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Ring graph induced by a ring endomorphism

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Published/Copyright: October 14, 2025

Abstract

Given a commutative ring R with unity, and an endomorphism f on R , the graph G ( R , f ) assigned to R is a simple graph with vertex set R , and two distinct vertices r and s are adjacent if r f ( s ) = 0 or s f ( r ) = 0 . This graph generalizes Beck’s zero-divisor graph G ( R ) . This extension enables a deeper insight into both the algebraic structure and graph-theoretic properties. We explore the properties of G ( R , f ) , such as connectivity, completeness, and cycles. We also determine the values for the diameter, girth, independence number, clique number, chromatic number, and domination number, sometimes under algebraic conditions on R or f . The study also reveals how algebraic properties of R and f relate to the graph-theoretic properties of G ( R , f ) . Applications to the ring Z n are provided with illustrative examples. The ties between G ( R , f ) and Beck’s graph G ( R ) are also presented offering new methods to tackle problems in the study of zero-divisor graphs.

MSC 2020: 05C07; 05C25; 05C38; 05C40; 05C69

1 Introduction

Algebra and graph theory are two key areas of mathematics, each with its own focus and methods, but they come together in interesting and useful ways. This connection happens when algebra helps us understand graphs, or when graphs help us solve problems in algebra. This crossover has led to the creation of “algebraic graph theory,” where algebraic ideas are used to study graphs, and graph-based methods are applied to explore algebraic structures. By combining the two, we gain new ideas and ways to better understand both fields and how they can be applied.

One of the most important co-study between graphs and algebraic structures is the zero-divisor graph of commutative rings. The concept of zero-divisor graphs was first introduced by Beck [1]. Beck’s interest was in the computation of the chromatic number of the graph by using the algebraic properties of the commutative ring. Following Beck’s introduction, the concept of zero-divisor graphs was further refined [2], expanded [3,4], and generalized to other algebraic structures as one can see for example in [5,6].

Building on the extensive work on generalizing the zero-divisor graphs, we introduce an interesting generalization of Beck’s zero-divisor graph of commutative rings. This graph is defined by taking into account the action of a ring endomorphism on the ring. More precisely, given a commutative ring R and a ring endomorphism f on R , our graph G ( R , f ) is a simple graph with vertex set R and two distinct vertices r and s are adjacent if r f ( s ) = 0 or s f ( r ) = 0 . This graph generalizes Beck’s zero-divisor graph G ( R ) since G ( R , i d ) = G ( R ) , where i d denotes the identity ring endomorphism on R . Throughout the study, we highlight the relationship between G ( R , f ) and G ( R ) . Also, we study the interaction between the algebraic properties of R and the graph properties of G ( R , f ) . We also provide formulas for graph characteristics like the clique number, independence number, chromatic number, girth, and so on. The results in this study can help solve coloring and optimization problems, as demonstrated in [7].

In Section 2, we cover the basics needed for this study. In Section 3, we explore and analyze G ( R , f ) , looking at aspects like connectivity, completeness, bipartiteness, and cycles. We also determine specific values or provide rules for calculating the diameter, girth, independence number, clique number, chromatic number, domination number, and vertex degree, sometimes with restrictions on the ring or the endomporphism. In Section 4, we provide examples and applications of the results obtained in Section 3 to the ring Z n .

Numerous additional properties of this new graph remain to be investigated. These aspects are left for future research and interested scholars.

2 Background

This section provides a review of the fundamental concepts related to rings and graphs. All the results presented here are borrowed from [8,9]. Throughout the study, we assume that R is a non-zero commutative ring with unity 1 0 . The set of units of R is denoted by U ( R ) . We begin by outlining some preliminaries from Ring theory.

Definition 2.1

A proper ideal I of a ring R is said to be maximal if I is not contained in another proper ideal of R .

Definition 2.2

A proper ideal I of R is prime if whenever x y I , then x I or y I .

Definition 2.3

Given an ideal I of R , the annihilator of I in R is defined to be the set Ann R ( I ) = { r R : r I = 0 } .

The set Ann R ( I ) is an ideal of R . Also, if I and J are ideals of R such that I J , then Ann R ( I ) Ann R ( J ) .

Definition 2.4

Let f be a ring endomorphism on R . We define Fix ( f ) to be the set of all fixed points of f , i.e., Fix ( f ) = { r R : f ( r ) = r } .

The set Fix ( f ) is a subring of R .

Next we present some background of graph theory concerning undirected graphs. In this section, G denotes an undirected graph. The number of vertices in G is referred to as the order of the graph. The set of vertices in G is denoted by V ( G ) . If two vertices x and y are adjacent, we write x y .

Definition 2.5

Let v be a vertex in G . The neighborhood N ( v ) of v is the set of all vertices adjacent to v .

If G is a simple undirected graph, then v N ( v ) . If N ( v ) = , then v is said to be an isolated vertex.

Definition 2.6

The degree of a vertex v of G is the number of edges incident to v . The degree of v is denoted by d e g G ( v ) (or deg ( v ) if there is no confusion with the underlined graph).

The minimum vertex degree is denoted by δ ( G ) , while the maximum vertex degree (called the degree of the graph) is denoted by Δ ( G ) . When G is a simple graph, then deg ( v ) = N ( v ) , where means the cardinality. Hence, v is isolated if and only if deg ( v ) = 0 .

Definition 2.7

Let v and u be two vertices of G . A path from u to v is a sequence of edges joined end to end such that u is the first vertex and v is the final vertex. If u = v , the path is called a closed path. A simple path is a path whose edges are not repeated. A simple closed path whose vertices are not repeated except for the initial vertex is called a cycle. The distance d ( u , v ) between v and u is the length of a shortest path between them. The diameter of G , denoted by diam ( G ) , is defined to be the supremum of the set { d ( u , v ) : u , v V ( G ) } .

