Abstract

Let be a simple graph of order and . A mapping is called a -colouring of if whenever the vertices and are adjacent in . The number of distinct -colourings of , denoted by , is called the chromatic polynomial of . The domination polynomial of is the polynomial , where is the number of dominating sets of of size . Every root of and is called the chromatic root and the domination root of , respectively. Since chromatic polynomial and domination polynomial are monic polynomial with integer coefficients, its zeros are algebraic integers. This naturally raises the question: which algebraic integers can occur as zeros of chromatic and domination polynomials? In this paper, we state some properties of this kind of algebraic integers.

1. Introduction

Let be a simple graph and . A mapping is called a -colouring of if whenever the vertices and are adjacent in . The number of distinct -colourings of , denoted by , is called the chromatic polynomial of . A zero of is called a chromatic zero of . For a complete survey on chromatic polynomial and chromatic root, see [1].

For any vertex , the open neighborhood of is the set , and the closed neighborhood is the set . For a set , the open neighborhood of is , and the closed neighborhood of is . A set is a dominating set if , or, equivalently, every vertex in is adjacent to at least one vertex in . An -subset of is a subset of of cardinality . Let be the family of dominating sets of which are -subsets and let . The polynomial is defined as domination polynomial of [2, 3]. A root of is called a domination root of . We denote the set of all roots of by . For more information and motivation of domination polynomial and domination roots, refer to [26].

We recall that a complex number is called an algebraic number (respectively, algebraic integer) if it is a zero of some monic polynomial with rational (resp., integer) coefficients (see [7]). Corresponding to any algebraic number , there is a unique monic polynomial with rational coefficients, called the minimal polynomial of (over the rationals), with the property that divides every polynomial with rational coefficients having as a zero. (The minimal polynomial of has integer coefficients if and only if is an algebraic integer.)

Since the chromatic polynomial and domination polynomial are monic polynomial with integer coefficients, its zeros are algebraic integers. This naturally raises the question: which algebraic integers can occur as zeros of chromatic and domination polynomials?

In Sections 2 and 3, we study algebraic integers as chromatic roots and domination roots, respectively.

As usual, we denote the complete graph of order and the complement of , by and , respectively.

2. Algebraic Integers as Chromatic Roots

Since chromatic polynomial is monic polynomial with integer coefficients, its zeros are algebraic integers. An interval is called a zero-free interval for a chromatic (domination) polynomial, if has no chromatic (domination) zero in this interval. It is well known that and are two maximal zero-free intervals for chromatic polynomials of the family of all graphs (see [8]). Jackson [8] showed that is another maximal zero-free interval for chromatic polynomials of the family of all graphs and the value is best possible.

For chromatic polynomials clearly those roots lying in are forbidden. Tutte [9] proved that , where is the golden ratio, cannot be a chromatic zero. Salas and Sokal in [10] extended this result to show that the numbers for and , with coprime to , are never chromatic zeros. For they showed the weaker result that and are not chromatic zeros of any plane near-triangulation.

Alikhani and Peng [11] have obtained the following theorem.

Theorem 2.1. , where is the golden ratio, cannot be zeros of any chromatic polynomials.

Also they extended this result to show that and all their natural powers cannot be chromatic zeros, where is called -annaci constant [12].

For some times it was thought that chromatic roots must have nonnegative real part. This is true for graphs with fewer than ten vertices. But Sokal showed the following.

Theorem 2.2 (see [13]). Complex chromatic roots are dense in the complex plane.

Theorem 2.3. The set of chromatic roots of a graph is not a semiring.

Proof. The set of chromatic roots is not closed under either addition or multiplication, because it suffices to consider and , where is nonreal and close to the origin.

Theorem 2.4. Suppose that are rational numbers, is an integer that is not a perfect square, and . Then is not the root of any chromatic polynomial.

Proof. If is a root of some polynomial with integer coefficients (e.g., a chromatic polynomial), then so is . But or cannot belong to , a contradiction.

Corollary 2.5. Let be a rational number, and let be a positive rational number such that is irrational. Then cannot ba a root of any chromatic polynomial.

We know that for every graph with edge , , where is the graph obtained from by contracting and and removing any loop. By applying this recursive formula repeatedly, we arrive at where ’s are some constants and Let us recall the definition of join of two graphs. The join of two graphs and , denoted by , is a graph with vertex set and edge set and .

Theorem 2.6 (see [14]). Let and be any two graphs with expressed in factorial form, . Then where is called umbral product, and acts as powers (i.e., .

