# Monomial ideals of weighted oriented graphs

Yuriko Pitones<EMAIL_ADDRESS>Departamento de Matemáticas
Centro de Investigación y de Estudios Avanzados del Instituto Politécnico
Nacional
Apartado Postal 14–740, Ciudad de México
07000 México Enrique Reyes<EMAIL_ADDRESS>Departamento de
Matemáticas
Centro de Investigación y de Estudios Avanzados del Instituto Politécnico
Nacional
Apartado Postal 14–740, Ciudad de México
07000 México Jonathan Toledo<EMAIL_ADDRESS>Departamento de Matemáticas
Centro de Investigación y de Estudios Avanzados del Instituto Politécnico
Nacional
Apartado Postal 14–740, Ciudad de México
07000 México

( )

###### Abstract

Let $I=I(D)$ be the edge ideal of a weighted oriented graph $D$. We determine
the irredundant irreducible decomposition of $I$. Also, we characterize the
associated primes and the unmixed property of $I$. Furthermore, we give a
combinatorial characterization for the unmixed property of $I$, when $D$ is
bipartite, $D$ is a whisker or $D$ is a cycle. Finally, we study the Cohen-
Macaulay property of $I$.

Keywords: Weighted oriented graphs, unmixed property, irreducible
decomposition, Cohen-Macaulay property.

##  1 Introduction

A weighted oriented graph is a triplet $D=(V(D),E(D),w)$, where $V(D)$ is a
finite set, $E(D)\subseteq V(D)\times V(D)$ and $w$ is a function
$w:V(D)\to\mathbb{N}$. The vertex set of $D$ and the edge set of $D$ are
$V(D)$ and $E(D)$, respectively. Some times for short we denote these sets by
$V$ and $E$ respectively. The _weight_ of $x\in V$ is $w(x)$. If $e=(x,y)\in
E$, then $x$ is the tail of $e$ and $y$ is the head of $e$. The underlying
graph of $D$ is the simple graph $G$ whose vertex set is $V$ and whose edge
set is $\\{\\{x,y\\}|(x,y)\in E\\}$. If $V(D)=\\{x_{1},\ldots,x_{n}\\}$, then
we consider the polynomial ring $R=K[x_{1},\ldots,x_{n}]$ in $n$ variables
over a field $K$. In this paper we introduce and study the edge ideal of $D$
given by $I(D)=(x_{i}x_{j}^{w(x_{j})}:(x_{i},x_{j})\in E(D))$ in $R$, (see
Definition 3.1).

In Section 2 we study the vertex covers of $D$. In particular we introduce the
notion of strong vertex cover (Definition 2.6) and we prove that a minimal
vertex cover is strong. In Section 3 we characterize the irredundant
irreducible decomposition of $I(D)$. In particular we show that the minimal
monomial irreducible ideals of $I(D)$ are associated with the strong vertex
covers of $D$. In Section 4 we give the following characterization of the
unmixed property of $I(D)$.

$I(D)$ is unmixed$G$ is unmixed and $D$ has the minimal-strong property All
strong vertex covers have the same cardinality All minimal vertex covers have
the same cardinality All strong vertex covers are minimals

Furthermore, if $D$ is bipartite, $D$ is a whisker or $D$ is a cycle, we give
an effective (combinatorial) characterization of the unmixed property. Finally
in Section 5 we study the Cohen-Macaulayness of $I(D)$. In particular we
characterize the Cohen-Macaulayness when $D$ is a path or $D$ is complete.
Also, we give an example where this property depend of the characteristic of
the field $K$.

##  2 Weighted oriented graphs and their vertex covers

In this section we define the weighted oriented graphs and we study their
vertex covers. Furthermore, we define the strong vertex covers and we
characterize when $V(D)$ is a strong vertex cover of $D$. In this paper we
denote the set $\\{x\in V\mid w(x)\neq 1\\}$ by $V^{+}$.

######  Definition 2.1.

A vertex cover $C$ of $D$ is a subset of $V$, such that if $(x,y)\in E$, then
$x\in C$ or $y\in C$. A vertex cover $C$ of $D$ is minimal if each proper
subset of $C$ is not a vertex cover of $D$.

######  Definition 2.2.

Let $x$ be a vertex of a weighted oriented graph $D$, the sets
$N_{D}^{+}(x)=\\{y:(x,y)\in E(D)\\}$ and $N_{D}^{-}(x)=\\{y:(y,x)\in E(D)\\}$
are called the out-neighbourhood and the in-neighbourhood of $x$,
respectively. Furthermore, the neighbourhood of $x$ is the set
$N_{D}(x)=N_{D}^{+}(x)\cup N_{D}^{-}(x)$.

