Almost graded multiplication and almost graded comultiplication modules

Example 1.1

Consider R=Z[i] and G=Z2. Then R is G-graded by R0=Z and R1=iZ. Now, I=〈1+i〉 is an ideal of R with 1+i∈I. If I is G-graded, then 1∈I, so 1=a(1+i) for some a∈R, i.e., 1=(x+iy)(1+i) for some x,y∈Z. Thus, 1=x−y and 0=x+y, i.e., 2x=1 and hence x=12 a contradiction. So, I is not G-graded.

Lemma 1.2

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

1. If N and K are graded R-submodules of M, thenN+KandN∩Kare graded R-submodules of M.

2. If N is a graded R-submodule of M,r∈h(R), x∈h(M)and I is a graded ideal of R, then Rx, IN and rN are graded R-submodules of M. Moreover,(N:RM)={r∈R:rM⊆N}is a graded ideal of R.

Proposition 1.3

Let M be a G-graded R-module and N a graded R-submodule of M. If I is a graded ideal of R, then(N:MI)={m∈M:Im⊆N}is a graded R-submodule of M.

Proof

Clearly, (N:MI) is an R-submodule of M. Let m∈(N:MI). Then Im⊆N. Now, m=∑g∈Gmg, where mg∈Mg for all g∈G. Let x∈I. Then xg∈I for all g∈G since I is graded. Assume that h∈G. Then xhmg∈Mhg⊆h(M) for all g∈G such that ∑g∈Gxhmg=xh∑g∈Gmg=xhm∈N. Since N is graded, xhmg∈N for all g∈G which implies that ∑h∈Gxhmg∈N for all g∈G, and then xmg∈N for all g∈G. So, Img⊆N for all g∈G, and hence mg∈(N:MI) for all g∈G. Therefore, (N:MI) is a graded R-submodule of M.□

Proposition 1.4

Let M be a G-graded R-module anda∈h(R). Then(0:Ma)={m∈M:am=0}is a graded R-submodule of M.

Definition 2.1

A graded R-module M is said to be almost graded multiplication if whenever a∈h(R) satisfies AnnR(aM)=AnnR(M), then (0:Ma)={0}.

Proposition 2.2

Every graded multiplication module is almost graded multiplication.

Proof

Let M be a graded multiplication R-module. Suppose that a∈h(R) such that AnnR(aM)=AnnR(M). Then (0:Ma) is a graded R-submodule of M, and then there exists a graded ideal I of R such that (0:Ma)=IM. It follows that I⊆AnnR(aM). So, I⊆AnnR(M) and hence (0:Ma)={0}. Thus, M is an almost graded multiplication R-module.□

Example 2.3

Consider R=Z, M=(Z2⊕Z2)[i] and G=Z2. Then R is trivially G-graded by R0=R and R1={0}. Also, M is G-graded by M0=Z2⊕Z2 and M1=i(Z2⊕Z2). Clearly, AnnR(M)=2Z. Let a∈h(R). If a is even, then AnnR(aM)=AnnR({0})=R=Z≠AnnR(M). If a is odd, then AnnR(aM)=2Z=AnnR(M), and (0:Ma)={0}. Hence, M is almost graded multiplication. On the other hand, N=Z2⊕{0} is a graded R-submodule of M such that (N:RM)M={0}≠N, so M is not graded multiplication.

Proposition 2.4

Let M be a graded R-module. If(0:Ma)=((0:Ma):RM)Mfor eacha∈h(R), then M is an almost graded multiplication R-module.

Proof

Let a∈h(R) such that AnnR(aM)=AnnR(M). Then (0:Ma)=((0:Ma):RM)M=AnnR(aM)M=AnnR(M)M={0}. Therefore, M is an almost graded multiplication R-module.□

Proposition 2.5

Let M be a graded R-module. If(Rx:RM)=AnnR(M)impliesRx={0}for everyx∈h(M), then M is an almost graded multiplication R-module.

