M|0(+)K. Similarly, 0(+)M|0(+)N, and hence 0(+)M|H. So H ⊆ 0(+)M and this means that H = 0(+)G for some submodule G of M. Now, 0(+)G|0(+)K implies that 0(+)K = H′ (0(+)G) = (H′ + 0(+)M) (0(+)G) for some ideal H′ of R (M). Assume H′ +0(+)M = A(+)M for some ideal A of R. Then 0(+)K = 0(+)AG, and hence K = AG, that is G|K. Similarly, G|N. Suppose G′ is any submodule of M such that G′|K and G′|N. Then K = BG′ and N = CG′ for some ideals B and C of R. This implies that 0(+)K = 0(+)BG′ = (B(+)M) (0(+)G′) and 0(+)N = (C (+)M) (0(+)G′). So 0(+)G′|0(+)K and 0(+)G′|0(+)N, and this gives that 0(+)G′|0(+)G. If 0(+)G = H′′ (0(+)G′) for some ideal H′′ of R (M), then 0(+)G = (H′′ + 0(+)M) (0(+)G′) = (D(+)M) (0(+)G′) for some ideal D of R. Hence G = DG′ and hence G′|G. So G = GCD(K, N) and hence gcd (0(+)K, 0(+)N) = 0(+)GCD (K, N). Conversely, let GCD (K, N) = G. Then G|K and G|N and hence 0(+)G|0(+)K and 0(+)G|0(+)N. Assume H is an ideal of R (M) such that H|0(+)K and H|0(+)N. Then 0(+)K ⊆ H and hence annH ⊆ ann (0(+)K) = annK (+)M. Since M is faithful and prime, we infer that annH ⊆ 0(+)M and hence annH = 0(+)L for some submodule L of M. This implies that H (0(+)L) = 0 and hence H ⊆ ann (0(+)L) = annL(+)M = 0(+)M. So H = 0(+)G′ for some submodule G′ of M. As 0(+)G′|0(+)K and 0(+)G′|0(+)N, we get that G′|K and G′|N. Hence G′|G and this implies that 0(+)G′|0(+)G. So 0(+)G = gcd (0(+)K, 0(+)N).

(2) Suppose lcm (0(+)K, 0(+)N) = H. Then 0(+)K|H and 0(+)N|H. So H ⊆ 0(+)(K ∩ N), and hence H = 0(+)L for some submodule L ⊆ K ∩N. So 0(+)K|0(+)L and 0(+)N|0(+)L. This implies that K|L and N|L. Suppose L′ is any submodule of M such that K|L′ and N|L′. Then 0(+)K|0(+)L′ and 0(+)N|0(+)L′. So 0(+)L|0(+)L′ and hence L|L′ and L = LCM (K, N). This also shows that lcm (0(+)K, 0(+)N) = 0(+)L = 0(+)LCM (K, N). The converse is now obvious.

(3) For all ideals of A of R, if A(+)AM is an invertible ideal of R (M) then A is invertible and the converse is true if M is torsion-free, [4, Theorem 2]. Suppose M is divisible and torsion-free. Then for any ideal A of R, if A_{v} = (A^{−1})^{−1} is invertible then

(A(+)AM)_{v} = ((A(+)AM)^{−1})^{−1} = (A^{−1}(+)M)^{−1} = A_{v} (+)M
= A_{v} (+)A_{v}M,

[21, Theorem 25.10] and [17, Theorem 3.9]. Suppose gcd (I, J) exists. It follows by, [12, Theorem 2.1], that gcd (I, J) = (I + J)_{v}. Since (I + J)_{v} is invertible and M divisible, we infer that (I + J)_{v} (+)(I + J)_{v} M is an invertible ideal of R (M), see [4, Theorem 5]. Moreover,

(I (+)IM + J (+)JM)_{v} = ((I + J) (+) (I + J) M)_{v} = (I + J)_{v} + (I + J)_{v} M.

Hence (I (+)IM + J (+)JM)_{v} is an invertible of R (M) and hence gcd(I (+)IM, J (+)JM) exists and

gcd (I (+)IM, J (+)JM) = (I + J)_{v} (+) (I + J)_{v} M = gcd (I, J) (+) gcd (I, J) M.

Conversely, suppose gcd (I (+)IM, J (+)JM) exists. Then ((I + J) (+) (I + J) M)_{v} = (I + J)_{v} (+)M is an invertible ideal of R (M) . It follows by [4, Theorem 2] that (I + J)_{v} is an invertible ideal of R and gcd (I, J) exists and is (I (+)J)_{v}. Since (I + J)_{v} is invertible and M divisible (I + J)_{v} (+)M = (I + J)_{v} (+)(I + J)_{v} M = (I (+)IM + J (+)JM)_{v}. Hence gcd (I (+)IM, J (+)JM) = gcd (I, J)(+)gcd (I, J) M.

(4) Suppose M is divisible, I and J are finitely generated faithful multiplication ideals of R. It follows by, [4, Theorem 5], that I (+)N and J (+)K are finitely generated multiplication ideals of R (M). Since M is multiplication, we infer from, [1, Theorem 9] and [3, Theorem 3], that N and K are multiplication submodules of M. If lcm (I, J) exists then lcm (I, J) = I ∩ J is a finitely generated faithful multiplication ideal of R, [11, Lemma 5]. It follows that (I ∩ J (+) (N ∩ K)) is a finitely generated multiplication ideal of R (M) and consequently K ∩ N is a multiplication submodule of M, [4, Theorem 5] and [1, Theorem 9]. Using the fact that for all submodules X and Y of M if X ∩ Y is multiplication then LCM (X_{,}Y) exists and LCM (X_{,}Y) = X ∩ Y, we get that lcm (I (+)N, J (+)K) = I (+)N ∩ J (+)K = (I ∩ J) + (N ∩ K) = lcm (I, J)(+)LCM (N, K). For the converse, suppose M is faithful multiplication. Then I (+)N and J (+)K are finitely generated faithful multiplication ideals of R (M). Suppose lcm (I (+)N, J (+)K) exists. It follows by [11, Lemma 5] that lcm (I (+)N, J (+)K) = (I ∩ J)(+)(N ∩ K) is a finitely generated faithful multiplication ideal of R (M). Hence I ∩ J is multiplication and hence lcm (I, J) exists and is I ∩ J. Since M is multiplication, N ∩ K is multiplication and hence lcm (N, K) exists and is N ∩ K. Hence

lcm (I (+)N, J (+)K) = lcm (I, J) (+)LCM (N, K) .