Definition 2.8

A graph G is connected if there is a path between any two distinct vertices of G .

Definition 2.9

A tree is a connected simple graph with no cycles inside it.

Definition 2.10

By the girth of G , we mean the length of a shortest cycle in G . The girth of G is denoted by g ( G ) . If G has no cycles, then we write g ( G ) = .

Definition 2.11

A graph is said to be complete if it is a simple graph and every two distinct vertices are adjacent. The complete graph on n vertices is denoted by K n .

Definition 2.12

A subgraph of G , which is a complete graph, is called a clique of G . The order of a clique with the largest number of vertices is called the clique number of G and it is denoted by ω ( G ) .

Definition 2.13

A dominating set D of G is a nonempty subset of V ( G ) such that each vertex of G is either in D or adjacent to a vertex in D . The infimum of the set { D : D is a dominating set of G } is called the domination number of G and is denoted by γ ( G ) .

Definition 2.14

A subset S of vertices of a graph G is called independent if no two vertices in S are adjacent. The cardinality of largest independent set is called the independence number of G and is denoted by α ( G ) .

Definition 2.15

A simple graph G is called bipartite if we can partition V ( G ) into two disjoint nonempty independent subsets (each subset is called a part). A complete bipartite graph is a bipartite graph where each vertex in one part is adjacent to each vertex in the other part. A complete bipartite graph is denoted by K m , n or K n , m , where m is the cardinality of one part and n is the cardinality of the other part. A K 1 , m graph is called a star graph.

Definition 2.16

The chromatic number χ ( G ) of a graph G is the minimal number of colors needed to color the vertices in such a way that no two adjacent vertices have the same color. Mathematically, this is equivalent to say the minimum number of independent sets that can partition the graph.

For example, χ ( K n ) = n , while if G is bipartite or C 2 n , then χ ( G ) = 2 . However χ ( C 2 n + 1 ) = 3 .

Notation 2.17

If H V ( G ) , then G H represents the subgraph of G whose vertices are V ( G ) H and whose edges are those edges in G connecting them.

In the next definition, we remind the reader with Beck’s zero divisor graph, which we generalize in this study.

Definition 2.18

Two simple graphs G 1 and G 2 are said to be isomorphic if there exists a bijection f : V ( G 1 ) V ( G 2 ) such that if { u , v } is an edge in G 1 , then { f ( u ) , f ( v ) } is an edge in G 2 . Such a bijection f is called a graph isomorphism.

Definition 2.19

[1] Let R be a commutative ring with unity 1. The “Beck” zero divisor graph of R , denoted by G ( R ) , is an undirected graph whose vertex set is R , and two vertices r and s are adjacent if r s = 0 .

3 Ring graph induced by a ring endomorphism

We introduce in this section the graph G ( R , f ) induced by a ring endomorphism on a commutative ring R . Then, we investigate many aspects of this graph through several graphic notions and properties, as well as compute various graphic values of this graph.

Definition 3.1

Let R be a commutative ring with unity, and f : R R be a ring homomorphism. The simple graph G ( R , f ) induced by f is the graph whose vertices are the elements of R and two distinct vertices r and s are adjacent if r f ( s ) = 0 or s f ( r ) = 0 .

One can see that if f is the identity isomorphism on R , then G ( R , f ) = G ( R ) , where G ( R ) stands for Beck’s zero-divisor graph.

Example 3.2

Let R = Z 2 Z 2 . We consider all possible endomorphisms on R .

  1. If f ( m , n ) = ( n , m ) , then G ( R , f ) is

  2. which is a claw graph.

  3. If f ( m , n ) = ( m , n ) , then G ( R , f ) = G ( R ) as shown in the figure

  4. If f = 0 , then G ( R , f ) is

  5. which is a complete graph.

  6. If f ( m , n ) = ( 0 , n ) then G ( R , f ) is

  7. If f ( m , n ) = ( m , 0 ) then G ( R , f ) is

Now, we present some facts about G ( R , f ) .

Proposition 3.3

Let R be a commutative ring with unity, and f a ring isomorphism on R. Then, G ( R , f ) = G ( R , f 1 ) .

Proof

Let r and s be two distinct elements of R . Then, r s in G ( R , f ) r f ( s ) = 0 or s f ( r ) = 0 f 1 ( r ) s = 0 or f 1 ( s ) r = 0 r s in G ( R , f 1 ) .□

Proposition 3.4

Let R and T be commutative rings with unity, and f and h be ring endomorphisms on R and T, respectively. If μ : R T is a ring isomorphism such that the diagram

commutes, then μ is a graph isomorphism and therefore, G ( R , f ) and G ( T , h ) are isomorphic graphs.

Proof

Let r , s R . Then,

r s in G ( R , f ) s f ( r ) = 0 or r f ( s ) = 0 μ ( s f ( r ) ) = 0 or μ ( r f ( s ) ) = 0 μ ( s ) μ ( f ( r ) ) = 0 or μ ( r ) μ ( f ( s ) ) = 0 μ ( s ) h ( μ ( r ) ) = 0 or μ ( r ) h ( μ ( s ) ) = 0 μ ( r ) μ ( s ) in G ( T , h ) .

The proof of the following corollary follows easily from Proposition 3.4.

Corollary 3.5

Let R and T be commutative rings with unity, and f and h be ring endomorphisms on R and T, respectively. If μ : R T is a ring isomorphism, then G ( R , f ) is isomorphic to G ( T , μ f μ 1 ) .

