On Properties of Graded Rings with respect to Group Homomorphisms

Abstract

Let be a group and be a -graded ring with non-zero unity. The goal of our article is reconsidering some well-known concepts on graded rings using a group homomorphism . Next is to examine the new concepts compared to the known concepts. For example, it is known that is weak if whenever such that , then . In this article, we also introduce the concept of -weakly graded rings, where is said to be -weak whenever such that , and . Note that if is abelian, then the concepts of weakly and -weakly graded rings coincide with respect to the group homomorphism . We introduce an example of non-weakly graded ring that is -weak for some . Similarly, we establish and examine the concepts of -non-degenerate, -regular, -strongly, -first strongly graded rings, and -weakly crossed product.

1. Introduction

Let be a group and be a ring with unity 1. If additive subgroups of such that and exist for all , then is said to be -graded. A -graded ring is denoted by . The elements of are called homogeneous of degree and . The identity component of is a subring of with . For , it can be written uniquely as where is the component of in and except finitely many.

The set of all homogeneous elements of is and is denoted by . The support of is defined by . Let be a graded ring and let be an ideal of . If , i.e., if whenever , then for all . Then is said to be a graded ideal. An ideal of a graded ring is not necessarily graded (see [1, 2]).

The concepts of faithful, non-degenerate, regular, and strongly graded rings have been introduced and examined in [2]; is said to be faithful if for all , , we have and . It is clear if is faithful, then . is said to be non-degenerate if for all , , we have and . Otherwise, is said to be degenerate. is said to be regular if for all , , we have . Evidently, every regular graded ring is considered non-degenerate graded, and every faithful graded ring is said to be non-degenerate. is said to be strong if for all . Undoubtedly, every strongly graded ring is faithful. Firstly, strongly graded rings have been introduced and investigated in [3]; is said to be first strongly if for all . Therefore, every strongly graded ring is first strong. We can certainly say that if is strong, then . On the other hand, if is first strong, then is a subgroup of . Surely, is first strong if and only if is a subgroup of and for all . Additionally, in [3], the concept of second strongly graded rings was introduced; is second strong if is a monoid in and for all . Apparently, every strongly graded ring is second strong. Furthermore, every first strongly graded ring is second strong. In fact, if is second strong and is a subgroup of , then is first strong. Moreover, is first strong if and only if is second strong and non-degenerate. Furthermore, is strong if and only if is second strong and faithful. In [4], the concept of weakly graded rings was proposed and investigated; is said to be weak if with , and . Every non-degenerate graded ring is weak, and if is a subgroup of , then is weak. Therefore, we say that every first strongly graded ring is weak, and every strongly graded ring is weak. On the other hand, a second strongly graded ring is not necessarily weak, and a weakly graded ring is not necessarily second strong (see [4]). Following [2], is said to be a crossed product if contains a unit for all . On the other hand, the concept of weakly crossed product (crossed product over the support) has been proposed in [5] and was investigated in [4] where is said to be a weakly crossed product if contains a unit for all . It is clear that if is a crossed product, then is a weakly crossed product. However, the converse is not necessarily true. If , then the trivial graduation of ( and otherwise) is a weakly crossed product, but it is not a crossed product since for , does not contain units. For more terminology, see [6–9].

In this article, we also reconsider some of the aforementioned concepts using a group homomorphism and examine the new concepts compared to the aforementioned concepts. In Section 2, we introduce the concepts of -non-degenerate and -regular graded rings and examine them in comparison to non-degenerate and regular graded rings. is said to be -non-degenerate. If for all , , we have and . Otherwise, is said to be -degenerate. Clearly, if is abelian, then the concepts of non-degenerate and -non-degenerate graded rings coincide regarding the group homomorphism . However, Example 1 introduces a degenerate graded ring that is -non-degenerate for some . In Section 3, we establish the concepts of -strongly and -first strongly graded rings and study them in comparison to strongly and first strongly graded rings. If , for all , then is said to be -strong. It is clear that every strongly graded ring is -strong, for every . However, if is -strong, then is not necessarily strong, as we see in Example 3 where is said to be -first strong if , for all . Clearly, every -strongly graded ring is -first strong. However, the converse is not necessarily true, as we see in Example 4.

Additionally, if is first strong and , then is -first strong. However, if is -first strong, then is not necessarily first strong, even if , as mentioned in Example 3; is -first strong since it is -strong, and , but is not first strong since is not a subgroup of . In Section 4, we present the concepts of -weakly graded rings and -weakly crossed products and study them in comparison to weakly graded rings and weakly crossed products. is said to be -weak if whenever such that , then . It is clear that if is abelian, then the concepts of weakly and -weakly graded rings coincide regarding the group homomorphism . However, Example 5 is a non-weakly graded ring that is -weak for some . is said to be -weakly crossed product if contains a unit for all . The concepts of weakly crossed product and -weakly crossed product coincide with respect to the identity homomorphism. However, Example 8 is on non-weakly crossed product that is -weakly crossed product for some .

