The intersection graph of graded submodules of a graded module

Definition 1.1

Let R be a G -graded ring and M be a G -graded left R -module. The intersection graph of G -graded submodules of M , denoted by Γ ( G , R , M ) , is defined to be an undirected simple graph whose set of vertices is h S ∗ ( M ) and two vertices N and K are adjacent if N ∩ K ≠ { 0 } .

Lemma 2.1

[20, Lemma 2.1] Let R be a G -graded ring and M be a G -graded R -module.

1. If I and J are G -graded ideals of R , then I + J and I ⋂ J are G -graded ideals of R .

2. If N and K are G -graded R -submodules of M , then N + K and N ⋂ K are G -graded R -submodules of M .

3. If N is a G -graded R -submodule of M , r ∈ h ( R ) , x ∈ h ( M ) and I is a G -graded ideal of R , then R x , I N , and r N are G -graded R -submodules of M . Moreover, ( N : R M ) = { r ∈ R : r M ⊆ N } is a G -graded ideal of R .

Theorem 2.2

[8, Theorem 2.1] Let M be an R -module. Then, the graph G ( M ) is disconnected if and only if M is a direct sum of two simple R -modules.

Corollary 2.3

[8, Corollary 2.3] Let M be an R -module. Then, the graph G ( M ) is disconnected if and only if it is null graph with at least two vertices.

Theorem 2.4

Let R be a G -graded ring and M be a G -graded R -module such that ∣ Γ ( G , R , M ) ∣ ≥ 2 . Then, the followings are equivalent:

1. Γ ( G , R , M ) is disconnected.

2. Γ ( G , R , M ) is a null graph.

3. Every nontrivial G -graded R -submodule of M is G -graded maximal as well as G -graded simple.

4. M is a direct sum of two G -graded simple (or maximal) R -modules.

Proof

( 1 ) ⇒ ( 2 ) Suppose that Γ ( G , R , M ) is disconnected. For a contradiction, assume N and K are two adjacent vertices. So N , K , and N ∩ K belong to the same component of Γ ( G , R , M ) . Since Γ ( G , R , M ) is disconnected, there is a vertex L that is not connected to any of the vertices N , K , and N ∩ K . If ( N ∩ K ) + L ≠ M , then ( N ∩ K ) ↔ ( ( N ∩ K ) + L ) ↔ L is a path connecting N ∩ K and L , a contradiction. So ( N ∩ K ) + L = M . Now let a ∈ N . Then, a = t + c for some t ∈ N ∩ K and c ∈ L . So a − t = c ∈ N ∩ L = { 0 } ; consequently, a = t ∈ N ∩ K . This implies that N = N ∩ K . Similarly, we obtain K = N ∩ K . Hence, we have N = K , a contradiction. Therefore, Γ ( G , R , M ) contains no edges, and hence, it is a null graph.

( 2 ) ⇔ ( 3 ) Straightforward.

( 3 ) ⇒ ( 4 ) Let N and K be G -graded maximal as well as G -graded simple R -submodules of M . Then, N + K = M and N ∩ K = { 0 } . Hence, N and K are G -graded simple R -modules and M = N ⊕ K .

( 4 ) ⇒ ( 1 ) Suppose M = N ⊕ K , where N and K are G -graded simple R -modules. Then, N and K are G -graded simple R -submodules. Also, they are G -graded maximal because N ≅ M K and K ≅ M N . Thus, N and K are isolated vertices. Therefore, Γ ( G , R , M ) is disconnected.□

Corollary 2.5

Let M be a G -graded R -module. Then, Γ ( G , R , M ) is disconnected if and only if G M is a complete graph with at least two vertices.

Proof

The result follows by Theorem 2.4 and [6, Theorem 2.2].□

Corollary 2.6

Let M be a G -graded R -module. If Γ ( G , R , M ) is connected, then every pair of G -graded maximal R -submodules intersect nontrivially.

Theorem 2.7

Let M be a G -graded R -module. If Γ ( G , R , M ) is connected, then diam ( Γ ( G , R , M ) ) ≤ 2 .