Corollary 3.6

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then, G ( R ker ( f ) , f * ) G ( I m f , f ) , where f * is the natural homomorphism induced by f on R ker ( f ) .

Proof

The proof comes from Proposition 3.4 by observing the commutativity of the diagram

where μ is the isomorphism, which maps r + ker ( f ) to f ( r ) , for every r R .□

Remark 3.7

It should be noted that not every ring isomorphism is a graph isomorphism. As illustrated in Example 3.2, the identity isomorphism on Z 2 Z 2 is not a graph isomorphism between G ( R ) and G ( R , f ) , where f is the ring endomorphism on Z 2 Z 2 defined by f ( m , n ) = ( n , m ) . So, the commutativity of the diagram included in Proposition 3.4 is necessary to ensure that the ring isomorphism turns out to be a graph isomorphism.

The next lemma has a great impact in this study.

Lemma 3.8

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then,

  1. k r for every k ker ( f ) and every r R such that k r .

  2. If r s , then α r β s for every α , β , r , s R is provided that α r β s .

  3. If r s , then f m ( r ) f m ( s ) for every m N is provided that f m ( r ) f m ( s ) .

  4. If f ( 1 ) = 1 , u U ( R ) , and r R such that r u , then u r if and only if r ker ( f ) .

  5. If r Fix ( f ) and s R such that r s , then r s if and only if r s ker ( f ) . Moreover, if r , s Fix ( f ) such that r s , then r s if and only if r s = 0 . That is, G ( Fix ( f ) , f Fix ( f ) ) = G ( Fix ( f ) ) , where f Fix ( f ) denotes the restriction of f to Fix ( f ) .

Proof

  1. Let r R and k ker ( f ) such that r k . Since f ( k ) = 0 , we have r f ( k ) = 0 and hence r k .

  2. Let r , s R such that r s . If r s , then

    r f ( s ) = 0 or s f ( r ) = 0 α f ( β ) r f ( s ) = 0 or f ( α ) β s f ( r ) = 0 α r f ( β ) f ( s ) = 0 or β s f ( α ) f ( r ) = 0 α r f ( β s ) = 0 or β s f ( α r ) = 0 α r β s ,

    provided that α r β s .

  3. Let r , s R such that r s and m N . If r s , then

    r f ( s ) = 0 or s f ( r ) = 0 f m ( r f ( s ) ) = 0 or f m ( s f ( r ) ) = 0 f m ( r ) f m + 1 ( s ) = 0 or f m ( s ) f m + 1 ( r ) = 0 f m ( r ) f ( f m ( s ) ) = 0 or f m ( s ) f ( f m ( r ) ) = 0 f m ( r ) f m ( s ) ,

    provided that f m ( r ) f m ( s ) .

  4. Assume f ( 1 ) = 1 . Let u U ( R ) and r R such that u r . Then, f ( u ) U ( R ) . Since f 0 , and hence ker ( f ) is a proper ideal of R , we obtain

    u r u f ( r ) = 0 or r f ( u ) = 0 f ( r ) = 0 or r = 0 r ker ( f ) .

  5. Assume r Fix ( f ) and s R such that r s . Then,

r s r f ( s ) = 0 or s f ( r ) = 0 f ( r ) f ( s ) = 0 or s r = 0 f ( r s ) = 0 or r s = 0 r s ker ( f ) . Now, if both r , s Fix ( f ) , then

r s in G ( Fix ( f ) , f Fix ( f ) ) r f ( s ) = 0 or s f ( r ) = 0 r s = 0 or s r = 0 r s = 0 r s in G ( Fix ( f ) ) .□

3.1 Completeness, connectivity, and girth

We begin our study in this subsection by investigating the completeness of G ( R , f ) .

Lemma 3.9

Let p be a prime number, and f a ring endomorphism on Z p . Then,

G ( Z p , f ) = K p , if f = 0 K 1 , p , if f 0 .

Proof

If f = 0 , then ker ( f ) = Z p . Lemma 3.8 implies the completeness of G ( Z p , f ) . If f 0 , then ker ( f ) = 0 and hence, f is an isomorphism. Since Z p is a field, we obtain that r s for every distinct elements r , s R 0 . We conclude that G ( Z p , f ) is a star graph with center 0.□

Theorem 3.10

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then, G ( R , f ) is complete if and only if f = 0 .

Proof

If R = Z p , then the result follows by Lemma 3.8. So, assume R Z p . Suppose G ( R , f ) is complete. Then, 1 r for every r R { 0,1 } . So, r f ( 1 ) = 0 or 1 f ( r ) = 0 . Since Z p is a field, we obtain r = 0 or f ( r ) = 0 . But r 0 , therefore f ( r ) = 0 . Thus, ker ( f ) = R { 1 } . Now let r R { 0,1 } . We have f ( 1 ) = f ( 1 + r r ) = f ( 1 r ) + f ( r ) = 0 + 0 = 0 . So, 1 ker ( f ) and hence ker ( f ) = R , which means that f = 0 . For the converse, assume f = 0 . Then, ker ( f ) = R . By Lemma 3.8, G ( R , f ) is a complete graph.□

The reader should note that when f 0 , then the subsets ker ( f ) , U ( R ) , and R ( ker ( f ) U ( R ) ) form a partition of R . This fact leads to the following theorem.

Theorem 3.11

Let R be a commutative ring with unity, f 0 a ring endomorphism on R, and r R . Then,

  1. deg ( r ) = R 1 , if r ker ( f ) .

  2. deg ( r ) = ker ( f ) , if r U ( R ) .

  3. ker ( f ) deg ( r ) R U ( R ) , if r ker ( f ) U ( R ) .