This completes the proof of the theorem. □

2. Annihilator conditions via idealization In [2] we discussed two annihilator conditions, namely (a.c.) and property (A) via idealization. A ring R satisfies Property (A) if each finitely generated ideal I ⊆ Z (R) has a nonzero annihilator, where Z (R) is the set of zero divisors of R. A condition closely intertwined with Property (A) is the annihilator condition; abbreviated as (a.c.): A ring R satisfies (a.c.) if for each pair of elements a and b in R, there exists c in R such that ann (Ra + Rb) = ann (c). More generally, we say that an R-module M satisfies (a.c.) if for each pair of elements m and n in M, there exists k in M such that ann (Rm + Rn) = ann (k), [21]. It is proved that if M is flat and R(M) has property (A) then so too has R and the converse is true if M is finitely generated. On the other hand, let R(M) be homogeneous. If M is flat and each of R and M satisfies the (a.c.), then so too does R(M) and the converse is true if M is finitely generated faithful and multiplication, [2, Theorem 16]. In this paper we investigate another two annihilator conditions on a ring R. Recall that R is a quasi-Frobenius ring, shortly QF ring, if R is Artinian and ann (annI) = I for each ideal I of R. And an R-module M is called a QF module if M is Artinian and [0 :_{M} annN] = N for each submodule N of M, [23, p. 336]. A ring R is called distinguished if annI ≠ 0 for each proper ideal I of R and M is distinguished if [0 :_{M} I] ≠ 0 for each proper ideal I of R, [26]. A ring R is QF if and only if every ideal of R is projective, equivalently, every ideal of R is injective, [23, Theorem 13.6.1]. Since QF rings are Artinian rings, we first give necessary and sufficient conditions for a ring R (M) to be Artinian. The proof of the next result is given in [17, Theorem 4.8] however we give here different proof which may be of some interest.

Proposition 10. Let R be a ring and M an R-module. Then R (M) is Artinian (resp. Noetherian) if and only if R is Artinian (resp. Noetherian) and M is Artinian (resp. Noetherian).

Proof. Suppose R (M) is Artinian. Let I_{1} ⊇ I_{2} ⊇ ⋯ be a descending chain of ideals of R. Then I_{1}(+)I_{1}M ⊇ I_{2}(+)I_{2}M ⊇ ⋯ is a descending chain of ideals of R (M) and hence there exists a positive integer k such that I_{k}(+)I_{k}M = I_{k+1}(+)I_{k+1}M = · · ·. This implies that I_{k} = I_{k+1} = ⋯, and hence R is Artinian. Next, let N_{1} ⊇ N_{2} ⊇ ⋯ be a descending chain of submodules of M then 0(+)N_{1} ⊇ 0(+)N_{2} ⊇ ⋯ is a descending chain of ideals of R (M). There exists a positive integer l such that 0(+)N_{l} = 0(+)N_{l+1} = ⋯, and hence N_{l} = N_{l+1} = · · ·. Hence M is Artinian. Conversely, suppose R is an Artinian ring and M an Artinian module. Let H_{1} ⊇ H_{2} ⊇ ⋯ be a descending chain of ideals of R (M). Then H_{1} + 0(+)M ⊇ H_{2} + 0(+)M ⊇ ⋯ and H_{1} ∩ 0(+)M ⊇ H_{2} ∩ 0(+)M ⊇ ⋯ are descending chains of ideals of R (M). Assume H_{i} + 0(+)M = I_{i}(+)M and H_{i} ∩ 0(+)M = 0(+)N_{i} for some ideals I_{i} of R and some submodules N_{i} of M. This gives that I_{1} ⊇ I_{2} ⊇ ⋯ is a descending chain of ideals of R and N_{1} ⊇ N_{2} ⊇ ⋯ is a descending chain of submodules of M. Hence there exist positive integers k and l such that I_{k} = I_{k+1} = ⋯ and N_{l} = N_{l+1} = · · ·. It follows that H_{k} + 0(+)M = H_{k+1} + 0(+)M = ⋯, and H_{l} ∩ 0(+)M = H_{l+1} ∩ 0(+)M = · · ·. If k < l, then H_{k} + 0(+)M = H_{k+1} + 0(+)M and H_{k} ∩ 0(+)M = H_{k+1} ∩ 0(+)M. So by using the modular law, one gets that

H_{k} = (H_{k} + 0(+)M) ∩ H_{k} = (H_{k+1} + 0(+)M) ∩ H_{k}
= H_{k+1} + (H_{k} ∩ 0(+)M) = H_{k+1} + (H_{k+1} ∩ 0(+)M)
= H_{k+1}.

Similarly, if k ≥ l then H_{l} = H_{l+1} = ⋯ and this shows that R (M) is an Artinian ring. The Noetherian case is now obvious by replacing the descending chain of ideals (submodules) by ascending chain of ideals (submodules). Alternatively, let R (M) be Noetherian. If I is an ideal of R, then I (+)IM is a finitely generated ideal of R (M). It follows by [1, Theorem 7] that I is finitely generated. If N is a submodule of M then 0(+)N is a finitely generated ideal of R(M), and hence N is finitely generated [1, Theorem 2] and [14, Theorem 3.1]. So R is a Noetherian ring and M a Noetherian module. Conversely, suppose R is a Noetherian ring and M a Noetherian module. Suppose H is an ideal of R (M), and let H + 0(+)M = I (+)M for some ideal I of R and H ∩ 0(+)M = 0(+)N for some submodule N of M. Since I is finitely generated and M Noetherian (hence finitely generated), I (+)M = H + 0(+)M is finitely generated. Also N is finitely generated since M is Noetherian. So 0(+)N = H ∩ 0(+)M is finitely generated. The fact that H is finitely generated follows by [25, Example 2.3, p. 13]. Hence R (M) is Noetherian. □

We next determine when R (M) is a QF ring or a distinguished ring.

Theorem 11. Let R be a ring and M an R-module.

(1) If R (M) is a QF ring then M is a QF module. Assuming further that M is projective then R is a QF ring.

(2) Let M be faithful. If R is a QF ring and M a QF module then R (M) is a QF ring.

(3) If R (M) is a distinguished ring then R is a distinguished ring or M a distinguished module. The converse is true if M is faithful.

(4) Let M be faithful multiplication. If R (M) is a distinguished ring then R is a distinguished ring and M a distinguished module.

Proof. (1) If R (M) is a QF ring then R (M) is Artinian, so by Proposition 10, R is Artinian and M Artinian. Suppose N is a submodule of M. Then

0(+)N = ann (ann (0(+)N)) = ann (annN (+)M)
= ann (annN) ∩ annM (+) [0 :_{M} annN] ,

and hence N = [0 :_{M} annN]. This gives that M is a QF module. Suppose now M is projective and I an ideal of R. It follows by, [2, Lemma 5], that

I (+)IM = ann (ann (I (+)IM)) = ann (annI) (+)ann (ann (I)) M.

So I = ann (annI), and R is a QF ring.

