On Pure Vertex Decomposable Graphs and their Dominating ...€¦ · Moreover, the Cohen-Macaulay...

On Pure Vertex Decomposable Graphs and their DominatingShedding Vertices

Nuno Miguel Januario Alves

Thesis to obtain the Master of Science Degree in

Mathematics and Applications

Examination CommitteeChairperson: Prof. Pedro Manuel Agostinho Resende

Supervisor: Prof. Maria da Conceicao Pizarro de Melo Telo Rasquilha Vaz Pinto

Member of the Committee: Prof. Teresa Maria Jeronimo Sousa

October 2017

Acknowledgements

I would like to thank the Center for Mathematical Analysis, Geometry, and Dynamical Systems of the

University of Lisbon for the financial support I was provided throughout the project RD0447/CAMGSD with

reference MAT/04459, funded by FCT/MEC. Its support was crucial for maintaining me in Lisbon in order to

complete my degree.

I would like to thank my advisers for all the help they provided, but more importantly, for helping me face

some mathematical demons I possessed. More than a year ago, before starting the research that ultimately

became this thesis, I could not fathom the idea of studying polynomial rings or graphs. Those were exam-

ples of mathematical theories that I was afraid of due to my lack of knowledge of them. Now the fear is gone.

I am also really grateful for having had the opportunity to learn and get inspired by some teachers I encoun-

tered during my bachelors and masters degree. They shaped the way I understand mathematics, and from

that I became a better mathematician.

I want to thank my family, in particular my Mom, my Dad, and my Grandmother Teresa, for their eternal

unconditional support. Without them none of this would be possible.

Moreover, I feel really thankful for the friends I have made so far throughout my mathematical journey.

Unequivocally, they helped me became a better person. In particular, I want to emphasize the relevance of

Joao Paulos. He was one of the important triggers of my evolving as a human being.

Last, but not least, I want to thank Matilde for the caress, kindness and love she gave me over the past

two and a half years.

2

Resumo

Pretendemos com esta dissertacao introduzir alguma teoria de algebra comutativa, ideais monomiais,

grafos e complexos simpliciais de Cohen-Macaulay; mas principalmente estudar a condicao de decomposicao

por vertices, nas suas formas pura e nao-pura, de grafos e complexos simpliciais. Estabelecemos, no caso

de grafos, a relacao existente entre os casos puro e nao-puro dessa propriedade. Alem disso, definimos o

conjunto de vertices perdidos de um grafo puramente decomponıvel por vertices, e em particular provamos

que a vizinhana de um vertice simplicial e um subconjunto de vertices perdidos. Por fim, determinamos

algumas famılias de grafos puramente decomponıveis por vertices para os quais o conjunto de vertices

perdidos e um conjunto dominante, e terminamos com dois exemplos de famılias de grafos puramente

decomponıveis por vertices para os quais o conjunto de vertices perdidos nao e um conjunto dominante.

Palavras-chave: Grafos, Complexos Simpliciais, Condicao de Cohen-Macaulay,

Condicao de Decomposicao por Vertices, Vertices Perdidos, Conjunto Dominante.

3

Abstract

The aim of this dissertation is to introduce some theory of commutative algebra, monomial ideals, Cohen-

Macaulayness of graphs and simplicial complexes; however, the main focus is studying vertex decompos-

ability of graphs and simplicial complexes, in both its pure and non-pure forms. We establish, for graphs, a

relation between those two forms of vertex decomposability. Also, we define the set of shedding vertices for

pure vertex decomposable graphs, and we prove in particular that the neighbourhood of a simplicial vertex

is a subset of shedding vertices. Finally, we determine some families of pure vertex decomposable graphs

for which the set of shedding vertices is a dominant set, ending with two examples of families of pure vertex

decomposable graphs for which the set of shedding vertices is not a dominating set.

Keywords: Graphs, Simplicial Complexes, Cohen-Macaulay property,

Vertex Decomposability, Shedding Vertices, Dominating Set.

4

Contents

Introduction 8

0 Preliminaries 9

0.1 Basic notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

0.2 Abstract Commutative Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

0.3 Monomial ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1 Graphs and Simplicial Complexes 29

1.1 Simplicial Complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

1.2 Cohen-Macaulay Simplicial complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

1.3 Graph Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

1.4 Cohen-Macaulay Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2 Pure Vertex Decomposability 46

2.1 Pure Vertex Decomposable Simplicial Complexes . . . . . . . . . . . . . . . . . . . . . . . . . 46

2.2 Pure Vertex Decomposable Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2.3 Dominating Shedding Vertices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.4 Non-Dominating Shedding Vertices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

Bibliography 66

5

Introduction

Combinatorial commutative algebra is a relatively recent area of mathematics which uses methods of

commutative algebra to solve combinatorial problems. It lies at the intersection between these two theories.

In 1975 Richard Stanley used the theory of Cohen-Macaulay rings to prove affirmatively the upper bound

conjecture for spheres, observing that commutative algebra could supply basic methods in the algebraic

study of convex polytopes and simplicial complexes. Later, in 1983, he published a book that would become

a reference for the development of combinatorial commutative algebra ([17]).

Besides the contribution of Richard Stanley, we must also mention the importance of Rafael Villarreal's work

on the subject. He studied Cohen-Macaulay graphs (see [22]). More specifically, given a (finite simple)

graph G with vertex set V = {1, . . . , n} and edge set E, one considers the ring R = K[x1, . . . , xn]/I, where

I = ({xixj | {i, j} ∈ E}) and K is a field, and we set that G is Cohen-Macaulay if and only if R is a Cohen-

Macaulay ring. With this, combinatorial properties of a graph can be inferred using notions of commutative

algebra.

In addition to the Cohen-Macaulay condition, there are other properties of graphs, monomial edge ideals

and simplicial complexes that can illustrate some interactions between commutative algebra and combi-

natorics. Examples of those properties are: shellability, (pure and non-pure) vertex decomposability and

well-coveredness. These properties have been studied in ([3],[5], [6], [7],[8], [9], [10], [11], [12], [13], [14]),

and the following implications hold (see [3],[10], [13], [14])

pure vertex decomposable ⇒ pure shellable ⇒ Cohen-Macaulay ⇒ well-covered

Generally, the implications above are strict; however, there are families of graphs for which the pure vertex

decomposable property is equivalent to the Cohen-Macaulay property. Examples of such families are bipar-

tite graphs (see [6] and [7]), very-well covered graphs (see [9] and [15]), and graphs without 4-cycles and

5-cycles (see [16]).

From all the properties mentioned above, the ones that will be more important for us are the pure and

non-pure vertex decomposable properties. Pure vertex decomposability was first introduced by Provan and

Billera ([19]) for simplicial complexes. They studied the notion of k-decomposability for simplicial complexes,

and the most restrictive case, k = 0, is what we call pure vertex decomposability. The non-pure version

of vertex decomposability (what we will simply call vertex decomposability) was introduced by Bjorner and

6

Wachs ([20]).

This thesis is based on a recent paper by Adam Van Tuyl, Jonathan Baker and Kevin Vander Meulen

([21]). Its goal is to review and establish bridges between the existing literature on vertex decomposability.

In particular, we prove that if a graph is well-covered and vertex decomposable, then it is pure vertex de-

composable.

Moreover, the Cohen-Macaulay property is also introduced in this thesis, since it is necessary to understand

the motivation that led Adam Van Tuyl and the other authors to create the paper that served as base for

this study. Their paper was initially motivated by a conjecture of Villarreal ([22]) on Cohen-Macaulay graphs.

Based upon computer experiments on all graphs on six or less vertices, Villarreal proposed the following:

Conjecture ([22], Conjectures 1 and 2). Let G be a Cohen-Macaulay graph with vertex set V and let

D = {x ∈ V | G \ x is a Cohen-Macaulay graph}.

Then D 6= ∅ and D is a dominating set of G.

A dominating set D of a graph G with vertex set V is a subset of V such that every vertex of V \ D is

adjacent to a vertex of D. It is already known that the above conjecture is false. Earl, Kevin Vander Meulen

and Adam Van Tuyl ([28]) found an example of a circulant graph G on 16 vertices such that G is Cohen-

Macaulay, but there is no vertex x such that G \ x is Cohen-Macaulay.

Although the conjecture above is false in general, Villarreal’s work suggests that there may exist some nice

subset of Cohen-Macaulay graphs for which the conjecture still holds. Since pure vertex decomposable

graphs are Cohen-Macaulay, Jonathan Baker, Kevin Vander Meulen and Adam Van Tuyl ([21]) considered

the following variation of conjecture above:

Question. Let G be a pure vertex decomposable graph with vertex set V . Let

Shed(G) = {x ∈ V | G \ x is a pure vertex decomposable graph}.

Is Shed(G) a dominating set of G?

The set Shed(G) denotes the set of shedding vertices of a pure vertex decomposable graph G. It will

follow from the definition of pure vertex decomposable graphs that Shed(G) 6= ∅, and that is the reason

for not including that condition in the question. In the last chapter of this thesis, among other things, we

are going to explore some families of pure vertex decomposable graphs for which the question above is

answered positively and negatively.

In Chapter 0 we firstly introduce some basic commutative algebra, from Noetherian rings, passing

through primary decomposition and then some necessary dimension theory in order to reach the notion of

Cohen-Macaulay rings. Then we move our attention to monomial ideals, which take part in any introduction

to combinatorial commutative algebra. Monomial ideals are relevant for us since both the Stanley-Reisner

ideal of a simplicial complex and the monomial edge ideal of a graph are examples of monomial ideals.

7

Those ideals establish the bridge between the theory of graphs or simplicial complexes and commutative

algebra.

Chapter 1 is aimed to be a basic introduction to the theory of simplicial complexes and graph theory.

Various different families of graphs are introduced, such as chordal graphs, complete graphs, simplicial

graphs, bipartite graphs, Cameron-Walker graphs and well-covered graphs. Also, a short approach to the

Cohen-Macaulayness of simplicial complexes and graphs is done. In particular, we introduce the Stanley-

Reisner ideal of a simplicial complex and the monomial edge ideal of a graph.

In Chapter 2 we define (pure and non-pure) vertex decomposability separately for simplicial complexes

and graphs, and derive some properties out of it. Moreover, we make a connection between (pure and non-

pure) vertex decomposable simplicial complexes and (pure and non-pure) graphs via the independence

complex. Then we study some families of pure vertex decomposable graphs for which the set of shed-

ding vertices is a dominating set. Finally, we end up with the constructions of two families of pure vertex

decomposable graphs for which the set of shedding vertices is not a dominating set.

Whenever possible I tried to include illustrative examples of the definitions in order to complement the

understanding of it. Also, I tried to make this text as self-contained as possible.

8

Chapter 0

Preliminaries

0.1 Basic notations

• The set of real numbers is denoted by R, and the set of positive real numbers is denoted by R+.

• The set of integers is denoted by Z.

• The set of non-negative integers is denoted by N.

• Given a positive integer n, we set [n] = {1, . . . , n}.

• Given two sets X,Y, we write X ⊆ Y when X is a subset of Y, and we write X ( Y when X is a

proper subset of Y .

• All rings will be commutative and will have a unit.

0.2 Abstract Commutative Algebra

Definition 0.2.1. Let R be a ring. We say that R is a Noetherian ring if R satisfies the ascending chain

condition on ideals, that is, given any chain of ideals I0 ⊆ I1 ⊆ I2 ⊆ . . . , there exist a k ∈ N such that

Ik = Ik+1 = . . . . We also say that a chain with the property above satisfies the ascending chain condition.

Example 0.2.2. Every field is a Noetherian ring.

Proposition 0.2.3. Let R be a ring. Then the following statements are equivalent:

1. R is a Noetherian ring,

2. Every ideal of R is finitely generated,

3. Every non-empty family of ideals of R has a maximal element.

9

Proof. (1⇒ 2) Let I ⊆ R be an ideal. Let i1 ∈ I. If I = (i1), then I is finitely generated. If I 6= (i1), then there

exists i2 ∈ I such that i2 /∈ (i1). If I = (i1, i2), then I is finitely generated. If I 6= (i1, i2), then there exists

i3 ∈ I such that i3 /∈ (i1, i2). Since R is Noetherian this process eventually ends, otherwise we would obtain

a chain of ideals of R that would not satisfy the ascending chain condition, (i1) ( (i1, i2) ( (i1, i2, i3) ( . . . .

Therefore, there exist i1, . . . , in ∈ I such that I = (i1, i2, . . . , in), so I is finitely generated.

(2 ⇒ 1) Let I0 ⊆ I1 ⊆ I2 ⊆ . . . be an ascending chain of ideals of R. Let I =⋃j≥0 Ij . Since Ii ⊆ Ii+1

∀i ∈ N, we have that I is an ideal. By hypothesis I is finitely generated, so there exist i1, . . . , in ∈ I such

that I = (i1, . . . , in). For each ij there exists Ikjcontaining ij . Let K = max{kj | j = 1, . . . , n}. Then

I = (i1, . . . , in) ⊆ Ik ⊆ Ik+1 ⊆ . . . ⊆ I, so R satisfies the ascending chain condition on its ideals.

(1 ⇒ 3) Let X be a non-empty set of ideals of R. Since R is Noetherian, every chain on X has an upper

bound, so, by Zorn’s lemma, X has a maximal element.

(3 ⇒ 1) Let I0 ⊆ I1 ⊆ I2 ⊆ . . . be an ascending chain of ideals of R. Let X = {In | n ∈ N}. By hypothesis,

X has a maximal element, so there exists k ∈ N such that Ik is the maximal element of X. Therefore

Ik = Ik+1 = Ik+2 = . . ., since Ik ⊆ Ik+1 ⊆ Ik2 ⊆ . . . . Hence R is Noetherian.

Example 0.2.4. Every principal ideal domain is a Noetherian ring.

Given a ring R, recall that R[x1, . . . , xn] ∼= (R[x1, . . . , xn−1])[xn], where R[x] = {r0 +r1x+. . .+rnxn | ri ∈

R,n ≥ 0}.

Theorem 0.2.5 (Hilbert’s Basis Theorem). Let R be a ring. If R is Noetherian, then R[x1, . . . , xn] is Noethe-

rian.

Proof. It suffices to show that if R is Noetherian, then R[x] is Noetherian. Let I be an ideal of R[x]. We

are going to prove that I is finitely generated. Let f =∑nk=0 rkx

k ∈ R[x]. We define in(f) to be 0 if f = 0

and to be rn if f 6= 0, and the degree of f is deg(f) = max{k | rk 6= 0}. Let in(I) = {in(f) | f ∈ I} and

inj(I) = {in(f) | f ∈ I, deg(f) ≤ j}. Now we prove that inj(I) is an ideal of R. Since 0 ∈ I, 0 = in(0)

and deg(0) ≤ j, we have 0 ∈ inj(I). Let a, b ∈ inj(I) \ {0}, and f, g ∈ I be such that a = in(f) and

b = in(g), deg(f),deg(g) ≤ j. Without loss of generality, assume that m = deg(f) − deg(g) ≥ 0. Then

a + b = in(f + xmg) ∈ inj(I). Also note that if a ∈ inj(I) and r ∈ R, then ra ∈ inj(I), because if a = in(f)

then ra = in(rf) ∈ inj(I). Hence inj(I) is an ideal of R. Therefore we have the following ascending chain

of ideals of R

in0(I) ⊆ in1(I) ⊆ in2(I) ⊆ . . . .

Since R is Noetherian there exists N such that inN (I) = inN+1(I) = . . . , and inj(I) is finitely generated

for every j ∈ N, inj(I) = (rj1, . . . , rjnj ). For each rjl there exists fjl ∈ I, with deg(fjl) ≤ j and such that

rjl = in(fjl). We claim that I = (fjl)j=0,...,Nl=1,...,nj

. It is clear that I ⊇ (fjl)j=0,...,Nl=1,...,nj

since fjl ∈ I. Suppose that

I 6⊆ (fjl)j=0,...,Nl=1,...,nj

. Let f be the polynomial in I \ (fjl)j=0,...,Nl=1,...,nj

of minimum degree. Let m = deg(f). If m ≤ N

then in(f) ∈ inm(I) = (rm1, . . . , rmnm), hence in(f) =

∑nm

k=1 skrmk. Let g = f−∑nm

k=1 skfmkxm−deg(fmk) ∈ I.

Then deg(g) < deg(f) and g /∈ (fjl)j=0,...,Nl=1,...,nj

, which is impossible since f is the polynomial with minimum

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degree in I \ (fjl)j=0,...,Nl=1,...,nj

. If m > N then in(f) ∈ inm(I) = inN (I) = (rN1, . . . , rNnN). Using an analogous

argument as before we reach again a contradiction. Therefore I ⊆ (fjl)j=0,...,Nl=1,...,nj

. Hence I = (fjl)j=0,...,Nl=1,...,nj

is

finitely generated, so R[x] is Noetherian.

Example 0.2.6. Z[x1, . . . , xn] is a Notherian ring.

Proposition 0.2.7. Let R be a ring and I ⊆ R an ideal. If R is Noetherian, then R/I is Noetherian.

Proof. Let J/I be an ideal of R/I. Then J is an ideal of R that contains I. Since R is Noetherian, J is finitely

generated, so there exist a1, . . . , an ∈ J such that J = (a1, . . . , an). Therefore J/I = (a1 + I, . . . , an + I),

hence R/I is Noetherian.

Example 0.2.8. If R[x] is a Noetherian ring, then R ∼= R[x]/(x) is a Noetherian ring.

Definition 0.2.9. Let R be a ring and let M be an R-module. We define

AnnR(M) := {r ∈ R | rm = 0 ∀m ∈M}.

Note that AnnR(M) ⊆ R is an ideal. Moreover, if J ⊆ R is an ideal such that J ⊆ AnnR(M), then the

structure of M as R-module is the same as R/J-module.

Definition 0.2.10. Let R be a ring and M be a R-module. We say that M is Noetherian if M satisfies the

ascending chain condition on R-submodules, that is, given any chain of R-submodules M0 ⊆ M1 ⊆ M2 ⊆

. . . , there exist a k ∈ N such that Mk = Mk+1 = . . . .

Proposition 0.2.11. Let R be a ring and M an R-module. Then the following statements are equivalent:

1. M is Noetherian,

2. Every R-submodule of M is finitely generated,

3. Every non-empty family of R-submodules of M has a maximal element.

Proposition 0.2.12. Let R be a ring, M be a R-module and N be a R-submodule of M . If M is a Noetherian

R-module, then N and M/N are Noetherian R-modules.

Proof. It suffices to recall that every R-submodule of N is also an R-submodule of M , and every R-

submodule of M/N is of the form P/N where P is an R-submodule of M such that N ⊆ P .

Proposition 0.2.13. Let R be a ring and let 0 → Nf−→ M

g−→ P → 0 be a short exact sequence of

R-modules. Then M is a Noetherian R-module if and only if both N and P are Noetherian R-modules.

Proof. (⇒) Since f is injective, we have N ∼= Im(f) = f(N), so N is Noetherian because it is an R-

submodule of M and M is Noetherian by hypothesis. We also have that P = Im(g) ∼= M/ker(g) since g is

surjective, hence P is Noetherian because M is Noetherian and ker(g) is an R-submodule of M .

11

(⇐) Let M1 ⊆M2 ⊆ . . . be an ascending chain of R-submodules of M . Then M1 ∩ f(N) ⊆M2 ∩ f(N) ⊆ . . .

is an ascending chain of R-submodules of f(N) ∼= N . Since N is Noetherian, there exists k such that

Mk ∩ f(N) = Mk+1 ∩ f(N) = . . .. On the other hand, g(M1) ⊆ g(M2) ⊆ . . . is an ascending chain of

R-submodules of P , so there exists l such that g(Ml) = g(Ml+1) = . . . because P is Noetherian. Let

t = max{k, l}. Then Mt = Mt+1 = .... Indeed, if x ∈ Mt+n (n ≥ 1), then g(x) ∈ g(Mt+n) = g(Mt), so

g(x) = g(y) for some y ∈ Mt, hence x− y ∈ ker(g) = f(N). Therefore x− y ∈ Mt+n ∩ f(N) = Mt ∩ f(N).