Proof

Let a∈h(R) such that AnnR(aM)=AnnR(M). Suppose that x∈(0:Ma). Then xg∈(0:Ma) for all g∈G since (0:Ma) is graded. Let g∈G. Then (Rxg:RM)⊆((0:Ma):RM)=AnnR(aM)=AnnR(M). Thus, (Rxg:RM)=AnnR(M) as the reverse inclusion is obvious. Now by assumption, Rxg={0}, which implies that xg=0 for all g∈G as 1∈R. So, x=∑g∈Gxg=0. Hence, (0:Ma)={0}, so M is an almost graded multiplication R-module.□

Proposition 2.6

Let M be a G-graded R-module. If M is graded co-Hopfian, thenWG(M)=ZG(R/AnnR(M))if and only if M is almost graded multiplication.

Proof

Suppose that WG(M)=ZG(R/AnnR(M)). Let a∈h(R) such that AnnR(aM)=AnnR(M). If aM≠M, then a∈WG(M), and then by assumption, there exists b∈h(R)−AnnR(M) such that ab∈AnnR(M). Hence, b∈AnnR(aM)=AnnR(M), which is a contradiction. So, aM=M, and then (0:Ma)={0} as M is graded co-Hopfian. Therefore, M is almost graded multiplication. Conversely, certainly, ZG(R/AnnR(M))⊆WG(M). Let a∈WG(M). Then a∈h(R) such that aM≠M. Now, as M is graded co-Hopfian, (0:Ma)≠{0}. So, by assumption, we can choose b∈AnnR(aM)−AnnR(M). It follows that a∈ZG(R/AnnR(M)), as needed.□

Proposition 2.7

Let M be a G-graded R-module. Ifb∈h(R)−WG(M)such that bM is an almost graded multiplication R-module, then M is an almost graded multiplication R-module.

Proof

Let a∈h(R) such that AnnR(aM)=AnnR(M). Then ab∈h(R) such that AnnR(bM)=AnnR(abM), and then by assumption, we have that (0:bMa)={0}. Let x∈(0:Ma). Then ax=0. Since b∉WG(M), bM=M, and then x=by for some y∈M. Hence, aby=0, which implies that by∈(0:bMa)={0}. Therefore, x=by=0, and hence (0:Ma)={0}. Thus, M is an almost graded multiplication R-module.□

Proposition 2.8

Every graded prime R-module is an almost graded multiplication R-module.

Proof

Let a∈h(R) such that AnnR(aM)=AnnR(M). Suppose that x∈(0:Ma). Then ax=0. Since {0} is a graded prime R-submodule of M, either x=0 or a∈(0:RM). If a∈(0:RM), then aM={0}, and then AnnR(M)=AnnR(aM)=AnnR({0})=R, which means that RM={0}. Again since {0} is a graded prime R-submodule of M, either M={0} or R⊆(0:RM). If R⊆(0:RM), then R=(0:RM) which is impossible since (0:RM) is a graded prime ideal of R by [5, Lemma 2.1]. So, x=0, as needed.□

Remark 2.9

The converse of Proposition 2.8 is not true in general, because if it is true, then by Proposition 2.2, we have that every graded multiplication module is graded prime which is not true, as {0} is a graded multiplication module which is not a graded prime module.

Proposition 2.10

Let M be an almost graded multiplication R-module. If R is an integral domain and M is faithful, then M is a graded prime R-module.

Proof

Let a∈h(R) such that (0:Ma)≠M. Clearly, AnnR(M)⊆AnnR(aM). Now, let b∈AnnR(aM). Then ba∈AnnR(M)={0}. Since R is an integral domain and (0:Ma)≠M, we have that b∈AnnR(M). So, AnnR(aM)⊆AnnR(M). Thus, AnnR(aM)=AnnR(M). Hence, (0:Ma)={0} since M is almost graded multiplication. Therefore, M is a graded prime R-module.□

Proposition 2.11

Let M be an almost graded multiplication R-module. If M is graded second, then M is a graded prime R-module.

Proof

Let a∈h(R). Then by assumption, aM={0} or AnnR(M)=AnnR(M/(0:Ma))=AnnR(aM). Hence, aM={0} or (0:Ma)={0}, as needed.□

Definition 3.1

A graded R-module M is said to be almost graded comultiplication if whenever a∈h(R) satisfies AnnR(aM)=AnnR(M), then aM=M.

Proposition 3.2

Every graded comultiplication module is almost graded comultiplication.

