Graded coprime submodules

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.

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 Σ_g∈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, N^t is defined to be

$$\stackrel{t\text{times}}{\stackrel{⏟}{N\ast N\ast \dots \ast N}}.$$

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 $G=(\mathbb{Z},+)$ and $R=(\mathbb{Z},+,·)$. Clearly, R is a G-graded ring. Let $M=\mathbb{Z}\times \mathbb{Z}$. 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),

$$\begin{array}{ll}{N}_i\hfill & +(N_1\ast \dots \ast N_{i-1}\ast N_{i+1}\ast \dots \ast N_t)\hfill \\ =N_i+(N_1\cap \dots \cap N_{i-1}\cap N_{i+1}\cap \dots \cap N_t)=M.\hfill \end{array}$$

Furthermore,

$$N_1\ast N_2\ast \dots \ast N_t=N_1\cap N_2\cap \dots \cap N_t.$$

Proof

Since

$$N_1\ast \dots \ast N_{i-1}\ast N_{i+1}\ast \dots \ast N_t\subseteq N_1\cap \mathrm{...}\cap N_{i-1}\cap N_{i+1}\cap \mathrm{...}\cap N_t,$$

it is enough to show that

$$N_i+(N_1\ast \dots \ast N_{i-1}\ast N_{i+1}\ast \dots \ast N_t)=M.$$

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 x_1j ∈ N₁, x _j  ∈ N _j such that x_1j + x _j  = m. So,

$$\begin{array}{ll}{\left(RM\right)}^t\hfill & =R(x_{12}+x_2)\ast R(x_{13}+x_3)\ast \dots \ast R(x_{1t}+x_t)\hfill \\ =\mathrm{Rx}+(R{x}_2\ast R{x}_3\ast \dots \ast R{x}_t),\hfill \end{array}$$

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,

$$N_1\ast N_2\ast \dots \ast N_{t-1}=N_1\cap N_2\cap \dots \cap N_{t-1}.$$

Therefore,

$$N_1\ast N_2\ast \dots \ast N_{t-1}\ast N_t=(N_1\cap N_2\cap \dots \cap N_{t-1})\ast N_t.$$

But we showed that N₁∩N₂∩…∩N_t−1and N _t are graded coprime. Hence, by applying Proposition 2.3, we find that

$$(N_1\cap N_2\cap \dots \cap N_{t-1})\ast N_t=N_1\cap N_2\cap \dots \cap N_{t-1}\cap N_t.$$

Accordingly, □

$$N_1\ast N_2\ast \dots \ast N_t=N_1\cap N_2\cap \dots \cap N_t.$$

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 K^tare graded coprime for every positive integer t.

Proposition 2.6

With the assumption in Proposition 2.4, let φ : M → ${\pi }_{i=1}^t$(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 $\acute{P}$ be a graded prime submodule of M containing Pⁿ. So, by Theorem 1.3, $P\subseteq \acute{P}$ as P is grade maximal; $P=\acute{P}.$ □

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

$$M=RM=(I+J)M\subseteq IM+JM=N+K.$$

□

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

$$RM=M=N+K=IM+JM=(I+J)M.$$

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 (N^t) = 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

$$N+K=M\iff gra{d}_M(N+K)=M$$

$$\iff gra{d}_M\left(gra{d}_M\right(N)+gra{d}_M(K\left)\right)=M$$

$$\iff gra{d}_M\left(N\right)+gra{d}_M\left(K\right)=M.$$

□