Proof

Suppose N and K are distinct vertices in Γ ( G , R , M ) . If N and K are adjacent, then d ( N , K ) = 1 . If N and K are nonadjacent, then d ( N , K ) ≥ 2 . If N ⊕ K ≠ M , then we have the path N ↔ N ⊕ K ↔ K , and hence, d ( N , K ) = 2 . If N ⊕ K = M , then either N or K is not G -graded simple, say N . Let ( 0 ) ≠ L ⊊ N . Thus, we have the path N ↔ L ⊕ K ↔ K , and hence, d ( N , K ) = 2 . As a result, d ( N , K ) ≤ 2 .□

Theorem 2.8

Let M be a G -graded Artinian R -module such that Γ ( G , R , M ) is not a null graph. Then, the followings are equivalent:

1. Γ ( G , R , M ) is regular.

2. ∣ GMin ( M ) ∣ = 1 .

3. Γ ( G , R , M ) is complete.

Proof

( i ) ⇒ ( i i ) Suppose Γ ( G , R , M ) is regular. Assume that M contains two distinct G -graded simple R -submodules N and K . Clearly, N and K are nonadjacent. By Theorem 2.7, there is a G -graded R -submodule Y that is adjacent to both N and K . Hence, by minimality of N , we obtain N ⊊ Y . This implies that N ( N ) ⊊ N ( Y ) ; consequently, deg ( Y ) > deg ( N ) , a contradiction. Hence, M contains a unique G -graded simple R -submodule.

( i i ) ⇒ ( i i i ) Suppose M contains a unique G -graded simple R -submodule, say N . Since M is G -graded Artinian, N ⊆ K for all K ∈ h S ∗ ( M ) . Thus, Γ ( G , R , M ) is complete.

( i i i ) ⇒ ( i ) Straightforward.□

Remark 2.9

In a G -graded R -module M , a G -graded submodule N of M is called G -graded essential if N ∩ K ≠ ( 0 ) for all K ∈ h S ∗ ( M ) . The graded socle, Gsoc ( M ) , of M is the sum of all G -graded simple R -submodules of M . Equivalently Gsoc ( M ) equals the intersection of all G -graded essential R -submodules of M , see [21, page 48]. So, if M is G -graded Artinian and Γ ( G , R , M ) is complete, then every G -graded R -submodule is G -essential, and thus, by Theorem 2.8, GMin ( M ) = Gsoc ( M ) .

Theorem 2.10

If M is a G -graded R -module, then g r ( Γ ( G , R , M ) ) ∈ { 3 , ∞ } .

Proof

Assume g ( Γ ( G , R , M ) ) < ∞ and g ( Γ ( G , R , M ) ) ≥ 4 . This implies that every pair of distinct nontrivial G -graded submodules of M with nonzero intersection should be comparable, otherwise Γ ( G , R , M ) will have a cycle of length 3, a contradiction. Since g ( Γ ( G , R , M ) ) ≥ 4 , Γ ( G , R , M ) contains a path of length 3, say N ↔ L ↔ K ↔ P . Since any two submodules in this path are comparable and any chain of nontrivial G -graded submodules of length 2 induces a cycle of length 3 in Γ ( G , R , M ) , the only possible two cases are N ⊆ L , K ⊆ L , K ⊆ P or L ⊆ N , L ⊆ K , P ⊆ K . The first case yields K ⊆ L ∩ P , and hence, L ∩ P ≠ ( 0 ) . Thus, L ↔ K ↔ P ↔ L is a cycle of length 3 in Γ ( G , R , M ) , a contradiction. In the second case, we have ( 0 ) ≠ L ⊆ N ∩ K , and therefore, N ↔ L ↔ K ↔ N is a cycle of length 3 in Γ ( G , R , M ) , which again yields a contradiction. Therefore, g ( Γ ( G , R , M ) ) = 3 .□

Theorem 2.11

Let M be a G -graded Noetherian R -module such that Γ ( G , R , M ) is not a null graph with ∣ Γ ( G , R , M ) ∣ ≥ 2 . If g ( Γ ( G , R , M ) ) = ∞ , then M is a G -graded local module and Γ ( G , R , M ) is a star graph whose center is the unique G -graded maximal R -submodule of M .