Thus, x = x− y + y ∈Mt, and so Mt = Mt+1 = . . . .

Corollary 0.2.14. Let R be a ring and M1, . . . ,Mn be R-modules. We have:

1. M1 ⊕M2 is a Noetherian R-module if and only if both M1 and M2 are Noetherian R-modules,

2. M1 ⊕ · · · ⊕Mn is a Noetherian R-module if and only if M1, . . . ,Mn are Noetherian R-modules,

3. R is an Notherian ring if and only if Rn = R⊕ · · · ⊕R is a Noetherian R-module.

Theorem 0.2.15. Let R be a ring and let M an R-module. If R is a Noetherian ring and M is finitely

generated, then M is a Noetherian R-module.

Proof. By hypothesis there exist a1, . . . , an ∈M such that M = Ra1 + . . .+Ran. Let ϕ : Rn →M be given

by ϕ(r1, . . . , rn) = r1a1 + . . . + rnan. Since ϕ is a surjective R-homomorphism we have M ∼= Rn/ker(ϕ),

hence M is Noetherian.

Definition 0.2.16. Let R be a ring and I, J two ideals of R. The ideal I : J = {f ∈ R | fg ∈ I ∀g ∈ J} is

called the colon ideal of I with respect to J .

Definition 0.2.17. Let R be a ring and I ⊆ R be an ideal. The ideal√I = {f ∈ R | fk ∈ I for some k > 0}

is called the radical ideal of I.

Definition 0.2.18. Let R be a ring and I ⊆ R and ideal. We say that I is a radical ideal if√I = I.

Proposition 0.2.19. Let R be a ring and I ⊆ R an ideal. Then the radical ideal of I is the intersection of all

prime ideals of R containing I, that is,√I =

⋂I⊆P∈Spec(R)

P.

Proof. See ([1]).

Corollary 0.2.20. Let R be a ring and I ⊆ R an ideal. Then I is a radical ideal if and only if I is the

intersection of all prime ideals of R that contain I.

Definition 0.2.21. Let R be a ring and I ( R be an ideal. We say that I is irreducible if given ideals J,K of

R such that I = J ∩K, then I = J or I = K.

Example 0.2.22. Consider the ring R = R[x, y], and the ideals I = (x2, xy, y2), J = (x, y2), K = (x2, y). We

have that I is not an irreducible ideal since I = J ∩K, I ( J and I ( K.

12

Proposition 0.2.23. Let R be a Noetherian ring and I ⊆ R an ideal. Then I is a finite intersection of

irreducible ideals of R.

Proof. Let R be a Noetherian ring. Assume to the contrary that there exists a proper ideal of R that is not a

finite intersection of irreducible ideals. Then the set

Γ = {J | J is a proper ideal of R that is not a finite intersection of irreducible ideals}

is non-empty. Since R is Noetherian, it follows that Γ has a maximal element, say L. Note that L is not

irreducible, so L = L1 ∩ L2 with L ( L1 and L ( L2. Then L1, L2 /∈ Γ by the maximality of L. Thus L1 and

L2 are finite intersections of irreducible ideals. Therefore L is a finite intersection of irreducible ideals, that

is, L /∈ Γ, contradiction.

Proposition 0.2.24. Let R be a Noetherian ring and Q ⊆ R an ideal. If Q is irreducible, then√Q is a prime

ideal.

Proof. Let x, y ∈ R be such that xy ∈√Q and x /∈

√Q. We want to prove that y ∈

√Q. Consider the

following ascending chain of ideals of R:

Q ⊆ (Q : (x)) ⊆ (Q : (x2)) ⊆ . . .

Since R is Noetherian, there exists n ∈ N such that (Q : (xn)) = (Q : (xn+1)) = . . . .

Now we prove that Q = (Q : (xn)) ∩ (Q, xn). The inclusion Q ⊆ (Q : (xn)) ∩ (Q, xn) is evident. Let

a + bxn ∈ (Q : (xn)) ∩ (Q, xn), where a ∈ Q and b ∈ R. Then axn + bx2n ∈ Q, so bx2n ∈ Q since axn ∈ Q.

Therefore b ∈ (Q : (x2n)) = (Q : (xn)), hence bxn ∈ Q, and so a+ bxn ∈ Q.

Note that xn /∈ Q (otherwise x ∈√Q), hence Q ( (Q, xn). By hypothesis Q is irreducible, so it follows

that Q = (Q : (xn)). Since xy ∈√Q we have xmym ∈ Q for some m ≥ 1, so xnmynm ∈ Q. Therefore

ynm ∈ (Q : (xnm)) = (Q : (xn)) = Q, that is, y ∈√Q.

Definition 0.2.25. Let R be a ring and I ( R an ideal. We say that P is a minimal prime ideal of I if

1. P is a prime ideal of R,

2. I ⊆ P ,

3. If P ′ is a prime ideal of R such that I ⊆ P ′ ⊆ P , then P ′ = P .

The set of minimal prime ideals of an ideal I is denoted by Min(I).

Proposition 0.2.26. Let R be a Noetherian ring and I ( R an ideal. Then:

1. Min(I) is a finite non-empty set,

2. If Min(I) = {P1, . . . , Pk}, then ∃l ≥ 1 such that P l1 ∩ . . . ∩ P lk ⊆ I.

13

Proof. By proposition 0.2.23 we have that I = Q1 ∩ . . . ∩Qn, where each Qi is an irreducible ideal, and by

proposition 0.2.24 it follows that each√Qi = Pi is a prime ideal. Note that I = Q1 ∩ . . .∩Qn ⊆ Qi ⊆

√Qi =

Pi, for each i. Since R is Noetherian, each Pi is finitely generated, so for each i there exists ri ≥ 1 such that

P rii ⊆ Qi. Let l = max{r1, . . . , rn}. Then

P l1 ∩ . . . ∩ P ln ⊆ Q1 ∩ . . . ∩Qn = I.

Without loss of generality, let P1, . . . , Pk be the prime ideals among P1, . . . , Pn which are not comparable,

that is, Pi 6⊆ Pj and Pj 6⊆ Pi for i 6= j. Moreover, Pk+1 contains Pi for some i ∈ {1, . . . , k}, and the same

happens for Pk+2, . . . , Pn. Thus,

P l1 ∩ . . . ∩ P lk = P l1 ∩ . . . ∩ P ln ⊆ I.

We are going to prove that P1, . . . , Pk are the only minimal prime ideals of I. We already know that Pi is a

prime ideal of R and I ⊆ Pi. Let P be a prime ideal such that I ⊆ P ⊆ Pi. We have that

P1 . . . P1 . . . Pk . . . Pk = P l1 . . . Plk ⊆ P l1 ∩ . . . P ln ⊆ I ⊆ P ⊆ Pi.

Since P is a prime ideal there exists j such that Pj ⊆ P ⊆ Pi, hence i = j and Pj = P = Pi because Pi and

Pj are not comparable. Therefore Pi is a minimal prime ideal of I. Now let P be a minimal prime ideal of I.

Then

P1 . . . P1 . . . Pk . . . Pk = P l1 . . . Plk ⊆ I ⊆ P

and so there exists j such that Pj ∈ P . Note that I ⊆ Pj and P is minimal, hence Pj = P .

Definition 0.2.27. Let R be a ring. A presentation of an ideal I of R as an intersection I = Q1 ∩ . . .∩Qm of

ideals is called irredundant if none of the ideals Qi can be omitted in this presentation.

Lemma 0.2.28. Let R be a ring and I ⊆ R an ideal. If I has an irredundant presentation I = P1 ∩ . . . ∩ Pmas an intersection of prime ideals, then Min(I) = {P1, . . . , Pm}.

Proof. Suppose without loss of generality that P1 /∈ Min(I). Then there exists a prime ideal Q such that

I ⊆ Q ( P1 . Since I = P1 ∩ . . . ∩ Pm ⊆ Q and Q is a prime ideal, then some Pi is contained in Q, thus

Pi ⊆ Q ( P1, therefore the presentation I = P1∩ . . .∩Pm is not irredundant. Hence {P1, . . . , Pm} ⊆ Min(I).

On the other hand, let Q ∈ Min(I). Again, since I = P1 ∩ . . . ∩ Pm ⊆ Q and Q is a prime ideal,

some Pi is contained in Q, and since both Pi and Q are minimal prime ideals of I, Q = Pi . Hence,

Min(I) = {P1, . . . , Pm}, as desired.

Definition 0.2.29. Let R be a ring and M an R-module. A prime ideal P of R is called an associated prime

ideal of M if there exists v ∈M \ {0} such that AnnR(v) = {r ∈ R | rv = 0} = P .

Note that P is an associated prime ideal of M if and only if there exists an injective homomorphism

0 → R/P → M . Moreover, if P is an associated prime ideal of M = R/I, where I is some ideal of R, then

we say that P is an associated prime ideal of I.

We denote by Ass(M) the set of associated prime ideals of M .

14

Proposition 0.2.30. Let R be a Noetherian ring and M be a finitely generated R-module. Then P ∈ Ass(M)

if and only if PP ∈ Ass(MP ).

Proof. (⇒) Let P ∈ Ass(M). Then there exists v ∈M \ {0} such that AnnR(v) = P . We are going to prove

that PP = AnnRP( v1 ).

(⊇) Let ab ∈ AnnRP( v1 ). Then there exists c ∈ R \ P such that acv = 0, so ac ∈ AnnR(v) = P . Since P is

prime and c /∈ P we have a ∈ P , hence ab ∈ PP .

(⊆) Let ab ∈ PP , with a ∈ P . Then abv1 = av

b = 0b = 0

1 , hence ab ∈ AnnRP

( v1 ).

(⇐) Let PP ∈ Ass(MP ). Then there exists vd ∈MP \ { 0

1} such that AnnRP( vd ) = PP . Note that AnnRP

( vd ) =

AnnRP( v1 ) since d

1 is invertible, abvd = 0

1 ⇔abd1vd = 0

1 ⇔abv1 = 0

1 . Since R is Noetherian we have P =

(x1, . . . , xn) for some xi ∈ P . For each i = 1, . . . , n there exists si /∈ P such that sixiv = 0, becausexi

1 ∈ PP = AnnRP( v1 ). Let s = s1s2 . . . sn. We are going to prove that P = AnnR(sv).

(⊆) Let a ∈ P . Then a = r1x1 + . . .+ rnxn, so asv = r1sx1v + . . .+ rnsxnv = 0, hence a ∈ AnnR(sv).

(⊇) Let a ∈ AnnR(sv). Then asv = 0, so as1 ∈ AnnRP

( v1 ) = PP , hence as1 = d

b for some d ∈ P and

b ∈ R \ P . Therefore there exists c ∈ R \ P such that bcsa = cd ∈ P . Since b, c, s /∈ P and P is prime, it

follows that bcs /∈ P , and so a ∈ P .

Proposition 0.2.31. Let R be a Noetherian ring and M be a finitely generated R-module. If P is a minimal

prime ideal of AnnR(M), then P ∈ Ass(M).

Proposition 0.2.32. Let R be a ring and M an R-module. Then every maximal element of the set Σ =

{AnnR(v) | v ∈M \ {0}} is a prime ideal of R.

Proof. Let AnnR(v) be a maximal element of Σ = {AnnR(v) | v ∈ M \ {0}}, and let a, b ∈ R be such

that ab ∈ AnnR(v). Suppose that b /∈ AnnR(v). Then bv 6= 0, AnnR(v) ⊆ AnnR(bv) ∈ Σ, and since

AnnR(v) is a maximal element it follows that AnnR(bv) = AnnR(v). Consequently a ∈ AnnR(v), because

a ∈ AnnR(bv).

Corollary 0.2.33. Let R be a Noetherian ring and M an R-module. Then Ass(M) 6= ∅.

Proposition 0.2.34. Let R be a Noetherian ring and M,M1,M2 non-zero finitely generated R-modules.

Consider the following short exact sequence 0→M1f−→M

g−→M2 → 0. Then

Ass(M1) ⊆ Ass(M) ⊆ Ass(M1) ∪Ass(M2).

Proof. First we prove that Ass(M1) ⊆ Ass(M). Let P ∈ Ass(M1). Then there exists an exact sequence

0→ R/Ph−→M1, and so we have the exact sequence 0→ R/P

h−→M1f−→M . Therefore P ∈ Ass(M), since

fh is an injective homomorphism.

Now let P ∈ Ass(M), and let v ∈ M \ {0} be such that P = AnnR(v). We are going to prove that

P ∈ Ass(M1) ∪ Ass(M2). Note that M1 ∼= f(M1), and so if v ∈ f(M1), then P ∈ Ass(M1). Therefore

suppose that v /∈ f(M1). Let u = v + f(M1) ∈ M/f(M1) ∼= M/ker(g) ∼= M2. We have that Pv = 0 so

15

Pu = 0, hence P ⊆ AnnR(u). If P = AnnR(u), then P ∈ Ass(M2). Therefore assume that P ( AnnR(u),

so there exists s ∈ AnnR(u) such that s /∈ P . We have that sv ∈ M \ {0} (because s /∈ P = AnnR(u)), and

su = 0⇔ sv + f(M1) = 0 + f(M1)⇔ sv ∈ f(M1). We are going to prove that P = AnnR(sv).

(⊆) P ⊆ AnnR(sv) since Pv = 0.

(⊇) Let t ∈ AnnR(sv). Then tsv = 0, so ts ∈ P . Since s /∈ P and P is prime it follows that t ∈ P .

Thus, P ∈ Ass(M1).

Corollary 0.2.35. Let R be a Noetherian ring and M1,M2 non-zero finitely generated R-modules. Then

Ass(M1 ⊕M2) = Ass(M1) ∪Ass(M2).

Proposition 0.2.36. Let R be a Noetherian ring and M a finitely generated R-module. Then |Ass(M)| <∞.

Proof. See ([1]).

Lemma 0.2.37. Let R be a Noetherian ring and M a non-zero finitely generated R-module. Then

Z(M) =⋃

P∈Ass(M)

P

where Z(M) = {r ∈ R | ∃v ∈M \ {0} : rv = 0}.

Proof. (⊇) Let r ∈ P for some P ∈ Ass(M). Then there exists m ∈ M \ {0} such that P = AnnR(m). In

particular, rm = 0, hence r ∈ Z(M).

(⊆) Let r ∈ Z(M). Then there exists m ∈M \ {0} such that rm = 0. Let Λ = {AnnR(sm) | s ∈ R, sm 6= 0}.

Note that Λ 6= ∅ since AnnR(1m) ∈ Λ. Therefore Λ has a maximal element, Q = AnnR(sm) for some s ∈ R

such that sm 6= ∅, since R is Noetherian. Note that r ∈ Q because rsm = srm = 0. Now we prove that Q is

prime. Let a, b ∈ R be such that ab ∈ Q and a /∈ Q. Then asm 6= 0, so AnnR(asm) ∈ Λ. We also have that

Q = AnnR(sm) ⊆ AnnR(asm), and therefore Q = AnnR(asm) by its maximality. Then basm = (ab)(sm) = 0

(ab ∈ Q), and so b ∈ AnnR(asm) = Q. Thus, r ∈⋃P∈Ass(M) P.

The set Z(M) is called the set of zero divisors of M .

Proposition 0.2.38. Let R be a Noetherian ring and I ⊆ R an ideal. Then Min(I) ⊆ Ass(R/I).

Proof. Let P ∈ Min(I). Then PP ∈ Min(IP ), thus PP is the only prime ideal containing IP . Since all associ-

ated prime ideals of RP /IP must contain IP , it follows that Ass(RP /IP ) ⊆ {PP }, and since Ass(RP /IP ) 6= ∅,

then Ass(RP /IP ) = {PP }. On the other hand, RP /IP and (R/I)P are isomorphic RP -modules, and so

Ass((R/I)P ) = {PP }. Hence P ∈ Ass(R/I).

Definition 0.2.39. Let R be a ring and Q ⊆ R an ideal. We say that Q is a primary ideal of R if for every

a, b ∈ R such that ab ∈ Q we have a ∈ Q or b ∈√Q.

16

Example 0.2.40. Consider R = Z and the ideal Q = 4Z. We have that Q is a primary ideal of R that is not

a prime ideal. Indeed, Q is not prime because 2 × 2 = 4 ∈ Q but 2 /∈ Q. Moreover, Q is primary since for

nm = 4k = 2 × 2k ∈ Q, if n /∈√Q = 2Z, then 2 divides m and so m = 2l. Hence 2nl = 4k ⇔ nl = 2k and

because n /∈ 2Z we have l = 2j. Therefore m = 4j ∈ Q .

In the previous example we noted that there are primary ideals that are not prime ideals. However, the

radical of any primary ideal is a prime ideal.

Proposition 0.2.41. Let R be a ring and Q ⊆ R an ideal. If Q is primary, then√Q is prime.

Proof. Let a, b ∈ R such that ab ∈√Q. Then anbn ∈ Q for some n ≥ 1. Since Q is primary we have that

an ∈ Q or bn ∈√Q. Hence an ∈ Q or bnm ∈ Q for some m ≥ 1, so a ∈

√Q or b ∈

√Q. Consequently

√Q is

prime.

Definition 0.2.42. Let R be a Noetherian ring, Q ⊆ R an ideal and P ⊆ R a prime ideal. We say that Q is

P -primary if Q is primary and moreover√Q = P .

Proposition 0.2.43. Let R be a Noetherian ring and Q ⊆ R an ideal. Then Q is P -primary if and only if

Ass(R/Q) = {P}.

Proof. (⇒) Let Q be a P -primary ideal of the Noetherian ring R. We want to prove that Ass(R/Q) = {P}.

(⊇) Observe that Q ⊆√Q = P =

⋂Q⊆L∈Spec(R) L, so P is a minimal prime ideal of Q. Since Q =

AnnR(R/Q), it follows, by proposition 0.2.31, that P ∈ Ass(R/Q).

(⊆) Let Q ∈ Ass(R/Q). Then Q = AnnR(y+Q), where y ∈ R \Q. Observe that Qy ⊆ Q ⊆√Q = P . We

want to prove thatQ = P . Let x ∈ Q. Then xy ∈ Q. Since y /∈ Q andQ is primary it follows that x ∈√Q = P .

On the other hand, if x ∈ P =√Q, then xn ∈ Q for some n ≥ 1, so xn(y + Q) = xny + Q = 0 + Q, that is,

xn ∈ Q. Since Q is prime it follows that x ∈ Q.

(⇐) Suppose that Ass(R/Q) = {P}. Then Q 6= R because P 6= R, and there exists an injective homomor-

phism 0 → R/P → R/Q. First we prove that√Q = P . We have that

√Q =

⋂Q⊆L∈Spec(R) L =

⋂L∈Min(Q).

If L is a minimal prime ideal of Q, then P is a minimal prime ideal of AnnR(R/Q), so L ∈ Ass(R/Q), hence

L = P . Thus√Q = P .

Now we prove that Q is primary. Let a, b ∈ R such that ab ∈ Q. Suppose that a /∈ Q. Then b(a + Q) =

0 +Q⇒ b ∈ Z(R/Q) =⋃L∈Ass(R/Q) L = P =

√Q.

Proposition 0.2.44. Let R be a Noetherian ring and Q ( R an ideal. If Q is irreducible, then Q is primary.

Proof. Let a, b ∈ R such that ab ∈ Q. Suppose that a /∈√Q and consider the following ascending chain of

ideals of R

(Q : (a)) ⊆ (Q : (a2)) ⊆ . . .

Since R is Noetherian there exists n ≥ 1 such that (Q : (an)) = (Q : (an+1)) = . . . . Recalling the proof of

proposition 0.2.24 we have that Q = (Q, an) ∩ (Q : (an)). Because a /∈ Q we have Q ( (Q, an). Since Q is

irreducible it follows that Q = (Q : (an)). Thus ab ∈ Q⇒ b ∈ (Q : (a)) ⊆ (Q : (an)) = Q.