######  Definition 2.3.

Let $C$ be a vertex cover of a weighted oriented graph $D$, we define

$L_{1}(C)=\\{x\in C\mid N_{D}^{+}(x)\cap C^{c}\neq\emptyset\\},$
$L_{2}(C)=\\{x\in C\mid\mbox{$x\notin L_{1}(C)$ and $N^{-}_{D}(x)\cap
C^{c}\neq\emptyset$}\\}\text{ and }$ $L_{3}(C)=C\setminus(L_{1}(C)\cup
L_{2}(C)),$

where $C^{c}$ is the complement of $C$, i.e. $C^{c}=V\setminus C$.

######  Proposition 2.4.

If $C$ is a vertex cover of $D$, then

$L_{3}(C)=\\{x\in C\mid N_{D}(x)\subset C\\}$.

###### Proof.

If $x\in L_{3}(C)$, then $N_{D}^{+}(x)\subseteq C$, since $x\notin L_{1}(C)$.
Furthermore $N^{-}_{D}(x)\subseteq C$, since $x\notin L_{2}(C)$. Hence
$N_{D}(x)\subset C$, since $x\notin N_{D}(x)$. Now, if $x\in C$ and
$N_{D}(x)\subset C$, then $x\notin L_{1}(C)\cup L_{2}(C)$. Therefore $x\in
L_{3}(C)$. ∎

######  Proposition 2.5.

If $C$ is a vertex cover of $D$, then $L_{3}(C)=\emptyset$ if and only if $C$
is a minimal vertex cover of $D$.

###### Proof.

$\Rightarrow)$ If $x\in C$, then by Proposition 2.4 we have
$N_{D}(x)\not\subset C$, since $L_{3}(C)=\emptyset$. Thus, there is $y\in
N_{D}(x)\setminus C$ implying $C\setminus\\{x\\}$ is not a vertex cover.
Therefore, $C$ is a minimal vertex cover.

$\Leftarrow)$ If $x\in L_{3}(C)$, then by Proposition 2.4, $N_{D}(x)\subseteq
C\setminus\\{x\\}$. Hence, $C\setminus\\{x\\}$ is a vertex cover. A
contradiction, since $C$ is minimal. Therefore $L_{3}(C)=\emptyset$.

######  Definition 2.6.

A vertex cover $C$ of $D$ is strong if for each $x\in L_{3}(C)$ there is
$(y,x)\in E(D)$ such that $y\in L_{2}(C)\cup L_{3}(C)$ with $y\in V^{+}$ (i.e.
$w(y)\neq 1$).

######  Remark 2.7.

Let $C$ be a vertex cover of $D$. Hence, by Proposition 2.4 and since
$C=L_{1}(C)\cup L_{2}(C)\cup L_{3}(C)$, we have that $C$ is strong if and only
if for each $x\in C$ such that $N(x)\subset C$, there exist $y\in
N^{-}(x)\cap(C\setminus L_{1}(C))$ with $y\in V^{+}$.

######  Corollary 2.8.

If $C$ is a minimal vertex cover of $D$, then $C$ is strong.

###### Proof.

By Proposition 2.5, we have $L_{3}(C)=\emptyset$, since $C$. Hence, $C$ is
strong. ∎

######  Remark 2.9.

The vertex set $V$ of $D$ is a vertex cover. Also, if $z\in V$, then
$N_{D}(z)\subseteq V\setminus z$. Hence, by Proposition 2.4, $L_{3}(V)=V$.
Consequently, $L_{1}(V)=L_{2}(V)=\emptyset$. By Proposition 2.5, $V$ is not a
minimal vertex cover of $D$. Furthermore since $L_{3}(V)=V$, $V$ is a strong
vertex cover if and only if $N_{D}^{-}(x)\cap V^{+}\neq\emptyset$ for each
$x\in V$.

######  Definition 2.10.

If $G$ is a cycle with
$E(D)=\\{(x_{1},x_{2}),\ldots,(x_{n-1},x_{n}),(x_{n},x_{1})\\}$ and
$V(D)=\\{x_{1},\ldots,x_{n}\\}$, then $D$ is called oriented cycle.

######  Definition 2.11.

$D$ is called unicycle oriented graph if it satisfies the following
conditions:

  1. 1)

The underlying graph of $D$ is connected and it has exactly one cycle $C$.

  2. 2)

$C$ is an oriented cycle in $D$. Furthermore for each $y\in V(D)\setminus
V(C)$, there is an oriented path from $C$ to $y$ in $D$.

  3. 3)

$w(x)\neq 1$ if $deg_{G}(x)\geq 1$.

######  Lemma 2.12.

If $V(D)$ is a strong vertex cover of $D$ and $D_{1}$ is a maximal unicycle
oriented subgraph of $D$, then $V(D^{\prime})$ is a strong vertex cover of
$D^{\prime}=D\setminus V(D_{1})$.

