Graphs over Graded Rings and Relation with Hamming Graph

Abstract

Hamming graph is known to be an important class of graphs, and it is a challenge to obtain algorithms that recognize whether a given graph G is a Hamming graph. Let G be a group and $S\subseteq G$ be a nonempty subset of G. The Cayley graph with respect to S is a graph whose vertex set is G and arcset is the set of pairs (u, v) such that $v=su$ for some element $s\in S$. This graph is denoted by $\mathrm{Cay}(G,S)$.Let $R=\oplus _{i}R_{i}$ be a graded ring, S be the set of homogeneous elements of R, $S'$ a subset of S, and $S''=\oplus _{i\ge k}R_{i}$. In this paper, with a different view, we study $\mathrm{Cay}(R, S')$ as a generalization of $\mathrm{Cay}(R, S)$ to obtain a new point of view to study Cartesian products of complete graphs (Hamming graph). In particular, we show that any Hamming graph over sets of prime power sizes is isomorphic to $\mathrm{Cay}(R,S')$ for some graded ring R and a subset $S'\subseteq S$. Also we study $\mathrm{Cay}(R, S'')$ as another Cayley graph over graded rings and obtain relations between this graph and total, cototal and counit graphs.

1 Introduction

Cartesian products of complete graphs are known as Hamming graphs [7]. In other words if $\Gamma _{1}, \ldots , \Gamma _{n}$ are complete graphs with vertex sets $G_{1}, \ldots , G_{n}$, then $H=\Gamma _{1}\times \cdots \times \Gamma _{n}$ is a graph with vertex set $V(H)=V(\Gamma _{1})\times \cdots \times V(\Gamma _{n})$ and two vertices $(u_{1}, \ldots , u_{n})$ and $(v_{1}, \ldots , v_{n})$ being adjacent if they differ in exactly one position. This graph is known as a Hamming graph over $G_{1}, \ldots , G_{n}$. Hamming graph is an important class of graphs and researchers have tried to obtain algorithms that recognize whether a given graph G is a Hamming graph; see [6].

In [8], Mahmoudi et al. introduced the Cayley graph $\mathrm{Cay}(R, S)$, where R is a graded ring and S is the set of homogeneous elements of R. When R is an Artinian graded ring, they showed that $\mathrm{Cay}(R, S)$ is isomorphic to a Hamming graph and conversely any Hamming graph is isomorphic to a subgraph of $\mathrm{Cay}(R, S)$ for some finite graded ring R. In this paper, we generalize this graph to obtain more results on Cartesian products of complete graphs. In particular, we show that for any Hamming graph H over sets of prime power sizes there exists a graded ring R such that H is isomorphic to $\mathrm{Cay}(R, S')$. Also, we study $\mathrm{Cay}(R, S'')$ as another Cayley graph over graded rings and obtain interesting results.

Let $R=\oplus _{i}R_{i}$ be a graded ring, S be the set of homogeneous elements of R, $S'$ a subset of S, and $S''=\oplus _{i\ge k}R_{i}$.

In Sect. 3, we study $\mathrm{Cay}(R, S')$ as a generalization of $\mathrm{Cay}(R, S)$ in [8]. We show that $\mathrm{Cay}(R, S')$ is isomorphic to a product of Cayley graphs over abelian groups, and any Cartesian product of Cayley graphs over abelian groups, in some special cases, can be seen as $\mathrm{Cay}(R, S')$ for some graded ring R. In particular, we prove that the Hamming graph is a generalized Cayley graph in more cases. Hence, we obtain a new point of view to study Cartesian products of Cayley graphs and Hamming graphs.

In Sect. 4, we study $\mathrm{Cay}(R, S'')$ and obtain important results about this graph. We show that there are more relations between this graph and total, cototal and counit graphs.

Among interesting results, we can refer to the following:

• Any Cartesian product of Cayley graphs over cyclic groups of prime power orders is isomorphic to $\mathrm{Cay}(R,S')$ for some graded ring R.

• For any Hamming graph H over sets of prime power sizes there exists a graded ring R such that H is isomorphic to $\mathrm{Cay}(R, S')$.

• Let T be a finite commutative local ring with the maximal ideal ${\mathfrak {m}}$. If $2\in {\mathfrak {m}}$, then the total graph of T is isomorphic to the Cayley graph $\mathrm{Cay}(R, S'')$ for some graded ring R and a subset $S''$.

2 Preliminaries

Let $\Gamma$ be a graph with the vertex set $V (\Gamma )$. The degree of a vertex x in a graph $\Gamma$ is the number of edges incident with x. The degree of a vertex x is denoted by $\deg (x)$. Let r be a nonnegative integer. The graph $\Gamma$ is said to be r-regular, if the degree of each vertex is r. If x and y are two adjacent vertices of $\Gamma$, then we write $x\sim y$. An edge from x to y is denoted by xy. The complete graph of order n, denoted by $K_{n}$, is a graph with n vertices in which any two distinct vertices are adjacent. A graph $\Gamma$ is connected if there is a path between every two distinct vertices. For distinct vertices x and y of $\Gamma$, let d(x, y) be the length of the shortest path from x to y and if there is no such a path we define $d(x, y) = \infty$. The girth of $\Gamma$, denoted by $\mathrm{gr}(\Gamma )$, is the length of a shortest cycle in $\Gamma$ ($\mathrm{gr}(\Gamma )=\infty$ if $\Gamma$ contains no cycle). The diameter of $\Gamma$, $\mathrm{diam}(\Gamma )$, is the supremum of the set $\{d(x, y) : x \ \text {and} \ y \ \text {are distinct vertices of} \ \Gamma \}$. A clique of $\Gamma$ is a complete subgraph of $\Gamma$ and the number of vertices in the largest clique of $\Gamma$, denoted by $\omega (\Gamma )$, is called the clique number of $\Gamma$. An independent set of $\Gamma$ is a subset of the vertices of $\Gamma$ such that no two vertices in the subset represent an edge of $\Gamma$. The independence number of $\Gamma$, denoted by $\alpha (\Gamma )$, is the cardinality of the largest independent set. Let $a\in \Gamma$ be a vertex, the set of all elements $b\in \Gamma$ where $d(a, b)=r$ is denoted by $N_{r}(a)$. If $T\subseteq V(\Gamma )$, then the induced subgraph by T is denoted by $\Gamma [T]$.

Now, we remind some facts of graded rings, Cayley graphs, Cartesian product of graphs and Hamming graphs.

Definition 1

A ring R is called graded (or more precisely, ${\mathbb {Z}}$-graded ) if there exists a family of subgroups $\{R_{i}\}_{i\in {\mathbb {Z}}}$ of R such that

• 1. $R=\oplus _{i}R_{i}$ (as abelian groups), and

• 2. $R_{i}R_{j}\subseteq R_{i+j}$ for all i, j.

A graded ring R is called nonnegatively graded if $R_{i}= 0$ for all $i <0$. A nonzero element $x \in R_{i}$ is called a homogeneous element of R of degree i. If $x=\sum x_{i}\in R$ is an element of R, then the number of homogenous elements $x_{i}$ of x is denoted by $s_{x}$.

Note that every ring R is trivially a graded ring by letting $R_{0} = R$ and $R_{i} =0$ for all $i \ne 0$. But there are rings with more interesting gradings.

Example 1

Let $R'$ be a ring and $x_{1}, \ldots , x_{d}$ indeterminates over $R'$. For $m=(m_{1}, \ldots , m_{d})\in N^{d}$, let $X^{m}=x^{m_{1}}_{1}\ldots x^{m_{d}}_{d}$. Then, the polynomial ring $R=R'[x_{1}, \ldots , x_{d}]$ is a graded ring, where

$$\begin{aligned} R_{i}=\{\sum _{m\in N^{d}}r_{m}X^{m}:\ r_{m}\in R', m_{1}+\cdots + m_{d}=i\} . \end{aligned}$$

Example 2

Let R be a ring and ${\mathcal {I}}=\{I_{i}\}^{\infty }_{i=0}$ a sequence of ideals of R. ${\mathcal {I}}$ is called a filtration of R if

• 1. $I_{0}=R$,

• 2. $I_{i}\supseteq I_{i+1}$ for all i, and

• 3. $I_{i}.I_{j}\subseteq I_{i+j}$ for all i, j.