Proof

The proof follows from Lemma 3.8.□

Theorem 3.12

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then, G ( R , f ) is connected with diameter at most 2.

Proof

Let r and s be distinct elements of R . If r s , then d ( r , s ) = 1 . If r s , then we have the path r 0 s of length 2.□

Next we move to investigate the cycles and the girth of G ( R , f ) . In the following lemma, recall that the identity homomorphism on Z 2 Z 2 is the only isomorphism on Z 2 Z 2 .

Lemma 3.13

Let R be a commutative ring with unity such that there exist distinct elements r , s R 0 such that Ann R ( r ) = { 0 , s } and Ann R ( s ) = { 0 , r } . Then, R Z 2 Z 2 .

Proof

By the assumptions, we have r , s 0 , where 0 is the nil-radical of R , and R 4 . Let α R 0 . Then, r ( α s ) = 0 . So, either α s = 0 or α s = s . If α s = 0 , then α = r , while if α s = s , then ( 1 α ) s = 0 . So, either 1 α = 0 which implies α = 1 or 1 α = r , implying α = 1 r . Consequently, R = { 0,1 , r , 1 r } . If R Z 4 , then, either r 2 = 0 or r is a unit, which is a contradiction. Thus, R Z 2 Z 2 .□

Theorem 3.14

Let R be a commutative ring with unity such that R 3 , and f a ring endomorphism on R. Then, G ( R , f ) has a cycle only in the following cases:

  1. f is not one-to-one.

  2. f is one-to-one and R Z 2 Z 2 .

  3. f is one-to-one, and R is not an integral domain such that R Z 2 Z 2 .

Moreover, when a cycle exists, we have g ( G ( R , f ) ) = 3 .

Proof

  1. Assume f is not one-to-one. If f = 0 , then by Theorem 3.10 G ( R , f ) is complete and hence, it contains a triangle. Assume f 0 . Then, there exists a nonzero and non-unit element k ker ( f ) . So, we have the triangle 0 k 1 0 .

  2. We have f = i d the identity isomorphism and therefore, G ( R , i d ) = G ( R ) , which contains a triangle.

  3. By Lemma 3.13, there is a zero divisor r R 0 such that Ann R ( r ) > 2 . Let t , s Ann R ( r ) 0 such that t s . Then, r s = r t = 0 . Thus, f ( r ) f ( s ) = f ( r ) f ( t ) = 0 . Definitely, either f ( s ) r or f ( t ) r (otherwise, f ( t ) = r = f ( s ) , and since f is one-to-one, then t = s which is a contradiction). Without loss of generality, suppose f ( s ) r . Then, r f ( s ) . Since f ( s ) 0 , we obtain the triangle r 0 f ( s ) r .

Finally, we show that G ( R , f ) has no cycles in the remaining case. Suppose that R is an integral domain and f is one-to-one. Then, for each distinct elements r , s R 0 , we have f ( r ) 0 and f ( s ) 0 . Moreover, r f ( s ) 0 and s f ( r ) 0 . By Lemma 3.8, G ( R , f ) is a star graph with center 0, and so no cycle exists in the graph.□

Next we investigate cycle-free G ( R , f ) . Precisely, we demonstrate that in G ( R , f ) , the notions of tree-graph, star-graph, and bipartite graph are equivalent.

Theorem 3.15

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then, G ( R , f ) is a star graph if and only if R is an integral domain and f is one-to-one.

Proof

( ) : The proof is the same as the final part of the proof of Theorem 3.14.

( ) : Assume G ( R , f ) is a star graph (hence a tree). By Lemma 3.8, the center of the star graph should be 0. Or else, we obtain a cycle, which contradicts that G ( R , f ) is a tree. On the other hand, by Part 1 of Theorem 3.14, f must be one-to-one. It remains to show that R is an integral domain. Let x 0 and y 0 be two distinct elements. Then, x y which means that x f ( y ) 0 and y f ( x ) 0 . We consider two cases

Case 1: If x = f ( x ) or y = f ( y ) , then we obtain x y 0 .

Case 2: If x f ( x ) and y f ( y ) , then x f ( x ) and y f ( y ) , which means that x f ( x ) 0 and y f ( y ) 0 . We have x f ( y ) x f ( x ) (otherwise, x f ( x y ) = 0 , which implies x x y , which implies in turn x = 0 or x = y which is a contradiction). Thus, x f ( x ) x f ( y ) . So, f ( x ) f 2 ( x ) x f ( y ) 0 and x f ( x 2 ) f 2 ( y ) 0 . The former non-equality implies f ( x y ) = f ( x ) f ( y ) 0 , which yields x y 0 .

In either case, we obtain that x y 0 and therefore, R is an integral domain.□

Theorem 3.16

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then, G ( R , f ) is bipartite if and only if G ( R , f ) is a star graph.

Proof

The “if” direction is trivial. For the “only if” direction, assume G ( R , f ) is bipartite. Since each vertex of ker ( f ) is adjacent to all other vertices of G ( R , f ) , then ker ( f ) = 0 . Thus, { { 0 } , R 0 } is the only bipartition of G ( R , f ) . So, G ( R , f ) = K 1 , R 1 , i.e., G ( R , f ) is a star graph.□

Theorem 3.17

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then, G ( R , f ) is a tree if and only if G ( R , f ) is a star graph.

Proof

The “if” direction is trivial. For the “only if” direction, assume G ( R , f ) is a tree. Since each vertex of ker ( f ) is adjacent to all other vertices of G ( R , f ) , then we obtain by Theorem 3.14 that ker ( f ) = 0 . Because 0 is adjacent to all nonzero vertices, and G ( R , f ) is a tree, we obtain that R 0 is an independent set. Thus, G ( R , f ) is a star graph.□

We can sum up the previous three theorems in one theorem as shown here.