(2) Suppose M is faithful. Since R is QF (hence Artinian) and M is a QF (hence Artinian) R-module, we infer from Proposition 10 that R (M) is Artinian. Let H be an ideal of R (M). Assume H + 0(+)M = I (+)M and H ∩ 0(+)M = 0(+)N for some ideal I of R and some submodule N of M. Then

H + 0(+)M = I (+)M = ann (annI) (+)M

⊇ ann(annI) ∩ annM (+) [0 :_{M} annI] + 0(+)M = ann (ann (I (+)M)) + 0(+)M = ann(ann (H + 0(+)M) + 0(+)M = ann (annH ∩ 0(+)M) + 0(+)M ⊇ ann (annH) + ann(0(+)M) + 0(+)M = ann (annH) + 0(+)M ⊇ H + 0(+)M.

On the other hand,

H ∩ 0(+)M = 0(+)N = 0(+) [0 :_{M} annN]

= ann (ann (0(+)N)) = ann (ann (H ∩ 0(+)M)) ⊇ ann(annH + 0(+)M) = ann(annH) ∩ 0(+)M ⊇ H ∩ 0(+)M.

The application of the modular law shows that H = ann (annH) and hence R (M) is a QF ring.

(3) Suppose R (M) is distinguished. Let I be a proper ideal of R. Then I (+)M is a proper ideal of R(M) and hence 0 ≠ ann (I (+)M) = annI ∩ annM (+)[0 :_{M} I]. If annI ∩ annM ≠ 0, then annI ≠ 0 and hence R is distinguished. If [0 :_{M} I] ≠ 0, then obviously M is distinguished. For the converse, assume R is distinguished and M is faithful and let H be a proper ideal of R (M). Assume H+0(+)M = I (+)M for some ideal I of R. Then

annH ⊇ ann (I (+)M) = annI ∩ annM (+) [0 :_{M} I]
= 0(+) [0 :_{M} I] ⊇ 0(+) (annI) M.

Since M is faithful and annI ≠ 0, (annI) M ≠ 0. So annH ≠ 0 and R (M) is distinguished. Suppose now that M is distinguished. Then annH ⊇ 0(+)[0 :_{M} I] ≠ 0 and clearly R (M) is distinguished.

(4) Suppose M is faithful multiplication. Let I be a proper ideal of R. Then 0 ≠ ann (I (+)IM) = 0(+)(annI) M. Hence (annI) M ≠ 0, and hence annI ≠ 0. So R is distinguished. Moreover, [0 :_{M} I]≠0 and this shows that M is distinguished. □

An R-module M is called finitely annihilated if annM = annN for some finitely generated submodule N of M, [30]. The next result gives necessary and sufficient conditions for a homogeneous ideal of R(M) to be finitely annihilated.

Proposition 12. Let R be a ring, M an R-module, I an ideal of R and N a submodule of M.

(1) N is finitely annihilated if and only if 0(+)N is a finitely annihilated ideal of R (M).

(2) Suppose M is finitely generated faithful multiplication and IM ⊆ N. If I (+)N is finitely annihilated then I is finitely annihilated and N finitely annihilated.

(3) Suppose M is faithful multiplication or projective. If I is finitely annihilated then I (+)IM is a finitely annihilated ideal of R (M). The converse is true if M is finitely generated faithful and multiplication.

(4) Suppose M is finitely generated faithful multiplication and IM ⊆ N. If I (+)N is finitely generated and I finitely annihilated then I (+)N is a finitely annihilated ideal of R (M).

(5) Suppose M is faithful multiplication and IM ⊆ N. If I is finitely annihilated and annI ⊆ annN then I (+)N is a finitely annihilated ideal of R (M).

(6) Suppose R has the property that annA = ann(annA) for each ideal A of R and M a faithful multiplication module such that IM ⊆ N. If I is a finitely annihilated ideal of R (M) then I (+)N is finitely annihilated.

Proof. (1) Suppose 0(+)N is finitely annihilated. There exists a finitely generated ideal H ⊆ 0(+)N of R (M) such that annH = ann(0(+)N). Assume H = 0(+)K for some finitely generated submodule K of N, [15, Theorem 3.1]. Then annK (+)M = annH = ann (0(+)N) = annN (+)M, and hence annK = annN. So N is finitely annihilated. The statement is reversible.

(2) Suppose M is finitely generated faithful multiplication. Let I (+)N be finitely annihilated. There exists a finitely generated ideal H of R (M) that is contained in I (+)N with annH = ann (I (+)N). Then annH = annN (+)(annI) M, and hence

ann(H + 0(+)M) = annH ∩ ann(0(+)M) = annH ∩ 0(+)M = 0(+) (annI) M.

Assume H + 0(+)M = J (+)M for some ideal J of R. Then

0(+)(annJ)M = ann (J (+)M) = 0(+) (annI) M.

This gives that (annJ)M = (annI)M and hence annJ = annI. Since H +0(+)M = J (+)M is finitely generated, it follows by, [1, Theorem 9], that J is finitely generated, and this shows that I is finitely annihilated. Alternatively, suppose

H = ∑_{i=1}^{n}R (M) (a_{i}, m_{i}). It follows that

annN (+) (annI) M = ann (I (+)N) = annH = ⋂_{i=1}^{n}ann (a_{i}, m_{i}) .

For all i,

ann (a_{i}, m_{i}) ⊇ ann (Ra_{i}(+)(Rm_{i} + a_{i}M))
= ann (a_{i}) ∩ ann (Rm_{i} + a_{i}M) (+)ann (a_{i}) M
= ann (Rm_{i} + a_{i}M) (+)ann (a_{i}) M.

Since M is faithful multiplication, it follows by [19, Corollary 1.7], that

annH ⊇ ⋂_{i=1}^{n}ann (Rm_{i} + a_{i}M) (+) ( ⋂_{i=1}^{n}ann (a_{i})) M

= ann (∑_{i=1}^{n}(Rm_{i} + a_{i}M)) (+) (∑_{i=1}^{n}ann (a_{i})) M.

Since M is finitely generated faithful multiplication, we infer from [29, Corollary

to Theorem 9] that annI ⊇ ann∑_{i=1}^{n}Ra_{i} ⊇ annI, so that annI = ann∑_{i=1}^{n}Ra_{i}, and

I is finitely annihilated. Also annN ⊇ ann (∑_{i=1}^{n}(Rm_{i} + a_{i}M)) ⊇ annN, so that

annN = ann (∑_{i=1}^{n}Rm_{i} + a_{i}M), and N is finitely annihilated.

(3) Assume M is faithful multiplication or projective. Let J ⊆ I be finitely generated ideal of R such that annI = annJ. It follows by, [1, Theorem 7], that J (+)JM is a finitely generated ideal of R(M) that is contained in I (+)IM. Moreover,

ann (I (+)IM) = annI (+) (annI) M = annJ (+) (annJ) M
= ann (J (+)JM) .

Hence I (+)IM is finitely annihilated. The converse follows by (2).