Let ${\mathcal {I}}=\{I_{i}\}^{\infty }_{i=0}$ be a filtration of R. As an R-module suppose

$$\begin{aligned} G({\mathcal {I}})=\oplus \frac{I_{i}}{I_{i+1}}=\frac{R}{I_{1}}\oplus \frac{I_{1}}{I_{2}} \oplus \frac{I_{2}}{I_{3}}\oplus \cdots . \end{aligned}$$

Let i and j be nonnegative integers and suppose $x_{i}+I_{i+1}$ and $x_{j}+I_{j+1}$ are elements of $G({\mathcal {I}})_{i}$ and $G({\mathcal {I}})_{j}$, respectively. Define the product on $G({\mathcal {I}})$ by

$$\begin{aligned} (x_{i}+I_{i+1})(x_{j}+I_{j+1})=x_{i}x_{j}+I_{i+j+1}. \end{aligned}$$

With this product, $G({\mathcal {I}})$ is a graded ring which is called the associated graded ring of ${\mathcal {I}}$.

Definition 2

Let $R= \oplus R_{i}$ be a graded ring. A subring $R'$ of R is called a graded subring of R if $R' = \sum _{i}( R_{i}\cap R')$. Equivalently, $R'$ is graded if for every element $r\in R'$ all the homogeneous components of r (as an element of R) are in $R'$.

Now, we remind some facts of Cayley graphs.

Definition 3

(Cayley graph over a group) Let G be a group and $S\subseteq G$ be a nonempty subset of G. The Cayley graph with respect to S is a graph whose vertex set is G and arcset is the set of pairs (u, v) (an arc from u to v) such that $v=su$ for some element $s\in S$. This graph is denoted by $\mathrm{Cay}(G,S)$.

Note that $\mathrm{Cay}(G,S)$ is a directed graph where (u, v) is an arc if and only if $vu^{-1}\in S$. But it is easy to see that $\mathrm{Cay}(G,S)$ is a graph (undirected graph) if and only if $1_{G}\notin S$ and $s^{-1}\in S$ for all $s\in S$.

Definition 4

(Cartesian product of graphs) The Cartesian product $\Gamma _{1}\times \cdots \times \Gamma _{n}$ of graphs $\Gamma _{1}, \ldots , \Gamma _{n}$ has the vertex set $V(\Gamma _{1})\times \cdots \times V(\Gamma _{n})$, two vertices $(u_{1}, \ldots , u_{n})$ and $(v_{1}, \ldots , v_{n})$ being adjacent if they differ in exactly one position, say in ith, and $u_{i}\sim v_{i}$ in $\Gamma _{i}$. If $\Gamma _{1}, \ldots , \Gamma _{n}$ are complete graphs then two vertices $(u_{1}, \ldots , u_{n})$ and $(v_{1}, \ldots , v_{n})$ are adjacent if they differ in exactly one position. A Hamming graph is the Cartesian product of complete graphs. In other words if $\Gamma _{i}$ is the complete graph with the vertex set $G_{i}$, then the Cartesian product $\Gamma _{1}\times \cdots \times \Gamma _{n}$ is called the Hamming graph over $G_{1}, \ldots , G_{n}$.

The following remark shows that Hamming graphs can be considered a product of Cayley graphs.

Remark 1

Let $\Gamma =\Gamma _{1}\times \cdots \times \Gamma _{n}$ be a Hamming graph over $G_{1}, \ldots , G_{n}$. For $1\le i\le n$, let $|G_{i}|=q_{i}$. Then, $\Gamma$ is isomorphic to

$$\begin{aligned} \mathrm{Cay}({\mathbb {Z}}_{q_{1}},{\mathbb {Z}}_{q_{1}}{\setminus } \{0\})\times \cdots \times \mathrm{Cay}({\mathbb {Z}}_{q_{n}},{\mathbb {Z}}_{q_{n}}{\setminus } \{0\}). \end{aligned}$$

By the following remark Hamming graph is isomorphic to a subgraph of a Cayley graph over a graded ring.

Remark 2

Mahmoudi et al. (2017) introduced a Cayley graph over graded rings as follows: let $R=\oplus _{n}R_{n}$ be a graded ring and $S=(\cup _{n}R_{n}){\setminus } \{0\}$ be the set of homogeneous elements of R. The Cayley graph $\mathrm{Cay}(R,S)$ is a graph, whose vertex set is R and two vertices $a, b \in R$ are adjacent whenever $b-a\in S$. They showed that any hamming graph is isomorphic to a subgraph of $\mathrm{Cay}(R,S)$, where $R={\mathbb {Z}}_{m}[X]/\langle X^{d}\rangle$ for some positive integers m and d; see [8].

Let R be a commutative ring and $\mathrm{Z}(R)$ be the set of zero-divisors of R. The Cayley sum graph $\mathrm{Cay}^{+}(R, \mathrm{Z}(R))$ is a graph whose vertex set is R and two vertices a and b are adjacent if and only if $a+b\in \mathrm{Z}(R)$.

Definition 5

(Cayley sum graph) Let G be an abelian group and S a subset of G. The Cayley sum graph $\mathrm{Cay}^{+}(G, S)$ on G with respect to S is the graph with vertex set G in which $x, y \in G$ are adjacent if and only if $x + y \in S$. Since G is abelian, this is an undirected graph of degree |S|, with a loop at x if $x + x \in S$. If S is required to be square-free (that is, $x + x \notin S$ for any $x \in \Gamma$), then $\mathrm{Cay}^{+}(G, S)$ is an undirected graph without loops. Cayley sum graphs are also known as addition Cayley graphs; see [5, 11].

The total graph and the unit graph are important examples of addition Cayley graphs.

Example 3

Let R be a commutative ring.

• 1. If $\mathrm{Z}(R)$ is the set of zero-divisors, then $\mathrm{Cay}^{+}(R, \mathrm{Z}(R))$ is called the total graph [1, 3].

• 2. If U(R) is the set of unit elements in R, then $\mathrm{Cay}^{+}(R, U(R))$ is called the unit graph [2].

Maimani et al. defined the cototal graph and used the term counit graph instead of the unitary graph; see [9, Definition 5.1]. The cototal graph of R is the graph obtained by setting all the elements of R to be the vertices and defining distinct vertices x and y to be adjacent if and only if $x - y \in \mathrm{Z}(R)$. Also the counit graph of R is the graph obtained by setting all the elements of R to be the vertices and defining distinct vertices x and y to be adjacent if and only if $x - y \in U(R)$.

Clearly R is an abelian group with respect to addition. Hence, by the definition 3, the cototal and the counit graphs are $\mathrm{Cay}(R,\mathrm{Z}(R){\setminus } \{0\})$ and $\mathrm{Cay}(R,U(R))$; respectively. In general case, we define this concept as follows.

Definition 6

(Coaddition graph over a ring) Let R be a ring and H be a subset of R such that $-h\in H$ if $h\in H$. In this case, we have the Cayley sum graph $\mathrm{Cay}^{+}(R, H)$ on R for some subset $H\subseteq R$. Then, we call the Cayley graph $\mathrm{Cay}(R, H{\setminus } \{0\})$, coaddition graph over R.

3 The Generalized Cayley Graph $\mathrm{Cay}(R,S')$ over Graded Rings

In this section, we introduce and study a Cayley graph $\mathrm{Cay}(R,S')$ over graded rings which is a generalization of $\mathrm{Cay}(R,S)$ in Remark 2. We show that the Cartesian product of Cayley graphs over cyclic groups of prime power orders is isomorphic to the generalized Cayley graph $\mathrm{Cay}(R,S')$ over a graded ring R. In particular, by Remark 1, we show that the Hamming graph over sets of prime power orders is isomorphic to the generalized Cayley graph $\mathrm{Cay}(R,S')$ for some graded ring R.

Definition 7

Let $R=\oplus _{i}R_{i}$ be a graded ring and S be the set of homogeneous elements of R. Suppose that $S'\subseteq S$ is a subset of S such that $-s\in S$ if $s\in S$. The Cayley graph $\mathrm{Cay}(R,S')$ is a graph, whose vertex set is R and two vertices $a, b \in R$ are adjacent if $b-a\in S'$.