17

Corollary 0.2.45. Let R be a Noetherian ring and I ( R an ideal. Then I is a finite intersection of primary

ideals.

Lemma 0.2.46. Let R be a Noetherian ring. If Q1, . . . , Ql are P -primary ideals of R, then Q1 ∩ . . . ∩Ql is a

P -primary ideal of R.

Proof. Consider the injective homomorphism

ϕ : R

Q1 ∩ . . . ∩Ql→ R

Q1⊕ . . .⊕ R

Ql

given by ϕ(r + Q1 ∩ . . . ∩Ql) = (r + Q1, . . . , r + Ql). Observe that RQ1∩...∩Ql

6= (0) since 1 /∈ Q1 ∩ . . . ∩Ql,

so RQ1⊕ . . .⊕ R

Ql6= (0) and Ass( R

Q1∩...∩Ql) 6= ∅. Thus

Ass( R

Q1 ∩ . . . ∩Ql) ⊆ Ass( R

Q1⊕ . . .⊕ R

Ql) =

l⋃i=1

Ass( RQi

) = {P}.

Hence Q1 ∩ . . . ∩Ql is P -primary.

Definition 0.2.47. Let R be a ring. A primary irredundant decomposition of an ideal I of R is an irredundant

decomposition I = Q1 ∩ . . . ∩Ql such that all ideals Qi are primary ideals.

Corollary 0.2.48. Let R be a Notherian ring and I ( R an ideal. Then I has a primary irredundant decom-

position.

Theorem 0.2.49. Let R be a Notherian ring and I ( R an ideal. Let I = Q1 ∩ . . . ∩ Qn be a primary

irredundant decomposition of I, where√Qi = Pi. Then Ass(R/I) = {P1, . . . , Pn}.

Proof. Consider the following exact sequence

0→ R/I → R/Q1 ⊕ . . .⊕R/Qn.

Then

Ass(R/I) ⊆ Ass(R/Q1 ⊕ . . .⊕R/Qn) =n⋃i=1

Ass(R/Qi) = {P1, . . . , Pn}.

On the other hand, since the decomposition of I is irredundant we have that I = Q1 ∩ Q2 ∩ . . . Qn (

Q2 ∩ . . . ∩ Qn, that is, we cannot preserve the intersection if we remove Q1 from it. Let y ∈ Q2 ∩ . . . ∩ Qn,

y /∈ I. Then Q1y ⊆ Q1 ∩ . . . ∩Qn = I, that is, Q1 ⊆ (I : (y)). Since P1 =√Q1 and P1 is finitely generated,

there exists l ≥ 1 such that P l1 ⊆ Q1. Thus P l1 ⊆ (I : (y)), so P l1y ⊆ I. Let k be the smallest positive integer

such that P k1 y ⊆ I. Then P k−11 y 6⊆ I. Let z ∈ P k−1

1 y, z /∈ I. Now we prove that P1 = AnnR(z + I). On the

one hand P1z ⊆ P k1 y ⊆ I ⇒ P1(z + I) = {0 + I}. On the other hand, let s ∈ AnnR(z + I) and suppose that

s /∈ P1 =√Q1. Then sz ∈ I = Q1 ∩ . . . ∩ Qn ⊆ Q1, and so z ∈ Q1 because Q1 is primary and s /∈

√Q1.

But z ∈ P k−11 y ⊆ Q2 ∩ . . . ∩Qn, hence z ∈ Q1 ∩Q2 ∩ . . . ∩Qn = I, contradiction. Therefore P1 ∈ Ass(R/I).

Analogously we can conclude that P2, . . . , Pn ∈ Ass(R/I).

18

Now we present some dimension theory and some necessary material in order to reach the notion of

Cohen-Macaulay ring.

Definition 0.2.50. Let R be a ring. The Krull dimension of R is

dim(R) = sup{n ∈ N | there exists an ascending chain of prime ideals P0 ( . . . ( Pn ( R}.

Example 0.2.51. If K is a field then dim(K) = 0 since the only prime ideal of K is (0).

Definition 0.2.52. Let R be a Noetherian ring and P ⊆ R a prime ideal. The height of P is

ht(P ) = sup{n ∈ N | there exists an ascending chain of prime ideals P0 ( . . . ( Pn = P}.

Note that ht(P ) = ht(PP ) = dim(RP ).

Definition 0.2.53. Let R be a Noetherian ring and I ⊆ R an ideal. The height of I is

ht(I) = min{ht(P ) | P ∈ Min(I)}.

Theorem 0.2.54. Let R be a Noetherian ring and P ⊆ R a prime ideal. If P is a minimal prime ideal of

(x1, . . . , xn) ( R then ht(P ) ≤ n.

Proof. See ([2]).

Corollary 0.2.55. Let R be a Noetherian ring and I = (x1, . . . , xn) ( R. Then ht(I) ≤ n.

Proposition 0.2.56. Let I be an ideal of K[x1, . . . , xn], where K is a field. Then

ht(I) + dim(R/I) = n.

Proof. See ([13]).

Definition 0.2.57. Let R be a ring and M an R-module. We say that r ∈ R is M -regular, or regular on M , if

rv = 0 implies v = 0 for every v ∈ M . In the case M = R we simply say that an element is regular instead

of R-regular.

Note that the set Z(M) = {r ∈ R | ∃v ∈ M \ {0} : rv = 0} of non-regular elements on M is precisely

the set of zero divisors of M . Recall, by lemma 0.2.37, that Z(M) =⋃P∈Ass(M) P .

Proposition 0.2.58. Let R be Noetherian a ring. Let P be a prime ideal of R and r ∈ P be regular. Then

ht(P ) ≥ ht(P/(r)) + 1.

Proof. Let P0/(r) ( . . . ( Pn/(r) = P/(r) be a maximal ascending chain of prime ideals of R/(r). Then

r ∈ P0 ( . . . ( Pn = P is an ascending chain of prime ideals of R. If P0 ∈ Min((0)) then P0 ∈ Ass(R), and

so r ∈ Z(R), contradiction. Therefore, there exists a prime ideal L of R such that L ( P0 ( . . . ( Pn = P .

Consequently ht(P ) ≥ n+ 1 = ht(P/(r)) + 1.

19

Corollary 0.2.59. Let R be a Noetherian ring. If r ∈ R is regular, then dim(R) ≥ dim(R/(r)) + 1.

Proof. Since r ∈ R is regular, for any given prime ideal P of R containing r we have dim(R) ≥ ht(P ) ≥

ht(P/(r)) + 1, hence dim(R) ≥ dim(R/(r)) + 1.

Definition 0.2.60. Let R be a ring and M an R-module. We say that x1, . . . , xn ∈ R is an M -regular

sequence, or a regular sequence on M , if the following conditions hold:

1. M 6= (x1, . . . , xn)M ,

2. x1 is M -regular, and xi is regular on M/(x1, . . . , xn−1)M for i = 2, . . . , n.

If M = R we simply say regular sequence instead of R-regular sequence.

An M -regular sequence x1, . . . , xn is a maximal M -regular sequence if x1, . . . , xn, y is not a regular

sequence on M , for every y ∈ R.

Example 0.2.61. Consider the case R = M = Z[x1, . . . , x3]. We have that x1, . . . , x3 is a regular sequence

on Z[x1, . . . , x3]. Indeed, if x1f = 0 then f = 0, so x1 is M -regular. Also, if x2f ∈ (x1), then f ∈ (x1)

since (x1) is a prime ideal and x2 /∈ (x1), so x2 is regular on M/(x1)M . Moreover, if x3f ∈ (x1, x2), then

f ∈ (x1, x2) since (x1, x2) is a prime ideal and x3 /∈ (x1, x2).

Example 0.2.62. Now consider the case R = M = K[x, y, z], where K is a field. We are going to show that

x, y(1− x), z(1− x) is an M -regular sequence, and y(1− x), z(1− x), x is not an M -regular sequence.

If xf = 0, then f = 0, so x is M -regular. If y(1 − x)f ∈ (x) then yf ∈ (x), hence f ∈ (x), and so

y(1 − x) is regular on K[x, y, z]/(x)K[x, y, z]. If z(1 − x)f ∈ (x, y(1 − x)) then zf ∈ (x, y(1 − x)) = (x, y),

hence f ∈ (x, y) = (x, y(1−x)), and so z(1−x) is regular on K[x, y, z]/(x, y(1−x))K[x, y, z]. Consequently,

x, y(1− x), z(1− x) is a regular sequence on K[x, y, z].

Now we prove that y(1 − x), z(1 − x), x is not a regular sequence on K[x, y, z]. It suffices to show that

z(1 − x) is not regular on K[x, y, z]/(y(1 − x))K[x, y, z]. To do so, just note that z(1 − x)y ∈ (y(1 − x)) and

y /∈ (y(1− x)).

Theorem 0.2.63 (Rees). Let R be a ring, I ( R an ideal, and M a finitely generated R-module such that

M 6= IM . Then every two maximal M -regular sequences contained in I have the same cardinality.

Proof. See ([3]).

Definition 0.2.64. Let R be a ring, I ( R an ideal, and M a finitely generated R-module such that M 6= IM .

The cardinality of any maximal M -regular sequence contained in I is denoted by grade(I,M). If M = R

we set grade(I) = grade(I,R). If (R,m) is a local ring, then the depth of M is depth(M) = grade(m,M).

Moreover, if (R,m) is a local ring and M = R, then the depth of R is depth(R) = grade(m) = grade(m,R).

Proposition 0.2.65. Let R be a ring, I ( R an ideal, and M a finitely generated R-module such that

M 6= IM . Let r ∈ I be M -regular. Then grade(I,M/(r)M) = grade(I,M)− 1.

20

Proof. On the one hand, if r1, . . . , rn ∈ I is a maximal (M/(r)M)-regular sequence, then r, r1, . . . , rn ∈ I

is an M -regular sequence, because r is M -regular. Hence grade(I,M) ≥ n + 1 = grade(I,M/(r)M) + 1.

On the other hand, since r ∈ I is M -regular, there is a maximal M -regular sequence r, r1, . . . , rn ∈ I, so

r1, . . . , rn ∈ I is a (M/(r)M)-regular sequence. Consequently grade(I,M/(r)M) ≥ n = grade(I,M) − 1.

Thus grade(I,M/(r)M) = grade(I,M)− 1.

Proposition 0.2.66. Let R be a Noetherian ring, and P a prime ideal of R. Then grade(P ) ≤ ht(P ).

Proof. We proceed by induction on n = grade(P ). The case n = 0 is trivial, so suppose that n ≥ 0. Let

r ∈ P be regular. Then P/(r) is a prime ideal of R/(r), and grade(P/(r)) = n− 1 = grade(P )− 1. Therefore

ht(P ) ≥ ht(P/(r)) + 1 = grade(P ).

Corollary 0.2.67. Let R be a Noetherian ring, and I an ideal of R. Then grade(I) ≤ ht(I).

Proof. Let P ∈ Min(I) be such that ht(P ) = ht(I). Then

grade(I) ≤ grade(P ) ≤ ht(P ) = ht(I).

Definition 0.2.68. Let R be a Noetherian ring and I an ideal of R. If I is generated by a regular sequence

we say that I is a complete intersection.

Proposition 0.2.69. Let R be a Noetherian ring and I ⊆ R a complete intersection. Then grade(I) = ht(I).

Proof. Let r1, . . . , rn ∈ I be a regular sequence such that I = (r1, . . . , rn). Then grade(I) = n and ht(I) ≤ n,

hence n = grade(I) ≤ ht(I) ≤ n.

Definition 0.2.70. Let (R,m) be a Noetherian local ring. We say that R is a Cohen-Macaulay local ring if

depth(R) = dim(R).

Definition 0.2.71. Let R be a Noetherian ring. We say that R is a Cohen-Macaulay ring if Rm is a Cohen-

Macaulay local ring for every maximal ideal m of R.

Example 0.2.72. The rings Z and K[x1, . . . , xn], where K is a field, are two examples of Cohen-Macaulay

rings.

0.3 Monomial ideals

Monomial ideals are ideals in polynomial rings that can be described in combinatorial and geomet-

ric terms. In spite of their simplicity, monomial ideals are powerful tools. For example, in combinatorial

commutative algebra they are used to attach algebraic invariants to graphs and, more generally, simplicial

complexes. These invariants have led to the solutions of several important problems in combinatorics.

21

Let K be a field and consider the the polynomial ring in n variables over K, K[x1, . . . , xn].

Monomials form a natural K-basis for K[x1, . . . , xn]. A monomial ideal I also has a K-basis of monomials.

Consequently, a polynomial f ∈ K[x1, . . . , xn] belongs to I if and only if all monomials in f with a non-zero

coefficient also belong to I. This is one of the reasons why algebraic operations with monomial ideals are

easy to perform.

Definition 0.3.1. Let S = K[x1, . . . , xn]. A monomial in S is a product xa = xa11 . . . xan

n , with a = (a1, . . . , an) ∈

Nn.

The monomials inK[x1, . . . , xn] correspond bijectively to the lattice points in Rn+. Moreover, xaxb = xa+b

holds for every a,b ∈ Nn.

If I is an ideal of K[x1, . . . , xn], then the set of monomials of I is denoted by , lMon(I).

Definition 0.3.2. Let S = K[x1, . . . , xn]. An ideal I ⊆ S is called a monomial ideal if its set of generators

consist of monomials.

Recall that K[x1, . . . , xn] is a Noetherian ring, that is, every ideal of K[x1, . . . , xn] is finitely generated.

As it would be expected, every monomial ideal of K[x1, . . . , xn] has a finite system of monomial generators.

Proposition 0.3.3. Let S = K[x1, . . . , xn] and I ⊆ S a monomial ideal. Then I has a finite system of

monomial generators.

Proof. Let I be a non-zero monomial ideal of S. Let

Σ = {L ⊆ S | L is an ideal with finite system of monomial generators in I}.

Since S is Notherian and Σ 6= ∅ (because (0) ∈ Σ), it follows that Σ has a maximal element, J . Suppose

I 6= J . Since I is a monomial ideal, then there exists u ∈ Mon(I) \ J , and so J + (u) is an ideal with a

finite system of monomial generators in I which strictly contains J , a contradiction. Hence I = J has a finite

system of monomial generators.

The set Mon(S) is a K-basis of S = K[x1, . . . , xn], that is, every polynomial f ∈ S is a unique finite

K-linear combination of monomials

f =∑

u∈Mon(S)

auu, with au ∈ K.

Definition 0.3.4. The support of f is supp(f) = {u ∈ Mon(S) | au 6= 0}.

Theorem 0.3.5. Let S = K[x1, . . . , xn] and I ⊆ S a monomial ideal. Then the setN of monomials belonging

to I is a K-basis of I.

Proof. First observe that the elements of N are linearly independent, since N ⊆ Mon(S). To show that

N generates the K-vector space I, it suffices to show that supp(f) ⊆ N for any f ∈ I. If f ∈ I, then

22

there exist monomials u1, . . . , um ∈ I and polynomials f1, . . . , fm ∈ S such that f =∑mi=1 fiui, hence

supp(f) ⊆⋃mi=1 supp(fiui). Since each monomial in supp(fiui) is of the form wui with w ∈ supp(fi), it

follows that supp(fiui) ⊆ N for every 1 ≤ i ≤ m, hence supp(f) ⊆ N , as desired.

Corollary 0.3.6. Let S = K[x1, . . . , xn] and I ⊆ S an ideal. Then I is a monomial ideal if and only if

supp(f) ⊆ I for every f ∈ I.

Proof. Suppose I is a monomial ideal and let f ∈ I. By theorem 0.3.5, there exist monomials u1, . . . , um ∈ I

and constants a1, . . . , am ∈ K such that f = a1u1 + . . .+ amum, hence supp(f) ⊆ {u1, . . . , um} ⊆ I.

Suppose supp(f) ⊆ I for every f ∈ I and let f1, . . . , fm ∈ I be a set of generators of I. Since supp(fi) ⊆

I for every i, it follows that⋃mi=1 supp(fi) is a set of monomial generators of I and thus I is a monomial

ideal.

Definition 0.3.7. Let S = K[x1, . . . , xn]. For monomials xa = xa11 . . . xan

n and xb = xb11 . . . xbnn of S, we say

that xb divides xa if bi ≤ ai for each i. In this case we write xb | xa.

The set of monomials which belong to a monomial ideal I of K[x1, . . . , xn] can be described as follows.

Proposition 0.3.8. Let S = K[x1, . . . , xn], I ⊆ S a monomial ideal, and {u1, . . . , um} be a monomial system

of generators of I. Then a monomial v belongs to I if and only if there exists a monomial w such that v = wui

for some i ∈ [m], that is, ui | v.

Proof. Suppose v ∈ I. Then there exist polynomials f1, . . . , fm ∈ S such that v = f1u1 + . . . + fmum,

therefore v ∈ supp(fiui) for some i ∈ [m], that is, v = wui for some w ∈ supp(fi).

Proposition 0.3.9. Let S = K[x1, . . . , xn] and I ⊆ S a monomial ideal. A monomial set of generators G of

I is minimal if and only if there is no pair of distinct monomials u, v ∈ G such that u | v.

Proof. We will show that G is not minimal if and only if there exists two distinct monomials u, v ∈ G such

that u | v.

Suppose G is not minimal. Then there exists a monomial set of generators G′ of I such that G′ ( G. Let

v ∈ G \G′. By proposition 0.3.8 there exists u ∈ G′ such that u | v.

Suppose there exists two distinct monomials u, v ∈ G such that u | v. Then G \ {v} is a monomial set of

generators of I strictly contained in G, hence G is not minimal.

Proposition 0.3.10. Let S = K[x1, . . . , xn] and I ⊆ S a monomial ideal. Then I has a unique minimal

monomial set of generators.

Proof. Let G1 = {u1, . . . , ur} and G2 = {v1, . . . , vs} be two minimal sets of generators of the monomial ideal

I. Let i ∈ [r]. Since ui ∈ I and I is monomial ideal, it follows that there exists vj such that ui = w1vj for some

monomial w1. Similarly there exists uk and a monomial w2 such that vj = w2uk. Then we have ui = w1w2uk.

Since G1 is a minimal set of generators of i, we conclude that k = i and w1w2 = 1. In particular, w1 = 1 and

hence ui = vj ∈ G2. This shows that G1 ⊆ G2. By symmetry we also have G2 ⊆ G1.

23

The unique minimal set of monomial generators of a monomial ideal I of K[x1, . . . , xn] is denoted by

G(I).

Definition 0.3.11. Let S = K[x1, . . . , xn] andM be a non-empty subset of Mon(S). A monomial xa ∈M is

said to be a minimal element ofM with respect to divisibility if whenever xb | xa with xb ∈M, then xb = xa.

The set of minimal elements ofM is denoted byMmin.

Theorem 0.3.12 (Dickson’s lemma). Let S = K[x1, . . . , xn] andM be a non-empty subset of Mon(S). Then

Mmin is a finite non-empty set.

Proof. Let I ⊆ S be the ideal with generator set M. Since G(I) is the only minimal set of monomial

generators of I, we have that G(I) ⊆M. It is enough to show that G(I) =Mmin, since G(I) is a finite set.

(⊆) Let u ∈ G(I), and let v ∈ M be such that v | u. Then v ∈ Mon(I) and by proposition 0.3.8 there

exists w ∈ G(I) such that w | v. Since w | v and v | u, we have w | u, and since u,w ∈ G(I), it follows that

w = u and so v = u. Hence u ∈Mmin.

(⊇) Let u ∈ Mmin. Then u ∈ I, therefore there exists v ∈ G(I) such that v | u. But then v ∈ M, and

since v | u, it follows that u = v ∈ G(I).

It is obvious that sums and products of monomial ideals are again monomial ideals. More precisely, if

I and J are monomial ideals, then I + J and IJ are monomial ideals, with G(I + J) ⊆ G(I) ∪ G(J) and

G(IJ) ⊆ G(I)G(J).