###### Proof.

We take $x\in V(D^{\prime})$. Thus, by Remark 2.9, there is $y\in
N_{D}^{-}(x)\cap V^{+}(D)$. If $y\in D_{1}$, then we take
$D_{2}=D_{1}\cup\\{(y,x)\\}$. Hence, if $C$ is the oriented cycle of $D_{1}$,
then $C$ is the unique cycle of $D_{2}$, since $deg_{D_{2}}(x)=1$. If $y\in
C$, then $(y,x)$ is an oriented path from $C$ to $x$. Now, if $y\notin C$,
then there is an oriented path $\mathcal{L}$ form $C$ to $y$ in $D_{1}$.
Consequently, $\mathcal{L}\cup\\{(y,x)\\}$ is an oriented path form $C$ to
$x$. Furthermore, $deg_{D_{2}}(x)=1$ and $w(y)\neq 1$, then $D_{2}$ is an
unicycle oriented graph. A contradiction, since $D_{1}$ is maximal. This
implies $y\in V(D^{\prime})$, so $y\in N^{-}_{D^{\prime}}(x)\cap
V^{+}(D^{\prime})$. Therefore, by Remark 2.9, $V(D^{\prime})$ is a strong
vertex cover of $D^{\prime}$. ∎

######  Lemma 2.13.

If $V(D)$ is a strong vertex cover of $D$, then there is an unicycle oriented
subgraph of $D$.

###### Proof.

Let $y_{1}$ be a vertex of $D$. Since $V=V(D)$ is a strong vertex cover, there
is $y_{2}\in V$ such that $y_{2}\in N^{-}(y_{1})\cap V^{+}$. Similarly, there
is $y_{3}\in N^{-}(y_{2})\cap V^{+}$. Consequently, $(y_{3},y_{2},y_{1})$ is
an oriented path. Continuing this process, we can assume there exist
$y_{2},y_{3},\ldots,y_{k}\in V^{+}$ where $(y_{k},y_{k-1},\ldots,y_{2},y_{1})$
is an oriented path and there is $1\leq j\leq k-2$ such that $(y_{j},y_{k})\in
E(D)$, since $V$ is finite. Hence, $C=(y_{k},y_{k-1},\ldots,y_{j},y_{k})$ is
an oriented cycle and $\mathcal{L}=(y_{j},\ldots,y_{1})$ is an oriented path
form $C$ to $y_{1}$. Furthermore, if $j=1$, then $w(y_{1})\neq 1$. Therefore,
$D_{1}=C\cup\mathcal{L}$ is an unicycle oriented subgraph of $D$. ∎

######  Proposition 2.14.

Let $D=(V,E,w)$ be a weighted oriented graph, hence $V$ is a strong vertex
cover of $D$ if and only if there are $D_{1},\ldots,D_{s}$ unicycle oriented
subgraphs of $D$ such that $V(D_{1}),\ldots,V(D_{s})$ is a partition of
$V=V(D)$.

###### Proof.

$\Rightarrow)$ By Lemma 2.13, there is a maximal unicycle oriented subgraph
$D_{1}$ of $D$. Hence, by Lemma 2.12, $V(D^{\prime})$ is a strong vertex cover
of $D^{\prime}=D\setminus V(D_{1})$. So, by Lemma 2.13, there is $D_{2}$ a
maximal unicycle oriented subgraph of $D^{\prime}$. Continuing this process we
obtain unicycle oriented subgraphs $D_{1},\ldots,D_{s}$ such that
$V(D_{1}),\ldots,V(D_{s})$ is a partition of $V(D)$.

$\Leftarrow)$ We take $x\in V(D)$. By hypothesis there is $1\leq j\leq s$ such
that $x\in V(D_{j})$. We assume $C$ is the oriented cycle of $D_{j}$. If $x\in
V(C)$, then there is $y\in V(C)$ such that $(y,x)\in E(D_{j})$ and $w(y)\neq
1$, since $deg_{D_{j}}(y)\geq 2$ and $D_{j}$ is a unicycle oriented subgraph.
Now, we assume $x\notin V(C)$, then there is an oriented path
$\mathcal{L}=(z_{1},\ldots,z_{r})$ such that $z_{1}\in V(C)$ and $z_{r}=x$.
Thus, $(z_{r-1},x)\in E(D)$. Furthermore, $w(z_{r-1})\neq 1$, since
$deg_{D_{j}}(z_{r-1})\geq 2$. Therefore $V$ is a strong vertex cover. ∎