Theorem 3.18

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then, the following statements are equivalent:

  1. G ( R , f ) is a star graph.

  2. G ( R , f ) is a tree.

  3. G ( R , f ) is bipartite.

  4. R is an integral domain and f is one-to-one.

3.2 Domination number, independence number, clique number, and chromatic number

This subsection starts with the determination of the domination number followed by the discussion of the independence number.

Theorem 3.19

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then, γ ( G ( R , f ) ) = 1 .

Proof

By Lemma 3.8, { 0 } is a dominating set.□

Proposition 3.20

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then,

  1. If f = 0 , then α ( G ( R , f ) ) = 0 .

  2. If f 0 , then R ker ( f ) α ( G ( R , f ) ) U ( R ) .

  3. If f is one-to-one and R is an integral domain, then α ( G ( R , f ) ) = R 1 .

Proof

For Parts 1 and 2, apply Theorem 3.10 and Lemma 3.8, respectively. Part 3 is a consequence of Theorem 3.15.□

In the sequel work, f 2 means f f .

Theorem 3.21

Let R be a commutative ring with unity, and f a ring endomorphism on R. If ker ( f ) is a prime ideal, then R ker ( f 2 ) is an independent set, and so α ( G ( R , f ) ) R ker ( f 2 ) . Moreover, if ker ( f ) = ker ( f 2 ) , then α ( G ( R , f ) ) = R ker ( f ) .

Proof

Let r , s ker ( f 2 ) such that r s . Assume for contrary that r f ( s ) = 0 , then r f ( s ) ker ( f ) . Since ker ( f ) is prime, either r ker ( f ) ker ( f 2 ) or f ( s ) ker ( f ) . Thus, either r ker ( f 2 ) or s ker ( f 2 ) , which is a contradiction. So, r f ( s ) 0 . Similarly, we obtain s f ( r ) 0 . We obtain r s . Consequently, R ker ( f 2 ) is an independent set. Hence, α ( G ( R , f ) ) R ker ( f 2 ) . The rest follows by Lemma 3.8 and Part 2 of Proposition 3.20.□

Remark 3.22

Note that Part 3 of Proposition 3.20 follows from Theorem 3.21. On the other hand, the condition ker ( f ) = ker ( f 2 ) in Theorem 3.21 holds for example when f is one-to-one or f is an idempotent element in the ring End ( R ) of endomorphisms on R .

In order to give the independence number of G ( R , f ) a more algebraic flavor, we shall discuss the adjacency of special cosets. The next lemma summarizes this discussion.

Lemma 3.23

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then,

  1. If r s ( mod ker ( f ) ) , and r s in G ( R , f ) , then r + ker ( f ) s + ker ( f ) in G R ker ( f ) , f * . Similarly, if r s ( mod ker ( f 2 ) ) , and r s in G ( R , f ) , then r + ker ( f 2 ) s + ker ( f 2 ) in G R ker ( f 2 ) , f * .

  2. If ker ( f ) is a prime ideal, then for every x ker ( f 2 ) , then x + ker ( f ) is an independent set in G ( R , f ) . Further, the cosets x + ker ( f 2 ) , where x ker ( f 2 ) form a partition of R ker ( f 2 ) .

Proof

1. Assume r s ( mod ker ( f ) ) . Then,

r s in G ( R , f ) r f ( s ) = 0 or s f ( r ) = 0 r f ( s ) ker ( f ) or s f ( r ) ker ( f ) ( r + ker ( f ) ) ( f ( s ) + ker ( f ) ) = ker ( f ) or ( s + ker ( f ) ) ( f ( r ) + ker ( f ) ) = ker ( f ) ( r + ker ( f ) ) f * ( s + ker ( f ) ) = ker ( f ) or ( s + ker ( f ) ) f * ( r + ker ( f ) ) = ker ( f ) r + ker ( f ) s + ker ( f ) in G R ker ( f ) , f * .

The remaining part is proved in the same manner.

2. Assume ker ( f ) is a prime ideal, h ker ( f ) , and x ker ( f 2 ) . The argument in the proof of Theorem 3.21 guarantees that R ker ( f 2 ) is an independent set. Since f ( x + h ) = f ( x ) , we obtain that x + h lies outside ker ( f 2 ) . Therefore, the coset x + ker ( f ) R ker ( f 2 ) , which implies it is an independent set. It is easy to see that the cosets x + ker ( f 2 ) , where x ker ( f 2 ) form a partition of R ker ( f 2 ) .□

Theorem 3.24

Let R be a commutative ring with unity, and f a ring endomorphism on R such that ker ( f ) is a prime ideal. Then,

ε α ( G ( R , f ) ) ε + α ( G ( ker ( f 2 ) , f ) ker ( f ) ) ,

where ε = R ker ( f 2 ) = ker ( f ) R ker ( f 2 ) { ker ( f 2 ) } . Further, if R is finite, then ε = ker ( f ) R ker ( f 2 ) 1 . On the other hand, if ker ( f ) = ker ( f 2 ) , then α ( G ( R , f ) ) = ε , which is equal to R ker ( f ) when R is also finite.