Given two monomials u and v, we denote by gcd(u, v) their greatest common divisor and by lcm(u, v)

their least common multiple.

For the intersection of monomial ideals we have the following.

Proposition 0.3.13. Let S = K[x1, . . . , xn], and I ⊆ S and J ⊆ S monomial ideals. Then I∩J is a monomial

ideal with set of generators given by

{lcm(u, v) : u ∈ G(I), v ∈ G(J)}.

Proof. Let f ∈ I ∩ J . By corollary 0.3.6, supp(f) ⊆ I ∩ J . Now, applying corollary 0.3.6 again, it follows that

I ∩ J is a monomial ideal, since f ∈ I ∩ J is arbitrary,

Let w ∈ Mon(I ∩ J). Then there exist u ∈ G(I) and v ∈ G(J) such that u | w and v | w, hence lcm(u, v) |

w. It is clear that {lcm(u, v) : u ∈ G(I), v ∈ G(J)} ⊆ I ∩ J , therefore {lcm(u, v) : u ∈ G(I), v ∈ G(J)} is a set

of generators of I ∩ J .

Proposition 0.3.14. Let S = K[x1, . . . , xn], and I ⊆ S and J ⊆ S monomial ideals. Then I : J is a monomial

ideal such that

I : J =⋂

v∈G(J)

I : (v).

Moreover, {u/ gcd(u, v) : u ∈ G(I)} is a set of generators of I : (v).

24

Proof. Let f ∈ I : J . Then fv ∈ I for every v ∈ G(J). By corollary 0.3.6 we have that supp(f)v =

supp(fv) ⊆ I. This implies that supp(f) ⊆ I : J . Since f ∈ I : J is arbitrary, corollary 0.3.6 yields that I : J

is a monomial ideal.

The given interpretation of I : J as an intersection is obvious, and it is clear that {u/ gcd(u, v) : u ∈

G(I)} ⊆ I : (v). So now let w ∈ Mon(I : (v)). Then vw ∈ I, therefore there exists u ∈ G(I) such that u | vw,

hence u/ gcd(u, v) | w, as desired.

Proposition 0.3.15. Let S = K[x1, . . . , xn] and I ⊆ S an ideal. If I is a monomial ideal, then its radical√I

is also a monomial ideal.

Proof. Let f ∈√I. Then fk ∈ I for some k ≥ 1 and by corollary 0.3.6 one has supp(fk) ⊆ I since I is a

monomial ideal. Let supp(f) = {xa1 , . . . ,xar}. We may assume that a1 is a vertex of the convex hull of the

set {a1, . . . ,ar}, so a1 does not belong to the convex hull of the set {a2, . . . ,ar}.

Assume (xa1)k = (xa1)k1(xa2)k2 . . . (xar )kr , with k = k1 + . . . + kr and k1 < k. Then a1 =∑ri=2

ki

k−k1ai

with∑ri=2

ki

k−k1= 1, so a1 belongs to the convex hull of {a2, . . . ,ar}, a contradiction. It follows that the

monomial (xa1)k cannot cancel against other terms in fk and hence (xa1)k belongs to supp(fk), which is a

subset of I. Therefore xa1 ∈√I and f − cxa1 ∈

√I, where c is the coefficient of f in the monomial xa1 . By

induction on the cardinality of supp(f) we conclude that supp(f) ⊆√I. Thus corollary 0.3.6 implies that

√I

is a monomial ideal.

Definition 0.3.16. Let S = K[x1, . . . , xn]. A monomial xa11 . . . xan

n ∈ S is called square-free if a1, . . . , an ∈

{0, 1}.

For u = xa11 . . . xan

n we set√u =

∏i:ai 6=0 xi. One has u =

√u if and only if u is square-free.

Proposition 0.3.17. Let S = K[x1, . . . , xn] and I ⊆ S a monomial ideal. Then {√u | u ∈ G(I)} is a set of

generators of√I.

Proof. Obviously {√u | u ∈ G(I)} ⊆

√I. Since

√I is a monomial ideal, it suffices to show that each

monomial v ∈√I is a multiple of some

√u with u ∈ G(I). If v ∈

√I, then vk ∈ I for some integer k ≥ 1 and

therefore u | vk for some u ∈ G(I). This yields the desired conclusion.

Definition 0.3.18. A monomial ideal I of S = K[x1, . . . , xn] is called a square-free monomial ideal if it is

generated by square-free monomials.

Corollary 0.3.19. A monomial ideal I of S = K[x1, . . . , xn] is a radical ideal if and only if it is a square-free

monomial ideal.

Theorem 0.3.20. Let S = K[x1, . . . , xn] and I ⊆ S a monomial ideal. Then I =⋂mi=1Qi, where each Qi is

generated by pure powers of the variables. In other words, each Qi is of the form (xa1i1, . . . , xak

ik). Moreover,

an irredundant presentation of this form is unique.

25

Proof. Let G(I) = {u1, . . . , ur}, and suppose some ui, say u1, is not a pure power of a variable. Then

we can write u1 = vw where v, w ∈ Mon(S) are such that gcd(v, w) = 1 and v 6= 1 6= w. We claim that

I = I1 ∩ I2, where I1 = (v, u2, . . . , ur) and I2 = (w, u2, . . . , ur). Obviously, I is contained in the intersection.

Conversely, let u ∈ Mon(I1 ∩ I2). If u is the multiple of some ui, with 2 ≤ i ≤ r, then u ∈ I. If not, then u is a

multiple of both v and w, and since gcd(v, w) = 1, u is a multiple of u1. In any case, u ∈ I.

If either G(I1) or G(I2) contains an element which is not a pure power, we repeat the argument and,

after a finite number of steps, we obtain a presentation of I as an intersection of monomial ideals generated

by pure powers. By omitting those ideals which contain the intersection of the others we end up with an

irredundant presentation.

Let Q1 ∩ . . .∩Qr = Q′1 ∩ . . .∩Q′s be two such irrendundant presentations of I. We will show that for each

i ∈ [r] there exists j ∈ [s] such that Q′j ⊆ Qi. By symmetry we then also have that for each k ∈ [s] there

exists l ∈ [r] such that Ql ⊆ Q′k. This will then imply that r = s and {Q1, . . . , Qr} = {Q′1, . . . , Q′s}.

Let i ∈ [r]. We may assume that Qi = (xa11 , . . . , xak

k ). Suppose that Q′j 6⊆ Qi for all j ∈ [s]. Then for each

j there exists xbj

lj∈ Q′j \ Qi. It follows that either lj 6∈ [k] or bj < alj . Let u = lcm(xb1l1 , . . . , x

bs

ls). We have

u ∈⋂sj=1Q

′j = I ⊆ Qi. Thus there exists i ∈ [k] such that xai

i divides u. But this is obviously impossible.

Example 0.3.21. Let I = (x21x2, x

21x

23, x

22, x2x

23) ⊆ K[x1, x2, x3]. Then:

I = (x21x2, x

21x

23, x

22, x2x

23)

= (x21, x

21x

23, x

22, x2x

23) ∩ (x2, x

21x

23, x

22, x2x

23)

= (x21, x

22, x2x

23) ∩ (x2, x

21x

23)

= (x21, x

22, x2) ∩ (x2

1, x22, x

23) ∩ (x2, x

21)

= (x21, x

22, x

23) ∩ (x2, x

21) ∩ (x2, x

23).

Definition 0.3.22. A monomial ideal I of K[x1, . . . , xn] is called irreducible if it cannot be written as proper

intersection of two other monomial ideals. It is called reducible if it is not irreducible.

Corollary 0.3.23. A monomial ideal I of K[x1, . . . , xn] is irreducible if and only if it is generated by pure

powers of the variables.

Proof. Let Q = (xa1i1, . . . , xak

ik) and suppose Q = I ∩ J , where I and J are monomial ideals properly contain-

ing Q. By theorem 0.3.20 we have I =⋂ri=1Qi and J =

⋂sj=1Q

′j , where the ideals Qi and Q′j are generated

by pure powers. Therefore we get the presentation

Q =r⋂i=1

Qi ∩s⋂j=1

Q′j .

By ommiting suitable ideals in the intersection on the right hand side, we obtain an irredundant presen-

tation of Q. The uniqueness of the irredundant presentation implies that Q = Qi for some i ∈ [r] or Q = Q′j

for some j ∈ [s], a contradiction.

26

Conversely, if G(Q) contains a monomial u = vw with gcd(v, w) = 1 and v 6= 1 6= w, then, as in the proof

of theorem 0.3.20, Q can be written as the intersection of monomial ideals properly containing Q.

If I is a square-free monomial ideal, the above procedure yields that the irreducible monomial ideals

appearing in the intersection of I are all of the form (xi1 , . . . , xik ). These monomial ideals are precisely the

monomial prime ideals.

Corollary 0.3.24. A square-free monomial ideal I of K[x1, . . . , xn] is the intersection of monomial prime

ideals.

Combining this corollary with lemma 0.2.28 we obtain:

Corollary 0.3.25. Let S = K[x1, . . . , xn] and I ⊆ S a square-free monomial ideal. Then I =⋂P∈Min(I) P ,

and each P ∈ Min(I) is a monomial prime ideal.

Proposition 0.3.26. The irreducible ideal (xa1i1, . . . , xak

ik) is (xi1 , . . . , xik )-primary.

Proof. Let Q = (xa1i1, . . . , xak

ik) and P = (xi1 , . . . , xik ). By proposition 0.2.43, Q is P -primary if and only

if Ass(S/Q) = {P}. Notice that Min(Q) = {P}. Since P ∈ Min(Q), by proposition 0.2.38 it follows that

P ∈ Ass(S/Q). Suppose there exists another P ′ ∈ Ass(S/Q).

Let g ∈ S \Q such that P ′ = Ann(g+Q) and consider its decomposition as a finite K-linear combination

of monomials. Removing from such K-linear combination the monomials which are multiples of some

monomial in the set {xa1i1, . . . , xak

ik}, we get a polynomial g′ ∈ S \Q whose support does not intersect Mon(Q)

and such that g+Q = g′ +Q. Hence we can suppose without loss of generality that supp(g)∩Mon(Q) = ∅.

Since Q ⊆ P ′ and Min(Q) = {P}, then P ( P ′.

Pick f ∈ P ′ \ P and consider its decomposition as a finite K-linear combination of monomials. Re-

moving from such K-linear combination the monomials which are multiples of some monomial in the set

{xi1 , . . . , xik}, we get a polynomial in P ′ \ P whose support does not intersect Mon(P ). Hence we can

suppose without loss of generality that supp(f) ∩Mon(P ) = ∅.

Since f 6= 0 and g 6= 0, then fg 6= 0. Let w ∈ supp(fg). Then there exist u ∈ supp(f) and v ∈ supp(g)

such that w = uv. Since none of the monomials xi1 , . . . , xik divide u and none of the monomials xa1i1, . . . , xak

ik

divide v, it follows that none of the monomials xa1i1, . . . , xak

ikdivide w. Since w ∈ supp(fg) is arbitrary, then

supp(fg) ∩Mon(Q) = ∅, hence fg 6∈ Q. But since f ∈ P ′ and P ′ = Ann(g + Q), by definition it follows that

fg ∈ Q, a contradiction.

Corollary 0.3.27. The irredundant decomposition of a monomial ideal as the intersection of monomial ideals

generated by pure powers is in fact a primary irredundant decomposition.

Corollary 0.3.28. The associated prime ideals of a monomial ideal I of K[x1, . . . , xn] are monomial prime

ideals.

27

Even though a primary irredundant decomposition of a monomial ideal I may not be unique, the primary

decomposition, obtained from an irredundant intersection of irreducible ideals as described above, is unique.

We call it the standard primary decomposition of I.

Example 0.3.29. As we saw in example 0.3.21, if I = (x21x2, x

21x

23, x

22, x2x

23), then

I = (x21, x

22, x

23) ∩ (x2, x

21) ∩ (x2, x

23)

is the standard primary decomposition of I and in particular Ass(R/I) = {(x1, x2), (x2, x3), (x1, x2, x3)}.

Notice that, in this particular case, Min(I) 6= Ass(R/I).

We end this section by stating a result about the height of a monomial prime ideal P of S, and also about

the Krull dimension of S/P .

Proposition 0.3.30. Let S = K[x1, . . . , xn] and P ⊆ S a monomial prime ideal generated by k variables.

Then

ht(P ) = k and dim(S/P ) = n− k.

Proof. Let P = (xi1 , . . . , xik ). Since P is generated by k elements of S, theorem 0.2.54 implies that ht(P ) ≤

k. On the other hand, (0) ( (xi1) ( (xi1 , xi2) ( . . . ( (xi1 , . . . , xik−1) ( P is an ascending chain of prime

ideals of length k and so ht(P ) = k. By proposition 0.2.56, ht(P ) + dim(S/P ) = n and so dim(S/P ) =

n− k.

28

Chapter 1

Graphs and Simplicial Complexes

We introduce basic concepts related to simplicial complexes and graphs and also the Cohen-Macaulay

property of these mathematical objects.

Section 1.1 is devoted to simplicial complexes. In particular we define pure simplicial complexes.

In section 1.2 we begin to define the Stanley-Reisner ideal of a simplicial complex in order to present the

notion of Cohen-Macaulay simplicial complexes.

In section 1.3 we give a brief introduction to graph theory. We only consider finite simple graphs and we

present various different families of such mathematical object.

In section 1.4 we define the monomial edge ideal of a graph and then we introduce the Cohen-Macaulay

property for graphs.

1.1 Simplicial Complexes

Definition 1.1.1. A simplicial complex ∆ on [n] is a collection of subsets of [n] that contains all one-element

subsets and such that if F ∈ ∆ and F ′ ⊆ F , then F ′ ∈ ∆. In this case, we say that [n] supports ∆, or ∆ is

supported by [n].

We can replace [n] by any finite set in the definition above. We chose the set [n] for simplicity. Moreover,

in the cases where the finite set, that supports the simplicial complex, is not important for the onward

discussion, we are going to omit it.

The simplicial complex supported by the empty set is called the empty complex and denoted by ∅.

Example 1.1.2. The collection

∆ = {{1, 2, 3}, {2, 3, 4}, {1, 2}, {1, 3}, {2, 3}, {2, 4}, {3, 4}, {1}, {2}, {3}, {4}}

is a simplicial complex on [4].

29

Definition 1.1.3. A face of a simplicial complex ∆ is just an element of ∆.

Definition 1.1.4. The dimension of a face F of a simplicial complex ∆ is |F | − 1.

Definition 1.1.5. A vertex of a simplicial complex ∆ is a face of ∆ of dimension 0.

Definition 1.1.6. A facet of a simplicial complex ∆ is a maximal face of ∆ with respect to inclusion.

The set of facets of ∆ is denoted by F(∆).


∆ = {{1, 2, 3}, {1, 3, 4}, {1, 2}, {1, 3}, {2, 3}, {1, 4}, {3, 4}, {1, 5}, {1}, {2}, {3}, {4}, {5}}

is a simplicial complex on [5] with 13 faces (of which five are vertices) and F(∆) = {{1, 2, 3}, {1, 3, 4}, {1, 5}}.

The facets {1, 2, 3} and {1, 3, 4} have dimension 2, and the facet {1, 5} has dimension 1.

It is clear that F(∆) determines ∆. When F(∆) = {F1, . . . , Fm}, we write ∆ = 〈F1, · · · , Fm〉. More

generally, given a set {G1, · · · , Gs} of faces of ∆, we denote by 〈G1, · · · , Gs〉 the subcomplex of ∆ consisting

of those faces of ∆ which are contained in some Gi.

Definition 1.1.8. A non-face of ∆ is a subset F of [n] such that F 6∈ ∆. A minimal non-face of ∆ is a

non-face F of ∆ such that for every G ( F one has G ∈ ∆.

The set of minimal non-faces of ∆ is denoted by N (∆).

Example 1.1.9. Let ∆ be the simplicial complex in example 1.1.2. Then the set of minimal non-faces of ∆

is

N (∆) = {{1, 4}}.

Definition 1.1.10. A simplex is a simplicial complex with only one facet or the empty complex.

Example 1.1.11. The collection ∆ = {{1, 2, 3}, {1, 2}, {1, 3}, {2, 3}, {1}, {2}, {3}} is a simplicial complex on

[3] that is a simplex.

Definition 1.1.12. Let ∆ be a simplicial complex. We say that ∆ is pure if all its facets have the same

dimension. More precisely, if such dimension is d, we say ∆ is a pure d-dimensional simplicial complex, and

d is called the dimension of ∆.


∆1 = {{1, 2, 3}, {2, 3, 4}, {1, 3, 4}, {1, 2}, {1, 3}, {2, 3}, {2, 4}, {3, 4}, {1, 4}, {1}, {2}, {3}, {4}}

is a pure simplicial complex on [4], since all its facets have dimension 2; and the collection

∆2 = {{1, 2, 3}, {3, 4}, {1, 2}, {1, 3}, {2, 3}, {1}, {2}, {3}, {4}}

is a non-pure simplicial complex on [4], since {1, 2, 3} is a facet with dimension 2 and {3, 4} is a facet with

dimension 1.

30

Definition 1.1.14. Let ∆ be a simplicial complex and H a face of ∆. The link of H is defined as follows

lk∆(H) = {F ∈ ∆ | F ∩H = ∅, F ∪H ∈ ∆}.

Example 1.1.15. Let ∆ be the simplicial complex in example 1.1.2. We have

lk∆({1, 2}) = {3} and lk∆({3}) = {{1, 2}, {2, 4}, {1}, {2}, {4}}.

Proposition 1.1.16. Let ∆ be a simplicial complex. If F ∈ ∆ and H ⊆ F , then F \H ∈ lk∆(H).

Proof. Let F ∈ ∆ and H ⊆ F . Note that F \H ∈ ∆, since F \H ⊆ F and F ∈ ∆. Also, (F \H)∪H = F ∈ ∆

and (F \H) ∩H = ∅. Hence F \H ∈ lk∆(H).

On the other hand, by definition, it immediately follows that G ∈ lk∆(H) implies H∪G ∈ ∆. This provides

a natural bijection between the faces of lk∆(H) and the faces of ∆ containing H. Moreover, such bijection

provides a natural bijection between the facets of lk∆(H) and the facets of ∆ containing H.

Definition 1.1.17. Let ∆ be a simplicial complex and H a face of ∆. The deletion of H is defined as follows

del∆(H) = {F ∈ ∆ | F ∩H = ∅}.

Example 1.1.18. Let ∆ be the simplicial complex in example 1.1.7. We have

del∆({1, 5}) = {{2, 3}, {3, 4}, {2}, {3}, {4}},

and

del∆({4}) = {{1, 2, 3}, {1, 2}, {1, 3}, {2, 3}, {1, 5}, {1}, {2}, {3}, {5}}.

IfH = {x} is a vertex of ∆, we simplify the notation and write lk∆(x) and del∆(x), instead of lk∆({x}) and

del∆({x}), respectively. In this case lk∆(x) = {F ∈ ∆ | x /∈ F, F ∪{x} ∈ ∆} and del∆(x) = {F ∈ ∆ | x /∈ F}.

Proposition 1.1.19. Let ∆ be a simplicial complex and H a face of ∆. Then both lk∆(H) and del∆(H) are

simplicial complexes, and lk∆(H) ⊆ del∆(H).

Proof. Let ∆ be a simplicial complex on [n] and H a face of ∆. From the definitions of lk∆(H) and del∆(H)

we can easily conclude that lk∆(H) ⊆ del∆(H). Now note that both lk∆(H) and del∆(H) are a collection

of subsets of [n] \H, since for every F ⊆ [n] belonging to lk∆(H) or del∆(H) we have F ∩H = ∅. Now let

F ∈ lk∆(H) and F ′ ⊆ F , then F ′ ∩ H = (F ′ ∩ F ) ∩ H = F ′ ∩ (F ∩ H) = ∅. Moreover, F ′ ∪ H ∈ ∆ since

F ′ ⊆ F ∪ H, F ∪ H ∈ ∆ and ∆ is a simplicial complex. Therefore F ′ ∈ lk∆(H), so we can conclude that

lk∆(H) is a simplicial complex on [n] \H. From what was said before we can also easily infer that del∆(H)

is a simplicial complex on [n] \H.