Note that $S=\cup _{i} R_{i}{\setminus } \{0\}$. Hence, $S'=\cup _{i} R'_{i}{\setminus } \{0\}$ for some subsets $R'_{i}$ of $R_{i}$.

Remark 3

If $S'=S$, then $\mathrm{Cay}(R,S')=\mathrm{Cay}(R,S)$. Hence, $\mathrm{Cay}(R,S')$ is a generalization of $\mathrm{Cay}(R,S)$ which was defined in Remark 2.

Let R be a ring and H a subset of R. Also let $\Gamma$ be a coaddition graph with respect to H. Then, by letting $R_{0} = R$ and $R_{i} =0$ for all $i \ne 0$, R is trivially a graded ring where $S=R{\setminus } \{0\}$ is the set of homogenous elements. Hence, by letting $S'=H{\setminus } \{0\}$, $\Gamma =\mathrm{Cay}(R, S')$. Therefore, any coaddition graph over a ring is a generalized Cayley graph where the ring is considered a graded ring trivially. Thus, the generalized Cayley graph over graded rings is a generalization of coaddition graphs.

The following theorem gives some properties of $Cay(R,S')$.

Theorem 1

Let $\Gamma =Cay(R,S')$ be the generalized Cayley graph; then,

• 1. $\Gamma$ is connected if and only if $S'$ generates R.

• 2. $\Gamma$ is an $|S'|$-regular graph.

• 3. If a, b are two vertices of $\Gamma$ and d(a, b) denotes the distance of a and b, then $d(a,b)\ge s_{b-a}$. If $S'=S$, then $d(a,b)= s_{b-a}$.

• 4. The clique number of $\Gamma$ is equal to $\omega (\Gamma )=\max \{|R'_{i}|:\ i\in {\mathbb {Z}}\}$.

Proof

• 1. Clearly, R is an abelian group with respect to addition and $\Gamma$ is a Cayley graph over this abelian group. By elementary properties of Cayley graphs over abelian graphs, we have the result.

• 2. Let $x=\sum _{i} x_{i}\in R$ be a vertex of $\Gamma$. Consider the vertex $y=\sum _{i} y_{i}$ that $y\sim x$. Then, there exists $\mathrm{beg}(R)\le j\le \mathrm{end}(R)$, such that $y-x\in R'_{j}{\setminus } \{0\}$. Hence $y_{j}\ne x_{j}$ and $y_{i}=x_{i}$ for all $i\ne j$ where $y_{j}\in R'_{j}$. The number of such elements is equal to $|R'_{j}{\setminus } \{0\}|$. By induction on $\mathrm{beg}(R)\le j\le \mathrm{end}(R)$, $\mathrm{deg}(x)=\sum _{j} |R'_{j}{\setminus } \{0\}|=|S'|$.

• 3. If there is no path between a and b, we have the result trivially; $d(a,b)=\infty > s_{b-a}$. Now, let $a, c_{1},\ldots ,c_{s},b$ be the shortest path between a and b. By part (2), for all $1\le i\le s$, $c_{i}$ and $c_{i+1}$ are different exactly in one component. Let $c_{ij}\ne c_{(i+1)j}$ in $R'_{j}$ and $c_{(i+1)k}\ne c_{(i+2)k}$ in $R'_{k}$. If $c_{ik}=c_{(i+2)k}$, then $c_{i}$ and $c_{i+2}$ are adjacent; a contradiction. Hence $c_{i}$ and $c_{i+2}$ are different in at least two components. By induction, $c_{i}$ and $c_{i+j}$ are different in at least j components. Thus, $d(a,b)> s_{b-a}$. If $S'=S$, we have the result by Theorem 2.3 in [8].

• 4. It follows by the same argument as Theorem 2.4 in [8].

$\square$

Theorem 2

Let $\Gamma [S']$ be the induced subgraph of $\Gamma =Cay(R,S')$ by $S'=\cup R'_{i}{\setminus } \{0\}$, and $\Gamma [R'_{i}]$ be the induced subgraph by $R'_{i}$. If $R'_{i}$ is an additive group for all i, then

• 1. $\Gamma [R'_{i}]$ is a complete graph and hence is a clique of $\Gamma$;

• 2. $\Gamma [S]=\bigcup _{i\in {\mathbb {Z}}}\Gamma [R'_{i}{\setminus } \{0\}]$.

Proof

Let x, y be two distinct vertices of $R'_{i}$. Since $R'_{i}$ is an additive group, $x-y\in R'_{i}{\setminus } \{0\}\subseteq S$ and hence x, y are adjacent. This completes the proof of part (1). Part (2) is followed by the same argument as Theorem 2.6 in [8]. $\square$

Now let $R=\oplus _{i\ge 0} R_{i}$ be a positively graded ring of finite Krull’s dimension d. If $\lambda _{R_{0}}(R_{i})$ denotes the length of $R_{i}$ as an $R_{0}$-module, then the Hilbert function $H_{R}:{\mathbb {Z}}\longrightarrow {\mathbb {Z}}$ of R is defined by $H_{R}(i)=\lambda _{R_{0}}(R_{i})$ for all $i\ge 0$. By Theorem 4.1.3 in [4], there exists a unique polynomial $Q_{R}(x)\in {\mathbb {Q}}[x]$ of degree $d-1$ such that $H_{R}(i)=Q_{R}(i)$ for all sufficiently large integers i. This polynomial is called the Hilbert polynomial. The following proposition gives the clique number and independence number of some generalized Cayley graphs over graded rings.

Proposition 1

Let $R'=\oplus _{i\ge 0} R'_{i}$ be a graded subring of a positively graded ring $R=\oplus _{i\ge 0} R_{i}$. If $S'=\cup _{i}R'_{i}{\setminus } \{0\}$, $R'_{0}$ is a finite field and $\Gamma =Cay(R,S')$, then

• 1. $\omega (\Gamma )=\max \{ |R'_{0}|^{H_{R'}(i)}:\ i\ge 0\}$;

• 2. If $Q_{R'}(x)\in {\mathbb {Q}}[x]$ is the polynomial function of $R'$, then for sufficiently large integers i, $\Gamma$ has cliques of size $|R'_{0}|^{Q_{R'}(i)}$;

• 3. If $Q_{R'}(x)\ne 0$, then the independence number $\alpha (\Gamma )$ is infinite.

Proof

Since $R'$ is a graded subring of R, $S'=\cup _{i}R'_{i}{\setminus } \{0\}= \cup _{i}(R_{i}\cap R'){\setminus } \{0\}$. Hence, $S'$ is a subset of S, and so we have the generalized Cayley graph $\Gamma =Cay(R,S')$. Now, $R_{0}$ is a finite field; hence, $H_{R'}(i)=\lambda _{R'_{0}}(R'_{i})=\dim _{R'_{0}}R'_{i}$. Thus $|R'_{i}|=|R'_{0}|^{H_{R'}(i)}$. Now, we have part (1) by part (4) of Theorem 1. Also by the above theorem, $\Gamma [R'_{i}]$ is a clique of $\Gamma$. By above discussion and part (1), $|R'_{i}|=|R'_{0}|^{H_{R'}(i)}=|R'_{0}|^{Q_{R'}(i)}$ for sufficiently large integers i. This completes the proof of part (2). To prove part (3), consider the integer t such that $H_{R'}(i)=Q_{R'}(i)$ for all $i \ge t$. Then, $R'_{i}\ne 0$ for all $i\ge t$. For every $i \ge t$, let $x_{i }\in R'_{i}{\setminus } \{0\}$. In this case, $X= \{x_{t}, x_{t+1}, \ldots \}$ is an infinite independent set. This shows that $\alpha (\Gamma )$ is infinite. $\square$

Definition 8

Let R and T be graded rings and $f : R \longrightarrow T$ be a ring homomorphism. Then f is called a graded ring homomorphism of degree d if $f(R_{i})\subseteq T_{i+d}$ for all i. The homogeneous homomorphism is a graded homomorphism of degree $d=0$. Moreover, two graded rings R and T are said to be isomorphic as graded rings if there exists a homogeneous ring isomorphism between them.