Proof

By Theorem 2.4, Γ ( G , R , M ) is connected. Suppose that N 1 and N 2 are two distinct G -graded maximal R -submodules of M . By Theorem 2.7, d ( N 1 , N 2 ) ≤ 2 . If N 1 ∩ N 2 ≠ ( 0 ) , then N 1 ↔ ( N 1 ∩ N 2 ) ↔ N 2 ↔ N 1 is a 3-cycle, a contradiction. So N 1 ∩ N 2 = ( 0 ) . Since N 1 and N 2 are G -graded maximal R -submodules, we obtain M = N 1 ⊕ N 2 . Thus, Γ ( G , R , M ) is null, which contradicts the assumption that Γ ( G , R , M ) is not null. Therefore, M is G -graded local module. Let N be the G -graded maximal submodule of M . It is easy to see that every proper graded submodule of a G -graded Noetherian module is contained in a G -graded maximal submodule. So N ∩ K = K ≠ ( 0 ) for all K ∈ h S ∗ ( M ) . However, since Γ ( G , R , M ) has no cycles, we conclude that Γ ( G , R , M ) is a star graph.□

Lemma 2.12

Let M be a G -graded R -module. If ω ( Γ ( G , R , M ) ) < ∞ , then M is G -graded Artinian and G -graded Noetherian.

Proof

Members of any ascending or descending chain of G -graded R -submodules form a clique in Γ ( G , R , M ) , and hence, the chain is finite.□

Corollary 2.13

Let M be a G -graded R -module. If Γ ( G , R , M ) is connected, then ∣ GMax ( M ) ∣ ≤ ω ( Γ ( G , R , M ) ) .

Proof

If Γ ( G , R , M ) is connected, then by Corollary 2.6, GMax ( M ) is a clique, and thus, ∣ GMax ( M ) ∣ ≤ ω ( Γ ( G , R , M ) ) .□

Theorem 2.14

Let M be a G -graded R -module that contains a G -graded maximal submodule. Then, γ ( Γ ( G , R , M ) ) ≤ 2 . Furthermore, if M is G -graded indecomposable, then γ ( Γ ( G , R , M ) ) = 1 .

Proof

Let N be a G -graded maximal R -submodule of M . If there exists K ∈ h S ∗ ( M ) such that N ∩ K = ( 0 ) , then N + K = M , and hence, M = N ⊕ K . So the set { N , K } is a dominating set, and thus, γ ( Γ ( G , R , M ) ) ≤ 2 . This proves the first part.

If M is G -graded indecomposable, then N ∩ K ≠ ( 0 ) for all K ∈ h S ∗ ( M ) . Consequently, { N } is a dominating set, and hence, γ ( Γ ( G , R , M ) ) = 1 .□

Theorem 3.1

Let M be a G -graded R -module. If for some σ ∈ G , G ( M σ ) is connected with at least two vertices, then Γ ( G , R , M ) is connected, and hence, G ( M ) is connected.

Proof

Since G ( M σ ) is connected, it must contain an edge. Let N σ , K σ be two adjacent vertices in G ( M σ ) . Then, R N σ and R K σ are vertices in Γ ( G , R , M ) . Moreover, R N σ ∩ M σ = N σ and R K σ ∩ M σ = K σ , and so, R N σ ≠ R K σ . In addition, we have { 0 } ≠ N σ ∩ K σ ⊆ R N σ ∩ R K σ . Therefore, Γ ( G , R , M ) is not null, and hence, it is connected. The last part follows from Corollary 2.3 because Γ ( G , R , M ) is a subgraph of G ( M ) .□

Remark 3.2

The converse of Theorem 3.1 needs not to be true in general. Let R = Z 6 with trivial Z -grading; that is, R 0 = Z 6 , and R k = 0 , for all k ≠ 0 , and choose M = Z 6 [ x ] as Z 6 -module with grading M k = Z 6 x k , k ≥ 0 , and M k = 0 , k < 0 . The Z -graded Z 6 -submodules Z 6 and Z 6 + Z 6 x are adjacent in Γ ( Z , Z 6 , Z 6 [ x ] ) , and by Theorem 2.4, we have Γ ( Z , Z 6 , Z 6 [ x ] ) is connected. On the other hand, for each k ≥ 0 , ⟨ 2 x k ⟩ and ⟨ 3 x k ⟩ are the only Z 6 -submodules of Z 6 x k , and their intersection is ( 0 ) . So G ( Z 6 x k ) is disconnected for all k ≥ 0 .

Lemma 3.3

[21, Proposition 2.6.3] A G -graded R -module M is σ -faithful for some σ ∈ G if and only if N ∩ M σ ≠ { 0 } for all N ∈ h S ∗ ( M ) .

Theorem 3.4

Let M be a G -graded R -module such that Γ ( G , R , M ) is not null graph. Then, M is σ -faithful for some σ ∈ G if and only if the map ϕ σ : Γ σ ( G , R , M ) ⟶ Γ ( G , R , M ) defined by ϕ ( N ) = N is a graph isomorphism.