(4) Since M is finitely generated and I (+)N is finitely generated, it follows by, [1, Therorem 9], that N is finitely generated. Let J ⊆ I be a finitely generated ideal of R such that annI = annJ. Since JM ⊆ IM ⊆ N, J (+)N is a homogeneous ideal of R(M) that is finitely generated. As

ann (I (+)N) = annN (+) (annI) M = annN (+) (annJ) M
= ann (J (+)N) ,

we infer that I (+)N is finitely annihilated.

(5) Using the same arguments of parts (3) and (4) and the fact that annI ⊆annN, there exists a finitely generated ideal J (+)JM that is contained in I (+)N with the property ann (I (+)N) = ann (J (+)JM). So I (+)N is finitely annihilated.

(6) Assume that J ⊆ I is a finitely generated ideal of R such that annI = annJ and K ⊆ N is a finitely generated submodule of M such that annK = annN. Since M is faithful and multiplication, we infer that

ann (I (+)N) = annN (+) (annI) M = annK (+) (annJ) M.

Since annK = annN ⊆ annI = annJ, we have that (ann[K : M])J = ann(K)J = 0, and hence J ⊆ ann(ann([K : M])) = [K : M]. This gives that JM ⊆ K and hence J (+)K is a finitely generated homogeneous ideal of R(M) that is contained in I (+)N and ann (I (+)N) = ann (J (+)K). So I (+)N is finitely annihilated. □

Let R be a ring and M an R-module. Define

T (M)={m∈M : ann (m) ≠ 0} and Si (M)={m∈M : ann (m) is large in R} .

T (M) is called the torsion submodule of M and Si (M) the singular submodule of M, [23]. If R is an integral domain then T (M) is closed in M, [23, p. 139].

The next result characterizes the torsion and singular subideals of homogeneous ideals of R (M).

Proposition 13. Let R be a ring, M an R-module and I (+)N a homogeneous ideal of R (M).

(1) If M is torsion-free then T (I (+)N) = T (I)(+)N.

(2) If M is faithful multiplication and R (M) a homogeneous ring then Si (I (+)N) = Si(I)(+)N.

Proof. (1) Let (a, m) ∈ T (I (+)N). Then ann (a, m) ≠ 0. Since M is torsionfree, we infer that ann (a) ≠ 0. Hence a ∈ T (I) and hence (a, m) ∈ T (I)(+)N. So T (I (+)N) ⊆ T (I)(+)N. For the reverse inclusion, if (a, m) ∈ T (I)(+)N, then a ∈ T (I). Hence ann (a) ≠ 0, and hence ann (a, m) ≠ 0. So (a, m) ∈ T (I (+)N) and hence T (I)(+)N ⊆ T (I (+)N). This shows that T (I (+)N) = T (I)(+)N.

(2) Suppose (a, m) ∈ Si (I (+)N). Then ann (a_{,}m) is a large ideal of R (M). Since R (M) is homogeneous, it follows by, [17, Theorem 3.3], that

ann (a_{,}m) = ann (Ra(+)Rm + aM)
= ann (a) ∩ ann (m) (+)ann (a) M.

By [3, Proposition 17], ann (a) M is a large submodule of M. Since M is faithful multiplication, we infer from [6, Proposition 12] that ann (a) is a large ideal of R. Hence a ∈ Si(I) and hence (a_{,}m) ∈ Si (I)(+)N. If (a_{,}m) ∈ Si (I)(+)N then a ∈ Si (I) and hence ann (a) is a large ideal of R. It follows that ann (a) M is a large submodule of M, [6, Proposition 12]. [3, Proposition 14] shows that ann (a, m) = ann (a) ∩ ann (m) (+)ann (a) M is a large ideal of R (M). So (a_{,}m) ∈ Si (I (+)N) and hence Si(I (+)N) = Si (I)(+)N. □

3. Cancellation modules Generalizing the case for ideals, an R-module M is defined to be cancellation (resp. weak cancellation) if IM = JM for ideals I and J of R then I = J (resp. I + annM = J + annM), [15] and [28]. Examples of cancellation modules include invertible ideals, free modules [28] and finitely generated faithful multiplication modules, [29, Corollary to Theorem 9]. If M is a finitely generated faithful multiplication R-module (hence cancellation), then it is easily verified that [IN : M] = I [N : M] for all ideals I of R and all submodules N of M. Anderson, [15], defined an R-module M to be restricted cancellation if IM = JM ≠ 0 for some ideals I and J of R then I = J. Every cancellation module is restricted cancellation but the converse is not true in general, [15]. An R-module M is a cancellation module if and only if it is a faithful weak cancellation module. It is also shown, [15, Theorem 2.5], that M is restricted cancellation if and only if it is weak cancellation and annM is comparable to every ideal of R. A submodule N of M is called join principal if [(IN + K) : N] = I + [K : N] for all submodules K of M and all ideals I of R. Setting K = 0, N becomes weak cancellation. A submodule N is join principal if and only if each of its homomorphic images is weak cancellation, [15, Theorem 2.2]. An R-module M is called \frac{1}{2} cancellation (resp. \frac{1}{2} weak cancellation) if for all ideals I of R, IM = M implies I = R (resp. I +annM = R), [5] and [27]. M is a \frac{1}{2} cancellation module if and only if M ≠ P M for each maximal ideal P of R. In [5], we introduced and investigated the concept of \frac{1}{2} join principal submodules. A submodule N of M is \frac{1}{2} join principal if for all ideals I of R and all submodules K of M, N = IN + K implies R = I + [K : N].

