Abstract
Abstract
Let G be a group. Let R be a G-graded commutative ring with identity, and let M be a G-graded module over R. Two graded submodules N and K of graded module M are called graded coprime whenever N + K = M. In this paper, some properties of graded coprime submodules are discussed. For example, we show that if M is a graded finitely generated module, then two graded submodules N and K of M are graded coprime if and only if grad M (N) and grad M (K) are graded coprime.
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Introduction
We define a G-graded ring R and a G-graded module over R in the same way as in [1] and [2]. Let G be a group with identity e and R be a commutative ring. Then, R is a G-graded ring if there exist additive subgroups R g of R indexed by the elements g ∈ G such that R = ⊕g∈GR g and R g R h ⊆ R gh for all g, h ∈ G; here, R g R h denotes the additive subgroup of R consisting of all finite sums of elements r g k h with r g ∈ R g and k h ∈ R h . Moreover, R e is a subring of R and 1 R ∈ R e . We denote this by (R, G). The elements of R g are called homogeneous of degree g. If x ∈ R, then x can be written uniquely as Σg∈Gx g , where x g is the component of x in R g . Also, we write h(R) = ∪g∈GR g .
Let R be a G-graded ring and M be an R-module. We say that M is a graded R-module if there exists a family of submodules {M g }g∈G of M such that M = ⊕g∈GM g and R g M h ⊆ M gh for all g, h ∈ G, and we write h(M) = ∪g∈GM g .
Throughout this paper, G is a group, R is a G-graded commutative ring with identity, and M is a G-graded module over R. Also, for basic properties of coprime ideals, one may refer to [3].
The concept of multiplication module has been studied by various authors (see, for example, [4, 5]). Also, the notion of the product of two submodules of a multiplication module has been studied in [6].
We define a graded multiplication module and the product of two graded submodules of a graded multiplication module in the same way as in [7].
Let R be a graded ring. A graded R-module M is said to be a graded multiplication module if for every graded submodule N of M, there exists a graded ideal I of h(R)such that N = IM. Assume that M is a graded multiplication R-module. If N and K are graded submodules of M, then there exist graded ideals I and J of h(R) such that N = IM and K = JM. Then, the product of N and K is defined to be (IJ)M and is denoted by N ∗ K. In fact, IJ is a graded ideal of R by [7, Lemma 1.1], and N ∗ K is well-defined and is independent of the choices of I and J by [6, Theorem 3.4], and [2, Theorem 4]. Also, for every positive integer t, Nt is defined to be
Lemma 1.1
[7, Lemma 1.2], Let R be a graded ring and M be a graded R-module.
(i) If N and K are graded submodules of M, then N + K and N ∩ K are graded submodules of M.
(ii) If a is an element of h(R) and x is an element of h(M), then aM and Rx are graded submodules of M.
(iii) If N and K are graded submodules of M, then (N: R K) is a graded ideal of R.
Definition 1.2
Let R be a graded ring and M be a graded module over R. Let P be a proper graded submodule of M.
(i) P is called a graded prime submodule of M whenever am ∈ P implies that m ∈ P or a ∈ (P: R M) where a ∈ h(R) and m ∈ h(M).
(ii) P is called a graded semiprime submodule of M whenever InK ⊆ P implies that IK ⊆ P where I ⊆ h(R) and K ⊆ h(M).
(iii) P is called a graded maximal submodule of M if there is no graded submodule K of M such that P ⊂ K ⊂ M.
Theorem 1.3
[2, Theorem 5], Let R be a graded ring and M be a graded multiplication R-module. Let N be a proper graded submodule of M. Then, N is graded prime if and only if K ∗ L ⊆ N implies that K ⊆ N or L ⊆ N for graded submodules K and Lof M.
Theorem 1.4
[7, Theorem 2.1], Let R be a graded ring and M be a graded multiplication R-module. Let N be a graded submodule of M. Then, N is graded semiprime if and only if (Rx)n ⊆ N implies that x ∈ N for each x ∈ h(M) and positive integer n.
The graded radical of a graded submodule N of a graded module M is the intersection of all graded prime submodules of M containing N and is denoted by gra d M (N). If there is no graded prime submodule of M containing N, then we say gra d M (N) = M. Also, gra d M (M) = M. It is easy to show that if M is a graded multiplication module, then gra d M (N) is the set of all elements m of h(M) such that (Rm)k ⊆ N for some positive integer k.