Theorem 3

Let R and T be graded rings and $f : R \longrightarrow T$ be a graded ring homomorphism of degree d. If $S_{R}$ and $S_{T}$ are the homogenous sets of R and T, respectively, then

• 1. $f : \mathrm{Cay}(R, S'_{R}) \longrightarrow \mathrm{Cay}(T, S'_{T})$ is a graph homomorphism if and only if $f(S'_{R})\subseteq S'_{T}$;

• 2. If f is a graded ring isomorphism, then $\mathrm{Cay}(R, S'_{R})$ and $\mathrm{Cay}(T, f(S'_{R}))$ are isomorphic.

Proof

Let f be a graph homomorphism and $x\in S'_{R}$. Then, $x\sim 0_{R}$ and $f(x)\sim f(0_{R})$ . Hence, $f(x)=f(x)-0_{T}=f(x)-f(0_{R})\in S'_{T}$. This shows that $f(S'_{R})\subseteq S'_{T}$. Conversely, let $x\sim y$ in $\mathrm{Cay}(R, S'_{R})$. Then, $f(x)-f(y)=f(x-y)\in f(S'_{R})\subseteq S'_{T}$ and hence $f(x)\sim f(y)$. Thus, f is a graph homomorphism. This completes the proof of part (1). To prove part (2), note that f is a graded homomorphism and hence $f(S'_{R})\subseteq S'_{T}$. Now, we have the result by part (1). $\square$

By Remark 3, any coaddition graph over a ring is a generalized Cayley graph where the ring is considered a graded ring trivially. In particular, counit graph can be seen as a generalized Cayley graph. In the following proposition, we give relations between counit and generalized graphs in the case that grading is not trivial.

Proposition 2

Let $R=\oplus R_{i}$ be a graded ring, S be the set of homogeneous elements of R and $\Gamma$ be the counit graph over R. Then,

• 1. If R is a graded domain and U is the set of unit elements, then $\Gamma$ is the generalized graph $\mathrm{Cay}(R,S')$ where $S'=U$.

• 2. If $R_{0}= \mathrm{k}$ is a field and either $R = \mathrm{k}$ or $R = \mathrm{k}[t, t^{-1}]$ for some homogeneous element $t \in R$ of positive degree which is transcendental over $\mathrm{k}$, then for any subset $S'$ of homogeneous elements, the generalized Cayley graph $\mathrm{Cay}(R,S')$ is a subgraph of the counit graph over R.

Proof

• 1. By [10, Exercise 1.1], all units in a graded domain are homogeneous. Hence, $U\subseteq S$ which shows that $\Gamma$ is the generalized graph.

• 2. By [4, Lemma 1.5.7], $R_{0}= \mathrm{k}$ is a field and either $R = \mathrm{k}$ or $R = \mathrm{k}[t, t^{-1}]$ if and only if every nonzero homogeneous element is invertible. Thus, $S\subseteq U$ and hence $\mathrm{Cay}(R,S)$ is a subgraph of the counit graph $\mathrm{Cay}(R,U)$.

$\square$

Theorem 4

Let $R=R_{0}\oplus \cdots \oplus R_{d-1}$ be a positively finite graded ring. If $\mathrm{Cay}(R, S')$ denotes the generalized Cayley graph where $S'=\cup ^{d-1}_{i=0} R'_{i}$, then

• 1. If $n= |\{i:\ R'_{i}\ne 0\}|$, then $\mathrm{diam}(\mathrm{Cay}(R, S'))=n$;

• 2. If $0\le r\le d$ and $a\in R$, then

  $$\begin{aligned} |\mathrm{N}_{r}(a)|=\underset{0\le i_{1}< \cdots < i_{r}\le d}{\sum }(|R'_{i_{1}}|-1)\cdots (|R'_{i_{r}}|-1). \end{aligned}$$

Proof

See Theorem 3.2 for $\mathrm{Cay}(R, S)$ in [8]. The proofs are the same. $\square$

Example 4

Let $R= {\mathbb {F}}_{q}[x,y,z]/ \langle x^3,x^2y^3,y^4\rangle$ and $R'={\mathbb {F}}_{q}[x,y]/ \langle x^3,x^2y^3,y^4\rangle$. Let S and $S'$ be the set of homogenous elements of R and $R'$; respectively. It is easy to see that $R'$ is a graded subring of R. Hence, $S'\subseteq S$. Let $\Gamma =\mathrm{Cay}(R, S')$. Then,

• 1. R is an infinite ring and hence $\Gamma$ is an infinite graph.

• 2. The Hilbert function of $R'$ is defined as follows: $H(0) = 1, H(1) = 2, H(2) = 3, H(3) = 3, H(4) = 2, H(t) = 0$ for $t \ge 5$.

• 3. $|S'|=\sum ^{i=4}_{i=0}|R_{0}|^{H(i)}=q+2 q^{2}+2q^{3}$. Hence, $\Gamma$ is a $(q+2 q^{2}+2q^{3})$-regular graph.

• 4. $\omega (\Gamma )=\max \{|R_{0}|^{H(i)}:\ i=0,1,2,3,4\}=q^{3}$.

• 5. By the above theorem, $\mathrm{diam}(\Gamma )=5$.

Now, we give relations between Cartesian product of Cayley graphs over abelian groups and the generalized Cayley graphs over graded rings.

Theorem 5

Let $R=\oplus _{i}R_{i}$ be a graded ring. Then,

• 1. $\mathrm{Cay}(R, S')\cong \prod _{i}\mathrm{Cay}(R_{i}, S'\cap R_{i})$.

• 2. If R is an Artinian graded ring, then there are homogeneous components $R_{i_{1}}, \ldots , R_{i_{n}}$ such that $Cay(R,S')$ is isomorphic to $\prod ^{n}_{j=1}\mathrm{Cay}(R_{i_{j}}, S'\cap R_{i_{j}})$.

Proof

• 1. Define $f: R\longrightarrow \prod _{i} R_{i}$ by $f(\sum _{i} x_{i})=(x_{0}, \ldots , x_{i},\ldots )$. Clearly, f is bijective function. Now, let $x=\sum x_{i}$ and $y=\sum y_{i}$ be adjacent; then, $x-y\in S'$. Hence, $x-y=x_{j}-y_{j}\in R_{j}$ for some j. Thus, $x_{j}\sim y_{j}$ in $\mathrm{Cay}(R_{j}, S'\cap R_{j})$ for some j, and $x_{i}=y_{i}$ for all i. Hence, $f(x)\sim f(y)$ in $\prod _{i}\mathrm{Cay}(R_{i}, S'\cap R_{i})$. The converse part is proved by the same argument.

• 2. By Corollary 4.2 in [10], $R_{j}=0$ for all but finitely many j. Let $R_{i_{1}}, \ldots , R_{i_{n}}$ be the nonzero homogeneous components of R. By part (1), $Cay(R,S')$ is isomorphic to $\prod ^{n}_{j=1}\mathrm{Cay}(R_{i_{j}}, S'\cap R_{i_{j}})$.

$\square$

By the above theorem, the generalized Cayley graph $\mathrm{Cay}(R, S')$ is isomorphic to the product of Cayley graphs over abelian groups. Now, we show that any Cartesian product of Cayley graphs over cyclic groups of prime power orders is isomorphic to a generalized Cayley graph over a graded ring.