Let R be a ring and M an R-module. An ideal I of R is called M-cancellation (resp. M-weak cancellation) if for all submodules K and N of M, IK = IN implies K = N (resp. K + [0 :_{M} I] = N + [0 :_{M} I]). Equivalently, [IN :_{M} I] = N (resp. [IN :_{M} I] = N + [0 :_{M} I]) for all submodules N of M. I is said to be an M-restricted cancellation ideal if for all submodules K and N of M, if IK = IN ≠ 0 then K = N. An ideal I of R is called M-join principal if [(IK + N) :_{M} I] = K + [N :_{M} I] for all submodules K and N of M. Setting N = 0, I becomes an M-weak cancellation ideal, [7]. Every M-cancellation ideal is M-restricted cancellation and every M-restricted cancellation is M-weak cancellation, and the three concepts coincide if both I and M are faithful. It is proved that an ideal I of R is M-restricted cancellation if and only if I is M-weak cancellation and [0 :_{M} I] is comparable to every submodule N of M, [7, Proposition 1.5]. Several facts on M-cancellation properties are given in [7]. Motivated by the terminology of \frac{1}{2} join principal and \frac{1}{2} weak (cancellation) modules, we introduce the following definitions. Let R be a ring and M an R-module. An ideal I of R is called an M-\frac{1}{2} join principal ideal of R if for all submodules K and N of M, IM = IK + N implies M = K + [N :_{M} I]. It is easily verified that I is M-\frac{1}{2} join principal if IM ⊆ IK + N then M = K + [N :_{M} I]. Setting N = 0, we define I to be M-\frac{1}{2} weak cancellation. In particular, I is called an M-\frac{1}{2} cancellation ideal of R if for all submodules K of M, IM = IK implies M = K. If N is a proper pure submodule of M then [N : M] is not M-\frac{1}{2} cancellation. For, if N is pure in M, then JN = JM ∩ N for all ideals J of R. Take J = [N : M] then [N : M] N = [N : M] M ∩N = [N : M] M. M-cancellation ideals are M-\frac{1}{2} cancellation but not conversely. For a prime p, the Prüfer p-group M = ℤ_{p^{∞}} is a faithful but not multiplication ℤ-module. The only non-zero submodules of M are ℤ_{p} ⊆ ℤ_{p^{2}} ⊆ ⋯ ℤ_{p^{∞}}. For each k > 1, pℤ_{p^{k}} = ℤ_{p^{k−1}}. Hence p^{2}ℤ_{p^{2}} = p^{2}ℤ_{p} = 0, and hence p^{2}ℤ is not M-cancellation. On the other hand, for each k > 1, p^{2}ℤ_{p^{∞}} ≠ p^{2}ℤ_{p^{k}}, and hence p^{2}ℤ is an M-\frac{1}{2} cancellation ideal of R. Otherwise, if p^{2}ℤ_{p^{∞}} = p^{2}ℤ_{p^{k}}, then p^{k}ℤ_{p^{∞}} = 0, and hence p^{k}ℤ ⊆ [0 : ℤ_{p^{∞}}] = 0, a contradiction. Neither M-\frac{1}{2} cancellation ideals are \frac{1}{2} cancellation nor \frac{1}{2} cancellation ideals are M-\frac{1}{2} cancellation. To explain the last statement, recall the ideal associated with an R-module M, θ (M) = ∑_{m∈M} [Rm : M] , which proved useful in studying multiplication modules.

A finitely generated module M is multiplication if and only if θ (M) = R. If M is multiplication then M = θ (M) M from which it follows that for any submodule N of M, N = θ (M) N . If M is a faithful multiplication module then θ (M) is a pure ideal of R, that is multiplication and idempotent, [16, Theorem 2.6]. Let M be a faithful multiplication R-module but not finitely generated. Then θ (M) is M-\frac{1}{2} cancellation because for each submodule N of M, if θ (M) M = θ (M) N, then M = N, [16, Lemma 1.1]. But θ (M) is not \frac{1}{2} cancellation, since otherwise θ (M) = (θ (M))^{2} implies θ (M) = R and hence M is finitely generated. Let R = ℤ/4ℤ and I = 2ℤ/4ℤ. Then I is a \frac{1}{2} cancellation ideal of R, [5]. Consider M = ℤ/2ℤ as an R-module, then IM = 0 and this shows that I is not an M-\frac{1}{2} cancellation ideal of R.

The following two results give some properties of M − \frac{1}{2} (weak) cancellation ideals. First let us recall the trace ideal of an R-module M, Tr (M) = ∑_{f∈Hom(M,R)}f (M).

If M is a projective module then M = Tr (M) M, annM = annTr (M) and Tr(M) is a pure ideal of R, [20, Proposition 3.30]. For any submodule N of M, N = Tr (M) N, [8, Theorem 3.1]. If M is faithful multiplication then Tr (M) = θ (M), [16, Theorem 2.3].

Proposition 14. Let R be a ring, M an R-module and I an ideal of R.

(1) I is M-\frac{1}{2} cancellation if and only if IM ≠ IQ for each maximal submodule Q of M.

(2) Let M be faithful multiplication or projective. If I is a faithful M-\frac{1}{2} weak cancellation ideal then I is M-\frac{1}{2} cancellation.

(3) Let M be finitely generated faithful multiplication. If I is projective or faithful multiplication and contains an M-\frac{1}{2} cancellation ideal of R then I is M-\frac{1}{2} cancellation.

(4) If I is a pure ideal of R and contains an M-\frac{1}{2} cancellation ideal of R then I is M-\frac{1}{2} cancellation.

Proof. (1) Obviously, if I is M-\frac{1}{2} cancellation then IM ≠ IQ for each maximal submodule Q of M. Conversely, suppose I is not M-\frac{1}{2} cancellation. There exists a proper submodule N of M with IM = IN, and hence there exists a maximal submodule Q of M with N ⊆ Q. So IM = IN ⊆ IQ ⊆ IM, so that IM = IQ, a contradiction.

(2) Let N be a submodule of M with IM = IN. Then M = N + [0 :_{M} I]. Since M is faithful multiplication or projective, [0 :_{M} I] = (annI) M = 0. So M = N, and I is M-\frac{1}{2} cancellation.

(3) Let J ⊆ I be M-\frac{1}{2} cancellation and N a submodule of M with IM = IN. Since M is finitely generated faithful and multiplication, I = I [N : M] , and hence Tr (I) = Tr (I) [N : M]. This implies that Tr (I) M = Tr (I) N. If I is projective, then J = JTr (I) , [8, Theorem 3.1], and hence JM = JN from which it follows that M = N, and I is M-\frac{1}{2} cancellation. Alternatively, if I is not M-\frac{1}{2} cancellation then IM = IQ for some maximal submodule Q of M. Hence Tr (I) M = Tr (I) Q and hence JM = JTr (I) M = JTr (I) Q = JQ, a contradiction. If I is faithful and multiplication then Tr (I) = θ (I) and J = Jθ (I) , [16, Lemma 1.1]. The result is now clear.

(4) Let N be a submodule of M with IM = IN. Suppose J ⊆ I is M-\frac{1}{2} cancellation and I is pure, then J = JI and hence JM = JN. So M = N, and hence I is M-\frac{1}{2} cancellation. □

Proposition 15. Let R be a ring, M an R-module and I an ideal of R.

(1) Let M be finitely generated, faithful and multiplication. Then I is \frac{1}{2} join principal (resp. \frac{1}{2} weak cancellation) if and only if I is M-\frac{1}{2} join principal (resp. M-\frac{1}{2} weak cancellation).

(2) If M is finitely generated faithful multiplication and I is \frac{1}{2} cancellation then I is M-\frac{1}{2} cancellation. The converse is true if M is \frac{1}{2} cancellation.

(3) If M is multiplication and IM is a \frac{1}{2} join principal (resp. \frac{1}{2} weak cancellation) submodule of M then I is M-\frac{1}{2} join principal (resp. M-\frac{1}{2} weak cancellation). The converse is true if we assume further that M is finitely generated.

(4) If M is multiplication and IM is a \frac{1}{2} cancellation submodule of M then I is M-\frac{1}{2} cancellation. The converse is true if M is \frac{1}{2} cancellation.