Proof

Let M be a graded comultiplication R-module. Assume that a∈h(R) such that AnnR(aM)=AnnR(M). Now, aM is a graded R-submodule of M, so as M is graded comultiplication, we have that aM=(0:MI) for some graded ideal of R, and then IaM={0}, which means that I⊆AnnR(aM)=AnnR(M). Let x∈M. Then Ix={0}, and then x∈(0:MI)=aM. So, M⊆aM, but clearly, aM⊆M, so aM=M. Hence, M is an almost graded comultiplication R-module.□

Example 3.3

Consider the details of Example 2.3. Similarly, one can prove that M is almost graded comultiplication. On the other hand, N=Z2⊕{0} is a graded R-submodule of M such that (0:MAnnR(N))=M≠N, so M is not graded comultiplication.

Example 3.4

Consider R=Z, M=(Z⊕Z)[i] and G=Z2. Then R is trivially G-graded by R0=R and R1={0}. Also, M is G-graded by M0=Z⊕Z and M1=i(Z⊕Z). Now, for each a∈h(R), AnnR(aM)=AnnR(M)={0} and (0:Ma)={0}, so we have M is an almost graded multiplication R-module. On the other hand, a=2∈h(R) such that aM≠M, which implies that M is not an almost graded comultiplication R-module.

Example 3.5

Consider R=Z, M=(Zp∞⊕Zp∞)[i] and G=Z2. Then R is trivially G-graded by R0=R and R1={0}. Also, M is G-graded by M0=Zp∞⊕Zp∞ and M1=i(Zp∞⊕Zp∞). Now, for each a∈h(R), AnnR(aM)=AnnR(M)={0} and aM=M, so we have M is an almost graded comultiplication R-module. On the other hand, p∈h(R) such that (0:Mp)≠{0}, which implies that M is not an almost graded multiplication R-module.

Proposition 3.6

Let M be a graded R-module. IfaM=(0:MAnnR(aM))for eacha∈h(R), then M is an almost graded comultiplication R-module.

Proof

Let a∈h(R) such that AnnR(aM)=AnnR(M). Then we have aM=(0:MAnnR(aM))=(0:MAnnR(M))=M. Hence, M is an almost graded comultiplication R-module.□

Corollary 3.7

Every graded comultiplication module is almost graded comultiplication.

Proof

Let M be a graded comultiplication R-module. Then as aM is a graded R-submodule of M for all a∈h(R), we have aM=(0:MAnnR(aM)) for each a∈h(R), and then by Proposition 3.6, M is an almost graded comultiplication R-module.□

Remark 3.8

Let N and L be two graded R-submodules of M. To prove that N⊆L, it is enough to prove that if K is a graded completely irreducible R-submodule of M such that L⊆K, then N⊆K.

Proposition 3.9

Let M be a graded R-module. IfAnnR(K)=AnnR(M)implies thatK=Mfor every graded completely irreducible R-submodule K of M, then M is an almost graded comultiplication R-module.

Proof

Let a∈h(R) such that AnnR(aM)=AnnR(M). Let K be a graded completely irreducible R-submodule of M such that aM⊆K. Then AnnR(K)⊆AnnR(aM)=AnnR(M), and then AnnR(K)=AnnR(M). So, by assumption, K=M. Therefore, aM=M by Remark 3.8. Thus, M is an almost graded comultiplication R-module.□

Proposition 3.10

Let M be a graded R-module. Ifb∈h(R)−ZG(M)such that bM is an almost graded comultiplication R-module, then M is an almost graded comultiplication R-module.

Proof

Let a∈h(R) such that AnnR(M)=AnnR(aM). Then AnnR(aM)=AnnR(abM). So, by assumption, bM=abM. Now, let x∈M. Then bx=aby for some y∈M. So, b(x−ay)=0. Since b∉ZG(M), x=ay and so M⊆aM, as required.□

Proposition 3.11

Let M be an almost graded comultiplication R-module. If R is an integral domain and M is faithful, then M is a graded second R-module.

Proof

Let a∈h(R) such that aM≠{0}. Since R is an integral domain and M is faithful, we have AnnR(aM)=AnnR(Ra)=AnnR(M)={0}, which implies that aM=M as M is almost graded comultiplication.□

Proposition 3.12

Let M be an almost graded comultiplication R-module. If M is graded prime, then M is a graded second R-module.

Proof

Let a∈h(R) such that aM≠{0}. Then AnnR(M)=AnnR(aM), and then M=aM, as needed.□