Proof

It is enough to verify the right inequality because the left inequality holds by Theorem 3.21. Once again by Theorem 3.21 and Lemma 3.23, we have

α ( G ( R , f ) ) = α ( G ( R , f ) ker ( f ) ) R ker ( f 2 ) + α ( G ( ker ( f 2 ) , f ) ker ( f ) ) = x [ x + ker ( f 2 ) ] + α ( G ( ker ( f 2 ) , f ) ker ( f ) ) = x x + ker ( f 2 ) + α ( G ( ker ( f 2 ) , f ) ker ( f ) ) = ker ( f ) x x + α ( G ( ker ( f 2 ) , f ) ker ( f ) ) = ker ( f ) R ker ( f 2 ) { ker ( f 2 ) } + α ( G ( ker ( f 2 ) , f ) ker ( f ) ) ,

where the union and the sum run over all different representatives x of the cosets of ker ( f ) in R except x = 0 . Now, if R is finite, it is easy to see that ε = ker ( f ) R ker ( f 2 ) 1 . If ker ( f 2 ) = ker ( f ) , then α ( G ( ker ( f 2 ) , f ) ker ( f ) ) = 0 and hence, α ( G ( R , f ) ) = ε . If in addition R is finite, we obtain α ( G ( R , f ) ) = ε = R ker ( f ) .□

Remark 3.25

Note that the result of Theorem 3.21 agrees with the result of Theorem 3.24.

The next work is devoted to studying the clique number of G ( R , f ) and finding an algebraic formula that calculates it. Recall that by a maximal clique, we mean a clique that is not included in another clique. The number of vertices of a maximal clique with the largest number of vertices is the clique number of the underlying graph. We begin with the following lemma.

Lemma 3.26

Let R be a commutative ring with unity, and f a ring endomorphism on R. If Λ is a clique of G ( R , f ) , then Λ ker ( f ) is a clique. So, ω ( G ( R , f ) ) Λ ker ( f ) . Moreover, if 1 Fix ( f ) and u U ( R ) , then ker ( f ) { u } is a maximal clique of G ( R , f ) .

Proof

Let Λ a clique of G ( R , f ) . By Lemma 3.8, every element in ker ( f ) is adjacent to every element in R . Therefore, Λ ker ( f ) is a clique (possibly a larger clique than Λ but not necessarily with the largest cardinality). So, obviously we obtain that ω ( G ( R , f ) ) Λ ker ( f ) . By Part 4 of Lemma 3.8, if f ( 1 ) = 1 , then the units of R are adjacent to only the elements of ker ( f ) . Thus, if u U ( R ) , then ker ( f ) { u } is a maximal clique.□

Theorem 3.27

Let R be a commutative ring with unity, and f a ring endomorphism on R. If ker ( f ) is a prime ideal, then ω ( G ( R , f ) ) ker ( f ) + 1 . Moreover, if ker ( f ) = ker ( f 2 ) , then ω ( G ( R , f ) ) = ker ( f ) + 1 .

Proof

By Theorem 3.21, R ker ( f 2 ) is an independent set. Therefore, if x ker ( f ) , then ker ( f ) { x } is a clique and hence ω ( G ( R , f ) ) ker ( f ) { x } = ker ( f ) + 1 . Now, if ker ( f 2 ) = ker ( f ) , then ker ( f ) { x } , where x ker ( f ) are the only maximal cliques in G ( R , f ) . From the last statement, we conclude that ω ( G ( R , f ) ) = ker ( f ) + 1 .□

Lemma 3.28

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then, every maximal clique of G ( R , f ) must contain ker ( f ) .

Proof

Let Λ be a maximal clique of G ( R , f ) . Suppose, for contrary, that ker ( f ) Λ . By Lemma 3.26, Λ ker ( f ) is a larger clique than Λ , which contradicts the maximality of Λ .□

Theorem 3.29

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then,

ω ( G ( R , f ) ) = ker ( f ) + ω ( G ( R , f ) ker ( f ) ) .

Moreover, if ker ( f ) is a prime ideal, then either

ω ( G ( R , f ) ) = ker ( f ) + ω ( G ( ker ( f 2 ) , f ) ker ( f ) ) ,

or

ω ( G ( R , f ) ) = ker ( f ) + ω ( G ( ker ( f 2 ) , f ) ker ( f ) ) + 1 .

Furthermore, if ker ( f ) = ker ( f 2 ) , then ω ( G ( R , f ) ) = ker ( f ) + 1 .

Proof

Let Λ be a maximal clique in the subgraph G ( R , f ) ker ( f ) such that Λ = ω ( G ( R , f ) ker ( f ) ) . Then, by Lemma 3.26, Λ ker ( f ) is a clique of G ( R , f ) . As a matter of fact, Λ ker ( f ) is a maximal clique with the largest number of vertices in G ( R , f ) . To see this, assume it is not, then there exists a maximal clique Λ with a larger number of vertices in G ( R , f ) than Λ ker ( f ) . Since ker ( f ) Λ by Lemma 3.28, we deduce that Λ has less number of vertices than the clique Λ ker ( f ) in G ( R , f ) ker ( f ) , which contradicts that Λ is a clique of G ( R , f ) ker ( f ) with the largest number of vertices in G ( R , f ) ker ( f ) . Thus, we obtain ω ( G ( R , f ) ) = Λ ker ( f ) = Λ + ker ( f ) = ker ( f ) + ω ( G ( R , f ) ker ( f ) ) . Next assume ker ( f ) is a prime ideal. By Theorem 3.21, R ker ( f 2 ) is an independent set. If none of the elements of R ker ( f 2 ) is adjacent to all vertices of a clique of order ω ( G ( ker ( f 2 ) , f ) ker ( f ) ) , then ω ( G ( R , f ) ker ( f ) ) = ω ( G ( ker ( f 2 ) , f ) ker ( f ) ) . However, if there is one element in R ker ( f 2 ) (it cannot be more than one element of R ker ( f 2 ) ) that is adjacent to all vertices of a clique of order ω ( G ( ker ( f 2 ) , f ) ker ( f ) ) , then ω ( G ( R , f ) ker ( f ) ) = ω ( G ( ker ( f 2 ) , f ) ker ( f ) ) + 1 where the number 1 stands for that unique element of R ker ( f 2 ) . Finally, if ker ( f ) = ker ( f 2 ) , then ω ( G ( ker ( f 2 ) , f ) ker ( f ) ) = 0 and noting that ker ( f ) is adjacent to all elements of R ker ( f ) , then we have ω ( G ( R , f ) ) = ker ( f ) + 1 .□

Remark 3.30

Theorem 3.29 agrees with Theorem 3.27 when ker ( f ) = ker ( f 2 ) .