Proof. (1) Suppose I is \frac{1}{2} join principal. Let K and N be submodules of M such that IM = IK + N. Then IM = (I [K : M] + [N : M]) M, and hence I = I [K : M] + [N : M]. This implies that R = [K : M] + [[N : M] : I] = [K : M] + [N : IM], and hence M = [K : M] M + [N : IM] M = K + [N :_{M} I]. So I is M-\frac{1}{2} join principal. Conversely, let I be M-\frac{1}{2} join principal and let A and B be ideals of R with I = AI + B. Then IM = I (AM) + BM, and hence M = AM + [BM :_{M} I] = (A + [BM : IM]) M. It follows that R = A + [B : I], and I is \frac{1}{2} join principal. The \frac{1}{2} weak cancellation case is obvious by taking N = 0 and B = 0 respectively.

(2) Obvious.

(3) We only discuss the \frac{1}{2} join principal case. Let M be multiplication and IM a \frac{1}{2} join principal submodule of M. Let K and N be submodules of M with IM = IK +N. Then IM = [K : M] IM +N, and hence R = [K : M]+[N : IM]. So M = K + [N :_{M} I], and I is an M-\frac{1}{2} join principal ideal of R. Conversely, let M be finitely generated multiplication and I an M-\frac{1}{2} join principal. Suppose J is an ideal of R and K a submodule of M with IM = J (IM) + K. Then M = JM +[K :_{M} I] = (J + [K : IM]) M, and hence R = J +[K : IM]+annM = J + [K : IM]. This shows that IM is \frac{1}{2} join principal.

(4) Obvious. □

The following result is now straightforward.

Corollary 16. Let R be a ring and M a finitely generated faithful multiplication R-module. The following statements are equivalent for a submodule N of M.

(1) N is \frac{1}{2} join principal (resp. \frac{1}{2} weak cancellation, \frac{1}{2} cancellation).

(2) [N : M] is a \frac{1}{2} join principal (resp. \frac{1}{2} weak cancellation, \frac{1}{2} cancellation) ideal of R.

(3) [N : M] is an M-\frac{1}{2} join principal (resp. M-\frac{1}{2} weak cancellation, M-\frac{1}{2} cancellation) ideal of R.

The next theorem gives some relationships between M-join principal (resp. Mcancellation) and M-\frac{1}{2} join principal (resp.M-\frac{1}{2}cancellation) submodules (ideals) and their products.

Proposition 17. Let R be a ring, M an R-module and I, J ideals of R.

(1) If I is M-join principal and J is M-\frac{1}{2} join principal (resp. M-\frac{1}{2} weak cancellation) then IJ is M-\frac{1}{2} join principal (resp. M-\frac{1}{2} weak cancellation).

(2) If I is M-cancellation and J is M-\frac{1}{2} weak cancellation (resp. M-\frac{1}{2} cancellation) then IJ is M-\frac{1}{2} weak cancellation (resp. M-\frac{1}{2} cancellation).

(3) If I is M-cancellation and IJ is M-\frac{1}{2} join principal (resp. M-\frac{1}{2} weak cancellation) then J is M-\frac{1}{2} join principal (resp. M-\frac{1}{2} weak cancellation).

Proof. (1) We only do the join principal case. Let K and N be submodules of M such that IJM = IJK + N. Since I is M-join principal, we infer that JM ⊆ [IJM :_{M} I] = [(IJK + N) :_{M} I] = JK + [N :_{M} I]. But J is M-\frac{1}{2} join principal. Thus M = K + [[N :_{M} I] :_{M} J] = K + [N :_{M} IJ], and hence IJ is M-\frac{1}{2} join principal.

(2) Suppose N is a submodule of M such that IJM = IJN. Then JM = JN. Since J is M-\frac{1}{2} weak cancellation, M = N + [0 :_{M} J] ⊆ N + [0 :_{M} IJ] , so that M = N + [0 :_{M} IJ] and IJ is M-\frac{1}{2} weak cancellation. The \frac{1}{2} cancellation case is now obvious.

(3) Suppose K and N are submodules of M such that JM = JK + N. Then IJM = IJK + IN, and hence

M = K + [IN :_{M} IJ] = K + [[IN :_{M} I] :_{M} J] = K + [N :_{M} J] ,

and J is an M-\frac{1}{2} join principal ideal of R. The \frac{1}{2} weak cancellation modules case follows by taking N = 0. □

The following result shows how properties such as M-\frac{1}{2} join principal (resp. M − \frac{1}{2} cancellation) are related to 0(+)M-\frac{1}{2} join principal (resp. 0(+)M − \frac{1}{2} cancellation).

Proposition 18. Let R be a ring, M an R-module and I an ideal of R.

(1) I is M-\frac{1}{2} join principal ideal if and only if I (+)IM is an 0(+)M-\frac{1}{2} join principal ideal of R (M).

(2) I is M-\frac{1}{2} weak cancellation ideal if and only if I (+)IM is an 0(+)M-\frac{1}{2} weak cancellation ideal of R (M).

(3) I is M-\frac{1}{2} cancellation ideal if and only if I (+)IM is an 0(+)M-\frac{1}{2} cancellation ideal of R (M).

Proof. (1) Suppose I is M − \frac{1}{2} join principal. Let K and N be submodules of M and let (0(+)M) (I (+)IM) = (0(+)K) (I (+)IM) + 0(+)N. Then 0(+)IM = 0(+)(IK + N) , and hence IM = IK + N. It follows that M = K + [N :_{M} I] , and this gives that

0(+)M = 0(+) (K + [N :_{M} I]) = 0(+)K + 0(+) [N :_{M} I]
= 0(+)K + [0(+)N :_{0(+)M} I (+)IM] ,

and I (+)IM is an M-\frac{1}{2} join principal ideal of R (M). The converse is now clear.

(2) The M-\frac{1}{2} weak cancellation case follows from (1) by letting N = 0.

(3) Clear. □

The next result gives some conditions under which the ideal I (+)IM becomes weak cancellation (resp. restricted cancellation). Compare with [3, Proposition 10 and Theorem 11].

Proposition 19. Let R be a ring, M an R-module and I an ideal of R.

(1) If I (+)IM is a cancellation ideal of R(M) then I is cancellation and the converse is true if M is torsion-free.

(2) If I (+)IM is a weak cancellation ideal of R(M) then I is weak cancellation and the converse is true if M is finitely generated, faithful and multiplication.

(3) If I (+)IM is a restricted cancellation ideal of R(M) then I is restricted cancellation and IM ≠ 0 implies I is faithful. The converse is true if M is finitely generated, faithful and multiplication.

Proof. (1) [5, Proposition 11].