Lemma 1.1.20. Let ∆ be a pure simplicial complex. Then lk∆(x) is pure for every vertex x of ∆.

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Proof. Let ∆ be a pure simplicial complex and x a vertex of ∆. We want to prove that all facets of lk∆(x)

have the same dimension. Let M be a facet of lk∆(x). We claim that M ∪ {x} is a facet of ∆. Suppose not,

then there is G ∈ ∆ such that M ∪ {x} ( G. We can write G = F ∪ {x}, where F ∈ ∆ and x /∈ F . Therefore

F ∈ lk∆(x) and M ( F , which contradicts the maximality of M in lk∆(x). Thus, M ∪ {x} is a facet of ∆ for

every facet M of lk∆(x), and since ∆ is pure it follows that lk∆(x) is pure.

1.2 Cohen-Macaulay Simplicial complexes

Let S = K[x1, . . . , xn] be the polynomial ring in n variables over a field K. For each F ⊆ [n] we set

xF = Πi∈Fxi, PF = (xi | i ∈ F ), and F = [n] \ F.

Definition 1.2.1. Let ∆ be a simplicial complex on [n]. The Stanley-Reisner ideal of ∆ is the ideal I∆ of S

which is generated by those squarefree monomials xF with F /∈ ∆.

Proposition 1.2.2. Let ∆ be a simplicial complex on [n]. Then the Stanley-Reisner ideal of ∆ is given by

I∆ = (xF ∈ S | F ∈ N (∆)).

Lemma 1.2.3. Let ∆ be a simplicial complex on [n]. The standard primary decomposition of its Stanley-

Reisner ideal I∆ is given by

I∆ =⋂

F∈F(∆)

PF .

Proof. First we show that I∆ ⊆ PF for every F ∈ F(∆). Let G be a non-face of ∆. Note that G 6⊆ F ,

otherwise G would be a face of ∆. Take i ∈ G, i /∈ F . Then xi ∈ PF and xi|xG, hence xG ∈ PF . Since G is

an arbitrary non-face of ∆, it follows that I∆ ⊆ PF .

Now we show that⋂F∈F(∆) PF ⊆ I∆. Let u be a monomial such that u ∈

⋂F∈F(∆) PF , that is, u ∈

PF ∀F ∈ F(∆). Then for each F ∈ F(∆) there exists iF /∈ F such that xiF |u, hence lcm(xiF | F ∈ F(∆))|u.

Now suppose that {iF | F ∈ F(∆)} is a face of ∆. Then there is a facet H of ∆ such that {iF | F ∈ F(∆)} ⊆

H. In particular, iH ∈ H, contradiction. So {iF | F ∈ F(∆)} is a non-face of ∆, consequently u ∈ I∆.

Since the Stanley-Reisner ideal of a simplicial complex is uniquely determined by its facets, the primary

decomposition I∆ =⋂F∈F(∆) PF turns out to be irredundant.

Definition 1.2.4. Let ∆ be a simplicial complex on [n]. We say that ∆ is a Cohen-Macaulay simplicial

complex over a field K if S/I∆ is a Cohen-Macaulay ring, where I∆ is the Stanley-Reisner ideal of ∆.

Proposition 1.2.5. Let ∆ be a simplicial complex. If ∆ is Cohen-Macaulay, then ∆ is pure.

Proof. This is (Lemma 8.1.5, [4]).

Proposition 1.2.6. Let ∆ be a simplicial complex and F ∈ ∆. If ∆ is Cohen-Macaulay, then lk∆(F ) is

Cohen-Macaulay.

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Proof. This is (Corollary 8.1.8, [4]).

Definition 1.2.7. Let ∆ be a simplicial complex. We say that ∆ is connected if each pair of facets F,G

can be connected by a sequence of facets F = F0, F1, . . . , Fk = G that satisfies Fi−1 ∩ Fi 6= ∅ for every

i = 1, . . . , k.

Example 1.2.8. Let ∆ be the simplicial complex on [5] given by

∆ = {{1}, {2}, {3}, {4}, {5}, {1, 2}, {2, 3}, {2, 5}, {3, 4}, {3, 5}, {4, 5}, {2, 3, 5}, {3, 4, 5}}.

We have F(∆) = {{1, 2}, {2, 3, 5}, {3, 4, 5}}. Note that every pair of facets of ∆ is connected by a

sequence of facets satisfying the condition of the previous definition. Hence ∆ is connected.

Proposition 1.2.9. Let ∆ be a simplicial complex. If ∆ is Cohen-Macaulay, then ∆ is connected.

Proof. This is (Corollary 8.1.7, [4]).

Definition 1.2.10. Let ∆ be a pure d-dimensional simplicial complex. We say that ∆ is strongly connected

if each pair of facets F,G can be connected by a sequence of facets F = F0, F1, . . . , Fk = G such that the

dimension of Fi−1∩Fi is d−1 for all i = 1, . . . , k. In this case we say that such a sequence strongly connects

F to G.

If d > 0, then every d-dimensional strongly connected simplicial complex is clearly connected. However,

0-dimensional non-singular simplicial complexes are not connected even though they are strongly connected

(recall that, by definition, ∅ is a face with dimension −1).

Example 1.2.11. The collection ∆ given by

∆ = {{1}, {2}, {3}, {4}, {5}, {1, 2}, {2, 3}, {1, 3}, {1, 4}, {2, 4}, {1, 5}, {2, 5}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}}

is a pure 2-dimensional strongly connected simplicial complex on [5], since every intersection of a pair of

facets of ∆ is a face of ∆ of dimension 1.

Proposition 1.2.12. Let ∆ be a simplicial complex. If ∆ is Cohen-Macaulay, then ∆ is strongly connected.

Proof. Let ∆ be a Cohen-Macaulay simplicial complex. By proposition 1.2.5, ∆ is pure.

We proceed by induction on d, the dimension of ∆.

If d = 0, then ∆ is strongly connected, since the dimension of the empty face is −1. Therefore assume

that d > 0. Let F,G ∈ F(∆). By proposition 1.2.9, ∆ is connected, so there exist a sequence of facets of

∆, F = F0, F1, . . . , Fk = G such that Fi−1 ∩ Fi 6= ∅ for i = 1, . . . , k. For each i ∈ [k], pick yi ∈ Fi−1 ∩ Fiand consider the simplicial complex lk∆(yi). By lemma 1.1.20 and proposition 1.2.6, each lk∆(yi) is a

pure (d−1)-dimensional Cohen-Macaulay simplicial complex. Hence, by induction hypothesis, each lk∆(yi)

is strongly connected. Since there is a bijection between the facets of ∆ containing yi and the facets of

lk∆(yi), for each i ∈ [k], there exist a sequence of facets of ∆, Fi−1 = Hi0, Hi1, . . . ,Hiki = Fi with all Hij

containing yi and such that the dimension of Hi(j−1)∩Hij is d−1. Combining all these sequences we obtain

a sequence of facets of ∆ that strongly connects F to G. Consequently ∆ is strongly connected.

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1.3 Graph Theory

Definition 1.3.1. A graph G is an ordered pair of disjoint finite sets (V,E) such that E is a subset of the set

of unordered pairs of V .

The set V is the set of vertices of the graph and the set E is called the set of edges of the graph. An

edge {x, y}, with x, y ∈ V , is said to join the vertices x and y. We also say that the edge {x, y} is incident

with x and y, or that vertices x and y are adjacent or neighbouring vertices of G, or that vertices x and y are

the endpoints of the edge {x, y}.

We call a graph with empty vertex set and empty edge set the empty graph, and denote it by ∅. More

generally, a graph is called totally disconnected if its edge set is empty.

Example 1.3.2. The graph with set of vertices given by V = {a1,a2,a3,a4,a5,a6,a7,a8} and set of edges

given by E = {{a1,a2}{a2,a3}{a3,a4}, {a4,a5}, {a6,a7} can be represented as below.

Definition 1.3.3. The neighbourhood of a vertex x is N(x) = {y ∈ V | {x, y} ∈ E}, and the closed

neighbourhood of x is N [x] = N(x) ∪ {x}.

Definition 1.3.4. The order of a graph is its number of vertices.

The vertex set and the edge set of G are often denoted by V (G) and E(G), respectively.

Definition 1.3.5. The degree deg(x) of a vertex x is the number of edges incident with x.

Definition 1.3.6. A vertex with degree zero is called an isolated vertex.

Definition 1.3.7. A vertex with degree one is called a leaf.

An edge is called a pendant edge if it is incident to a leaf.

Example 1.3.8. In the graph below the vertex a is a leaf and the vertex b is an isolated vertex.

Definition 1.3.9. Two graphs G and H are isomorphic if there exists a bijective map φ from V (G) to V (H)

such that {x, y} ∈ E(G) if and only if {φ(x), φ(y)} ∈ E(H).

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Example 1.3.10. The two graphs below are isomorphic with bijective map φ given by φ(ai) = bi for i =

1, . . . , 12.

Definition 1.3.11. Let G and H be two graphs. Then H is called a subgraph of G if V (H) ⊆ V (G) and

E(H) ⊆ E(G).

Definition 1.3.12. Let G be a graph, V its vertex set and E its edge set. For any W ⊆ V , the induced

subgraph of G on W , denoted by G[W ], is the graph with vertex set W and edge set {e ∈ E | e ⊆ W}. In

this case we say that the vertices belonging to W induce that subgraph G[W ].

For S ⊆ V , we let G \ S denote the subgraph of G obtained by removing all the vertices of S and their

incident edges, that is, G \ S = G[V \ S]. If S = {x} we simplify the notation writing G \ x instead of G \ {x}.

Definition 1.3.13. Given two graphs G and H such that V (G) ∪ V (H) ⊆ [n], their intersection is the graph

G ∩H such that V (G ∩H) = V (G) ∩ V (H) and E(G ∩H) = E(G) ∩ E(H).

Definition 1.3.14. Given two graphs G and H such that V (G) ∪ V (H) ⊆ [n], their union is the graph G ∪H

such that V (G ∪H) = V (G) ∪ V (H) and E(G ∪H) = E(G) ∪ E(H).

Definition 1.3.15. Let G be a graph. A walk of length n in G is a sequence of vertices and edges

x0, z1, x1, . . . , xn−1, zn, xn such that, for each 1 ≤ i ≤ n, zi = {xi−1, xi} ∈ E(G). If x0 = xn, the walk

is called a closed walk. A walk may also be written x0, . . . , xn with the edges understood.

Example 1.3.16. In the graph below a1,a2,a3,a4,a5,a3 is a walk of length 5.

Definition 1.3.17. A path of length n is a walk x0, . . . , xn whose vertices are all distinct.

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Definition 1.3.18. Let G be a graph and x, y two vertices of G. The distance between x and y, d(x, y), is

the minimum of the lengths of all possible paths joining x and y. If there is no path joining x and y, then

d(x, y) =∞.

Definition 1.3.19. A graph G is connected if for every pair of vertices of G there is a path in G joining them.

Given a graph G, note that G has a vertex disjoint decomposition

G =r⋃i=1

Gi

where G1, . . . , Gr are the maximal (with respect to inclusion) connected subgraphs of G. The Gi are called

the connected components of G.

Definition 1.3.20. We say that a graph G with V (G) = {x0, . . . , xn−1} and E(G) = {{xi−1, xi} | 1 ≤ i ≤ n}

(with xn = x0) is a cycle of length n or a n-cycle, denoted by Cn.

Example 1.3.21. The graph below is a cycle of length 6.

By an induced cycle of G we mean an induced subgraph of G, G[W ], that is a cycle, where W is some

subset of V (G).

If a graph G has no induced cycles we say that G has infinite girth.

Definition 1.3.22. A cycle is even (respectively, odd) if its length is even (respectively, odd), that is, if it has

an even (respectively, odd) number of vertices.

Definition 1.3.23. A chord of a cycle C in the graph G is an edge of G joining two non-adjacent vertices of

C.

Example 1.3.24. In the graph below, we see a cycle of length six with two chords.

Definition 1.3.25. A graph G is called chordal if every cycle of G of length greater than 3 has a chord in G.

Example 1.3.26. The graph below is a chordal graph.

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Proposition 1.3.27. Any induced subgraph of a chordal graph is chordal as well.

Definition 1.3.28. The complete graph Kn is the graph such that every pair of its n vertices is adjacent.

Example 1.3.29. The graph below is a complete graph on four vertices.

Definition 1.3.30. A clique of a graph G is a set of vertices that induces a complete subgraph.

We will also call a complete subgraph of G a clique.

Definition 1.3.31. A vertex x ∈ V is a simplicial vertex if the induced graph on N(x) is a clique; equivalently

the vertex x appears in exactly one maximal clique of the graph.

Example 1.3.32. The vertex a in the graph below is simplicial.

Definition 1.3.33. A simplex of a graph is a clique containing at least one simplicial vertex of the graph.

Definition 1.3.34. A graph G is simplicial if every vertex of G is a simplicial vertex or adjacent to one.

Example 1.3.35. The graph below is a simplicial graph. The simplicial vertices are a1,a2,a6,a7 and each

vertex is either a simplicial vertex or adjacent to one.

Definition 1.3.36. A graph G is bipartite if its vertex set V (G) can be bipartitioned into two disjoint subsets

V1 and V2 such that every edge of G has one vertex in V1 and one vertex in V2.

The pair (V1, V2) is called a bipartition of G.

Example 1.3.37. The graph below is a bipartite graph with bipartition (V1, V2), where V1 = {a1,a2,a3} and

V2 = {b1,b2,b3,b4}.

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Proposition 1.3.38. A graph G is bipartite if and only if all the (induced) cycles of G are even.

Proof. (⇒) Let G be a bipartite graph with bipartition (V1, V2). If {x0, . . . , xn} is a cycle of G, (x0 = xn),

one may assume x0 ∈ V1. Note that x1 ∈ V2, and so xi ∈ V1 if and only if i is even. Thus, n is even since

xn = x0 ∈ V1.

(⇐) Let G be a graph such that all its cycles are even. We want to prove that G is bipartite. It suffices to

prove that each connected component of G is bipartite, thus one may assume that G is connected. Pick a

vertex x0 ∈ V (G). Set

V1 = {x ∈ V (G) | d(x, x0) is even} and V2 = V (G) \ V1.

It follows that no two vertices of Vi are adjacent for i = 1, 2, otherwise G would contain an odd cycle.

Therefore G is bipartite.

Definition 1.3.39. A bipartite graph G with bipartition (V1, V2) is called a complete bipartite graph if for every

two vertices x ∈ V1 and y ∈ V2, {x, y} is an edge of G. A complete bipartite graph with partitions of size

|V1| = m and |V2| = n, is denoted Km,n.

Example 1.3.40. The graph below is a complete bipartite graph with bipartition (V1, V2), where V1 =

{a1,a2,a3} and V2 = {b1,b2,b3,b4}.

Definition 1.3.41. A star is the complete bipartite graph K1,k.

Example 1.3.42. The graph G below is a star with bipartition (V1, V2), where V1 = {a} and V2 = V (G) \ V1.

Definition 1.3.43. Let n ≥ 1 and S ⊆ {1, . . . , bn2 c}. The circulant graph on S, denoted by Cn(S), is the

graph on the vertex set {0, . . . , n− 1} with all edges {a, b} that satisfy |a− b| ∈ S or n− |a− b| ∈ S.

Example 1.3.44. The graph below on the left is the circulant graph C8({1, 3}), and the graph below on the

right is the circulant graph C8({2, 3, 4}).

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Definition 1.3.45. Let G = (V,E) be a graph. A clique vertex partition of V is a set π = {W1, . . . ,Wt} of

disjoint subsets that partition V such that, for 1 ≤ i ≤ t, G[Wi] is a clique.

Example 1.3.46. In the graph G below we have a clique vertex partition π = {W1,W2,W3} given by W1 =

{a1,a2}, W2 = {b1,b2,b3,b4} and W3 = {c1, c2, c3, c4, c5}. The induced graphs G[W1], G[W2], G[W3] are

complete graphs on 2, 4 and 5 vertices, respectively, hence each G[Wi] is a clique.

Definition 1.3.47. Let G = (V,E) be a graph, where V = {x1, . . . , xn}. A clique-whiskered graph Gπ

constructed from the graph G with clique vertex partition π = {W1, . . . ,Wt} is the graph with V (Gπ) =

{x1, . . . , xn, w1, . . . , wt} and E(Gπ) = E ∪ {{x,wi} | x ∈Wi}. In order words, for each clique in the partition

π, we add a new vertex wi, and join wi to all the vertices in the clique.

Example 1.3.48. The graph below is the graph Gπ, where G and π are the graph and the clique vertex

partition, respectively, of the previous example.

Definition 1.3.49. Let G = (V,E) be a graph. A subset M of E is called a matching of G if for e and e′

belonging to M with e 6= e′, one has e ∩ e′ = ∅.

39

In other words, a matching of a graph is a subset of edges of the graph that do not share any common

endpoints. A matching is perfect if the set of vertices in the edges of the matching are all of the vertices of

the graph.

The matching number m(G) of G is the number of edges of a matching of G of largest size.

Example 1.3.50. The graph below has a matching M given by M = {{a2,a4}, {a5,a6}, {a3,a7}}, and a

perfect matching P given by P = {{a1,a2}, {a3,a4}, {a5,a6}, {a7,a8}}.

Definition 1.3.51. A matching M of a graph G is called an induced matching of G if, for e and e′ belonging

to M with e 6= e′, there is no f ∈ E(G) with f ∩ e 6= ∅ and f ∩ e′ 6= ∅.

The induced matching number im(G) of G is the number of edges of an induced matching of G of largest

size.

Example 1.3.52. The graph below has an induced matchingM given byM = {{a1,a2}, {a5,a6}, {a7,a8} {a10a11}}.

Notice that M is not a perfect matching.

Definition 1.3.53. A star triangle is a set of triangles that intersect at a common vertex.

Example 1.3.54. Both graphs below are star triangle graphs.

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Definition 1.3.55. A connected graph G is called a Cameron-Walker graph if im(G) = m(G) and if G is

neither a star nor a star triangle.

Example 1.3.56. The graph G below is a Cameron-Walker graph with im(G) = m(G) = 2.

Definition 1.3.57. Let G be a graph. A subset C of V (G) is called a vertex cover of G if every edge of G is

incident with at least one vertex of C.

Definition 1.3.58. Let G be a graph. A vertex cover C of G is called a minimal vertex cover of G if there is

no vertex cover D of G such that D is a proper subset of C.

Example 1.3.59. If G is a bipartite graph with bipartition (V1, V2), then both V1 and V2 are vertex covers of

G.

Definition 1.3.60. Let G be a graph. A subset W of V (G) is called an independent set of G if no two vertices

of W are adjacent.

Definition 1.3.61. Let G be a graph. An independent set W of G is called a maximal independent set of G

if there is no independent set U of G such that W is a proper subset of U .

We denote by Ind(G) the set of independent sets of a graph G.

Example 1.3.62. The set of independent sets of the graph G below is given by

Ind(G) = {{a1,a4}, {a2,a4}, {a3,a5}, {a1}, {a2}, {a3}, {a4}, {a5}}.

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Definition 1.3.63. Let G be a graph. The vertex covering number of G, denoted by β(G), is the number of

vertices in any vertex cover of G of smallest size.

Definition 1.3.64. Let G be a graph. The vertex independence number of G, denoted by α(G), is the

number of vertices in any independent set of G of largest size.

Proposition 1.3.65. Let G be a graph and W ⊆ V (G). Then W is an independent set of G if and only if

C = V (G) \W is a vertex cover of G.