Proof

Suppose M is σ -faithful for some σ ∈ G . Clearly, ϕ is a set bijection. Let N , K ∈ h S ∗ ( M ) such that N ∩ K ≠ { 0 } . Since M is σ -faithful, by Lemma 3.3, N ∩ K ∩ M σ ≠ { 0 } , which implies that N and K are adjacent in Γ σ ( G , R , M ) . Therefore, ϕ σ is a graph isomorphism. For the converse, suppose that there exists N ∈ h S ∗ ( M ) such that N ∩ M σ = { 0 } . Then, N is an isolated vertex in Γ σ ( G , R , M ) , which implies that N is an isolated vertex in Γ ( G , R , M ) because ϕ σ is an isomorphism. So Γ ( G , R , M ) is null, a contradiction. Then, N ∩ M σ ≠ { 0 } for all N ∈ h S ∗ ( M ) . Hence, by Lemma 3.3, M is σ -faithful.□

Theorem 3.5

Let M be a σ -faithful G -graded R -module. If R M σ = M , then the following assertions hold:

1. Γ ( G , R , M ) is connected if and only if G ( M σ ) is connected.

2. γ ( Γ ( G , R , M ) ) = γ ( G ( M σ ) ) .

3. ω ( Γ ( G , R , M ) ) < ∞ if and only if ω ( G ( M σ ) ) < ∞ , and for each N σ ∈ S ∗ ( M σ ) , the set β N σ = { N ∈ h S ∗ ( M ) ∣ N ∩ M σ = N σ } is finite.

Proof

(i) The “if” part is Theorem 3.1. For the “only if” part, assume Γ ( G , R , M ) is connected and let N σ and K σ be two distinct vertices in G ( M σ ) . If R N σ ∩ R K σ ≠ { 0 } , then by Theorem 3.4, R N σ ∩ R K σ ∩ M σ ≠ { 0 } . So we have N σ ∩ K σ = R N σ ∩ M σ ∩ R K σ ∩ M σ = R N σ ∩ R K σ ∩ M σ ≠ { 0 } , and hence, N σ ↔ K σ is a path. Assume R N σ ∩ R K σ = { 0 } . By Theorem 2.7, there is Y ∈ h S ∗ ( M ) such that R N σ ∩ Y ≠ { 0 } and R K σ ∩ Y ≠ { 0 } . Then, R N σ ∩ Y ∩ M σ ≠ { 0 } and R K σ ∩ Y ∩ M σ ≠ { 0 } . Since ϕ σ is a graph isomorphism, N σ ∩ ( Y ∩ M σ ) and K σ ∩ ( Y ∩ M σ ) are nontrivial. Moreover, Y ∩ M σ ≠ M σ because R M σ = M . Hence, we obtain a path connecting N σ and K σ in G ( M ) . Therefore, G ( M ) is connected.

(ii) Let S ⊆ S ∗ ( M σ ) be a minimal dominating set in G ( M σ ) , and let S = { R N σ ∣ N σ ∈ S } . Clearly, ∣ S ∣ = ∣ S ∣ . Let K ∈ h S ∗ ( M ) such that K ∉ S . By Lemma 3.3, we have K ∩ M σ ≠ { 0 } , and hence, K ∩ M σ ∩ N σ ≠ { 0 } for some N σ ∈ S . So we have R N σ ∈ S and K is adjacent to R N σ in Γ ( G , R , M ) . Hence, S is a dominating set in Γ ( G , R , M ) . Therefore, γ ( Γ ( G , R , M ) ) ≤ γ ( G ( M σ ) ) . Now assume S is a minimal dominating set in Γ ( G , R , M ) , and let S = { N ∩ M σ ∣ N ∈ S } . Let K σ ∈ S ∗ ( M σ ) such that K σ ∉ S . If R K σ ∈ S , then R K σ ∩ M σ ∈ S . We have K σ ∩ ( R K σ ∩ M σ ) = K σ ≠ ( 0 ) . So K σ is adjacent to R K σ ∩ M σ ∈ S . Now assume R K σ ∉ S . So there exists N ∈ S such that R K σ ∩ N ≠ ( 0 ) . Hence, by Theorem 3.4, we obtain ( 0 ) ≠ R K σ ∩ N ∩ M σ ⊆ K σ ∩ ( N ∩ M σ ) . Thus, K σ ∩ ( N ∩ M σ ) ≠ ( 0 ) , and so S is a dominating set in G ( M σ ) . So γ ( G ( M σ ) ) ≤ ∣ S ∣ ≤ ∣ S ∣ = γ ( Γ ( G , R , M ) ) .