##  3 Edge ideals and their primary decomposition

As is usual if $I$ is a monomial ideal of a polynomial ring $R$, we denote by
$\mathcal{G}(I)$ the minimal monomial set of generators of $I$. Furthermore,
there exists a unique decomposition,
$I=\mathfrak{q}_{1}\cap\cdots\cap\mathfrak{q}_{r}$, where
$\mathfrak{q}_{1},\ldots,\mathfrak{q}_{r}$ are irreducible monomial ideals
such that $I\neq\bigcap_{i\neq j}\mathfrak{q}_{i}$ for each $j=1,\ldots,r$.
This is called the irredundant irreducible decomposition of $I$. Furthermore,
$\mathfrak{q}_{i}$ is an irreducible monomial ideal if and only if
$\mathfrak{q}_{i}=(x_{i_{1}}^{a_{1}},\ldots,x_{i_{s}}^{a_{s}})$ for some
variables $x_{i_{j}}$. Irreducible ideals are primary, then a irreducible
decomposition is a primary decomposition. For more details of primary
decomposition of monomial ideals see [6, Chapter 6]. In this section, we
define the edge ideal $I(D)$ of a weighted oriented graph $D$ and we
characterize its irredundant irreducible decomposition. In particular we prove
that this decomposition is an irreducible primary decomposition, i.e, the
radicals of the elements of the irredundant irreducible decomposition of
$I(D)$ are different.

######  Definition 3.1.

Let $D=(V,E,w)$ be a weighted oriented graph with
$V=\\{x_{1},\ldots,x_{n}\\}$. The edge ideal of $D$, denote by $I(D)$, is the
ideal of $R=K[x_{1},\ldots,x_{n}]$ generated by
$\\{x_{i}x_{j}^{w(x_{j})}\mid(x_{i},x_{j})\in E\\}$.

######  Definition 3.2.

A source of $D$ is a vertex $x$, such that $N_{D}(x)=N_{D}^{+}(x)$. A sink of
$D$ is a vertex $y$ such that $N_{D}(y)=N_{D}^{-}(y)$.

######  Remark 3.3.

Let $D=(V,E,w)$ be a weighted oriented graph. We take
$D^{\prime}=(V,E,w^{\prime})$ a weighted oriented graph such that
$w^{\prime}(x)=w(x)$ if $x$ is not a source and $w^{\prime}(x)=1$ if $x$ is a
source. Hence, $I(D)=I(D^{\prime})$. For this reason in this paper, we will
always assume that if $x$ is a source, then $w(x_{i})=1$.

######  Definition 3.4.

Let $C$ be a vertex cover of $D$, the irreducible ideal associated to $C$ is
the ideal

$I_{C}=\bigl{(}L_{1}(C)\cup\\{x_{j}^{w(x_{j})}\mid x_{j}\in L_{2}(C)\cup
L_{3}(C)\\}\bigr{)}.$

######  Lemma 3.5.

$I(D)\subseteq I_{C}$ for each vertex cover $C$ of $D$.

###### Proof.

We take $I=I(D)$ and $m\in\mathcal{G}(I)$, then $m=xy^{w(y)}$, where $(x,y)\in
D$. Since $C$ is a vertex cover, $x\in C$ or $y\in C$. If $y\in C$, then $y\in
I_{C}$ or $y^{w(y)}\in I_{C}$. Thus, $m=xy^{w(y)}\in I_{C}$. Now, we assume
$y\notin C$, then $x\in C$. Hence, $y\in N_{D}^{+}(x)\cap C^{c}$, so $x\in
L_{1}(C)$. Consequently, $x\in I_{C}$ implying $m=xy^{w(y)}\in I_{C}$.
Therefore $I\subseteq I_{C}$. ∎

######  Definition 3.6.

Let $I$ be a monomial ideal. An irreducible monomial ideal $\mathfrak{q}$ that
contains $I$ is called a minimal irreducible monomial ideal of $I$ if for any
irreducible monomial ideal $\mathfrak{p}$ such that
$I\subseteq\mathfrak{p}\subseteq\mathfrak{q}$ one has that
$\mathfrak{p}=\mathfrak{q}$.

######  Lemma 3.7.

Let $D$ be a weighted oriented graph. If
$I(D)\subseteq(x_{i_{1}}^{a_{1}},\ldots,x_{i_{s}}^{a_{s}})$, then
$\\{x_{i_{1}},\ldots,x_{i_{s}}\\}$ is a vertex cover of $D$.

###### Proof.

We take $J=(x_{i_{1}}^{a_{1}},\ldots,x_{i_{s}}^{a_{s}})$. If $(a,b)\in E(D)$,
then $ab^{w(b)}\in I(D)\subseteq J$. Thus, $x_{i_{j}}^{a_{j}}|ab^{w(b)}$ for
some $1\leq j\leq s$. Hence, $x_{i_{j}}\in\\{a,b\\}$ and
$\\{a,b\\}\cap\\{x_{i_{1}}\ldots x_{i_{s}}\\}\neq\emptyset$. Therefore
$\\{x_{i_{1}},\ldots,x_{i_{s}}\\}$ is a vertex cover of $D$. ∎

######  Lemma 3.8.

Let $J$ be a minimal irreducible monomial ideal of $I(D)$ where
$\mathcal{G}(J)=\\{x_{i_{1}}^{a_{s}},\ldots,x_{i_{s}}^{a_{s}}\\}$. If
$a_{j}\neq 1$ for some $1\leq j\leq s$, then there is $(x,x_{i_{j}})\in E(D)$
where $x\notin\mathcal{G}(J)$.