Proof. Let G be a graph and W ⊆ V (G). Then W ∈ Ind(G) if and only if {i, j} /∈ E(G) for every i, j ∈W . In

other words, W ∈ Ind(G) if and only if every edge of G has at least one vertex in C = V (G) \W , that is, C

is a vertex cover of G.

Corollary 1.3.66. Let G be a graph with |V (G)| = n. Then α(G) + β(G) = n.

Proof. Since a subset of V (G) is an independent set if and only if its complement is a vertex cover, it follows

that a subset of V (G) is a maximal independent set if and only if its complement is a maximal vertex cover.

Hence the result follows.

Proposition 1.3.67. Let G be a graph. Then Ind(G) is a simplicial complex on the set V (G).

We call Ind(G) the independence complex of G.

Definition 1.3.68. Let G be a graph. We say that G is well-covered or unmixed if all of its maximal indepen-

dent sets have the same cardinality. Equivalently, G is well-covered or unmixed if all of its minimal vertex

covers have the same cardinality.

Example 1.3.69. The graph below is a well-covered graph with maximal independent sets having cardinality

2.

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Proposition 1.3.70. Let G be a graph. Then G is well-covered if and only if Ind(G) is pure.

Proof. Simply note that the facets of Ind(G) are the maximal faces of Ind(G), that is, the facets of Ind(G)

are the maximal independent sets of G with respect to inclusion and so, G is well-covered if and only if all

maximal independent sets of G have the same cardinality, which is equivalent to all facets of Ind(G) having

the same cardinality, that is, Ind(G) is pure.

Definition 1.3.71. A very well-covered graph is a well-covered graph in which every maximal independent

set has cardinality |V |2 .

Example 1.3.72. The graph below is a very well-covered graph of 8 vertices, with maximal independent

sets having cardinality 4.

Lemma 1.3.73. Let G be a graph and x one of its vertices. We have:

1. Ind(G \ x) = delInd(G)(x),

2. Ind(G \N [x]) = lkInd(G)(x).

Proof.

1. (⊆) Let S ∈ Ind(G \ x) ⊆ Ind(G). Then x /∈ S and so S ∈ delInd(G)(x).

(⊇) Let S ∈ delInd(G)(x). Then x /∈ S and so S ⊆ V (G \ x). Hence S ∈ Ind(G \ x).

2.(⊆) Let S ∈ Ind(G \N [x]). We have N [x] ∩ S = ∅ and consequently x /∈ S and S ∪ {x} ∈ Ind(G). Thus,

S ∈ lkInd(G)(x).

(⊇) Let S ∈ lkInd(G)(x). We have x /∈ S and S ∪ {x} ∈ Ind(G) and so N [x] ∩ S = ∅. Therefore

S ∈ Ind(G \N [x]).

The next and last definition of this section is the notion of dominating set of a graph. This definition will

play an important role on the last section of this thesis.

Definition 1.3.74. Let G be a graph and D a subset of V = V (G). We say that D is a dominating set of G

if every vertex x ∈ V \D is adjacent to a vertex of D.

Example 1.3.75. The set of vertices {a,b, c} of the graph below is a dominating set.

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Proposition 1.3.76. Let G be a graph and D a subset of V (G). If D is a dominating set, then every subset

of V (G) that contains D is also a dominating set.

Proof. Let F be a subset of V (G) such that D ⊆ F . Let x ∈ V (G) \ F ⊆ V (G) \D. Since D is a dominating

set, x is adjacent to some vertex in D ⊆ F , and so F is a dominating set.

1.4 Cohen-Macaulay Graphs

Let S = K[x1, . . . , xn] be the polynomial ring in n variables over a field K. For each F ⊆ [n] we set

xF = Πi∈Fxi, PF = (xi | i ∈ F ), and F = [n] \ F.

Definition 1.4.1. Let G be a graph with V (G) = [n]. The monomial edge ideal of G is the ideal I(G) of

S generated by all square-free monomials xixj such that {i, j} ∈ E(G). If G is a discrete graph, we set

I(G) = (0).

Proposition 1.4.2. Let G be a graph with V (G) = [n], and C ⊆ [n]. Then C is a vertex cover of G if and

only if the monomial prime ideal PC contains I(G).

Proof. (⇒) Let C be a vertex cover of G. Let xixj ∈ I(G). Since C is a vertex cover of G, we have i ∈ C or

j ∈ C, that is, xi ∈ PC or xj ∈ PC . Hence xixj ∈ PC .

(⇐) Let C ⊆ [n] be such that I(G) ⊆ PC . For every xixj ∈ I(G) we have xi ∈ PC or xj ∈ PC , so for every

{i, j} ∈ E(G) we have i ∈ C or j ∈ C, hence C is a vertex cover of G.

Corollary 1.4.3. Let G be a graph with V (G) = [n], and C ⊆ [n]. Then C is a minimal vertex cover of G if

and only if PC is a minimal prime ideal of I(G).

Corollary 1.4.4. Let G be a graph with V (G) = [n]. Then β(G) = ht(I(G)).

Proof. By proposition 0.3.30 we have ht(PC) = |C| for every C ∈ [n]. Hence, proposition 1.4.3 implies that

ht(I(G) = min{ht(PC) | PC is a minimal prime ideal of I(G)

= min{|C| | C is a minimal vertex cover of G}

= β(G).

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Corollary 1.4.5. Let G be a graph with V (G) = [n]. Then α(G) = dim(S/I(G)).

Proof. By corollary 1.3.66 we have α(G)+β(G) = n, by proposition 0.2.56 we have ht(I(G))+dim(S/I(G)) =

n, and by corollary 1.4.4 we have β(G) = ht(I(G)). Thus,

α(G) = α(G) + β(G)− β(G) = ht(I(G)) + dim(S/I(G))− ht(I(G)) = dim(S/I(G)).

Definition 1.4.6. Let G be a graph with V (G) = [n]. We say that G is Cohen-Macaulay if S/I(G) is a

Cohen-Macaulay ring.

Proposition 1.4.7. Let G be a graph with V (G) = [n]. Then G is Cohen-Macaulay if and only if its indepen-

dence complex Ind(G) is a Cohen-Macaulay simplicial complex on [n].

Proof. If suffices to show that I(G) = IInd(G), that is, the monomial edge ideal of G is equal to the Stanley-

Reisner ideal of Ind(G).

(⊆) Let xixj be one of the generators of I(G). Then {i, j} ∈ E(G), hence {i, j} /∈ Ind(G), and so xixj ∈

IInd(G).

(⊇) Let xF be one of the generators of IInd(G). Therefore F is a minimal non-face of Ind(G), in particular

F /∈ Ind(G), so there exist {i, j} ⊆ F such that {i, j} ∈ E(G). By the minimality of F it follows that

F = {i, j} ∈ E(G). Hence xF = x{i,j} ∈ I(G).

Corollary 1.4.8. Let G be a graph. If G is Cohen-Macaulay, then G is well-covered.

Proof. Let G be a Cohen-Macaulay graph. By the previous proposition we have that Ind(G) is a Cohen-

Macaulay simplicial complex, and so, by proposition 1.2.5, we have that Ind(G) is pure. Consequently, by

proposition 1.3.70, G is well-covered.

An important property of the family of Cohen-Macaulay graphs is their additivity with respect to connected

components.

Proposition 1.4.9. Let G be a graph. Then G is Cohen-Macaulay if and only if its connected components

are all Cohen-Macaulay.

Proof. This is (Proposition 7.3.8, [13]).

45

Chapter 2

Pure Vertex Decomposability

This is the main chapter of this dissertation. Here we explore in detail the notion of vertex decompos-

ability, both in its pure and non-pure forms, for simplicial complexes and graphs.

In section 2.1 we define vertex decomposability and pure vertex decomposability of simplicial complexes,

and its main result is that every link of a face of a pure vertex decomposable simplicial complex is itself a

pure vertex decomposable simplicial complex.

In section 2.2 we define vertex decomposability and pure vertex decomposability of graphs and relate

these two definitions. We prove that a well-covered vertex decomposable graph is pure vertex decompos-

able. Moreover, we define the set of shedding vertices for pure vertex decomposable graphs and we prove

that every neighbour of a simplicial vertex is a shedding vertex.

In section 2.3 we study some families of pure vertex decomposable graphs for which the set of shedding

vertices is a dominating set.

In section 2.4 we present two constructions of families of pure vertex decomposable graphs for which

the set of shedding vertices is not a dominating set.

2.1 Pure Vertex Decomposable Simplicial Complexes

Definition 2.1.1. Let ∆ be a simplicial complex. We say that ∆ is vertex decomposable if ∆ is a simplex, or

∆ contains a vertex x such that

1. both del∆(x) and lk∆(x) are vertex decomposable, and

2. no face of lk∆(x) is a facet of del∆(x).

Example 2.1.2. The simplicial complex ∆ on [4] given by ∆ = {{1, 3}, {1, 4}, {2, 4}, {1}, {2}, {3}, {4}} is

vertex decomposable. Indeed, ∆ is not a simplex, so we must find a vertex of ∆ that satisfies condition 2 of

46

the previous definition. Consider the vertex 2. We have

del∆(2) = {{1, 3}, {1, 4}, {1}, {3}, {4}} and lk∆(2) = {{4}}.

The only face of lk∆(2), {4}, is not maximal in del∆(2), hence 2 satisfies condition 2 of the previous definition.

Also note that lk∆(2) is a simplex, and therefore vertex decomposable. It remains to be shown that del∆(2)

is vertex decomposable. Consider the vertex 3. We have

deldel∆(2)(3) = {{1, 4}, {1}, {4}} and lkdel∆(2)(3) = {{1}}.

The only face of lkdel∆(2)(3), {1}, is not maximal in deldel∆(2)(3), hence 3 satisfies condition 2 of the pre-

vious definition. Moreover, both deldel∆(2)(3) and lkdel∆(2)(3) are simplices, hence vertex decomposable.

Therefore del∆(2) is vertex decomposable. It follows that ∆ is vertex decomposable.

Proposition 2.1.3. Let ∆ be a simplicial complex and x a vertex of ∆. The following conditions are equiva-

lent:

1. no face of lk∆(x) is a facet of del∆(x),

2. no facet of lk∆(x) is a facet of del∆(x),

3. any facet of del∆(x) is a facet of ∆.

Proof. Let ∆ be a simplicial complex and x a vertex of ∆.

(1)⇒ (2): Obvious.

(2) ⇒ (1): Let F be a face of lk∆(x). If F is a facet of lk∆(x) then, by hypothesis, F is not a facet of

del∆(x). Therefore, suppose that F is not a facet of lk∆(x), so that there exists a facet of lk∆(x), F ′, such

that F ( F ′. By hypothesis F ′ is not a facet of del∆(x), and so there exists a vertex y of del∆(x) such that

F ′ ∪ {y} ∈ del∆(x). Note that F ∪ {y} ∈ del∆(x), since F ∪ {y} ⊆ F ′ ∪ {y} ∈ ∆, x /∈ F and x 6= y. Thus, F

is not a facet of del∆(x).

(2) ⇒ (3): Let F be a facet of del∆(x). If F is not a facet of ∆ then F ( F ′, where F ′ is a facet of ∆. If

x /∈ F ′, then F ′ ∈ del∆(x); but this is a contradiction because F is a facet of del∆(x). Therefore, x ∈ F ′, and

since F ∪ {x} ⊆ F ′, we have F ∪ {x} ∈ ∆. This means that F ∈ lk∆(x). Note that, in this scenario, F is

also a facet of lk∆(x) (because lk∆(x) ⊆ del∆(x)), contradiction, since, by hypothesis, no facet of lk∆(x) is

a facet of del∆(x).

(3)⇒ (2): Let F be a facet of lk∆(x). Suppose that F is a facet of del∆(x). It follows, by hypothesis, that F

is a facet of ∆. However, since F ∈ lk∆(x), we have F ( F ∪ {x} ∈ ∆, which contradicts the maximality of

F in ∆. Hence F is not a facet of del∆(x).

Definition 2.1.4. Let ∆ be a simplicial complex. We say that ∆ is pure vertex decomposable if ∆ is pure

and either ∆ is a simplex, or ∆ contains a vertex x such that both del∆(x) and lk∆(x) are pure vertex

decomposable.

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Example 2.1.5. Let ∆ be a simplicial complex on [6] given by

∆ = {{1, 3}, {1, 4}, {2, 5}, {2, 6}, {4, 5}, {3, 6}, {1}, {2}, {3}, {4}, {5}, {6}}.

We are going to prove that ∆ is pure vertex decomposable. Clearly ∆ is pure and it is not a simplex, so we

must find a vertex of ∆ so that del∆(x) and lk∆(x) are pure vertex decomposable.

Consider the vertex 1. We have

del∆(1) = {{2, 5}, {2, 6}, {4, 5}, {3, 6}, {2}, {3}, {4}, {5}, {6}} and lk∆(1) = {{3}, {4}}.

Clearly lk∆(1) is pure vertex decomposable and del∆(1) is pure.

Consider the vertex 3 of del∆(1). We have

deldel∆(1)(3) = {{2, 5}, {2, 6}, {4, 5}, {2}, {4}, {5}, {6}} and lkdel∆(1)(3) = {{6}}.

Clearly lkdel∆(1)(3) is pure vertex decomposable (because it is a simplex), and deldel∆(1)(3) is pure.

Consider the vertex 6 of deldel∆(1)(3). We have

deldeldel∆(1)(3)(6) = {{2, 5}, {4, 5}, {2}, {4}, {5}} and lkdeldel∆(1)(3)(6) = {2}.

Clearly lkdeldel∆(1)(3)(6) is pure vertex decomposable (because it is a simplex), and deldeldel∆(1)(3)(6) is pure.

Consider the vertex 4 of deldeldel∆(1)(3)(6). We have

deldeldeldel∆(1)(3)(6)(4) = {{2, 5}, {2}, {5}} and lkdeldeldel∆(1)(3)(6)(4) = {{5}}.

Note that both deldeldeldel∆(1)(3)(6)(4) and lkdeldeldel∆(1)(3)(6)(4) are simplices, so pure vertex decomposable.

Therefore del∆(1) is pure vertex decomposable, so it follows that ∆ is pure vertex decomposable.

Lemma 2.1.6. Let ∆ be a simplicial complex and z, w two distinct vertices of ∆ such that {z, w} /∈ ∆. Then

lk∆(w) = lkdel∆(z)(w).

Proof.

(⊆) Let F ∈ lk∆(w). It suffices to show that z /∈ F . Suppose that z ∈ F . Then {z, w} ⊆ F ∪ {w} ∈ ∆, and

so {z, w} ∈ ∆, contradiction. Therefore z /∈ F and F ∈ lkdel∆(z)(w).

(⊇) It follows from the inclusion del∆(z) ⊆ ∆.

Lemma 2.1.7. Let ∆ be a simplicial complex and z, w two distinct vertices of ∆ such that {z, w} ∈ ∆. Then

1. dellk∆(w)(z) = lkdel∆(z)(w),

2. lklk∆(w)(z) = lk∆({z, w}) = lklk∆(z)(w).

Proof.

1. (⊆) Let F ∈ dellk∆(w)(z). It suffices to show that F ∪ {w} ∈ del∆(z). Since z /∈ F and z 6= w, we

48

have z /∈ F ∪ {w}, and so F ∪ {w} ∈ del∆(z). Therefore F ∈ lkdel∆(z)(w).

(⊇) Let F ∈ lkdel∆(z)(w). We have that F ∈ lk∆(w) (because del∆(z) ⊆ ∆), and since z /∈ F it follows that

F ∈ dellk∆(w)(z).

2. F ∈ lklk∆(w)(z)⇔ z /∈ F, w /∈ F, F ∪ {z, w} ∈ ∆⇔ F ∈ lk∆({z, w}).

Proposition 2.1.8. Let ∆ be a pure vertex decomposable simplicial complex. Then lk∆(x) is pure vertex

decomposable for every vertex x of ∆.

Proof. Let ∆ be a pure vertex decomposable simplicial complex and x a vertex of ∆. By lemma 1.1.20 we

already know that lk∆(x) is pure.

We proceed by induction on n, the number of vertices of ∆.

If n = 1, x is the only vertex of ∆, lk∆(x) = ∅, and therefore, pure vertex decomposable.

If n = 2, let x and z be the two vertices of ∆, then ∆ = {{x}, {z}} or ∆ = {{x}, {z}, {x, y}}, and so

lk∆(x) = ∅ or lk∆(x) = {{z}}, which is pure vertex decomposable in each one of the two cases.

Now assume that n ≥ 3. Suppose that ∆ is a simplex. Then lk∆(x) is a simplex: suppose not, and let F and

G be two distinct facets of lk∆(x). Recalling the proof of lemma 1.1.20 it follows that F ∪ {x} and G ∪ {x}

are facets of ∆, a contradiction. Since lk∆(x) is a simplex, it is pure vertex decomposable.

Therefore assume that ∆ is not a simplex, so that there is a vertex y of ∆ such that both del∆(y) and lk∆(y)

are pure vertex decomposable. Assume that y 6= x (otherwise lk∆(x) is pure vertex decomposable and we

are done).

Case 1. {x, y} /∈ ∆.

It follows from lemma 2.1.6 that lk∆(x) = lkdel∆(y)(x). By induction hypothesis we can conclude that

lkdel∆(y)(x) is pure vertex decomposable, hence lk∆(x) is pure vertex decomposable.

Case 2. {x, y} ∈ ∆.

We claim that dellk∆(x)(y) and lklk∆(x)(y) are pure vertex decomposable. Indeed, by lemma 2.1.7 one has

dellk∆(x)(y) = lkdel∆(y)(x) and lklk∆(x)(y) = lklk∆(y)(x), and by induction hypothesis both lkdel∆(y)(x) and

lklk∆(y)(x) are pure vertex decomposable. Therefore dellk∆(x)(y) and lklk∆(x)(y) are pure vertex decompos-

able, hence lk∆(x) is pure vertex decomposable.

2.2 Pure Vertex Decomposable Graphs

Definition 2.2.1. Let G be a graph. We say that G is vertex decomposable if G is totally disconnected, or G

contains a vertex x such that

1. both G \ x and G \N [x] are vertex decomposable, and

2. no independent set of G \N [x] is a maximal independent set of G \ x.

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Example 2.2.2. Both graphs below are vertex decomposable. The graph on the left is the complete graph

K8, and the graph on the right is a graph whose independence complex is the simplicial complex in example

2.1.2.

Proposition 2.2.3. Any complete graph is vertex decomposable.

Proof. Let G is a complete graph on n vertices. We proceed by induction on n.

n = 1: In this case G is totally disconnected, hence vertex decomposable.

n > 1: Let x be a vertex of G. Since G is a complete graph we have that G \ N [x] = ∅, so x satisfies

condition 2 of the definition 2.2.1and G \ N [x] is vertex decomposable. Moreover, since G \ x is also a

complete graph it follows, by induction hypothesis, that G \x is vertex decomposable. Therefore, G is vertex

decomposable.

Proposition 2.2.4. Let G be a graph. Then G is vertex decomposable if and only if the simplicial complex

Ind(G) is vertex decomposable.

Proof. It suffices to note that G is totally disconnected if and only if Ind(G) is a simplex and recall, by lemma

1.3.73, that Ind(G \ x) = delInd(G)(x) and Ind(G \N [x]) = lkInd(G)(x).

Proposition 2.2.5. Let G be a graph and x a vertex of G. The following conditions are equivalent:

1. no independent set of G \N [x] is a maximal independent set of G \ x,

2. no maximal independent set of G \N [x] is a maximal independent set of G \ x,

3. any maximal independent set of G \ x is a maximal independent set of G.

Proof. By lemma 1.3.73 we have Ind(G \ x) = delInd(G)(x) and Ind(G \N [x]) = lkInd(G)(x), so by applying

proposition 2.1.3 to the case ∆ = Ind(G) the result follows.