Proposition 3

Let $G_{0}, \ldots , G_{n-1}$ be cyclic groups such that $|G_{i}|\mid |G_{i-1}|$ for $1\le i\le n-1$. If $\mathrm{Cay}(G_{i}, H_{i})$ denotes a Cayley graph over $G_{i}$, then there exists a graded ring R with a subset $S'$ of homogeneous elements such that

$$\begin{aligned} \prod ^{n-1}_{i=0}\mathrm{Cay}(G_{i}, H_{i})\cong \mathrm{Cay}(R, S'). \end{aligned}$$

Proof

Let $|G_{i}|=m_{i}$ for $0\le i\le n-1$. Consider the ring ${\mathbb {Z}}_{m}$, where $m=m_{0}m_{1}\ldots m_{n-1}$. Define the ideals

$$\begin{aligned} I_{0}&=m_{0}{\mathbb {Z}}_{m},\\ I_{1}&=(m_{0}m_{1}){\mathbb {Z}}_{m},\\&\vdots \\ I_{n-1}&=(m_{0}m_{1}\cdots m_{n-1}){\mathbb {Z}}_{m} \end{aligned}$$

of ${\mathbb {Z}}_{m}$. Also let $I_{j}=m {\mathbb {Z}}_{m}$ for all $j\ge n$. It is easy to see that ${\mathcal {I}}=\{I_{j}\}^{\infty }_{j=0}$ is a filtration of ${\mathbb {Z}}_{m}$. Consider the associated graded ring $G({\mathcal {I}})$ of ${\mathcal {I}}$. Then,

$$\begin{aligned} G({\mathcal {I}}) =&\oplus \frac{I_{j}}{I_{j+1}}\\ =&\frac{{\mathbb {Z}}_{m}}{I_{0}}\oplus \frac{I_{0}}{I_{1}} \oplus \frac{I_{1}}{I_{2}}\oplus \cdots \\ =&\frac{{\mathbb {Z}}_{m}}{I_{0}}\oplus \frac{I_{0}}{I_{1}} \oplus \cdots \oplus \frac{I_{n-2}}{I_{n-1}}\\ :=&R_{0}\oplus R_{1} \oplus \cdots \oplus R_{n-1}. \end{aligned}$$

But

$$\begin{aligned} |I_{j}|=\frac{m}{\gcd (m, m_{0}m_{1}\cdots m_{j})}=m_{j+1}\cdots m_{n-1}. \end{aligned}$$

Hence,

$$\begin{aligned} |R_{j}|=\frac{|I_{j-1}|}{|I_{j}|}=\frac{m_{j}\cdots m_{n-1}}{m_{j+1}\cdots m_{n-1}}=m_{j}. \end{aligned}$$

Moreover, $R_{j}=\langle m_{0}\cdots m_{j-1}+ I_{j}\rangle$ as an additive group. Since cyclic groups of the same order are isomorphic, $G_{j} \cong R_{j}$ for $0\le j\le n-1$. Let $f_{j}:G_{j}\longrightarrow R_{j}$ be the isomorphism groups and $S'= \cup _{j} f_{j}(H_{j})$. We show that the following map is a graph isomorphism from $\prod ^{n-1}_{i=0}\mathrm{Cay}(G_{i}, H_{i})$ to $\mathrm{Cay}(G({\mathcal {I}}), S')$:

$$\begin{aligned} f: \prod ^{n-1}_{j=0} G_{j}\longrightarrow&G({\mathcal {I}})\\ (x_{j})\longmapsto&\sum ^{n-1}_{j=0}f_{j}(x_{j}). \end{aligned}$$

Clearly, this map is bijective. Now let $(x_{j})\sim (y_{j})$ in $\prod ^{n-1}_{i=0}\mathrm{Cay}(G_{i}, H_{i})$. Then $x_{t}\sim y_{t}$ in $\mathrm{Cay}(G_{t}, H_{t})$ for some $0\le t\le n-1$, and $x_{j}=y_{j}$ for all $j\ne t$. Hence $x_{t}y^{-1}_{t}\in H_{t}$. Therefore,

$$\begin{aligned} f(x_{0}, \ldots , x_{n-1})-f(y_{0}, \ldots , y_{n-1})=&\sum ^{n-1}_{j=0}f_{j}(x_{j})-f_{j}(y_{j})\\ =&f_{t}(x_{t})-f_{t}(y_{t})\\ =&f_{t}(x_{t}y^{-1}_{t})\in f_{t}(H_{t})\subseteq S'. \end{aligned}$$

Thus, $f(x_{0}, \ldots , x_{n-1})\sim f(y_{0}, \ldots , y_{n-1})$. This completes the proof. $\square$

Lemma 1

Let $G_{1}$ and $G_{2}$ be cyclic groups with $\gcd (|G_{1}|, |G_{2}|)=1$. If $\mathrm{Cay}(G_{1}, H_{1})$ and $\mathrm{Cay}(G_{2}, H_{2})$ are Cayley graphs over $G_{1}$ and $G_{2}$, then

$$\begin{aligned} \mathrm{Cay}(G_{1}, H_{1})\times \mathrm{Cay}(G_{2}, H_{2})\cong \mathrm{Cay}({\mathbb {Z}}_{m}, H), \end{aligned}$$

where $m=|G_{1}||G_{2}|$ and H is a subset of ${\mathbb {Z}}_{m}$.

Proof

Let $|G_{1}|=m_{1}$ and $|G_{2}|=m_{2}$; then, $G_{1}\cong {\mathbb {Z}}_{m_{1}}$ and $G_{2}\cong {\mathbb {Z}}_{m_{2}}$. Since $\gcd (m_{1}, m_{2})=1$, ${\mathbb {Z}}_{m_{1}}\times {\mathbb {Z}}_{m_{2}}\cong {\mathbb {Z}}_{m}$. Hence, we have the group isomorphism $f: G_{1}\times G_{2}\longrightarrow {\mathbb {Z}}_{m}$. Let $H=f(\{0\}\times H_{2})\cup f(H_{1}\times \{0\}){\setminus } \{{\overline{0}}\}$. We show that f is a graph isomorphism from $\mathrm{Cay}(G_{1}, H_{1})\times \mathrm{Cay}(G_{2}, H_{2})$ to $\mathrm{Cay}({\mathbb {Z}}_{m}, H)$. The vertices (a, b) and (c, d) are adjacent in $\mathrm{Cay}(G_{1}, H_{1})\times \mathrm{Cay}(G_{2}, H_{2})$ if and only if $a=c$ and $b\sim d$ in $\mathrm{Cay}(G_{2}, H_{2})$ (or $a\sim c$ in $\mathrm{Cay}(G_{1}, H_{1})$ and $b=d$). But we have this if and only if $a-c=0$ and $b-d\in H_{2}$, if and only if $f(a,b)-f(c,d)=f(0, b-d)\in f(\{0\}\times H_{2})\subseteq H$ (or $f(a,b)-f(c,d)=f(a-c, 0)\in f(H_{1}\times \{0\})\subseteq H$), if and only if $f(a,b)\sim f(c,d)$. This completes the proof. $\square$

The following theorem shows that any Cartesian product of Cayley graphs over cyclic groups of prime power orders is isomorphic to the generalized Cayley graph $\mathrm{Cay}(R,S')$ for some graded ring R.

Theorem 6

Let $G_{0}, \ldots , G_{n-1}$ be cyclic groups such that $|G_{i}|\ne |G_{j}|$ for $0\le i< j \le n-1$. Also let $|G_{i}|=p^{m_{i}}_{i}$ for $0\le i\le n-1$, where $p_{i}$ is a prime number and $m_{i}$ is a positive integer. If $\mathrm{Cay}(G_{i}, H_{i})$ denotes a Cayley graph over $G_{i}$, then there exists a graded ring R with a subset $S'$ of homogeneous elements such that

$$\begin{aligned} \prod ^{n-1}_{i=0}\mathrm{Cay}(G_{i}, H_{i})\cong \mathrm{Cay}(R, S'). \end{aligned}$$

Proof

Let $n_{0}=\mathrm{lcm}\{|G_{i}|: \ 0\le i\le n-1\}$ be the least common multiple of $|G_{0}|, \ldots , |G_{n-1}|$. For $j\ge 1$, define $n_{j}=\mathrm{lcm}\{|G_{i}|:\ \gcd (|G_{i}|,\frac{n_{j-1}}{|G_{i}|})\ne 1\}$. Then, there exists an integer $0\le t\le n-1$ which $\prod ^{n-1}_{i=0} |G_{i}|=\prod ^{t}_{j=0} n_{j}$. Now by above lemma, there are some subsets $H'_{j}\subseteq {\mathbb {Z}}_{n_{j}}$ such that

$$\begin{aligned} \prod ^{n-1}_{i=0}\mathrm{Cay}(G_{i}, H_{i})\cong \prod ^{t}_{j=0}\mathrm{Cay}({\mathbb {Z}}_{n_{j}}, H'_{j}). \end{aligned}$$

Since $n_{j}\mid n_{j-1}$ for $1\le j\le t$, by Proposition 3, there exists a graded ring R with a subset $S'$ of homogeneous elements such that

$$\begin{aligned} \prod ^{t}_{j=0}\mathrm{Cay}({\mathbb {Z}}_{n_{j}}, H'_{j})\cong \mathrm{Cay}(R, S'). \end{aligned}$$

This completes the proof. $\square$

The following corollary shows that any Hamming graph $\Gamma$ over sets of prime power sizes is isomorphic to the generalized Cayley graph $\mathrm{Cay}(R, S')$ for some graded ring R.