2. -Non-Degenerate and -Regular Graded Rings

In this section, we introduce the concepts of -non-degenerate and -regular graded rings and examine them compared to non-degenerate and regular graded rings.

Definition 1. Let be a -graded ring and be a group homomorphism. Then,(1) is added to be -non-degenerate if for all , , we have and . Otherwise, is said to be -degenerate.(2) is expressed to be -regular if for all , , we have .Clearly, if is abelian, then the concepts of non-degenerate and -non-degenerate (regular and -regular) graded rings coincide regarding the group homomorphism . However, the next example introduces a degenerate graded ring that is -non-degenerate for some .

Example 1. Let , where is a field, and . Then is -graded by , and . Consider the group homomorphism such that . Let and , ; then , and , for some . Since , , and is also a non-negative integer, , which implies that as . Similarly, . Hence, is -non-degenerate. On the other hand, is degenerate since with .

Remark 2. Clearly, every faithful graded ring is -non-degenerate, for every . However, if is -non-degenerate, then is not necessarily faithful as in Example 1, where is -non-degenerate, but is not faithful since .

Theorem 3. Every -regular graded ring is -non-degenerate.

Proof. Let be -regular. Suppose that and . This gives us as is -regular, which implies that and , and therefore is -non-degenerate.

Remark 4. If is -non-degenerate, then is not necessarily -regular as in Example 1, is -non-degenerate, and is not -regular since with .

Theorem 5. Let be -non-degenerate such that . If , for some , then .

Proof. Assume that . Then there is ; therefore, since is -non-degenerate, , which implies that , which is a contradiction. Hence, .

Theorem 6. Let be -non-degenerate such that . Suppose that . If will be a graded right (left) ideal of with , then , for all .

Proof. Let . Then . So, . Let . Then . If , then , and since is -non-degenerate, , which is a contradiction. So, , and therefore .

Corollary 7. Let be -non-degenerate such that . Suppose that , for all . If is a graded right (left) ideal of with , then , for all .

Proof. Let . Then , and so , which implies that , where , so , and hence the result holds by Theorem 6.

Corollary 8. Let be -non-degenerate, where is the identity homomorphism. Suppose that , for all . If will be a graded right (left) ideal of with , then .

Proof. By Corollary 7, , for all , which implies that .
The next example shows that the condition “ is -non-degenerate regarding the identity homomorphism” in Corollary 8 is necessary.

Example 2. Consider , where is a field, and that satisfies , for all .
Then is -graded by and .
Now, is a graded left ideal of that satisfiesNote that is -degenerate with respect to the identity homomorphism since with .

Theorem 9. Let be -non-degenerate, where is the identity homomorphism. Suppose that , for all . If is a field, then is graded simple.

Proof. Let be a graded ideal of . If , then by Corollary 8, . Suppose that . Then is a non-zero ideal of , and then . Assume that . Then . So, , for all , which implies that , and hence . Thus, is graded simple.

Corollary 10. Let be -non-degenerate, where is the identity homomorphism. Suppose that , for all . If is a field, then is graded Artinian and graded Noetherian.

Theorem 11. Consider , where is a field. If is -graded such that , then either is not -non-degenerate regarding the identity homomorphism or there is such that .

Proof. Since , is a commutative ring with unity. Let . Then , and then is unit in , and hence by ([4], Lemma 3.4). Hence, is a subfield of . Let be a non-constant homogeneous element of . Then is a graded ideal of with , but for all ; let and . Then for some and and then . Suppose that for some . Then for some and then, since is an integral domain, , a contradiction since and is non-constant. Therefore, there is no with for all , i.e., is not Artinian. Hence, by Corollary 10, either is not -non-degenerate regarding the identity homomorphism or there exists such that .

Remark 12. In Example 1, is graded where is a field, and one can prove that is -non-degenerate regarding the identity homomorphism. Absolutely, , for all .

Theorem 13. Let be a graded ring such that has no zero divisors. Suppose that is a group homomorphism and . If will be a graded right (or left) ideal of with , then , for all .

Proof. Let . Then . So, , and then either or . If , then . Hence, , for all .

Proposition 14. If is -non-degenerate, then .

Proof. Let . So there exists such that , and then there exists , and so , which implies that , and hence .

Remark 15. Example 2 shows that the converse of Proposition 14 is not necessarily true as , but is -degenerate.

Theorem 16. Let be a graded ring such that has no zero divisors and be a group homomorphism. Thus, if and only if is -non-degenerate.