Definition 2.2.6. Let G be a graph. We say that G is pure vertex decomposable if G is well-covered and

either G is totally disconnected, or G contains a vertex x such that both G \ x and G \N [x] are pure vertex

decomposable.

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Example 2.2.7. Consider the three graphs below. The graph on the left is not well-covered, so it is not pure

vertex decomposable. The graph on the right is well-covered, but it is not pure vertex decomposable since

by removing any one of its vertices, we obtain a graph that is not well-covered. The graph in the middle is a

pure vertex decomposable graph whose independence complex is the simplicial complex of example 2.1.5.

Proposition 2.2.8. Let G be a graph. Then G is pure vertex decomposable if and only if the simplicial

complex Ind(G) is pure vertex decomposable.

Proof. We have that G is totally disconnected if and only if Ind(G) is a simplex. Moreover, by proposition

1.3.70, we have that G is well-covered if and only if Ind(G) is pure. Now, using lemma 1.3.73, that is,

Ind(G \ x) = delInd(G)(x) and Ind(G \N [x]) = lkInd(G)(x), the result follows.

Proposition 2.2.9. Let G be a well-covered graph. If G is vertex decomposable, then G is pure vertex

decomposable.

Proof. Let G be a well-covered graph. Suppose that G is vertex decomposable. If G is totally disconnected,

then G is pure vertex decomposable. Therefore, assume that G is not totally disconnected so that there is

a vertex x such that G \ x and G \N [x] both are vertex decomposable, and every maximal independent set

of G \ x is a maximal independent set of G. We proceed by induction on n, the number of vertices of G. If

n = 2 (G is not totally disconnected) then G consists of only one egde, hence pure vertex decomposable.

So assume that n ≥ 3. Using proposition 1.3.70, lemma 1.3.73 and lemma 1.1.20 we can conclude that

G \ N [x] is well-covered, hence, by induction hypothesis, G \ N [x] is pure vertex decomposable. We also

note that G \ x is well-covered since every maximal independent set of G \ x is a maximal independent set

of G and G is well-covered. Therefore, by induction hypothesis, G \ x is pure vertex decomposable. Thus,

G is pure vertex decomposable.

Corollary 2.2.10. Any complete graph is pure vertex decomposable.

Proof. Let G be a complete graph. Then G is clearly well-covered. By proposition 2.2.3 we also have that

G is vertex decomposable. Now the previous proposition implies that G is pure vertex decomposable.

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Proposition 2.2.11. Let G1 and G2 be two disjoint graphs. Then G1 ∪ G2 is pure vertex decomposable if

and only if G1 and G2 are pure vertex decomposable.

Proof. Let G1 and G2 be two disjoint graphs. Let Vi = V (Gi) for i = 1, 2. We first note that G1 ∪ G2 is

well-covered if and only if G1 and G2 are well-covered, since every maximal independent set of G1 ∪ G2 is

of the form S1 ∪ S2 where Si is a maximal independent set of Gi, for i = 1, 2.

(⇒) Suppose that G1 ∪ G2 is pure vertex decomposable. If G1 ∪ G2 is totally disconnected, then both

G1 and G2 are totally disconnected, so G1 and G2 are pure vertex decomposable. Therefore, assume

that G1 ∪ G2 is not totally disconnected so that there exists x ∈ V (G1 ∪ G2) such that (G1 ∪ G2) \ x and

(G1 ∪ G2) \ N [x] are pure vertex decomposable. We proceed by induction on n, the number of vertices of

G1 ∪G2. If n = 3 then G1 ∪G2 consists of an isolated vertex and an edge, hence both G1 and G2 are pure

vertex decomposable. Now suppose that n > 3 and assume that x ∈ G1 (the case x ∈ G2 is completely

analogous). Thus (G1∪G2)\x = (G1 \x)∪G2 and (G1∪G2)\N [x] = (G1 \N [x])∪G2, and so, by induction

hypothesis, G1 \ x,G1 \ N [x] and G2 are pure vertex decomposable. Consequently, G1 and G2 are pure

vertex decomposable.

(⇐) Now suppose that G1 and G2 are pure vertex decomposable. If they are both totally disconnected, then

G1 ∪ G2 is totally disconnected, and so pure vertex decomposable. Therefore, assume that at least one of

them is not totally disconnected. By symmetry, we can assume that G1 is not totally disconnected. Thus,

there exist x ∈ V (G1) such that both G1 \ x and G1 \ N [x] are pure vertex decomposable. We proceed

by induction on n, the number of vertices of G1 ∪ G2. If n = 3 then G1 consists of only one edge and

G2 consists of an isolated vertex, hence G1 ∪ G2 is pure vertex decomposable. Now suppose that n > 3.

We have (G1 ∪ G2) \ x = (G1 \ x) ∪ G2 and (G1 ∪ G2) \ N [x] = (G1 \ N [x]) ∪ G2, which are pure vertex

decomposable by induction hypothesis. Hence G1 ∪G2 is pure vertex decomposable.

Now we present a construction that allows us to make pure vertex decomposable graphs from a given

graph, under certain conditions.

Definition 2.2.12. Let G be a graph and S ⊆ V (G). After relabelling let S = {x1, . . . , xs}. The whiskered

graph G∪W (S) is defined to be graph with vertex set V (G)∪{z1, . . . , zs} and edge set E(G)∪{{xi, zi} | i =

1, . . . , s}. In other words, the graph G ∪W (S) is obtained from G by adding leaves or whiskers to all the

vertices of S.

Example 2.2.13. The graph below on the right is the whiskered graph G ∪ W (S), where G is the graph

below on the left and S = {x1, x2, x3, x4}.

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The next result gives us sufficient conditions for a whiskered graph G ∪W (S) to be pure vertex decom-

posable.

Proposition 2.2.14. Let G be a graph and S ⊆ V (G). If the induced graph on V (G) \ S is a chordal graph

and if G ∪W (S) is well-covered, then G ∪W (S) is pure vertex decomposable.

Proof. By (Corollary 4.6, [23]) we have that Ind(G ∪W (S)) is vertex decomposable. Now, proposition 2.2.4

implies that G ∪W (S) is vertex decomposable. Since G ∪W (S) is well-covered, it follows, by proposition

2.2.9, that G ∪W (S) is pure vertex decomposable.

Definition 2.2.15. Let G be a pure vertex decomposable graph. The set of shedding vertices of G is defined

as follows:

Shed(G) := {x ∈ V | G \ x and G \N [x] are pure vertex decomposable}.

Woodroofe ([14]) defined shedding vertices in the context of (non-pure) vertex decomposability. In his

definition, a shedding vertex of a graph is a vertex that satisfies condition 2 of definition 2.2.1.

Proposition 2.2.16. Let G be a pure vertex decomposable graph. Then G \N [x] is pure vertex decompos-

able for every vertex x of G.

Proof. Let G be a pure vertex decomposable graph. Then, by proposition 2.2.8, Ind(G) is pure vertex

decomposable. Therefore, by proposition 2.1.8, lkInd(G)(x) is pure vertex decomposable for every vertex x

of G. Since lkInd(G)(x) = Ind(G \ N [x]), by lemma 1.3.73, it follows, by proposition 2.2.8, that G \ N [x] is

pure vertex decomposable.

Corollary 2.2.17. Let G be a pure vertex decomposable graph. Then

Shed(G) = {x ∈ V | G \ x is pure vertex decomposable}.

Observe that for a pure vertex decomposable graph G we always have that Shed(G) 6= ∅, if G is not the

empty graph. Indeed, if G is totally disconnected, then Shed(G) = V (G), and if G is not totally disconnected,

by definition of pure vertex decomposable graph it follows that there exists x ∈ Shed(G).

We can also state that if a graph G is pure vertex decomposable, then G is well-covered and there exists

a finite sequence of well-covered subgraphs of G, say Gk, k = 0, . . . , n, such that

G0 = G, Gk = (((G \ x1) \ x2) . . . \ xk),

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where xi ∈ Shed(Gi−1) for i = 1, . . . , n, and Gn is totally disconnected.

Lemma 2.2.18. Let G be a well-covered graph and x a simplicial vertex of G. Then, for every y ∈ N(x), the

graph G \ y is well-covered.

Proof. Let S be a maximal independent set of G \ y. Then S is also an independent set of G. Suppose that

S is not maximal in G. Then S ∪ {y} is an independent set of G. Therefore (N [x] \ {y}) ∩ S = ∅, and so

S ∪ {x} is an independent set of G \ y, which contradicts the maximality of S in G \ y. Thus, every maximal

independent set of G \ y is a maximal independent set of G, and since G is well-covered, it follows that G \ y

is well-covered.

Theorem 2.2.19. Let G be a pure vertex decomposable graph and x is a simplicial vertex of G. Then

N(x) ⊆ Shed(G).

Proof. Let N(x) = {y1, . . . , yn}. Due to the symmetry of every complete graph it is enough to prove that

y1 ∈ Shed(G), that is, G\y1 is pure vertex decomposable. By the previous lemma we have that G\y1 is well-

covered, so to prove that G\ y1 is pure vertex decomposable we must find a vertex z so that both (G\ y1)\ z

and (G \ y1) \N [z] are pure vertex decomposable. Consider z = y2. Note that x is still a simplicial vertex in

G \ y1 so by the previous result it follows that (G \ y1) \ y2 is well-covered. Also, (G \ y1) \N [y2] = G \N [y2],

hence pure vertex decomposable, by proposition 2.2.16. Analogously, ((G\ y1)\ y2)\ y3 is well-covered and

((G \ y1) \ y2) \ N [y3] is pure vertex decomposable. By continuing this reasoning, we can infer that G \ y1

is pure vertex decomposable if we prove that G \N(x) = ((G \ y1) \ . . .) \ yn is pure vertex decomposable.

To do so, we observe that G \ N(x) = (G \ N [x]) ∪ {x}, hence pure vertex decomposable by propositions

2.2.16 and 2.2.11. Consequently, N(x) ⊆ Shed(G).

2.3 Dominating Shedding Vertices

Based upon computer experiments on all graphs on six or less vertices, Villarreal proposed a two-part

conjecture:

Conjecture 1 ([22], Conjectures 1 and 2). Let G be a Cohen-Macaulay graph and let

D = {x ∈ V (G) | G \ x is a Cohen-Macaulay graph}.

Then

1. D 6= ∅,

2. D is a dominating set of G.

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It is already known that conjecture 1 is false. Earl, Kevin Vander Meulen and Adam Van Tuyl ([28]) found

an example of a circulant graph G on 16 vertices such that G is Cohen-Macaulay, but there is no vertex x

such that G \ x is Cohen-Macaulay.

Although conjecture 1 is false in general, Villarreal’s work suggests that there may exist some nice subset

of Cohen-Macaulay graphs for which conjecture 1 still holds. Since pure vertex decomposable graphs are

Cohen-Macaulay, Jonathan Baker, Kevin Vander Meulen and Adam Van Tuyl ([21]) considered the following

variation of conjecture 1:

Question 1. Let G be a pure vertex decomposable graph. Recall that

Shed(G) = {x ∈ V (G) | G \ x is a pure vertex decomposable graph}.

Is Shed(G) a dominating set of G?

The goal of this section is to find some families of pure vertex decomposable graphs for which the

question 1 is answered affirmatively. The next result summarizes what is going to be presented.

Theorem 2.3.1. Let G be a pure vertex decomposable graph. If G is

1. a bipartite graph, or

2. a chordal graph, or

3. a very well-covered graph, or

4. a circulant graph, or

5. a Cameron-Walker graph, or

6. a clique-whiskered graph.

then Shed(G) is a dominating set.

We first prove the previous result for the cases of circulant graphs and chordal graphs.

Theorem 2.3.2. Let G be a pure vertex decomposable circulant graph. Then Shed(G) is a dominating set.

Proof. Since G is pure vertex decomposable, there exists some vertex i such that G \ i is pure vertex

decomposable. By the symmetry of circulant graphs, it follows that G \ j is isomorphic to G \ i for all i 6= j.

This implies that Shed(G) = V (G), hence Shed(G) is a dominating set.

Next we give an important classification of pure vertex decomposable chordal graphs.

Lemma 2.3.3. Let G be a chordal graph. Then the following are equivalent:

1. G is pure vertex decomposable,

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2. G is well-covered,

3. Every vertex of G belongs to exactly one simplex of G.

Proof.

(1⇒ 2) If G is pure vertex decomposable, then it is well-covered by definition.

(2 ⇒ 1) Woodroofe (Corollary 7, [14]) showed that every chordal graph G is vertex decomposable. Now

proposition 2.2.9 implies that G is pure vertex decomposable.

(2⇔ 3) This is (Theorem 2, [24]).

Theorem 2.3.4. Let G be a pure vertex decomposable chordal graph. Then Shed(G) is a dominating set.

Proof. Given proposition 2.2.11, we can assume that G is connected and has at least two vertices. Since

G is pure vertex decomposable, by the previous lemma, the simplexes of G partition V (G), i.e., V (G) =

V1 ∪ · · · ∪ Vt where the induced graph on each Vi is a simplex. So, every Vi contains at least one simplicial

vertex.

For each i = 1, . . . , t, let xi ∈ Vi be a simplicial vertex. Note that this means that N(xi) = Vi \ {xi} for each

i = 1, . . . , t. By theorem 2.2.19, N(xi) ⊆ Shed(G). So N(x1) ∪ . . . ∪ N(xt) ⊆ Shed(G). This implies that

Shed(G) is a dominating set. Indeed, if x 6= xi for any i, then x is a neighbour of some xj , and so is in

Shed(G). If x = xi for some i, then all of its neighbours belong to Shed(G).

Now we consider a construction of Hibi, Higashitani, Kimura, and O’Keefe ([25]) that builds a pure vertex

decomposable graph by appending a clique at each vertex. This construction is going to allow us to prove

theorem 2.3.1 for the case of Cameron-Walker graphs. More precisely, let G be a graph with vertex set

V (G) = {x1, . . . , xn} and edge set E(G). Let k1, . . . , kn be n positive integers with ki ≥ 2 for i = 1, . . . , n.

We now construct a graph G with

V (G) = {x1,1, x1,2, . . . , x1,k1 , x2,1, . . . , x2,k2 , . . . , xn,1, . . . , xn,kn}

and edge set

E(G) = {{xi,1, xj,1} | {xi, xj} ∈ E(G)} ∪n⋃i=1{{xi,j , xi,l} | 1 ≤ j < l ≤ ki}.

That is, G is the graph obtained from G by attaching a clique of size ki at the vertex xi.

Example 2.3.5. Let G be the graph of example 1.3.62. The graph below is the graph G constructed from G

as above, with k1 = 3, k2 = k3 = k4 = 2 and k5 = 4.

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Proposition 2.3.6. Let G be a graph. Then the graph G constructed from G as above is pure vertex

decomposable.

Proof. This is (Theorem 1.1, [25]).

Proposition 2.3.7. Let G be a graph. Then the pure vertex decomposable graph G has the property that

Shed(G) is a dominating set.

Proof. First note that xi,ki 6= xi,1 for i = 1, . . . , n, since ki ≥ 2. The vertex xi,ki is a simplicial vertex, so

by theorem 2.2.19 xi,1 ∈ N(xi,ki) ⊆ Shed(G). Thus T = {x1,1, . . . , xn,1} ⊆ Shed(G), and since T is a

dominating set it follows, by proposition 1.3.76, that Shed(G) is a dominating set.

One of the most relevant results of ([25]) is the fact that a Cameron-Walker graph G is pure vertex

decomposable if and only if G = H for some graph H (with some hypothesis on the ki′s that appear in the

construction of H). Consequently, we have the following:

Theorem 2.3.8. Let G be a pure vertex decomposable Cameron-Walker graph. Then Shed(G) is a domi-

nating set.

The next result is theorem 2.3.1 for the case of clique-whiskered graphs. Let G be a graph on the

vertex set V = {x1, . . . , xn}. Note that if G is the graph obtained from G by appending cliques with k1 =

. . . = kn = 2, then G is isomorphic to the clique-whiskered graph Gπ using the clique vertex partition

π = {{x1}, . . . , {xn}}. Furthermore, Cook and Nagel (Theorem 3.3, [26]) showed that for any graph G and

any clique vertex partition π of G, the graph Gπ is always pure vertex decomposable.

Theorem 2.3.9. Let G be a graph with clique vertex partition π. Then the pure vertex decomposable graph

Gπ has the property that Shed(Gπ) is a dominating set.

Proof. If π = {W1, . . . ,Wt}, then V (Gπ) = {x1, . . . , xn, w1, . . . , wt}. Every vertex xi ∈ V (G) belongs to some

clique Wj , and so, in Gπ, the vertex xi is adjacent to wj . Note that wj is a simplicial vertex, because wj is

adjacent only to the vertices of Wj and Wj is a clique. Thus by theorem 2.2.19, xi ∈ N(wj) ⊆ Shed(Gπ).

Therefore {x1, . . . , xn} ⊆ Shed(Gπ), and since {x1, . . . , xn} is a dominating set of Gπ it follows, by 1.3.76,

that Shed(Gπ) is a dominating set.

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We now show theorem 2.3.1 for the case of very well-covered graphs. To do so, we need the following

structure result shown by Mahmoudi, Mousivand, Crupi, Rinaldo, Terai, and Yassemi in (Lemma 3.1 and

Theorem 3.2, [9]).

Lemma 2.3.10. Let G be a very well-covered graph with 2h vertices. Then the following are equivalent:

1. G is pure vertex decomposable,

2. There is a relabeling of the vertices V = X ∪ Y = {x1, . . . , xh} ∪ {y1, . . . , yh} such that the following

five conditions hold:

(a) X is a minimal vertex cover of G and Y is a maximal independent set of G,

(b) {x1, y1}, . . . , {xh, yh} ∈ E(G),

(c) if {zi, xj}, {yj , xk} ∈ E(G), then {zi, xk} ∈ E(G) for distinct i, j, k and for zi ∈ {xi, yi},

(d) if {xi, yj} ∈ E(G), then {xi, xj} /∈ E(G), and

(e) if {xi, yj} ∈ E(G), then i ≤ j.

Theorem 2.3.11. Let G be a pure vertex decomposable very well-covered graph. Then Shed(G) is a

dominating set.

Proof. Let G be a pure vertex decomposable very well-covered graph, with V (G) = V and E(G) = E.

Given the previous lemma we can assume that the vertices of G have been relabeled as V = X ∪ Y =

{x1, . . . , xh, y1, . . . , yh} so that the five conditions of the lemma hold.

We first note that y1 is a leaf. Indeed, by condition (a) we have that y1 is not adjacent to any other yj ; by

condition (b) we know that {x1, y1} ∈ E, and condition (e) implies {y1, xj} /∈ E for all j = 2, . . . , h. Now we

consider the set S = {N(z) | z is a leaf of G}, which is non-empty, since, as noted before, y1 is a leaf. By

theorem 2.2.19 we have that S ⊆ Shed(G) (every leaf is a simplicial vertex).

In order to prove that Shed(G) is a dominating set it suffices to show that S is a dominating set, by

proposition 1.3.76. Assume to the contrary that there is a vertex w ∈ V \S that is not adjacent to any vertex

in S (in particular, w is not a leaf), that is, assume that S is not a dominating set. We consider two cases.

Case 1. w ∈ X. In this case w = xi for some 1 ≤ i ≤ h, and actually we can assume that i is maximal in

the sense that for i < k ≤ h, xk is either in S or is adjacent to some vertex in S. Since xi is not a leaf and

{xi, yi} ∈ E, there is another vertex that is adjacent to xi.