Corollary 1

Let $\Gamma =\Gamma _{0}\times \cdots \times \Gamma _{n-1}$ be the Hamming graph over $G_{0}, \ldots , G_{n-1}$ such that $|G_{i}|\ne |G_{j}|$ for $0\le i< j \le n-1$. Also for all i, assume that $|G_{i}|=p^{m_{i}}_{i}$, where $p_{i}$ is a prime number and $m_{i}$ is a positive integer. Then, there exists a finite graded ring R with a subset $S'$ of homogeneous elements such that $\Gamma$ is isomorphic to the generalized Cayley graph $\mathrm{Cay}(R,S')$.

Proof

By Remark 1, the Hamming graph can be considered a product of Cayley graphs over cyclic groups. Now, by the above theorem, we have the result. $\square$

4 The Cayley Graph $\mathrm{Cay}(R,S'')$ over Graded Rings

Let $R=\oplus _{i}R_{i}$ be a graded ring. For any integer k, define $R_{\ge k}=\oplus _{i\ge k}R_{i}$ and $R_{\le k}=\oplus _{i\le k}R_{i}$. Also the beginning and ending of R is defined as follows:

$$\begin{aligned} \mathrm{beg}(R)= \inf \{i\in {\mathbb {Z}}:\ R_{i}\ne 0\}; \ \ \ \mathrm{end}(R)=\sup \{i\in {\mathbb {Z}}:\ R_{i}\ne 0\}. \end{aligned}$$

Definition 9

Let $R=\oplus _{i}R_{i}$ be a graded ring and $S''=R_{\ge k}{\setminus } \{0\}$ for some integer $\mathrm{beg}(R) \le k\le \mathrm{end}(R)$. The Cayley graph $\mathrm{Cay}(R,S'')$ is a graph, whose vertex set is R and two vertices $a, b \in R$ are adjacent whenever $b-a\in S''$.

In this section, we study $\mathrm{Cay}(R,S'')$ and give its relation with total, cototal and counit graphs.

Clearly,

$$\begin{aligned} R=R_{\le k-1}\oplus R_{\ge k}=\{x+x''; \ x\in R_{\le k-1}, x''\in R_{\ge k}\}. \end{aligned}$$

By this notation, two vertices $x+x''$ and $y+y''$ are adjacent if and only if $x=y$.

Theorem 7

Let $\Gamma =Cay(R,S'')$ be the above Cayley graph. Then,

• 1. $\Gamma$ is connected if and only if $k=\mathrm{beg}(R)$.

• 2. $\Gamma$ is an $|S''|$-regular graph.

• 3. If a, b are two vertices of $\Gamma$, then the distance of a and b is $d(a,b)\in \{1,\infty \}$.

• 4. $\Gamma$ is a complete graph or a non connected regular graph.

Proof

• 1. By the structure of Cayley graphs, $\Gamma$ is connected if and only if R is generated by $S''$ under addition. But R is generated by $S''$ if and only if $k=\mathrm{beg}(R)$.

• 2. Let $x+x''\in R$. We find all vertices $y+y''$ such that $x+x''\thicksim y+y''$. Let $x+x''\thicksim y+y''$; hence, $y+y''-(x+x'')=(y-x)+(y''-x'')\in R_{\ge k}{\setminus } \{0\}$ and hence $y-x=0$ and $y''-x''\in R_{\ge k}{\setminus } \{0\}$. Clearly, the number of such elements is equal to $|R_{\ge k}{\setminus } \{x''\}|=|S''|$. This completes the proof.

• 3. Let $a=x+x''$ and $b=y+y''$ be two elements of R. There are two cases; $x=y$ or $x\ne y$. If $x=y$, then $a\sim b$ and hence $d(a,b)=1$. Now, let $x\ne y$, and $a, a_{1},\ldots , a_{i}, \ldots ,a_{n}, b$ be a path between a and b where $a_{i}=x_{i}+x''_{i}$. Then, by induction, $x=x_{1}= \cdots =x_{i}=\cdots =x_{n}=y$; a contradiction. Hence, there is no path between a and b. In this case, $d(a,b)=\infty$.

• 4. It follows from parts (1) and (2).

$\square$

Theorem 8

Let $\Gamma =Cay(R,S'')$, then $\omega (\Gamma )=|S''|+1$.

Proof

Let $C\subseteq \Gamma$ be a clique and $x+x''$ be a vertex of C. By part (2) of above theorem, all vertices that are adjacent with $x+x''$ are

$$\begin{aligned} C_{x}=\{x+y'':\ y''\in R_{\ge k}{\setminus } \{x''\}\}. \end{aligned}$$

Hence, $C\subseteq C_{x}\cup \{x+x'' \}$. It is easy to see that $C_{x}\cup \{x+x'' \}$ is a clique of $\Gamma$. Therefore,

$$\begin{aligned} C=C_{x}\cup \{x+x'' \}=\{x+x'':\ x''\in R_{\ge k}\}. \end{aligned}$$

This completes the proof. $\square$

The following theorem shows that R is a union of complete graphs.

Theorem 9

Let $\Gamma =Cay(R,S'')$ and for $x\in R_{\le k-1}$ let $R_{x}=\{ x+y;\ \ y\in R_{\ge k}\}$. If $\Gamma [R_{x}]$ is the induced subgraph of $\Gamma$, then

• 1. $\Gamma [R_{x}]$ is a complete graph.

• 2. $\Gamma =\bigcup _{x\in R_{\le k-1}}\Gamma [R_{x}]$.

Proof

• 1. Let $x+y_{1}, x+y_{2}\in R_{x}$. Then, $(x+y_{2})-(x+y_{1})=y_{2}-y_{1}\in R_{\ge k}{\setminus } \{0\}$ and hence $(x+y_{1})\sim (x+y_{2})$. Thus, $\Gamma [R_{x}]$ is a complete graph.

• 2. Clearly, $R=\bigcup _{x\in R_{\le k-1}} R_{x}$. Now let $(x_{1}+y_{1}) (x_{2}+y_{2})$ be an edge of $\Gamma$. Then, $(x_{2}+y_{2})- (x_{1}+y_{1})\in R_{\ge k}$. Hence, $x_{1}=x_{2}$ which shows that $(x_{1}+y_{1})$ and $(x_{2}+y_{2})$ are vertices of $\Gamma [R_{x_{1}}]$. This completes the proof.

$\square$

Theorem 10

Let $\Gamma =Cay(R,S'')$. Then,

• 1. $\mathrm{diam}(\Gamma )= \left\{ \begin{array}{ll} \infty , &{} \hbox {k} >\mathrm{beg}\hbox {(R)}; \\ 1, &{} \hbox {k}=\mathrm{beg}\hbox {(R).} \end{array} \right.$

• 2. $\mathrm{gr}(\Gamma )\in \{3,\infty \}$, and $\mathrm{gr}(\Gamma )=3$ if and only if $|S''|\ge 2$.

• 3. $\Gamma$ is planar if and only if $|S''|\le 3$.

• 4. $\Gamma$ is a complete graph if and only if $k=\mathrm{beg}(R)$.

• 5. $\Gamma$ is a path if and only if $R\cong {\mathbb {Z}}_{2}$.

• 6. $\Gamma$ is a cycle if and only if $R\cong {\mathbb {Z}}_{3}$.