(2) Suppose I (+)IM is a weak cancellation ideal of R(M). By [15, Theorem 2.4], I (+)IM is a cancellation R(M)/ann(I (+)IM) ≅ R(M)/annI (+)[0 :_{M} I] ≅ R/annI (+)M/[0 :_{M} I]-module. It follows by (1) that I is a cancellation R/annImodule and hence I is a weak cancellation ideal of R, [15, Theorem 2.4]. Conversely, suppose M is finitely generated faithful multiplication and I is weak cancellation. Then I is a cancellation R/annI-module. Since M is a multiplication R-module, it is easily verified that M/(annI)M is a multiplication R/annImodule. Moreover, [0 :_{R/annI} M/(annI)M] = [(annI)M : M]/annI. Since M is finitely generated faithful and multiplication (hence cancellation), [0 :_{R/annI} M/(annI)M] = annI/annI = 0. That is M/(annI)M is a faithful multiplication R/annI-module, and hence it is a torsion free R/annI-module, [19, Lemma 4.1]. It follows by (1) that I (+)IM is a cancellation R/annI (+)M/(annI)M ≅ R(M)/ann(I (+)IM)-module. This implies that I (+)IM is a weak cancellation ideal of R (M).

(3) Suppose I (+)IM is a restricted cancellation ideal of R (M). Then I (+)IM is weak cancellation and by (2), I is a weak cancellation ideal of R. Let A be an ideal of R then A(+)AM ⊆ ann (I (+)IM) = annI (+)[0 :_{M} I] from which we get A ⊆ annI or ann(I (+)IM) ⊆ A(+)AM and this case gives that annI ⊆ A. By, [15, Theorem 2.5], I is restricted cancellation. Now suppose IM ≠ 0. Then 0 ≠ (I (+)IM)(0(+)M) = (I (+)IM) (annI (+)M), and hence annI = 0. Conversely, suppose M is finitely generated, faithful and multiplication. Let I be restricted cancellation. Then I is weak cancellation and by (2), I (+)IM is weak cancellation. If 0 ≠ IM, then I is faithful. Since M is faithful multiplication, I (+)IM is faithful and hence I (+)IM is restricted cancellation (in fact, I (+)IM is cancellation). If IM = 0, then M = [0 :_{M} I] = (annI)M. Since M is finitely generated, faithful and multiplication (hence cancellation), R = annI, that is I = 0, and this case shows that I (+)IM = 0 is again a restricted cancellation ideal of R(M). □

We close by a result that gives necessary and sufficient conditions for a homogeneous ideal of R (M) to be join principal.

Theorem 20. Let R be a ring, M an R-module and I (+)N a homogeneous ideal of R (M).

(1) If I (+)N is join principal then I is a weak cancellation ideal of R and N/IM is a weak cancellation R-submodule of M/IM.

(2) If I (+)N is join principal then I is a join principal ideal of R. The converse is true if H = [H :_{R(M )} 0(+)M] for any ideal H of R(M).

(3) Let M be cancellation. If I (+)N is weak cancellation then N is a weak cancellation submodule of M.

(4) Let M be cancellation. If I (+)N is cancellation then N is a cancellation submodule of M.

(5) Suppose I (+)N is restricted cancellation. If I^{2} ≠ 0 then I is a restricted cancellation ideal of R and if we assume further that M is restricted cancellation then N is a restricted cancellation submodule of M. If I^{2} = 0 then I (+)N is a nilpotent ideal of R (M) and if R is reduced or N faithful then I is restricted cancellation and N is restricted cancellation.

Proof. (1) [15, Theorem 2.2] says that if I (+)N is a join principal ideal of R (M) then I (+)N/0(+)N ≅ I is a weak cancellation ideal of R, and

I (+)N/I (+)IM = I (+)IM + 0(+)N/I (+)IM
≅ 0(+)N/I (+)IM ∩ 0(+)N
= 0(+)N/0(+)IM ≅ 0(+)N/IM

is a weak cancellation ideal of R (M) /I (+)IM ≅ R (M/IM). It follows that N/IM is a weak cancellation submodule of M/IM, [1, Theorem 2] and [15, Theorem 3.1].

(2) Suppose I (+)N is join principal. The fact that I is join principal follows by [3, Theorem 9]. Note that if I is join principal then I is weak cancellation and that confirms part (1). For the converse, let H_{1} and H_{2} be ideals of R(M) and let H_{1} + 0(+)M = A(+)M and H_{2} + 0(+)M = B(+)M for some ideals A and B of R. Then

[(H_{1} (I (+)N) + H_{2}) :_{R(M)} I (+)N]

⊆ [(H_{1} (I (+)N) + 0(+)IM) + H_{2} + 0(+)M) :_{R(M)} (I (+)N + 0(+)M)] = [((H_{1} + 0(+)M)(I (+)N) + H_{2} + 0(+)M) :_{R(M)} I (+)M] = [((A(+)M)(I (+)N) + B(+)M) :_{R(M)} I (+)M] = [(AI + B)(+)M :_{R(M)} I (+)M] = [(AI + B) : I](+)M = A + [B : I](+)M = A(+)M + [B : I](+)M = H_{1} + 0(+)M + [B(+)M :_{R(M)} I (+)M] = H_{1} + [H_{2} + 0(+)M :_{R(M)} I (+)M] ⊆ H_{1} + [H_{2}(0(+)M) :_{R(M)} 0(+)IM] ⊆ H_{1} + [(H_{2} ∩ 0(+)M) :_{R(M)} 0(+)IM] = H_{1} + [H_{2} :_{R(M)} 0(+)IM] = H_{1} + [H_{2} :_{R(M)} (0(+)M)(I (+)N)] ⊆ H_{1} + [H_{2} :_{R(M)} 0(+)M] :_{R(M)} I (+)N] = H_{1} + [H_{2} :_{R(M)} I (+)N].

The other inclusion is always true and hence

[H_{1}(I (+)N) + H_{2} :_{R(M)} I (+)N] = H_{1} + [H_{2} :_{R(M)} I (+)N].

This shows that I (+)N is join principal. Note that we used in the proof the fact that

0(+)M ⊆ B(+)M ⊆ [B(+)M :_{R(M)} I (+)M].

(3) Suppose M is cancellation and I (+)N a weak cancellation ideal of R (M). Let A and B be ideals of R such that AN = BN. Since AIM ⊆ AN = BN, we infer that 0(+)AIM ⊆ 0(+)BN, and hence

(0(+)AM) (I (+)N) ⊆ (B(+)BM) (0(+)N) ⊆ (B(+)BM) (I (+)N) .

This implies that

0(+)AM + (annI ∩ annN) (+) [0 :_{M} I] ⊆ B(+)BM + (annI ∩ annN) (+) [0 :_{M} I] ,

from which one gets that AM + [0 :_{M} I] ⊆ BM + [0 :_{M} I]. Hence AIM ⊆ BIM, and this gives that AI ⊆ BI. It follows that (A(+)M) (I (+)N) ⊆ (B(+)M) (I (+)N) , and hence

A(+)M + (annI ∩ annN) (+) [0 :_{M} I] ⊆ B(+)M (+) (annI ∩ annN) (+) [0 :_{M} I] .