Suppose that xi is adjacent to some vertex in X, say xj , with j 6= i. Since xi and xj do not belong to

S, we have that both yi and yj are not leaves. Thus yi is adjacent to some xp with p < i, and yj is

adjacent to some xq with q < j, by condition (e). Note that p 6= q, because otherwise we would have

{xi, xj}, {yj , xp} ∈ E implying that, by condition (c), {xi, xp} ∈ E, which contradicts condition (d) since

{xp, yi} ∈ E. So we have {xi, xj}, {yj , xq} ∈ E, and consequently, by condition (c), {xi, xq} ∈ E (note

that q 6= i, because otherwise we would have {xi, yj} ∈ E, implying that, by condition (d), {xi, xj} /∈ E,

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which is a contradiction). Similarly, because {xj , xi}, {yi, xp} ∈ E, we have {xj , xp} ∈ E and j 6= p. Finally,

because {xq, xi}, {yi, xp} ∈ E, we have {xq, xp} ∈ E by condition (c). Let Xq = {xi, xj , xp, xq}, and note

that xq /∈ S since xq is adjacent to xi = w. This implies that yq is not a leaf. Hence there exists a vertex xq1

adjacent to yq, with q1 < q. By condition (d) it follows that {xq1 , xq} /∈ E, and so, xq1 /∈ N(xq). Therefore

q1 /∈ {i, p}. Also, q1 6= j since q1 < q < j. Thus xq1 /∈ Xq. We also have that N(xq) ⊆ N(xq1) by

condition (c). Let Xq1 = {xi, xj , xp, xq, xq1}. Inductively, we can see that for each positive integer n ≥ 2,

xqn−1 /∈ S and hence there exists a vertex xqn∈ X, qn < qn−1, with xqn

adjacent to yqn−1 such that

xqn/∈ Xqn−1 = {xi, xj , xp, xq, xq1 , . . . , xqn−1} and N(xqn−1) ⊆ N(xqn

). Since the cardinality of the set of

vertices of G is finite, this is impossible. Therefore, w is not adjacent to any vertex in X.

Let j > i be such that w = xi is adjacent to yj . We have {xi, xj} /∈ E. Note that xj is not a leaf (since

otherwise w = xi would be adjacent to yj ∈ N(xj) ⊆ S), and so N(xj) \ {yj} 6= ∅. We have N(xj) ⊆ N(xi).

Indeed, if u ∈ N(xj) \ {yj}, since {u, xj}, {yj , xi} ∈ E, condition (c) implies that {u, xi} ∈ E, that is,

u ∈ N(xi). By our assumption on the maximality of i, xj is either in S, or xj is adjacent to a vertex in S. If

xj is adjacent to a vertex in S, then so is xi since N(xj) ⊆ N(xi), which contradicts our choice of xi. On

the other hand, if xj ∈ S then there exists a vertex with degree 1 adjacent to xj , but this is impossible since

every neighbourhood of xj is a neighbourhood of xi. So w cannot be adjacent to any vertex in Y . Therefore

we conclude that w /∈ X.

Case 2. w ∈ Y . In this case w = yi for some 1 ≤ i ≤ h. We assume that i is minimal in the sense that for

1 ≤ k < i, yk is either in S or is adjacent to some vertex in S. Note that i > 1 since we already observed that

x1 ∈ S (because y1 is a leaf). Since yi is not a leaf, there is some xj adjacent to yi, and j < i by condition

(e). By our choice of i, yj is either in S or yj is adjacent to some vertex in S. If yj ∈ S, then yj is adjacent to

a leaf xk. By condition (e), k ≤ j. Further, even though {xj , yj} ∈ E, xj is not a leaf since xj is also adjacent

to yi. Hence k < j. Now, since {xk, yj}, {xj , yi} ∈ E, we have, by condition (c), {xk, yi} ∈ E, which is a

contradiction since xk is a leaf. So yj /∈ S and hence yj is adjacent to a vertex in S. But then either xj ∈ S,

which means that w = yi is adjacent to an element of S, or yj is adjacent to some xl ∈ S with l < j, by

condition (e). But then xl is also adjacent to w = yi by condition (c), a contradiction. Therefore w /∈ Y .

These two cases show that every vertex of G is either in S or adjacent to a vertex in S, that is, S is a

dominating set, and we are done.

The next result is theorem 2.3.1 for the case of bipartite graphs. We are going to show in its proof that for

pure vertex decomposable graphs the class of very well-covered graphs contains the family of well-covered

bipartite graphs.

Theorem 2.3.12. Let G be a pure vertex decomposable bipartite graph. Then Shed(G) is a dominating set.

Proof. Let G be a pure vertex decomposable bipartite graph with vertex partition V (G) = V1 ∪ V2 =

{x1, . . . , xn} ∪ {y1, . . . , ym}. Due to proposition 2.2.11 we can assume that G is connected.

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By definition of bipartite graph we have that V1 and V2 are independent sets. In fact, they are actually

maximal independent sets. Indeed, if V1 is not maximal, then there is a vertex yj ∈ V2 such that V1 ∪ {yj} is

an independent set. This implies that yj is not adjacent to any other vertex of G, contradicting the fact that

G is connected. Analogously we can conclude that V2 is maximal.

Since G is pure vertex decomposable, G is well-covered, and so, for any maximal independent set W ,

we have |W | = |V1| = |V2| = n = m = |V (G)/2|. Therefore G is very well-covered, and consequently, by the

previous theorem, Shed(G) is a dominating set.

2.4 Non-Dominating Shedding Vertices

We end up with two constructions of a pure vertex decomposable graphs for which question 1 is an-

swered negatively, that is, families of pure vertex decomposable graphs for which the set of shedding ver-

tices is not a dominating set. Both constructions are due to Jonathan Baker, Kevin Vander Meulen and

Adam Van Tuyl ([21]).

The first construction is the following:

Let k1, . . . , km be m fixed integers such that ki ≥ 2 and k1 + . . . + km = n. We define Dn(k1, . . . , km) to

be the graph on the 5n vertices

V = X ∪ Y ∪ Z = {x1, . . . , x2n} ∪ {y1, . . . , y2n} ∪ {z1, . . . , zn}

with the edge set given by the following conditions:

1. the induced graph on Z is a complete graph on n vertices,

2. Y is an independent set,

3. the induced graph G[X] is Kk1,k1 t · · · tKkm,km where the vertices of G[X] are labeled so that the i-th

complete bipartite graph has bipartition

{x2w+1, x2w+3, . . . , x2(w+ki)−1} ∪ {x2w+2, x2w+4, . . . , x2(w+ki)}

with w =∑i−1l=1 kl where w = 0 if i = 1,

4. {xj , yj} are edges for 1 ≤ j ≤ 2n,

5. {zj , y2j} and {zj , y2j−1} are edges for 1 ≤ j ≤ n.

The graph Dn(k1, . . . , km) is formed by joining m complete bipartite graphs to a complete graph Kn by

first passing through an independent set of vertices Y .

Example 2.4.1. The graph below is the graph D5(3, 2).

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Now we present and prove some lemmas that are going to allow us to prove that the graphDn(k1, . . . , kn)

constructed as above is a pure vertex decomposable graph for which the set of shedding vertices is not a

dominating set.

Lemma 2.4.2. Let G = Dn(k1, . . . , kn) be constructed as above. Then G is well-covered.

Proof. Let G = Dn(k1, . . . , kn). We want to show that every maximal independent set of G has the same

cardinality. First note that we can partition V (G) into n sets of five vertices, namely, {x2i−1, x2i, y2i−1, y2i, zi}

for i ≤ i ≤ n. The induced graph on each set {x2i−1, x2i, y2i−1, y2i, zi} is a 5-cycle. Since every maximal

independent set of a 5-cycle has cardinality 2, it follows that every maximal independent set of G has

cardinality at most 2n.

Now we show that every maximal independent set of G has cardinality 2n. Assume to the contrary that there

is a maximal independent set H of G so that |H| < 2n.

Suppose that H ∩ Z = ∅. Then, there exists an i such that neither xi nor yi belongs to H. Indeed, if for all

j we had that xj or yj belonged to H, then we would have |H| = 2n, since there are 2n edges of the form

{xi, yi}; but |H| 6= 2n, by hypothesis. But then H ∪ {yi} is an independent set since yi is only adjacent to

some zj and to xi. Because H is a maximal independent set this is a contradiction.

So, there exists a zi ∈ H∩Z. Note that H∩Z = {zi}, since G[Z] is a clique. Thus, for j 6= 2i or j 6= 2i−1 we

have {xj , yj} ∩H 6= ∅, otherwise H ∪ {yj} would be an independent set containing H. Since |H| ≤ 2n− 1,

we have that neither x2i nor x2i−1 belong to H. Hence x2i is adjacent to some xl ∈ H, and x2i−1 is adjacent

to some xk ∈ H. Moreover, x2i−1, xl, x2i, xk all belong to the same induced bipartite graph. Then l must be

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odd since 2i is even, and k must be even since 2i − 1 is odd. This implies that xl and xk are adjacent but

this is impossible since xl, xk ∈ H and H is an independent set.

Thus H cannot be a maximal independent set if |H| < 2n, so every maximal independent set of G has

cardinality 2n. Therefore G is well-covered.

Lemma 2.4.3. Let G = Dn(k1, . . . , kn) be constructed as above. Then Gi = (((G \ z1) \ z2) . . . \ zi) is a

well-covered graph, for each i = 1, . . . , n.

Proof. Let G = Dn(k1, . . . , kn) and fix some i ∈ {1, . . . , n}. In the previous lemma we already concluded

that G is well-covered, so to prove that Gi = (((G \ z1) \ z2) . . . \ zi) is well-covered it suffices to prove that

every maximal independent set of Gi is a maximal independent set of G.

Let H be a maximal independent set of G1 = G \ z1. Suppose y1 /∈ H and y2 /∈ H. Since y1 is a

leaf in G1, y1 ∈ H, and H is a maximal independent set of G1, we must have x1 ∈ H. Similarly, x2 ∈ H.

But {x1, x2} is an edge in G1, and we have a contradiction. Therefore, y1 ∈ H or y2 ∈ H. Since, in G,

z1 ∈ N(y1) ∩N(y2) we conclude that H is also a maximal independent set of G.

Now let H be a maximal independent set of G2 = G1 \ z2. Repeating the argument, H is a maximal

independent set of G1, and therefore, a maximal independent set of G.

Continuing this reasoning, we can conclude that Gi is a well-covered graph, for each i = 1, . . . , n.

Lemma 2.4.4. Let G = Dn(k1, . . . , kn) be constructed as above. Then Gn = (((G \ z1) \ z2) . . . \ zn) is pure

vertex decomposable.

Proof. Let G = Dn(k1, . . . , kn) and Gn = (((G\z1)\z2) . . .\zn). Note that Gn = G[X∪Y ]. Also, Gn consists

of m disjoint connected components where the j-th component is the complete bipartite graph Kkj ,kj with

whiskers at every vertex.

Now we show that each such connected component is pure vertex decomposable. To do so, we are going to

use proposition 2.2.14. Take S = V (Kkj ,kj ), and note that every maximal independent set of Kkj ,kj ∪W (S)

has cardinality 2kj , hence Kkj ,kj ∪W (S) is well-covered. So Kkj ,kj ∪W (S) is pure vertex decomposable,

by proposition 2.2.14, which implies, by proposition 2.2.11, that Gn is pure vertex decomposable.

Lemma 2.4.5. Let G = Dn(k1, . . . , kn) be constructed as above. Then Ni = Gi−1 \ N [zi] is pure vertex

decomposable, for each i = 1, . . . , n, where Gj = (((G \ z1) \ z2) . . . \ zj) for j = 1, . . . , n, and G0 = G.

Proof. Let G = Dn(k1, . . . , kn) and Ni = Gi−1 \ N [zi]. Suppose that x2i−1 and x2i belong to the induced

complete bipartite graph Kkj ,kj . Note that Ni consists of m disjoint graphs, where m − 1 of these are the

complete bipartite graphs with whiskers at every vertex, and them-th graph is the graph Kkj ,kjwith whiskers

at every vertex except x2i−1 and x2i. The m− 1 graphs are pure vertex decomposable as noted in the proof

of the previous lemma. Now we show that the m-th graph is pure vertex decomposable. It is well-covered

since all of its maximal independent sets have cardinality 2kj . Moreover, if we let S = V (Kkj ,kj \{x2i−1, x2i})

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then the m-th graph is the graph Kkj ,kj ∪W (S), which is pure vertex decomposable by proposition 2.2.14.

Therefore, by proposition 2.2.11, Ni is pure vertex decomposable.

Theorem 2.4.6. Let G = Dn(k1, . . . , kn) be constructed as above. Then G is pure vertex decomposable.

Proof. Let G = Dn(k1, . . . , kn). By lemma 2.4.2 we have that G is well-covered. Note that if we prove that

G1 = G\z1 and N1 = G\N [z1] are both pure vertex decomposable we conclude, by definition, that G is pure

vertex decomposable. But G1 is pure vertex decomposable if G2 = ((G\z1)\z2) and N2 = ((G\z1)\N [z2])

are pure vertex decomposable. Continuing this reasoning, to show that G is pure vertex decomposable, it

suffices to show that Gn and N1, . . . , Nn are all pure vertex decomposable. But this was shown in lemma

2.4.3, lemma 2.4.4 and lemma 2.4.5. So G is pure vertex decomposable.

Theorem 2.4.7. Let G = Dn(k1, . . . , kn) be the pure vertex decomposable graph constructed as above.

Then Shed(G) = Z.

Proof. Let G = Dn(k1, . . . , kn), which is pure vertex decomposable. We showed that z1 ∈ Shed(G). By

symmetry we also have zj ∈ Shed(G), for j = 1, . . . , n, so Z ⊆ Shed(G). Now the idea is to prove that

Y ∩ Shed(G) = ∅ = X ∩ Shed(G).

Let y ∈ Y . After relabelling, assume that y = y2n. Then {y1, . . . , y2n−1, x2n} and {z1, x1, y3, . . . , y2n−2, x2n−1}

are maximal independent sets in G \ y of cardinality 2n and 2n − 1, respectively. Thus, G \ y is not well-

covered and so y /∈ Shed(G).

Let x ∈ X. After relabelling, assume that x = x1. Note that Y is a maximal independent set of G \ x of

cardinality 2n. Since k1 ≥ 2, x3 is adjacent to x2 and x4, so {z1, x3, y4, . . . , y2n} is a maximal independent

set of G \ x of cardinality 2n− 1. Hence G \ x is now well-covered, consequently x /∈ Shed(G).

Thus Shed(G) = Z.

Corollary 2.4.8. Let G = Dn(k1, . . . , kn) be the pure vertex decomposable graph constructed as above.

Then Shed(G) is not a dominating set.

Proof. Just observe that no vertex in X is adjacent to any vertex in Z.

Now we present the second and last construction.

Let n ≥ 1. We define the graph Ln to be the graph on 8n + 1 vertices with vertex set V (Ln) = X ∪

Y ∪ Z ∪ {w}, where X = {x1,1, x1,2} ∪ . . . ∪ {xn,1, xn,2}, Y = {y1,1, y1,2, y1,3} ∪ . . . ∪ {yn,1, yn,2, yn,3},

Z = {z1,1, z1,2, z1,3} ∪ . . . ∪ {zn,1, zn,2, zn,3}, and with edge set E(Ln) satisfying the following conditions:

1. For each i = 1, . . . , n the induced graph on {xi,1, xi,2, yi,1, yi,2, yi,3} is a 5-cycle with edges {yi,1, yi,2},

{yi,2, yi,3},{yi,3, xi,2},{xi,2, yi,1},{xi,1, yi,1},

2. {zi,j , yi,j} ∈ E(Ln) for i = 1, . . . , n and j = 1, 2, 3, and these edges form a matching between Y and

Z,

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3. The induced graph on Z ∪ {w} is the complete graph K3n+1.

Example 2.4.9. The graph below on the left is the graph L1 and the graph below on the right is the graph

L2.

Proposition 2.4.10. Let G = Ln be constructed as above. Then every maximal independent set of G has

cardinality 2n+ 1.

Proof. LetM be a maximal independent set ofG. Let z ∈ Z∪{w} be a vertex such thatM∩(Z∪{w}) = {z}.

Since G contains n 5-cycles and every maximal independent set of a 5-cycle has cardinality 2, it follows that

M has cardinality 2n+ 1.

Corollary 2.4.11. The graph Ln constructed as above is a well-covered graph.

Lemma 2.4.12. The graph L1 constructed as above is pure vertex decomposable.

Proof. By the previous corollary we already know that L1 is well-covered. Since L1 is not totally discon-

nected, we must find a vertex x such that L1 \x and L1 \N [x] are both pure vertex decomposable. Consider

the vertex z1,1. Observe that L1 \N [z1,1] is a path on four vertices, hence pure vertex decomposable. More-

over, L1 \ z1,1 is well-covered because every maximal independent set of L1 \ z1,1 is a maximal independent

set of L1 and L1 is well-covered. Now consider the vertex z1,2 of L1 \ z1,1. Note that (L1 \ z1,1) \ N [z1,2]

is a path on four vertices, hence pure vertex decomposable. Also, (L1 \ z1,1) \ z1,2 is well-covered since

every maximal independent set of (L1 \ z1,1) \ z1,2 is a maximal independent set of L1 and L1 is well-

covered. It suffices to prove that (L1 \ z1,1)\ z1,2 is pure vertex decomposable. To do so we just observe that

64

((L1 \z1,1)\z1,2)\z1,3 consists of an isolated vertex and a 5-cycle, and ((L1 \z1,1)\z1,2)\N [z1,3] consists of

a path on four vertices, hence both pure vertex decomposable. Thus L1 \ z1,1 is pure vertex decomposable,

and consequently L1 is pure vertex decomposable.

One may wonder why is it relevant the existence of the vertex w is the construction above. The reason

is the following: if we take the vertex w out of the construction of the graph Ln, then, in particular, the graph

(L1 \ z1,1) \ z1,2 would not be well-covered, therefore L1 would not be pure vertex decomposable.

Theorem 2.4.13. Let G = Ln be constructed as above. Then G is pure vertex decomposable.

Proof. We proceed by induction on n. The case n = 1 is the previous lemma.

Suppose n > 1. We set G1 = G \ zn,1, G2 = G1 \ zn,2, G3 = G2 \ zn,3, N1 = G \N [zn,1], N2 = G1 \N [zn,2]

and N3 = G2 \ N [zn,2]. Observe that for each j = 1, 2, 3, Nj is a path on four vertices, hence pure vertex

decomposable. Moreover, both G1 and G2 are well-covered since each maximal independent set of any of

them is a maximal independent set of G and G is well-covered. Now observe that G3 consists of Ln−1 and a

5-cycle. By induction hypothesis Ln−1 is pure vertex decomposable, henceG3 is pure vertex decomposable.

It follows that G = Ln is pure vertex decomposable.

Given the proof above we can conclude that Z ⊆ Shed(Ln).

Proposition 2.4.14. Let G = Ln be constructed as above. For any vertex x ∈ X ∪ Y the graph G \ x is not

well-covered.

Proof. It suffices to prove that G\x1,1 and G\y1,1 are not well-covered, since the other cases are analogous.

Consider I = {x2,1, y2,2}∪. . .∪{xn,1, yn,2}. First observe that A1 = {x1,2, y1,2, w}∪I and A2 = {z1,1, y1,3}∪I

are two maximal independent sets of G \ x1,1 with different cardinalities, hence G \ x1,1 is not well-covered.

Moreover, B1 = {x1,1, y1,3, w}∪ I and B2 = {z1,2, x1,2}∪ I are two maximal independent sets of G\y1,1 with

different cardinalities, hence G \ y1,1 is not well-covered.

Corollary 2.4.15. Let G = Ln be constructed as above. Then X ∩ Shed(G) = ∅ = Y ∩ Shed(G).

Theorem 2.4.16. Let G = Ln be constructed as above. Then Shed(G) is not a dominating set.

Proof. Suppose that Shed(G) is a dominating set of G. By the previous corollary it follows that X ∪ Y ⊆

V (G)\Shed(G). Then, in particular, x1,1 is adjacent to some vertex in Shed(G). This is a contradiction, since

x1,1 is only adjacent to x1,2 ∈ X and y1,1 ∈ Y . Therefore Shed(G) is not a dominating set.

65

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