###### Proof.

By contradiction suppose there is $a_{j}\neq 1$ such that if $(x,x_{i_{j}})\in
E(D)$, then $x\in M=\\{x_{i_{1}}^{a_{1}},\ldots,x_{i_{s}}^{a_{s}}\\}$. We take
the ideal $J^{\prime}=(M\setminus\\{x_{i_{j}}^{a_{j}}\\})$. If $(a,b)\in
E(D)$, then $ab^{w(b)}\in I(D)\subseteq J$. Consequently,
$x_{i_{k}}^{a_{k}}|ab^{w(b)}$ for some $1\leq k\leq s$. If $k\neq j$, then
$ab^{w(b)}\in J^{\prime}$. Now, if $k=j$, then by hypothesis $a_{j}\neq 1$.
Hence, $x_{i_{j}}^{a_{j}}|b^{w(b)}$ implying $x_{i_{j}}=b$. Thus,
$(a,x_{i_{j}})\in E(D)$, so by hypothesis $a\in
M\setminus\\{x_{i_{j}}^{a_{j}}\\}$. This implies $ab^{w(b)}\in J^{\prime}$.
Therefore $I(D)\subseteq J^{\prime}\subsetneq J$. A contradiction, since $J$
is minimal.

∎

######  Lemma 3.9.

Let $J$ be a minimal irreducible monomial ideal of $I(D)$ where
$\mathcal{G}(J)=\\{x_{i_{1}}^{a_{1}},\ldots,x_{i_{s}}^{a_{s}}\\}$. If
$a_{j}\neq 1$ for some $1\leq j\leq s$, then $a_{j}=w(x_{i_{j}})$.

###### Proof.

By Lemma 3.8, there is $(x,x_{i_{j}})\in E(D)$ with $x\notin
M=\\{x_{i_{1}}^{a_{1}},\ldots,x_{i_{s}}^{a_{s}}\\}$. Also,
$xx_{i_{j}}^{w(x_{i_{j}})}\in I(D)\subseteq J$, so
$x_{i_{k}}^{a_{k}}|xx_{i_{j}}^{w(x_{i_{j}})}$ for some $1\leq k\leq s$. Hence,
$x_{i_{k}}^{a_{k}}|x_{i_{j}}^{w(x_{i_{j}})}$, since $x\notin M$. This implies,
$k=j$ and $a_{j}\leq w(x_{i_{j}})$. If $a_{j}<w(x_{i_{j}})$, then we take
$J^{\prime}=(M^{\prime})$ where
$M^{\prime}=\\{M\setminus\\{x_{i_{j}}^{a_{j}}\\}\\}\cup\\{x_{i_{j}}^{w(x_{i_{j}})}\\}$.
So, $J^{\prime}\subsetneq J$. Furthermore, if $(a,b)\in E(D)$, then
$m=ab^{w(b)}\in I(D)\subseteq J$. Thus, $x_{i_{k}}^{a_{k}}|ab^{w(b)}$ for some
$1\leq k\leq s$. If $k\neq j$, then $x_{i_{k}}^{a_{k}}\in M^{\prime}$ implying
$ab^{w(b)}\in J^{\prime}$. Now, if $k=j$ then $x_{i_{j}}^{a_{j}}|b^{w(b)}$,
since $a_{j}>1$. Consequently, $x_{i_{j}}=b$ and $x_{i_{j}}^{w(x_{i_{j}})}|m$.
Then $m\in J^{\prime}$. Hence $I(D)\subseteq J^{\prime}\subsetneq J$, a
contradiction since $J$ is minimal. Therefore $a_{j}=w(x_{i_{j}})$. ∎

######  Theorem 3.10.

The following conditions are equivalent:

  1. 1)

$J$ is a minimal irreducible monomial ideal of $I(D)$.

  2. 2)

There is a strong vertex cover $C$ of $D$ such that $J=I_{C}$.