So A + (annI ∩ annN) ⊆ B + (annI ∩ annN). Since M is cancellation (hence faithful) and IM ⊆ N, we have that annN ⊆ ann (IM) = annI. This finally gives that A + annN ⊆ B + annN. Similarly, B + annN ⊆ A + annN. Hence N is weak cancellation.

(4) Suppose M is cancellation and I (+)N is cancellation. By (3), N is weak cancellation. By [1, Proposition 14] I is cancellation and hence IM is cancellation. It follows that annN ⊆ ann(IM) = 0, and hence N is cancellation.

(5) Suppose I (+)N is a restricted cancellation ideal of R (M). If I^{2} ≠ 0, then 0 ≠ I^{2}(+)IN = (I (+)N)^{2} = (I (+)N) (I (+)IM). So I (+)N = I (+)IM. The fact that I is restricted cancellation follows by Proposition 19. Suppose M is restricted cancellation. Since N = IM, it is easily verified that N is restricted cancellation. Next, let I^{2} = 0. Then I ⊆ annI and hence IM ⊆ (annI)M ⊆ [0 :_{M} I]. But I (+)N is weak cancellation, [15, Theorem 2.5]. Therefore (I (+)N)^{2} = (I (+)N) (I (+)IM) implies that

I (+)N + (annI ∩ annN) (+) [0 :_{M} I] = I (+)IM + (annI ∩ annN)(+) [0 :_{M} I]) .

So N + [0 :_{M} I] = IM + [0 :_{M} I] and hence IN = I^{2}M = 0. This gives that (I (+)N)^{2} = 0, and hence I (+)N is nilpotent. If R is reduced then I = 0 is restricted cancellation. Moreover, I (+)N = 0(+)N is restricted cancellation implies that N is restricted cancellation, see [1, Theorem 2] and [15, Theorem 3.1]. Finally, if N is faithful then IN = 0 implies I = 0 and again I is restricted cancellation and N is restricted cancellation (in fact, N is cancellation). □

References

[1] Ali, M. M.: Idealization and Theorems of D. D. Anderson. Comm. Algebra 34 (2006), 4479–4501. Zbl 1109.13010 [2] Ali, M. M.: Idealization and Theorems of D. D. Anderson II. Commun.

Algebra 35 (2007), 2767–2792. Zbl pre05207373 [3] Ali, M. M.: Multiplication modules and homogeneous idealization. Beitr. Al-

gebra Geom. 47 (2006), 249–270. Zbl 1107.13016 [4] Ali, M. M.: Multiplication modules and homogeneous idealization II. Beitr.

Algebra Geom. 48 (2007), 321–343. Zbl pre05191869 [5] Ali, M. M.: \frac{1}{2}Cancellation modules and homogeneous idealization. Commun.

Algebra 35 (2007), 3524–3543. Zbl pre05221141 [6] Ali, M. M.: Multiplication modules and tensor product. Beitr. Algebra Geom.

47 (2006), 305–327. Zbl 1112.13012 [7] Ali, M. M.: Residual submodules of multiplication modules. Beitr. Algebra

Geom. 46 (2005), 405–422. Zbl 1085.13003 [8] Ali, M. M.; Smith, D. J.: Some remarks on multiplication and projective

modules. Commun. Algebra 32 (2004), 3897–3909. Zbl 1065.13004 [9] Ali, M. M.: Some remarks on Generalized GCD domains. Commun. Algebra

36 (2008), 142–164. Zbl pre05241292 [10] Ali, M. M.; Smith, D. J: Generalized GCD modules. Beitr. Algebra Geom.

46 (2005), 447–466. Zbl 1084.13006 [11] Ali, M. M.; Smith, D. J.: Generalized GCD rings. Beitr. Algebra Geom. 42

(2001), 219–233. Zbl 0971.13002 [12] Ali, M. M.; Smith, D. J.: Generalized GCD rings II. Beitr. Algebra Geom.

44 (2003), 75–98. Zbl 1023.13003 [13] Ali, M. M.; Smith, D. J.: Multiplication modules and a theorem of

P. F. Smith. Period. Math. Hung. 44 (2002), 127–135. Zbl 1006.13001 [14] Ali, M. M.; Smith, D. J.: Finite and infinite collections of multiplication

modules. Beitr. Algebra Geom. 42 (2001), 557–573. Zbl 0987.13001 [15] Anderson, D. D.: Cancellation modules and related modules. Lect. Notes Pure

Appl. Math. 220 (2001), 13–25. Dekker, New York. Zbl 1037.13005 [16] Anderson, D. D.; Al- Shaniafi, Y.: Multiplication modules and the ideal θ(M ).

Commun. Algebra 30 (2002), 3383–3390. Zbl 1016.13002 [17] Anderson, D. D.; Winders, M.: Idealization of a module. Rocky Mountain J.

Math. (to appear).

[18] Barnard, A.: Multiplication modules. J. Algebra 71 (1981), 174–178. Zbl 0468.13011

[19] El-Bast, Z. A.; Smith, P. F.: Multiplication modules. Commun. Algebra, 16 (1988), 755–799. Zbl 0642.13002

[20] Faith, C.: Algebra I: Rings, Modules and Categories. Springer-Verlag, 1981. Zbl 0508.16001

[21] Huckaba, J. A.: Commutative Rings with Zero Divisors. Marcel Dekker, New York 1988. Zbl 0637.13001

[22] Kaplansky, I.: Commutative Rings. University of Chicago Press, 1974. Zbl 0296.13001 [23] Kasch, F.: Modules and Rings. Academic Press, London 1982.

Zbl 0523.16001

[24] Larsen, M. D.; McCarthy, P. J.: Multiplicative Theory of Ideals. Academic Press, New York 1971. Zbl 0237.13002

[25] Matsumura, H.: Commutative Ring Theory. Cambridge University Press, Cambridge 1986. Zbl 0603.13001

[26] Naoum, A. G.; Al-Shalgy, L. S.: Quasi-Frobenius modules. Beitr. Algebra Geom. 41 (2000), 257–266. Zbl 0976.16006

[27] Naoum, A. G.: \frac{1}{2}Cancellation modules. Kyungpook Math. J. 36 (1996), 97– 106. Zbl 0890.13003

[28] Naoum, A. G.; Mijbass, A. S.: Weak cancellation modules. Kyungpook Math. J. 37 (1997), 73–82. Zbl 0882.13002

[29] Smith, P. F.: Some remarks on multiplication modules. Arch. Math. 50 (1988), 223–235. Zbl 0615.13003

[30] Smith, P. F.; Woodward, A. R.: On finitely annihilated modules and Artinian rings. Mediterr. J. Math. 3 (2006), 301–311. Zbl 1119.16021

Received June 5, 2007