Proof. Suppose that . Let and . Then , and so , which implies that and . Hence, is -non-degenerate. The converse holds by Proposition 14.
Let and be two -graded rings. Then a ring homomorphism is said to be a graded homomorphism if , for all [2]. Note that if is a ring homomorphism and , then . So, being graded homomorphism is equivalent to for all . We have the following.

Lemma 17. If is a graded epimorphism, then for all .

Proof. Let . Then as is a graded homomorphism. Let . If , then . Suppose that . Since is onto, there exists such that . Suppose that , where , for . Then , where and for all . Since , . Thus, and hence and . Thus, , and hence .

Theorem 18. Let be a graded isomorphism. Then is -non-degenerate if and only if is -non-degenerate.

Proof. Suppose that is -non-degenerate. Let and . By Lemma 17, . So, there exists such that . Since is -non-degenerate, , and then . Similarly, , and hence is -non-degenerate. Conversely, let and . Assume that . Then , and then , which implies that . Similarly, , and hence is -non-degenerate.

3. -Strongly and -First Strongly Graded Rings

In this section, we establish the concepts of -strongly and -first strongly graded rings and study them in comparison to strongly and first strongly graded rings.

Definition 19. Let be a -graded ring and be a group homomorphism. Therefore, is said to be -strong if , for all .
It is clear that every strongly graded ring is -strong, for every . However, if is -strong, then is not necessarily strong, as we see in the following example.

Example 3. Let , where is a field, and . Then is -graded by , and otherwise. Consider the group homomorphism such that . Let be a non-negative integer. Then . Let be a negative integer. Then . So, , for all , and hence is -strong. On the other hand, is not considered strong since .

Proposition 20. Let be a graded ring. Then is strongly graded if and only if , for all .

Proof. Suppose that , for all . Let . Then , and . Hence, is strong. The converse is clear.
The next result is a consequence of Proposition 20.

Corollary 21. If is abelian, then strongly graded and -strongly graded rings coincide regarding the group homomorphism .

Theorem 22. Let be a -graded ring and be a group homomorphism. If , for all , then is strong, and hence is -strong. Moreover, is abelian.

Proof. Let . Then , and then , which implies that . So, . Now, let . Then . Thus, , which implies that , and thus is strong by Proposition 20, and hence is -strong. Moreover, since , for all and is homomorphism, then is abelian.

Definition 23. Let be a -graded ring and be a group homomorphism. Then is said to be -first strong if , for all .
Clearly, every -strongly graded ring is -first strong. However, the converse is not necessarily true, as we see in the following example.

Example 4. Let , where is a field, and and consider the identity homomorphism on . Then is -graded byBy easy calculations, we find that and . So, is -first strong. On the other hand, is not -strong since .

Remark 24. Clearly, if is a first strong and is a group homomorphism with , then is -first strong. However, if is -first strong, then is not necessarily first strong, even if , as in Example 2. is -first strong since it is -strong, and , but is not first strong since is not a subgroup of .

Proposition 25. Let be first strong. If is -non-degenerate, then is -first strong.

Proof. Apply Proposition 14 and Remark 24.

Proposition 26. Let be a graded ring. Then is first strong if and only if , for all .

Proof. Suppose that , for all . Then as , , for all , and hence is first strong. The converse is clear.
The next result is a consequence of Proposition 26.

Corollary 27. If is abelian, then the first strongly graded and -first strongly graded rings coincide regarding the group homomorphism .

Theorem 28. Let be a -graded ring and be a group homomorphism. If , for all , then is first strong, and hence is -first strong. Furthermore, is an abelian subgroup of .

Proof. Let . Then , and then , which suggests that . So, , and thus is first strong. So, is a subgroup of , and then , which implies that is -first strong by Remark 24. Since , for all and is homomorphism, then will be an abelian subgroup of .

Theorem 29. Let be a graded ring such that is an integral domain. Then , for all .

Proof. Let . Then , and then, since is commutative, . So, , which implies that .

Theorem 30. Let be a graded ring over a finite group . If is second strong, then is -first strong, for every .

Proof. Let be a group homomorphism. Assume that . Since is finite, , for some positive integer , and then since is a monoid, and hence is a subgroup of , and then since is second strong, is first strong. Since is finite, , for some positive integer , and then .
Hence, , which implies that is -first strong by Remark 24.

4. -Weakly Graded Rings and -Weakly Crossed Products

In this section, we present the concepts of -weakly graded rings and -weakly crossed products and study them in comparison to weakly graded rings and weakly crossed products.

Definition 31. Let be a -graded ring and be a group homomorphism. Then is said to be -weak if whenever such that , then .
Clearly, if is abelian, then the concepts of weakly and -weakly graded rings coincide regarding the group homomorphism . However, the following example is on non-weakly graded ring that is -weakly for some .

Example 5. Let and . Then is -graded by , and . Otherwise, consider the group homomorphism such that . Let be such that . Then is a negative number, and then is also a negative number, which implies that . Hence, is -weak. On the other hand, is not weak since but .