Results and discussion
Let and . Clearly, R is a G-graded ring. Let . So, M is a G-graded R-module. Let x, y ∈ h(M). Consider the graded submodules N = (Rx) × 0 and K = (Ry) × 0 of M. Then, N + K is the graded submodule generated by the greatest common factor of x and y.
Definition 2.1
Let R be a graded ring and M be a graded module over R; two graded submodules N and K of M are called graded coprime whenever N + K = M.
Clearly, two distinct graded maximal submodules of a graded module are graded coprime.
Proposition 2.2
Let R be a graded ring and M be a graded multiplication module over R. Let N1and K1be two graded coprime submodules of M. Let N2and K2be two graded submodules of M such that every element of N1(resp. K1) has a power in N2(resp. K2). Then, N2and K2are graded coprime.
Proof
Under the given hypothesis, every graded prime submodule which contains N2(resp. K2) contains N1(resp. K1). Let P be a graded prime submodule of M which contains both N2 and K2. So, P contains N1 and K1, that is, N1 + K1 ⊆ P, which is absurd. Hence, no graded prime submodule contains both N2 and K2. Therefore, N2 and K2 are graded coprime. □
Proposition 2.3
Let R be a graded ring and M be a graded multiplication module over R. Let N and K be graded coprime submodules of M. Then, N ∗ K = N ∩ K.
Proof
Similar to the proof of [6, Proposition 3.5]. □
Proposition 2.4
Let R be a graded ring and M be a graded multiplication module over R with this property that every graded submodule of M is graded semiprime. Let N1, N2, …, N t be graded submodules of M such that N i and N j are graded coprime whenever i ≠ j. Then, for each i(1 ≤ i ≤ t),
Furthermore,
Proof
Since
it is enough to show that
Without loss of generality, let i = 1. Suppose that 2 ≤ j ≤ t. Let m be an arbitrary element of h(M). So, there exist elements x1j ∈ N1, x j ∈ N j such that x1j + x j = m. So,
where x ∈ N1. Therefore, (Rm)t ⊆ N1 + (N2 ∗ N3 ∗ … ∗N t ). Now, N1 + (N2 ∗ N3 ∗ … ∗N t ) graded semiprime implies that m ∈ N1 + (N2 ∗ N3 ∗ … ∗N t ), as required.
We prove the next result by induction on t. When t = 1, there is nothing to prove, and when t = 2, the required result follows from Proposition 2.3. Now, assume that t > 2 and the relation has been proved for all smaller values of the inductive variable. So,
Therefore,
But we showed that N1∩N2∩…∩Nt−1and N t are graded coprime. Hence, by applying Proposition 2.3, we find that
Accordingly, □
The following result can be obtained by the above proposition easily.
Corollary 2.5
Let R be a graded ring and M be a graded multiplication module over R. Let N and K be graded coprime submodules of M. Then, N and Ktare graded coprime for every positive integer t.
Proposition 2.6
With the assumption in Proposition 2.4, let φ : M → (M / N i ) be a homomorphism by the rule φ(m) = (m + N1, … ,m + N t ). Then, φ is injective if and only if N1 ∗ N2 ∗ … ∗ N t = 0.
Proof
It is enough to show that kerφ = N1 ∗ N2 ∗ … ∗ N t . Let x ∈ kerφ. So, x ∈ N i for every i (1 ≤ i ≤ t). Thus, (Rx)t ⊆ N1 ∗ N2 ∗ … ∗ N t and N1 ∗ N2 ∗ ... ∗ N t graded semiprime implies that x ∈ N1 ∗ N2 ∗ ... ∗ N t , that is, kerφ ⊆ N1 ∗ N2 ∗ ... ∗ N t .
Conversely, let x ∈ N1 ∗ N2 ∗ … ∗ N t . Therefore, x ∈ N1 ∩ N2 ∩ … ∩ N t . Hence, x + N i = N i for every i (1 ≤ i ≤ t). Now, we conclude that φ(x) = (N1, …, N t ), that is, N1 ∗ N2 ∗ …∗ N t ⊆ kerφ. □
Proposition 2.7
Let R be a graded ring and M be a graded multiplication module over R. Let P be a graded maximal submodule of M. Then, for every positive integer n, the only graded prime submodule containing Pnis P.
Proof
Let be a graded prime submodule of M containing Pn. So, by Theorem 1.3, as P is grade maximal; □
Proposition 2.8
Let R be a graded ring and M be a graded module over R. Let N = IM and K = JM be graded submodules of M where I and J are graded coprime ideals of R. Then, N and K are graded coprime.