###### Proof.

2) $\Rightarrow$ 1) By definition $J=I_{C}$ is a monomial irreducible ideal.
By Lemma 3.5, $I(D)\subseteq J$. Now, suppose $I(D)\subseteq
J^{\prime}\subseteq J$, where $J^{\prime}$ is a monomial irreducible ideal. We
can assume
$\mathcal{G}(J^{\prime})=\\{x_{j_{1}}^{b_{1}},\ldots,x_{j_{s}}^{b_{s}}\\}$. If
$x\in L_{1}(C)$, then there is $(x,y)\in E(D)$ with $y\notin C$. Hence,
$xy^{w(y)}\in I(D)$ and $y^{r}\notin J$ for each $r\in\mathbb{N}$.
Consequently $y^{r}\notin J^{\prime}$ for each $r$, implyinig
$y\notin\\{x_{j_{1}},\ldots,x_{j_{s}}\\}$. Furthermore
$x_{j_{i}}^{b_{i}}|xy^{w(y)}$ for some $1\leq i\leq s$, since $xy^{w(y)}\in
I(D)\subseteq J^{\prime}$. This implies, $x=x_{j_{i}}^{b_{i}}\in J^{\prime}$.
Now, if $x\in L_{2}(C)$, then there is $(y,x)\in E(D)$ with $y\notin C$. Thus
$y\notin J$, so $y\notin\\{x_{j_{1}}^{b_{1}},\ldots,x_{j_{s}}^{b_{s}}\\}$.
Also, $x^{w(x)}y\in I(D)\subseteq J^{\prime}$, then
$x_{j_{i}}^{b_{i}}|x^{w(x)}y$ for some $1\leq i\leq s$. Consequently,
$x_{j_{i}}^{b_{i}}|x^{w(x)}$ implies $x^{w(x)}\in J^{\prime}$. Finally if
$x\in L_{3}(C)$, then there is $(y,x)\in E(D)$ where $y\in L_{2}(C)\cup
L_{3}(C)$ and $w(y)\neq 1$, since $C$ is a strong vertex cover. So,
$x^{w(x)}y\in I(D)\subseteq J^{\prime}$ implies $x_{j_{i}}^{b_{i}}|x^{w(x)}y$
for some $i$. Furthermore $y\notin J=I_{C}$, since $y\in L_{2}(C)\cup
L_{3}(C)$ and $w(y)\neq 1$. This implies $y\notin J^{\prime}$ so,
$x_{j_{i}}^{b_{i}}|x^{w(x)}$ then $x^{w(x)}\in J^{\prime}$. Hence,
$J=I_{C}\subseteq J^{\prime}$. Therefore, $J$ is a minimal monomial
irreducible of $I(D)$.

1) $\Rightarrow$ 2) Since $J$ is irreducible, we can suppose
$\mathcal{G}(J)=\\{x_{i_{1}}^{a_{1}},\ldots,x_{i_{s}}^{a_{s}}\\}$. By Lemma
3.9, we have $a_{j}=1$ or $a_{j}=w(x_{i_{j}})$ for each $1\leq j\leq s$. Also,
by Lemma 3.7, $C=\\{x_{i_{1}},\ldots,x_{i_{s}}\\}$ is a vertex cover of $D$.
We can assume
$\mathcal{G}(I_{C})=\\{x_{i_{1}}^{b_{1}},\ldots,x_{i_{s}}^{b_{s}}\\}$, then
$b_{j}\in\\{1,w(x_{i_{j}})\\}$ for each $1\leq j\leq s$. Now, suppose
$b_{k}=1$ and $w(x_{i_{k}})\neq 1$ for some $1\leq k\leq s$. Consequently
$x_{i_{k}}\in L_{1}(C)$. Thus, there is $(x_{i_{k}},y)\in E(D)$ where $y\notin
C$. So, $x_{i_{k}}y^{w(y)}\in I(D)\subseteq J$ and
$x_{i_{r}}^{a_{r}}|x_{i_{k}}y^{w(y)}$ for some $1\leq r\leq s$. Furthermore
$y\notin C$, then $r=k$ and $a_{k}=a_{r}=1$. Hence, $I_{C}\cap V(D)\subseteq
J\cap V(D)$. This implies, $I_{C}\subseteq J$, since
$a_{j},b_{j}\in\\{1,w(x_{i_{j}})\\}$ for each $1\leq j\leq s$. Therefore
$J=I_{C}$, since $J$ is minimal. In particular $a_{i}=b_{i}$ for each $1\leq
i\leq s$.