Lemma 3.31

Let R be a commutative ring with unity, and f a ring endomorphism on R. If x ker ( f 2 ) ker ( f ) and k ker ( f ) , then x + k ker ( f 2 ) ker ( f ) .

Proof

The proof follows from the fact that f ( x + k ) = f ( x ) .□

Corollary 3.32

Let R be a commutative ring with unity, and f a ring endomorphism on R such that ker ( f ) is a prime ideal of R. If ker ( f ) Ann R ( ker ( f 2 ) ) , then ω ( G ( R , f ) ) = ker ( f 2 ) + η , where η = 0 or η = 1 .

Proof

Assume ker ( f ) Ann R ( ker ( f 2 ) ) . Let x , y ker ( f 2 ) ker ( f ) be distinct elements. Then, f ( x ) ker ( f ) and so f ( x ) . y = 0 , which yields x y . We conclude that G ( ker ( f 2 ) , f ) is a complete subgraph of G ( R , f ) . It follows by Theorem 3.21 and Lemma 3.26, following a parallel argument as that of the proof of Theorem 3.29, that either ker ( f 2 ) is a maximal clique with the largest number of vertices or ker ( f 2 ) { y } for some y ker ( f 2 ) is a maximal clique with the largest number of vertices. Therefore, ω ( G ( R , f ) ) = ker ( f 2 ) + η , where η = 0 or η = 1 .□

In the final part of this subsection, we study the chromatic number of G ( R , f ) .

Theorem 3.33

Let R be a commutative ring with unity, and f a ring endomorphism on R. Then, χ ( G ( R , f ) ) = ker ( f ) + χ ( G ( R , f ) ker ( f ) ) . Further, if 1 Fix ( f ) , then χ ( G ( R , f ) ker ( f ) ) = χ ( G ( R , f ) ( ker ( f ) U ( R ) ) ) .

Proof

Assume that { S 1 , , S χ ( G ( R , f ) ker ( f ) ) } is a smallest partition of G ( R , f ) ker ( f ) consisting of independent sets. We know that m = χ ( G ( R , f ) ) ker ( f ) + χ ( G ( R , f ) ker ( f ) ) . Suppose for contrary that m < ker ( f ) + χ ( G ( R , f ) ker ( f ) ) . Let { T 1 , , T m } be a smallest partition of G ( R , f ) of independent sets. Since ker ( f ) is a clique whose vertices are adjacent to every vertex in R ker ( f ) , we assume without loss of generality that ker ( f ) = T 1 T ker ( f ) , where each T i is a singleton set containing one distinct vertex of ker ( f ) , for each 1 i ker ( f ) . Hence, R ker ( f ) = T ker ( f ) + 1 T m with m ker ( f ) < χ ( G ( R , f ) ker ( f ) ) , which contradicts that χ ( G ( R , f ) ker ( f ) ) is the chromatic number of G ( R , f ) ker ( f ) . Consequently, χ ( G ( R , f ) ) = ker ( f ) + χ ( G ( R , f ) ker ( f ) ) . Next, if 1 Fix ( f ) , then by Lemma 3.8 we have that every vertex of U ( R ) is adjacent to no vertex in R ker ( f ) . Therefore, χ ( G ( R , f ) ker ( f ) ) = χ ( G ( R , f ) ( ker ( f ) U ( R ) ) ) .□

Lemma 3.34

Let G be an undirected graph and L an independent set of vertices of G. Then, χ ( G ) = χ ( G L ) + η , where η = 0 or 1. Moreover, η = 1 if and only if L contains a vertex that is adjacent to χ ( G L ) vertices with different color each.

Proof

It is obvious that χ ( G L ) χ ( G ) . Let { L 1 , , L χ ( G L ) } be a smallest partition of G L by independent sets. If L L k is independent for some 1 k χ ( G L ) , then L 1 , , L k 1 , L k L , , L χ ( G L ) is a smallest partition of G by independent sets of G . Thus, χ ( G ) = χ ( G L ) . Assume L L k is not independent for all 1 k χ ( G L ) . This means that for every 1 k χ ( G L ) , there is a vertex in L , which is adjacent to a vertex in L k . If there is a vertex l in L that is adjacent to a vertex in each L k (i.e., l is adjacent to χ ( G L ) vertices with different color each), then { L , L 1 , , L χ ( G L ) } is a smallest partition of G by independent sets of G . Thus, χ ( G ) = χ ( G L ) + 1 . However, otherwise, we can distribute the vertices of L to the sets L 1 , , L χ ( G L ) such that the new family of sets forms a partition of G by independent sets. Hence, χ ( G ) = χ ( G L ) .

Next suppose η = 1 . Then, the vertices of L cannot be distributed to L 1 , , L χ ( G L ) . This yields the existence of a vertex in L that is adjacent to a vertex in each L k , or equivalently speaking, the existence of a vertex in L that is adjacent to χ ( G L ) vertices with different color each.□

Theorem 3.35

Let R be a commutative ring with unity, and f a ring endomorphism on R. If ker ( f ) is a prime ideal, then χ ( G ( R , f ) ) = χ ( G ( ker ( f 2 ) , f ) ) + η , where η = 0 or 1. If in addition ker ( f 2 ) = ker ( f ) , then χ ( G ( R , f ) ) = ω ( G ( R , f ) ) .