Proof
Since I and J are graded coprime ideals in R, we have
□
Definition 2.9
Let R be a graded ring and M be a graded module over R. Let I and J be graded ideals of R. Then, M is called graded cancelation whenever IM = JM gives I = J.
Theorem 2.10
Let R be a graded ring and M be a graded finitely generated cancelation module over R. Let N and K be graded submodules of M with graded presentation ideals I and J, respectively. Then, I and J are graded coprime in R if and only if N and K are graded coprime in M.
Proof
One direction is proved in Proposition 2.8.
Since N and K are graded coprime submodules in a graded finitely generated module M, we have
Now, M graded cancelation implies that R = I + J, as required. □
Corollary 2.11
Let M be a graded finitely generated module. Then, every proper graded submodule of M is contained in a graded prime submodule.
Proof
Similar to the proof of [8, Corollary 1]. □
Proposition 2.12
Let R be a graded ring and M be a graded module over R. Let N and K be graded submodules of M. then,
(i) N ⊆ gra d M (N).
(ii) grad M (grad M (N)) = grad M (N).
(iii) If M = N, then grad M (N) = M.
Moreover, if M is finitely generated, then grad M (N) = M if and only if N = M.
(iv) grad M (N + K) = grad M (grad M (N) + grad M (K)).
Also, if M is a graded multiplication module, then
(v) grad M (N ∗ K) = grad M (N ∩ K) = grad M (N) ∩ grad M (K).
(vi) Let N be graded prime. Then grad M (Nt) = N for every positive integer t.
Proof
□
(i,ii) (i) and (ii) are trivial by definition of grad M (N).
-
(iii)
Suppose that N = M. So, grad M (N) = grad M (M) = M. Also, let M be finitely generated and grad M (N) = M. Suppose to the contrary that N ≠ M. Hence, by Corollary 2.11, grad M (N) ⊆ P for some graded prime submodule P of M. Therefore, grad M (N) ≠ M, a contradiction.
-
(iv)
N ⊆ grad(N)and K ⊆ grad(K) implies that N + K ⊆ grad(N) + grad(K). Therefore, grad(N + K) ⊆ grad(grad(N) + grad(K)). Also, N, K ⊆ N + K, hence grad(N), grad(K) ⊆ grad(N + K). So, grad(N) + grad(K) ⊆ grad(N + K), and by (ii), grad(grad(N) + grad(K)) ⊆ grad(N + K).
-
(v)
We prove the first equality. Suppose that x ∈ grad(N ∗ K). Then, (Rx)t ⊆ N ∗ K⊆N ∩ K for some positive integer t. So, x ∈ grad M (N ∩ K).Conversely, suppose x ∈ grad M (N ∩ K). Then, (Rx)t ⊆ N and (Rx)t ⊆ K for some positive integer t. Hence, (Rx)2t = (Rx)t ∗ (Rx)t ⊆ N ∗ K, that is, x ∈ grad M (N ∗ K).Now, we prove the second equality. Let x ∈ grad M (N ∩ K). So, (Rx)t ⊆ N ∩ K ⊆ grad M (N) ∩ grad M (K) for some positive integer t. Thus, (Rx)t ⊆ grad M (N) and (Rx)t ⊆ grad M (K). Hence, x ∈ grad M (N) ∩ grad M (K).Conversely, if x ∈ grad M (N) ∩ grad M (K), then (Rx)n ⊆ N and (Rx)k ⊆ K for some positive integers n, k. Therefore, (Rx)nk ⊆ N ∩ K, giving us that in fact x ∈ grad M (N ∩ K).
-
(vi)
Note that N ⊆ grad M (N t). Now, suppose x ∈ grad M (N t). Then, (Rx)n ⊆ N t ⊆ N, for some positive integer n. Now, N graded prime implies that x ∈ N, as needed.
Proposition 2.13
Let R be a graded ring and M be a graded finitely generated module over R. Let N and K be graded submodules of M. Then, N and K are graded coprime if and only if grad M (N) and grad M (K) are graded coprime.
Proof
By using (iii) and (iv) above, we have
□
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Varmazyar, R. Graded coprime submodules. Math Sci 6, 70 (2012). https://doi.org/10.1186/2251-7456-6-70
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DOI: https://doi.org/10.1186/2251-7456-6-70