Now, assume $C$ is not strong, then there is $x\in L_{3}(C)$ such that if
$(y,x)\in E(D)$, then $w(y)=1$ or $y\in L_{1}(C)$. We can assume
$x=x_{i_{1}}$, and we take $J^{\prime}$ the monomial ideal with
$\mathcal{G}(J^{\prime})=\\{x_{i_{2}}^{a_{2}},\ldots,x_{i_{s}}^{a_{s}}\\}$. We
take $(z_{1},z_{2})\in E(D)$. If $x_{i_{j}}^{a_{j}}|z_{1}z_{2}^{w(z_{2})}$ for
some $2\leq j\leq s$, then $z_{1}z_{2}^{w(z_{2})}\in J^{\prime}$. Now, assume
$x_{i_{j}}^{a_{j}}\nmid z_{1}z_{2}^{w(z_{2})}$ for each $2\leq j\leq s$.
Consequently $z_{2}\notin\\{x_{i_{2}}\ldots x_{i_{s}}\\}$, since
$a_{j}\in\\{1,w(x_{i_{j}})\\}$. Also $z_{1}z_{2}^{w(z_{2})}\in I(D)\subseteq
J$, then $x_{i_{1}}^{a_{1}}|z_{1}z_{2}^{w(z_{2})}$. But $x_{i_{1}}\in
L_{3}(C)$, so $z_{1},z_{2}\in N_{G}[x_{i_{1}}]\subseteq C$. If
$x_{i_{1}}=z_{1}$, then there is $2\leq r\leq s$ such that $z_{2}=x_{i_{r}}$.
Thus $x_{i_{r}}^{a_{r}}\mid z_{1}z_{2}^{w(z_{2})}$. A contradiction, then
$x_{i_{1}}=z_{2}$, $z_{1}\in C$ and $(z_{1},x_{i_{1}})\in E(D)$. Then,
$w(z_{1})=1$ or $z_{1}\in L_{1}(C)$. In both cases
$z_{1}\in\mathcal{G}(I_{C})$. Furthermore $z_{1}\neq z_{2}$ since
$(z_{1},z_{2})\in E(D)$. This implies $z_{1}\in\mathcal{G}(J^{\prime})$. So,
$z_{1}z_{2}^{w(z_{2})}\in J^{\prime}$. Hence, $I(D)\subseteq J^{\prime}$. This
is a contradiction, since $J$ is minimal. Therefore $C$ is strong.

∎

######  Theorem 3.11.

If $\mathcal{C}_{s}$ is the set of strong vertex covers of $D$, then the
irredundant irreducible decomposition of $I(D)$ is given by
$I(D)=\bigcap_{C\in\mathcal{C}_{s}}I_{C}$.

###### Proof.

By [3, Theorem 1.3.1], there is a unique irredundant irreducible decomposition
$I(D)=\bigcap_{i=1}^{m}I_{i}$. If there is an irreducible ideal
$I_{j}^{\prime}$ such that $I(D)\subseteq I_{j}^{\prime}\subseteq I_{j}$ for
some $j\in\\{1,\ldots,m\\}$, then $I(D)=(\bigcap_{i\neq j}I_{i})\cap
I_{j}^{\prime}$ is an irreducible decomposition. Furthermore this
decomposition is irredundant. Thus, $I_{j}^{\prime}=I_{j}$. Hence,
$I_{1},\ldots,I_{m}$ are minimal irreducible ideals of $I(D)$. Now, if there
is $C\in\mathcal{C}_{s}$ such that $I_{C}\notin\\{I_{1},\ldots,I_{m}\\}$, then
there is $x_{j_{i}}^{\alpha_{i}}\in I_{i}\setminus I_{C}$ for each
$i\in\\{1,\ldots,m\\}$. Consequently,
$m=lcm(x_{j_{1}}^{\alpha_{1}},\ldots,x_{j_{m}}^{\alpha_{m}})\in\bigcap_{i=1}^{m}I_{i}=I(D)\subseteq
I_{C}$. Furthermore, if $C=\\{x_{i_{1}},\ldots,x_{i_{k}}\\}$, then
$I_{C}=(x_{i_{1}}^{\beta_{1}},\ldots,x_{i_{k}}^{\beta_{k}})$ where
$\beta_{j}\in\\{1,w(x_{i_{j}})\\}$. Hence, there is $j\in\\{1,\ldots,k\\}$
such that $x_{i_{j}}^{\beta_{j}}|m$. So, there is $1\leq u\leq m$ such that
$x_{i_{j}}^{\beta_{j}}\mid x_{j_{u}}^{\alpha_{u}}$. A contradiction, since
$x_{j_{u}}^{\alpha_{u}}\notin I_{C}$. Therefore
$I(D)=\bigcap_{C\in\mathcal{C}_{s}}I_{C}$ is the irredundant irreducible
decomposition of $I(D)$. ∎

######  Remark 3.12.

If $C_{1},\ldots,C_{s}$ are the strong vertex covers of $D$, then by Theorem
3.11, $I_{C_{1}}\cap\cdots\cap I_{C_{s}}$ is the irredundant irreducible
decomposition of $I(D)$. Furthermore, if $P_{i}=rad(I_{C_{i}})$, then
$P_{i}=(C_{i})$. So, $P_{i}\neq P_{j}$ for $1\leq i<j\leq s$. Thus,
$I_{C_{1}}\cap\cdots\cap I_{C_{s}}$ is an irredundant primary decomposition of
$I(D)$. In particular we have ${\rm Ass}(I(D))=\\{P_{1},\ldots,P_{s}\\}$.