Here we state and prove the following theorem.

Theorem 2.7. For any graph ,

Proof. It suffices to prove it for . Assume that . By Theorem 2.6,

Here we state and prove the following theorem.

Theorem 2.8. If is a chromatic root, then for any natural number , is a chromatic root.

Proof. Since we have the result.

By Theorem 2.1, and are not chromatic roots. However is a chromatic root (see Theorem 2.14). Therefore by Theorem 2.8 we have the following corollary:

Corollary 2.9. For every natural number , is a chromatic root.

There are the following conjectures.

Conjecture 2.10 (see [15]). Let be an algebraic integer. Then there exists a natural number such that is a chromatic root.

Conjecture 2.11 (see [15]). Let be a chromatic root. Then is a chromatic root for any natural number .

Definition 2.12 (see [15]). A ring of cliques is the graph whose vertex set is the union of complete subgraphs of sizes , where the vertices of each clique are joined to those of the cliques immediately preceding or following it mod.

Theorem 2.13 (see [15]). The chromatic polynomial of is a product of linear factors and the polynomial

We call the polynomial in Theorem 2.13 the interesting factor.

Theorem 2.14. is a chromatic root.

Proof. Consider the graph . Obviously this graph has eight vertices and by Theorem 2.13 its interesting factor is , with roots . Therefore the graph has as chromatic root.

Remark 2.15. We observed that is a chromatic root for every . Also we saw that is not a chromatic root, but we do not know whether is a chromatic root or not. Therefore this remains as an open problem.

3. Algebraic Integers as Domination Roots

For domination polynomial of a graph, it is clear that is zero-free interval. Brouwer [16] has shown that −1 cannot be domination root of any graph . For more details of the domination polynomial of a graph at −1 refer to [17]. We also have shown that every integer domination root is even [18].

Let us recall the corona of two graphs. The corona of two graphs and , as defined by Frucht and Harary in [19], is the graph formed from one copy of and copies of , where the th vertex of is adjacent to every vertex in the th copy of . The corona , in particular, is the graph constructed from a copy of , where for each vertex , a new vertex and a pendant edge are added.

Here we state the following theorem.

Theorem 3.1 (see [2]). Let be a graph. Then if and only if for some graph of order .

By above theorem there are infinite classes of graphs which have −2 as domination roots. Since −1 is not domination root of any graph, so we do not have result for domination roots similar to Theorem 2.8. Also we think that the following conjecture is correct.

Conjecture 3.2 (see [18]). If is an integer domination root of a graph, then or .

Now we recall the following theorem.

Theorem 3.3 (see [2]). Let be a connected graph of order . Then, , if and only if , for some graph . Indeed .

The following corollary is an immediate consequence of above theorem.

Corollary 3.4. All graphs of the form , have as domination roots.

The following theorem state that cannot be a domination root.

Theorem 3.5. cannot be a domination root.

Proof. Let be any graph. Since is a polynomial with integral coefficients, we have . But , a contradiction.

The following theorem is similar to Theorem 3.6 for domination roots.

Theorem 3.6. Suppose that are rational numbers, is an integer that is not a perfect square, and . Then is not the root of any domination polynomial.

Proof. If is a root of some polynomial with integer coefficients (e.g., a domination polynomial), then so is . But , a contradiction.

Corollary 3.7. Let be a rational number, and let be a positive rational number such that is irrational. Then cannot ba a root of any domination polynomial.

Here we will prove that for odd , cannot be a domination root. We need some theorems.

Theorem 3.8 (see [20]). For every natural number ,

Corollary 3.9. For every natural number

Proof. This follows from Theorem 3.8

Now, we recall the Cassini’s formula.

Theorem 3.10 (Cassini’s formula [20]). One has where .

Using this formula, we prove another property of golden ratio and Fibonacci numbers which is needed for the proof of Theorem 3.13.

Theorem 3.11.

Proof. Suppose that is even, therefore is odd, and by Corollary 3.9, we have Hence, and by multiplying in this inequality, we have Thus, By Theorem 3.10, we have Hence, for even , Similarly, the result holds when is odd.

Theorem 3.12 (see [20, page 78]). For every , .

Now we are ready to prove the following theorem.

Theorem 3.13. Let be an odd natural number. Then cannot be domination roots.

Proof. By Theorem 3.12, we can write Suppose that , that is Then (see page  187 in [21]), but we can write By Theorem 3.11, when is even and when is odd. Since is odd, we have . But we know that is zero-free interval for domination polynomial of any graph. Hence we have a contradiction.