Proof

• 1. If $k=\mathrm{beg}(R)$ then $\Gamma$ is a complete graph and hence $\mathrm{diam}(\Gamma )=1$. If $k >\mathrm{beg}(R)$, then by Theorem 7, $\Gamma$ is not connected. In this case, $\mathrm{diam}(\Gamma )=\infty$.

• 2. It follows from Theorem 8.

• 3. By Theorem 6.2.2 in [12], a graph is planar if and only if it contains no subdivision of $K_{5}$ or $K_{3,3}$. Let $\Gamma$ be planar. Then, $K_{5}$ is not a subgraph of $\Gamma$. Hence, $|S''|+1=\omega (\Gamma )\le 4$. Conversely, let $|S''|\le 3$; then by Theorem 8, $\omega (\Gamma )=|S''|+1\le 4$ which shows that $K_{5}$ is not a subgraph of $\Gamma$. Now, let

  $$\begin{aligned} \Lambda =\{x_{i}+x''_{i}:\ x_{i}\in R_{\le k-1}, x''_{i}\in R_{\ge k},i=1,..,6\} \end{aligned}$$

  contains $K_{3,3}$. Suppose $\{x_{1}+x''_{1}, x_{2}+x''_{2}, x_{3}+x''_{3}\}$ is a part of $K_{3,3}$. Then, $x_{i}+x''_{i}\sim x_{j}+x''_{j}$ for $i=1,2,3$ and $j=4,5,6$. Thus, $x_{i}=x_{j}$, and hence, $x_{1}=x_{4}=x_{2}=x_{5}=x_{3}=x_{6}$. This proves that $\Lambda$ is a clique of size 6, a contradiction. This completes the proof.

• 4. It is a result of Theorem 9.

• 5. If $R\cong {\mathbb {Z}}_{2}$, then $\Gamma =K_{2}$ is a path. Conversely, if $\Gamma$ is a path, then it is connected. By Theorem 7, $\Gamma$ is a complete graph. The only complete path is $K_{2}$ which proves that $R\cong {\mathbb {Z}}_{2}$.

• 6. It follows by the same argument as part (5).

$\square$

Theorem 11

Let T be a finite commutative local ring with the maximal ideal ${\mathfrak {m}}$. If $2\in {\mathfrak {m}}$, then the total graph of T is isomorphic to a Cayley graph $\mathrm{Cay}(R, S'')$ for some graded ring R and a subset $S''$.

Proof

T is a finite ring. Hence, every element of T is invertible or a zero divisor. Since T is local, $T{\setminus } {\mathfrak {m}}$ is the set of invertible elements. Hence, ${\mathfrak {m}}=\mathrm{Z}(T)$ is the set of zero divisors of T. By Lemma 2.3 in [3], the total graph $\mathrm{Cay}^{+}(T, \mathrm{Z}(T))$ is isomorphic to

$$\begin{aligned} \underbrace{K_{\alpha }\cup \cdots \cup K_{\alpha }}_{\beta \ \text {copies}}, \end{aligned}$$

where $|\mathrm{Z}(T)|=\alpha$ and $|T/\mathrm{Z}(T)|=\beta$.

Note that ${\mathcal {M}}=\{{\mathfrak {m}}^{i}\}^{\infty }_{i=0}$ is a filtration of T. Hence,

$$\begin{aligned} R=G({\mathcal {M}})=\frac{T}{{\mathfrak {m}}}\oplus \cdots \oplus \frac{{\mathfrak {m}}^{e-1}}{{\mathfrak {m}}^{e}} \end{aligned}$$

is a finite graded ring where e is the nilpotency index of ${\mathfrak {m}}$. In other words, ${\mathfrak {m}}^{e}=0$, and hence $|R|=|T|$. Now, if $S''=R_{\ge 1}$, then by Theorem 9,

$$\begin{aligned} \mathrm{Cay}(R, S'')=\bigcup _{x\in \frac{T}{{\mathfrak {m}}}}\Gamma [R_{x}] \end{aligned}$$

where $|\frac{T}{{\mathfrak {m}}}|=|T/\mathrm{Z}(T)|=\beta$ and $\Gamma [R_{x}]$ is a complete graph of size $|R_{x}|=|R_{\ge 1}|=\alpha$. This completes the proof. $\square$

Theorem 12

Let R and T be graded rings, $S''_{R}=R_{\ge k}{\setminus } \{0\}$, $S''_{T}=T_{\ge k}{\setminus }\{0\}$ and $f : R \longrightarrow T$ be a graded ring homomorphism of degree d. Then,

• 1. $f : \mathrm{Cay}(R, S''_{R}) \longrightarrow \mathrm{Cay}(T, S''_{T})$ is a graph homomorphism.

• 2. If f is a graded ring isomorphism, then $\mathrm{Cay}(R, S''_{R})$ and $\mathrm{Cay}(T, S''_{T})$ are isomorphic.

Proof

Let $x,y\in R$ and $x\sim y$. Then, $x-y\in R_{\ge k}$. Hence,

$$\begin{aligned} f(x)-f(y)=f(x-y)\in f(R_{\ge k})\subseteq T_{\ge k+d} \subseteq T_{\ge k}. \end{aligned}$$

Thus, $f(x)\sim f(y)$. Part (2) is a direct result of part (1) and Definition 8. $\square$

Consider $\Gamma =Cay(R,S'')$ with $S''=R_{\ge k}{\setminus } \{0\}$. We have

$$\begin{aligned} R=R_{\le k-1}\oplus R_{\ge k}=\{x+x''; \ x\in R_{\le k-1}, \ x''\in R_{\ge k}\}. \end{aligned}$$

By this notation, two vertices $x_{1}+x''_{1}$ and $x_{2}+x''_{2}$ are adjacent if and only if $x_{1}=x_{2}$. Hence, the complement of $\Gamma$, denoted by ${\overline{\Gamma }}$, is a graph with vertex set R, and two vertices $x_{1}+x''_{1}$ and $x_{2}+x''_{2}$ are adjacent if and only if $x_{1}\ne x_{2}$. The following theorem gives the structure of ${\overline{\Gamma }}$.

Theorem 13

Let $\Gamma =Cay(R,S'')$ with $S''=R_{\ge k}{\setminus } \{0\}$ and ${\overline{\Gamma }}$ be the complement of $\Gamma$. Then,

• 1. If $k=\mathrm{beg}(R)$, then ${\overline{\Gamma }}$ is totally disconnected.

• 2. If $k>\mathrm{beg}(R)$, then ${\overline{\Gamma }}$ is a complete $|R_{\le k-1}|$-partite graph where every part has $|S''|+1$ vertices.

• 3. For two vertices a, b in ${\overline{\Gamma }}$, $d(a,b)\in \{1,2,\infty \}$.

• 4. $\mathrm{diam}({\overline{\Gamma }})= \left\{ \begin{array}{ll} \infty , &{} \hbox {k} =\mathrm{beg}\hbox {(R)}; \\ 2, &{} \hbox {k} >\mathrm{beg}\hbox {(R)}. \end{array} \right.$

• 5. $\mathrm{gr}({\overline{\Gamma }})\in \{4, \infty \}$.

• 6. If $k>\mathrm{beg}(R)$, then ${\overline{\Gamma }}$ is planar if and only if $|R_{\ge k}|\le 2$ and $|R_{\le k-1}|\le 4$.

Proof

• 1. If $k=\mathrm{beg}(R)$, then by part (4) of Theorem 10, $\Gamma$ is a complete graph. Hence, ${\overline{\Gamma }}$ is totally disconnected.

• 2. Since $k> \mathrm{beg}(R)$, $|R_{\le k-1}|\ge 2$. Hence by Theorem 9,

  $$\begin{aligned} \Gamma =\bigcup _{x\in R_{\le k-1}}\Gamma [R_{x}] \end{aligned}$$

  where $R_{x}=\{ x+y:\ \ y\in R_{\ge k}\}$ and $\Gamma [R_{x}]$ is a complete graph. Thus, ${\overline{\Gamma }}$ is a complete $|R_{\le k-1}|$-partite graph where every part has $|R_{x}|=|R_{\ge k}|=|S''|+1$ vertices.

• 3. If $k=\mathrm{beg}(R)$, then by part (1), $d(a,b)=\infty$ for $a, b\in R$. If $k> \mathrm{beg}(R)$, then by part (2), $d(a,b)\in \{1,2\}$.

