Group Action on the Set of Nonunits in Rings

Abstract

Let be a ring, be the group of all units of , and . In this paper, we investigate for a ring , where is the set of all vertices of the zero-divisor graph of adjacent to . We also investigate the question on zero-divisor graphs posed in the literature such that when the equality holds in a commutative regular ring with identity. Here, is a nonzero idempotent of which is not the identity element of .

1. Introduction

In 1988, Beck introduced zero-divisor graphs for commutative rings [1]. The modified definition of a zero-divisor graph was given by Anderson and Livingston [2]. Let be a ring with identity , be the set of zero-divisors, and is the set of nonzero zero-divisors of . We define a zero-divisor graph with vertex set such that vertex is adjacent with vertex if and only if . This graph has been studied extensively by several authors like [3, 4]. The notion has been developed for noncommutative rings in [5]. The articles [6, 7] provided similar notions for the commutative semigroups. The research idea of regular group action in rings was introduced by Han (see [8–11]).

Throughout, will denote the set of all nonzero nonunit elements of , is the group of all units of , and is the Jacobson radical of . We consider a group action of on given by from , called the regular action. For each , we define the orbit of as under the given group action and is the set of all vertices of which are adjacent to , for all . In fact, . In [8], the authors showed that if is a local ring, and are a union of orbits under the regular action on by .

The first draft of this work appeared in [12] where, in Section 3, we investigate some properties of regular action in noncommutative rings with as a finite union of orbits under the regular action. In Section 4, we will investigate the question on zero-divisor graphs posed by Han in 2010 [13], that is, for any idempotent element () in a commutative regular ring with identity, when does the equality hold?

2. Preliminary Notes

In this section, we recall some lemmas.

Lemma 1 (see [9]). Let be a ring with identity 1 such that and is a union of orbits under the regular action on by . If there exists such that , then(i)(ii)is a local ring(iii), that is, is a nilpotent ideal of (iv) is one-dimensional vector space over

Lemma 2 (see [9]). If is a ring such that is a union of orbits under the regular action on by , then there exists with if and only if is a local ring and is the set of all right zero-divisors of.

A sufficient condition is given in [13], for a ring to be a division ring.

Lemma 3. Let be a regular ring with identity. If has no nontrivial idempotents, then is a division ring.

In [14], authors proved the following lemma.

Lemma 4. Let be a commutative local ring with identity 1. There are nontrivial rings and such that if and only if there exists a nontrivial idempotent . In this case, one can choose and .

In [15], authors proved the following lemma.

Lemma 5. Let be a local ring such that is a union of distinct orbits under the regular action of on . If, then the set of all ideals of is exactly

3. Regular Action in Rings

In this section, we investigate in commutative rings with identity.

Lemma 6. If there exists such that , then is a local ring.

Proof. Let such that , implies is a unique maximal ideal of . Therefore, and are a local ring.

Theorem 7. Let be ideal of and { is a nonunit} be the subset of . Then, is the group of all units of . Moreover, the action defined by is a regular action of on .

Proof. Let . Then, and such that . So, is the group consisting of all units of .
Let and , then . In addition, and . Thus, acts on .

Theorem 8. Let be a commutative ring with identity and is a union of orbits under the regular action of on . If is ideal of , then there exists , where , such that and , where is a nontrivial left ideal of .

Proof. Let . Since is an ideal of , then . There exists and such that . So, and . Thus, there exists , where , such that . Hence, .

Theorem 9. Let be a commutative ring with identity, , is a nontrivial ideal of and . Then, there exists an integer such that is a union of distinct orbits under the regular action of on.

Proof. Suppose and . Now, we have , , and is nonunit, . On the other hand, for each , there exist , and such that . Hence,Thus, . However, , for each . Consequently, .

Theorem 10. Let be a commutative ring with identity, , is a union of orbits under the regular action of on, then .

Proof. Let . We know that . We show that . The proof is done by induction on . For , let . So, and , where . Suppose . If , then and so there exists an element such that or . Therefore, . This implies, and , where . Thus, is the unique maximal ideal, which implies that is a local ring and (by Lemma 5), which is a contradiction. Thus, .
Now, suppose the assertion holds for and let such that . Without loss of generality, let . If for all , , then is a local ring, and by Lemma 5, , which is a contradiction. Thus, there exists an such that does not contain , where . Assume that does not contain , and . By Lemma 3, there exist an integer such that is a union of orbits under the regular action of on . By induction hypothesis, . Moreover, does not contain and . We consider following two cases.