Theorem 32. Let be -non-degenerate such that . So is -weak.

Proof. Apply Theorem 5.
The later result is a consequence of Theorem 32.

Corollary 33. Let be a graded ring and be a group homomorphism such that . If is faithful, then is -weak.

The following result is a consequence of Theorems 3 and 32.

Corollary 34. Let be a graded ring and be a group homomorphism such that . If is -regular, is -weak.

Theorem 35. Let be a graded such that has no zero divisors and be a group homomorphism such that . Then is -non-degenerate if and only if is -weak.

Proof. Suppose that is -weak. Let and . Then . If , then as is -weak, , which is a contradiction. So, , and thus and , and hence is -non-degenerate. The converse holds from Theorem 32.

Theorem 36. Let be a graded ring and be a group homomorphism such that . Then if and only if is -weak.

Proof. Suppose that . Let such that . If , , and then , which is a contradiction. So, , and therefore is -weak. Conversely, let . Then there is such that . If , then , which is a contradiction. So, , and then , and hence .

Proposition 37. Every strongly graded ring is -weak, for every .

Proof. Let be strong and be a group homomorphism. Then , and hence is -weak.
Similarly, one can prove the following.

Proposition 38. Every faithful graded ring is -weak, for every .

The next example shows that an -weakly graded ring is not necessarily strongly graded or faithful graded or even first strongly graded.

Example 6. Let and . Then is -graded byClearly, is -weak regarding the identity homomorphism, but is not strong and not faithful since . Also, is not first strong since is not a subgroup of .
Also, the next example is on a non-strongly graded ring that is -weakly graded, for every .

Example 7. Let and . Then is -graded by , and . Since , is -weak, for every . On the other hand, is not strong since as such that .

Definition 39. Let be a -graded ring and be a group homomorphism. Then is said to be -weakly crossed product if contains a unit for all .
Clearly, the concepts of weakly crossed product and -weakly crossed product coincide regarding the identity homomorphism. However, the following example is a non-weakly crossed product that is -weakly crossed product for some .

Example 8. Let and . Then is -graded by , and otherwise. Consider the group homomorphism such that , for all . Let . Then contains a unit. Hence, is -weakly crossed product. On the other hand, is not weakly crossed product since with does not contain any unit.

Proposition 40. Let be a -graded ring and be the group homomorphism , for all . Then is -weakly crossed product.

Proof. Let . Then contains a unit. Hence, is -weakly crossed product.

Proposition 41. Let be -weakly crossed product. Then .

Proof. Let . Then there is such that . Since is -weakly crossed product, contains a unit, and then , which implies that . Thus, .

Remark 42. The converse of Proposition 41 is not necessarily true; as in Example 6, is not -weakly crossed product regarding the identity homomorphism since with does not contain any unit, but .

Theorem 43. Let be a graded ring such that has no zero divisors. If is -weakly crossed product, then is -non-degenerate.

Proof. Let and . Then , and then contains a unit, which implies that . So, and . Hence, is -non-degenerate.
The next example introduces a -non-degenerate graded ring that is not a -weakly crossed product.

Example 9. Let , where is a field, and . Then is -graded byOtherwise, one can prove that is -non-degenerate regarding the group homomorphism , but is not -weakly crossed product since with does not contain any unit.

Theorem 44. If is -weakly crossed product and , then is -weak.

Proof. Let such that . If , then , and then contains a unit, which is a contradiction. So, , and hence is -weak.
The next example is on -weakly graded ring that is not -weakly crossed product.

Example 10. Let , where is a field, and . Then is -graded byClearly, is -weak, for every , but is not -weakly crossed product regarding the identity homomorphism since with does not contain any unit.

Proposition 45. If is -weakly crossed product, then for all , there exists a unit such that (i.e., is a cyclic -module).

Proof. Let . Then contains a unit, say . Let . Then . So, . Hence, .
Similarly, one can prove the following.

Proposition 46. If is -weakly crossed product, then for all , there exists a unit such that .

Theorem 47. If is -weakly crossed product, then is isomorphic to , as an -module, for all .

Proof. Let . Then by Proposition 46, , for some unit , and then such that is an -isomorphism.

5. Conclusion

Let be a group, be a group homomorphism, and be a -graded ring. In this article, we established and investigated the perceptions of -weakly, -non-degenerate, -regular, -strongly, -first strongly graded rings, and -weakly crossed product. We have indeed examined these notions comparable to weakly, non-degenerate, regular, strongly, first strongly graded rings, and weakly crossed product.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This study was supported by the Researchers Supporting Project (PNURSP2023R183), Princess Nourah bint Abdulrahman University, P.O. Box 84428, Riyadh 11671, Saudi Arabia.