• 4. It follows from part (3).

• 5. By parts (1) and (2), ${\overline{\Gamma }}$ is a complete $|R_{\le k-1}|$-partite graph or totally disconnected. Hence, $\mathrm{gr}({\overline{\Gamma }})$ is equal to 4 or $\infty$.

• 6. By part (2), ${\overline{\Gamma }}$ is a complete $|R_{\le k-1}|$-partite graph. Hence, it contains $K_{5}$ if and only if $|R_{\le k-1}|\ge 5$. Since every part of ${\overline{\Gamma }}$ has $|R_{\ge k}|$ vertices, it contains $K_{3,3}$ if and only if $|R_{\ge k}|\ge 3$. This completes the proof.

$\square$

Let $H_{R}:{\mathbb {Z}}\longrightarrow {\mathbb {Z}}$ be the Hilbert function of a graded ring R. Define the Poincare series (or Hilbert series) of R to be $P_{R}(t)=\Sigma _{i\in {\mathbb {Z}}}H_{R}(i)t^{i}$.

Corollary 2

Let $R=\oplus _{i\ge 0} R_{i}$ be a positively graded ring such that $R_{0}$ is a finite field. For $k\ge 1$, let $\Gamma =Cay(R,S'')$ with $S''=R_{\ge k}{\setminus } \{0\}$. If $m=\sum ^{k-1}_{i=0}H_{R}(i)$, then

• 1. $\alpha (\Gamma )=|R_{0}|^{m}$;

• 2. If $P_{R}(1)< \infty$, then $\omega (\Gamma )= |R_{0}|^{P_{R}(1)-m}$;

• 3. If $P_{R}(1)=\infty$, then $\Gamma$ is infinite graph with $\omega (\Gamma )=\infty$;

• 4. for all $i\ge k$, $\Gamma [R_{i}]$ is a clique of size $|R_{0}|^{H_{R}(i)}$. In particular, for sufficiently large integers i, $\Gamma [R_{i}]$ is a clique of size $|R_{0}|^{Q_{R}(i)}$ where $Q_{R}(x)\in {\mathbb {Q}}[x]$ is the Hilbert polynomial of R.

• 5. ${\overline{\Gamma }}$ is a complete $\alpha (\Gamma )$-partite graph where every part has $|\omega (\Gamma )|$ vertices.

Proof

Since $R_{0}$ is a finite field, $H_{R}(i)=\lambda _{R_{0}}(R_{i})=\dim _{R_{0}}R_{i}$ and hence

$$\begin{aligned} P_{R}(t)=\Sigma _{i\in {\mathbb {Z}}}H_{R}(i)t^{i}=\Sigma _{i\in {\mathbb {Z}}}\dim _{R_{0}}R_{i}t^{i}. \end{aligned}$$

Thus, $P_{R}(1)=\Sigma _{i\in {\mathbb {Z}}}\dim _{R_{0}}R_{i}$ is the dimension of R as an $R_{0}$-vector space. Therefore, $|R_{\le k-1}|=|R_{0}|^{m}$, $|R_{\ge k}|=|R_{0}|^{P_{R}(1)-m}$ and $|R|=|R_{0}|^{P_{R}(1)}$.

By Theorem 9, $\Gamma =\bigcup _{x\in R_{\le k-1}}\Gamma [R_{x}]$ where $R_{x}=\{ x+y;\ \ y\in R_{\ge k}\}$ and $\Gamma [R_{x}]$ is a complete graph. Hence, $\alpha (\Gamma )=|R_{\le k-1}|$ and $\omega (\Gamma )= |R_{\ge k}|$ which prove parts (1), (2) and (3). Also $\Gamma [R_{i}]$ is a induced subgraph of the complete graph $\Gamma [R_{x}]$ if $x=0\in R_{\le k-1}$. Hence, $\Gamma [R_{i}]$ is a complete graph of size $|R_{0}|^{H_{R}(i)}$. This completes the proof of part (4). Part (5) follows from Theorem 13 part (2). $\square$

Corollary 3

Let $R = {\mathbb {F}}[x_{1}, \ldots , x_{d}]$ be a polynomial ring over a field ${\mathbb {F}}$ and $\mathrm{deg}x_{i }= 1$ for $i = 1, \ldots , d$. Also let $\Gamma =Cay(R,S'')$ with $S''=R_{\ge k}{\setminus } \{0\}$. If $m=\sum ^{k-1}_{i=0}\left( {\begin{array}{c}i+d-1\\ d-1\end{array}}\right)$ and ${\overline{m}}=\sum _{i\ge 0}\left( {\begin{array}{c}i+d-1\\ d-1\end{array}}\right)$, then $\alpha (\Gamma )=|R_{0}|^{m}$ and $\omega (\Gamma )= |R_{0}|^{{\overline{m}}-m}$.

Proof

By Proposition 7.1 of [10], $H_{R}(i)=\left( {\begin{array}{c}i+d-1\\ d-1\end{array}}\right)$ for all $i\ge 0$. Now, we have the result by Corollary 2. $\square$

Example 5

Let $T= {\mathbb {F}}_{q}[x,y,z]$, $I = \langle x^{2},xy^{2},y^{4},z^{3}\rangle$ and $R=T/I$. Then, the Hilbert function of R is defined as follows: $H(0) = 1, H(1) = 3, H(2) = 5, H(3) = 5, H(4) = 3, H(5) = 1, H(t) = 0$ for $t \ge 6$. Also $P_{R}(t)=1 + 3t + 5t^{2} + 5t^{3} + 3t^{4} + t^{5}$. If $\Gamma =Cay(R,S'')$ with $S''=R_{\ge 3}{\setminus } \{0\}$, then $m=\sum ^{2}_{i=0}H(i)=1+3+5=9$ and $P_{R}(1)=18$. Hence,

• 1. $\alpha (\Gamma )=|R_{0}|^{m}=q^{9}$;

• 2. $\omega (\Gamma )= |R_{0}|^{P_{R}(1)-m}=q^{9}$;

• 3. ${\overline{\Gamma }}$ is a complete $q^{9}$-partite graph where every part has $q^{9}$ vertices.

The following theorem gives relations between the generalized Cayley graph, cototal and counit graphs.

Theorem 14

Let $R=\oplus _{i}R_{i}=R_{0}\oplus R_{+}$ be a positively Artinian graded ring and $S''=R_{+}{\setminus } \{0\}$. Then,

• 1. $R_{0}$ is a field if and only if $\mathrm{Cay}(R,S'')$ is the cototal graph $\mathrm{Cay}(R,\mathrm{Z}(R))$.

• 2. $R_{0}$ is a field if and only if the complement graph $\overline{\mathrm{Cay}(R,S'')}$ is the counit graph $\mathrm{Cay}^{+} (R,U(R))$.

Proof

Since R is Artinian, any element of R is invertible or a zero divisor and hence $U(R)=R{\setminus } \mathrm{Z}(R)$. Also by Theorem 3.11 in [8], $\mathrm{Z}(R)=\mathrm{Z}(R_{0})+ R_{+}$.

• 1. Let $R_{0}$ be a field, then $\mathrm{Z}(R)= R_{+}$. Thus $\mathrm{Cay}(R,S'')=\mathrm{Cay}(R,\mathrm{Z}(R){\setminus } \{0\})$ is the cototal graph. Conversely, if $\mathrm{Cay}(R,S'')=\mathrm{Cay}(R,\mathrm{Z}(R){\setminus } \{0\})$, then $\mathrm{Z}(R){\setminus } \{0\}=S''=R_{+}{\setminus } \{0\}$. Hence, $\mathrm{Z}(R_{0})=\{ 0\}$ which proves that $R_{0}$ is a field.

• 2. By above discussion, $R_{0}$ is a field if and only if

  $$\begin{aligned} \overline{\mathrm{Cay}(R,S'')}=\mathrm{Cay}(R, R{\setminus } S'')=\mathrm{Cay}(R, R{\setminus } \mathrm{Z}(R))=\mathrm{Cay}(R, U(R)) \end{aligned}$$

  is the counit graph.

$\square$