Case 11. . In this case, . Then, there exists an element and . Therefore, but . So and . On the one hand, implies . Therefore, and . On the other hand, ; hence, . Also, . Thus, does not contain . Consequently, , which is a contradiction.

Case 12. We first let , for every . Then, there exists such that . We show that is a local ring, and is the unique maximal left ideal of . Let be an ideal of such that and be an arbitrary element. Since , there exists an orbit such that . This yields the existence of an element such that . Therefore, and . So, . Therefore, is a local ring and (by Lemma 5).
Secondly, suppose that there is an element such that . Then, (follows similarly as in Case 1) and . Now, let . Then, by the same argument as in Case 1, the result follows.

Theorem 13. Let be a commutative local ring with identity 1 such that is a union of a finite number of orbits under the regular action on by , then .

Proof. Let be a commutative local ring with identity, then there exists an element , such that (by Lemma 2) and (by Lemma 1). Since for every i (i  = 1, 2, …, n) , then . So, we haveMoreover, for all , there exists an orbit , such that . Therefore, there exists such that and thus . Hence, .

Theorem 14. Let be a commutative ring with identity 1 and . If for every , there exists an element such that , and then, is a local ring.

Proof. Since , then there exists an orbit such that . So, there exists an element such that , also . Therefore, and , for every . Thus, is a maximal ideal, and this proves that R is a local ring.

4. When Does the Equality = Hold?

Before stating our main theorem in this section, we need to recall some definitions and lemmas [14].

Definition 15. Let be a finite commutative ring with identity, is called irreducible if it does not contain nontrivial idempotent elements. Therefore, is irreducible if and only if it is not isomorphic to the product of some other nontrivial commutative rings with identity.

Lemma 16. Let be a commutative regular ring with identity. If contains exactly two nontrivial idempotent elements, then contains exactly two orbits.

Proof. Since , then and are two idempotent elements in . We show that . Let , then there exists an element such thatTherefore, . Thus, is an inverse element of which is a contradiction. By Lemma 1 in [8], contains exactly two orbits.

Lemma 17. Let be a commutative regular ring with identity. Then, exactly contains two nontrivial idempotent elements if and only if , where and are fields.

Proof. Let . Then, contains exactly two nontrivial idempotent elements. Conversely, if contains two idempotent elements, we show that . It is clear that (see Lemma 4) and for each orbit which contains , there exist an idempotent element such that (by Lemma 17). Thus, for , there exists an element such that or . Hence,We show that is a commutative regular ring. For , there exists an element such that . In a similar way, is a regular commutative ring. Then, by Lemma 4, it follows that and are fields.

Corollary 18. Let be a regular commutative ring and has two nontrivial idempotent elements. Then, the zero-divisor graph is a complete bipartite graph.

Now, we have the following theorem.

Theorem 19. Let be a regular commutative ring. For any idempotent ∈ , if and only if , where and are fields.

Proof. If , then . Conversely, if is an idempotent element of , by Lemma 4, we get . If be another idempotent element of , then we show that does not contain . If , then . Therefore, . So there is an element such that . Thus, . So implies . Now, there is an element such thatWe conclude . In the same way, we can show that does not contain . Now, suppose that . Then, and . Since is idempotent and by the above argument, and . Therefore, . Then, . Thus, contains exactly two idempotent elements. So, by Lemma 17, .

5. Discussion and Conclusion

In this paper, we investigated the cardinality of the set equal to the cardinality of the set for a ring , where is the set of of which are adjacent to , for all , and is the orbit of under the group action. Furthermore, we discussed some properties of regular action in noncommutative rings with as a finite union of orbits under the regular action. In [12], we investigated the question on zero-divisor graphs, denoted by posed by Han in 2010 [13], when the equality holds for any idempotent element () in a commutative regular ring with identity. The researchers show that there are several notions of zero-divisor graphs for commutative rings in [1], noncommutative rings in [5], and commutative semigroups in [6, 7] which link between algebraic structure and graph theory and motivate others to focus on the same method.

Data Availability

No data were used to support this study.

Disclosure

This article was submitted as a preprint in the link https://assets.researchsquare.com/files/rs-2782623/v1-covered.pdf?c=1681360249.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

The authors contributed equally to this article. They have all read and approved the final manuscript.