Proof

Suppose that ker ( f ) is a prime ideal. By Theorem 3.21 L = R ker ( f 2 ) is an independent set. It follows by Lemma 3.34 that χ ( G ( R , f ) ) = χ ( G ( ker ( f 2 ) , f ) ) + η , where η = 0 or 1. Now, if in addition ker ( f ) = ker ( f 2 ) , then each vertex in R ker ( f ) is adjacent to each vertex of ker ( f ) by Lemma 3.8. Again, by Lemma 3.34, we obtain η = 1 and hence, by Theorem 3.29, we obtain χ ( G ( R , f ) ) = χ ( ker ( f ) , f ) + 1 = ker ( f ) + 1 = ω ( G ( R , f ) ) .□

4 Applications

This section offers different examples of and applications to the results found in Section 3.

Example 4.1

Let p > 2 be a prime number, and f a ring endomorphisms on Z p . Then, either f = 0 and hence, G ( R , f ) = K p or f = i d and hence G ( R , f ) = G ( R ) .

Example 4.2

Let { R i : i I } be a family of commutative rings with unities. Fix j I . Consider the endomorphism f j : i I R i i I R i defined by f j ( r i ) i I = ( x i ) i I , where x i = 0 for all i I { j } and x j = r j . Then, if ( r i ) i I and ( r i ) i I are distinct elements of i I R i , we obtain ( r i ) i I ( r i ) i I in G ( i I R i , f j ) if and only if r j r j = 0 (when r j r j , we can write r j r j in G ( R j ) ). Accordingly, the inclusion monomorphism g j : R j i I R i defined by g j ( r j ) = ( x i ) i I , where x i = 0 for all i I { j } and x j = r j , induces an embedding of G ( R j ) in G ( i I R i , f j ) .

Example 4.3

Let p and q be prime numbers. Consider the ring endomorphism f : Z p q Z p q defined by f ( m + n ) = m , where m Z p and n Z q . We have I m ( f ) = Z p and ker ( f ) = Z q is a prime ideal of Z p q . According to Example 4.2, m 1 + n 1 m 2 + n 2 in G ( Z p q , f ) if and only if m 1 m 2 = 0 (if m 1 m 2 this means that m 1 m 2 in G ( Z p q ) ). Since f 2 = f , by Theorem 3.24, α ( G ( Z p q , f ) ) = ε = p q q = q ( p 1 ) .

Example 4.4

Let p be a prime number and R = Z p Z p . To make things easy, the left Z p in R will be presented as Z p , while the right Z p in R will be presented as Z p + . Consider the ring endomorphism f : Z p Z p Z p Z p defined by f ( m , n ) = ( n , m ) . It is easy to see that f is an isomorphism. One can check out that ( m , n ) ( m , n ) in G ( Z p Z p , f ) if and only if m = n = 0 , m = m = 0 , n = n = 0 , or m = n = 0 . In other words, the adjacency in G ( Z p Z p , f ) can be translated as follows:

  1. ( 0 , 0 ) is adjacent to all nonzero elements.

  2. ( 0 , n ) ( 0 , n ) for every distinct elements n , n Z p .

  3. ( m , 0 ) ( m , 0 ) for every distinct elements m , m Z p .

Thus, Z p + and Z p are the largest cliques in the graph. This implies ω ( G ( Z p Z p , f ) ) = p . By the pigeonhole principle, any set of three nonzero vertices in Z p Z p + must contain at least two vertices that lie in Z p + or Z p (i.e., they are adjacent). Thus, α ( G ˆ ( Z p Z p + , f ) ) = 2 , where G ˆ ( Z p Z p + , f ) is the subgraph of G ( R , f ) whose vertex set is Z p Z p + . Besides, by Lemma 3.8, U ( R ) = { ( m , n ) : m , n Z p 0 } is an independent set with U ( R ) = ( p 1 ) 2 . Since R = ( Z p Z p + ) U ( R ) . We conclude that α ( G ( Z p Z p , f ) ) = ( p 1 ) 2 + 2 . The graph clique number ω has been computed by investigating the graph itself along with some combinatorial techniques. However, we can deduce the same value from our algebraic angle. Observing that f 2 = i d and ker ( f ) = ker ( f 2 ) = 0 , which is not a prime ideal, we obtain by Theorem 3.29,

ω ( G ( R , f ) ) = ker ( f ) + ω ( G ( R , f ) ker ( f ) ) .

We have V ( G ( R , f ) ker ( f ) ) = ( ( Z p Z p + ) 0 ) U ( R ) . Hence, ω ( G ( R , f ) ker ( f ) ) = p 1 . Thus, ω ( G ( R , f ) ) = 1 + p 1 = p , which is the same result obtained above. For the chromatic number, we should observe that each vertex of Z p (resp., Z p + ) is not adjacent to any of the vertices outside Z p (resp., Z p + ). Since each of Z p + and Z p are maximal cliques and the vertices of U ( R ) are only adjacent to 0, we obtain χ ( G ( R , f ) ) = p = ω ( G ( R , f ) ) .

  1. Funding information: The authors state no funding involved.

  2. Author contributions: Both authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results, and approved the final version of the manuscript. Both authors were involved throughout the preparation of the manuscript and contributed equally.

  3. Conflict of interest: The authors state no conflict of interest.

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Received: 2024-12-21
Revised: 2025-06-28
Accepted: 2025-08-20
Published Online: 2025-10-14

© 2025 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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