(iii) Suppose ω ( Γ ( G , R , M ) ) < ∞ . Let N σ ∈ S ∗ ( M σ ) . Since all elements of β N σ contain N σ , β N σ is a clique in Γ ( G , R , M ) . Hence, ∣ β N σ ∣ ≤ ω ( Γ ( G , R , M ) ) < ∞ . Let C be a clique in G ( M σ ) . Then, ∪ N σ ∈ C β N σ is a clique in Γ ( G , R , M ) . Thus, ∪ N σ ∈ C β N σ is finite, which yields C itself is finite. Therefore, ω ( G ( M σ ) ) < ∞ .

For the converse, suppose that ω ( G ( M σ ) ) < ∞ , and for each N σ ∈ S ∗ ( M σ ) , the set β N σ is finite. Let D be a clique in Γ ( G , R , M ) . Let Λ be the set of all N σ ∈ S ∗ ( M σ ) such that D ∩ β N σ is nonempty. Clearly, the collection { D ∩ β N σ ∣ N σ ∈ Λ } is a partition of D . So D = ∪ N σ ∈ Λ ( D ∩ β N σ ) . We note that, by the assumption for the converse, D ∩ β N σ is finite for all N σ ∈ Λ . Let N σ , K σ ∈ Λ . Then, there are N , K ∈ D such that N σ = N ∩ M σ and K σ = K ∩ M σ . Since D be a clique, N ∩ K ≠ ( 0 ) . In addition, because the grading is σ -faithful, by Lemma 3.3, we obtain that ( 0 ) ≠ ( N ∩ K ) ∩ M σ = N σ ∩ K σ . So Λ is a clique in G ( M σ ) , and hence, it is finite. This implies that D = ∪ N σ ∈ Λ ( D ∩ β N σ ) is finite. We just proved that every clique in Γ ( G , R , M ) is finite. Therefore, ω ( Γ ( G , R , M ) ) < ∞ .□

Corollary 3.6

Let M be a σ -faithful G -graded R -module such that R M σ = M . Then, M σ is a direct sum of two simple R e -modules if and only if M is a direct sum of two G -graded simple R -modules.

Proof

The proof follows directly from Theorems 2.2, 2.4, and Part (i) of Theorem 3.5.□

Lemma 3.7

[23, Fact 2.5] A grading ( R , G ) is first strong if and only if H = supp ( R , G ) ≤ G and ( R , H ) is strong.

Lemma 3.8

Let ( R , G ) be first strong grading and M be a G -graded R -module. If supp ( M , G ) ⊆ supp ( R , G ) , then Γ ( G , R , M ) ≅ G ( M σ ) for all σ ∈ supp ( M , G ) .

Proof

Fix σ ∈ supp ( M , G ) . We claim that if N is G -graded R -submodule of M , then N = R ( N ∩ M σ ) . Let 0 ≠ x ∈ N ∩ M τ for some τ ∈ G . Now σ , τ ∈ supp ( M , G ) ⊆ supp ( R , G ) . Thus, since supp ( R , G ) ≤ G , τ σ − 1 ∈ supp ( R , G ) . So R τ σ − 1 R σ τ − 1 = R e . This implies that 1 = ∑ i = 1 n r i s i for some r i ∈ R τ σ − 1 and s i ∈ R σ τ − 1 . Hence, x = ∑ i = 1 n r i s i x . Since x ∈ M τ and s i ∈ R σ τ − 1 , s i x ∈ R σ τ − 1 M τ ⊆ M σ , for all i . Also s i x ∈ N because N is an R -submodule. So x ∈ R ( N ∩ M σ ) . Hence, N ∩ M τ ⊆ R ( N ∩ M σ ) for all τ ∈ supp ( M , G ) . This implies that R ( N ∩ M σ ) ⊆ N = ⊕ τ ∈ G ( N ∩ M τ ) ⊆ R ( N ∩ M σ ) . Hence, N = R ( N ∩ M σ ) . From the claim, we conclude that M = R ( M ∩ M σ ) = R M σ and N ∩ M σ ≠ { 0 } for all N ∈ h S ∗ ( M ) . Therefore, the correspondence N → N ∩ M σ yields an isomorphism between Γ ( G , R , M ) and G ( M σ ) .□