######  Example 3.13.

Let $D$ be the following oriented weighted graph

$x_{3}$$x_{4}$$x_{5}$$x_{1}$$x_{2}$$5$$2$$2$$3$$4$

whose edge ideal is
$I(D)=(x_{1}^{3}x_{2},x_{2}^{4}x_{3},x_{3}^{5}x_{4},x_{3}x_{5}^{2},x_{4}^{2}x_{5})$.
From Theorem 3.10 and Theorem 3.11, the irreducible decomposition of $I(D)$
is:

$I(D)=(x_{1}^{3},x_{3},x_{4}^{2})\cap(x_{1}^{3},x_{3},x_{5})\cap(x_{2},x_{3},x_{4}^{2})\cap(x_{2},x_{3}^{5},x_{5})\cap(x_{2},x_{4},x_{5}^{2})\cap(x_{1}^{3},x_{2}^{4},x_{3}^{5},x_{5})\cap(x_{1}^{3},x_{2}^{4},x_{4},x_{5}^{2})\cap(x_{2},x_{3}^{5},x_{4}^{2},x_{5}^{2})\cap(x_{1}^{3},x_{2}^{4},x_{3}^{5},x_{4}^{2},x_{5}^{2})$.

######  Example 3.14.

Let $D$ be the following oriented weighted graph

$x_{1}$$x_{2}$$x_{3}$$x_{4}$$2$$5$$7$

Hence, $I(D)=(x_{1}x_{2}^{2},x_{2}x_{3}^{5},x_{3}x_{4}^{7})$. By Theorem 3.10
and Theorem 3.11, the irreducible decomposition of $I(D)$ is:

$I(D)=(x_{1},x_{3})\cap(x_{2}^{2},x_{3})\cap(x_{2},x_{4}^{7})\cap(x_{1},x_{3}^{5},x_{4}^{7})\cap(x_{2}^{2},x_{3}^{5},x_{4}^{7})$.

In Example 3.13 and Example 3.14, $I(D)$ has embedding primes. Furthermore the
monomial ideal $(V(D))$ is an associated prime of $I(D)$ in Example 3.13.
Proposition 2.14 and Remark 3.12 give a combinatorial criterion for to decide
when $(V(D))\in{\rm Ass}(I(D))$.

##  4 Unmixed weighted oriented graphs

Let $D=(V,E,w)$ be a weighted oriented graph whose underlying graph is
$G=(V,E)$. In this section we characterize the unmixed property of $I(D)$ and
we prove that this property is closed under c-minors. In particular if $G$ is
a bipartite graph or $G$ is a whisker or $G$ is a cycle, we give an effective
(combinatorial) characterization of this property.

######  Definition 4.1.

An ideal $I$ is unmixed if each one of its associated primes has the same
height.

######  Theorem 4.2.

The following conditions are equivalent:

  1. 1)

$I(D)$ is unmixed.

  2. 2)

Each strong vertex cover of $D$ has the same cardinality.

  3. 3)

$I(G)$ is unmixed and $L_{3}(C)=\emptyset$ for each strong vertex cover $C$ of
$D$.

###### Proof.

Let $C_{1},\ldots,C_{\ell}$ be the strong vertex covers of $D$. By Remark
3.12, the associated primes of $I(D)$ are $P_{1},\ldots,P_{\ell}$, where
$P_{i}=rad(I_{C_{i}})=(C_{i})$ for $1\leq i\leq\ell$.

$1)\Rightarrow 2)$ Since $I(D)$ is unmixed,
$|C_{i}|=ht(P_{i})=ht(P_{j})=|C_{j}|$ for $1\leq i<j\leq\ell$.

$2)\Rightarrow 3)$ If $C$ is a minimal vertex cover, then by Corollary 2.8,
$C\in\\{C_{1},\ldots,C_{\ell}\\}$. By hypothesis, $|C_{i}|=|C_{j}|$ for each
$1\leq i\leq j\leq\ell$, then $C_{i}$ is a minimal vertex cover of $D$. Thus,
by Lemma 2.5, $L_{3}(C_{i})=\emptyset$. Furthermore $I(G)$ is unmixed, since
$C_{1},\ldots,C_{\ell}$ are the minimal vertex covers of $G$.

$3)\Rightarrow 1)$ By Proposition 2.5, $C_{i}$ is a minimal vertex cover,
since $L_{3}(C_{i})=\emptyset$ for each $1\leq i\leq\ell$. This implies,
$C_{1},\ldots,C_{\ell}$ are the minimal vertex covers of $G$. Since $G$ is
unmixed, we have $|C_{i}|=|C_{j}|$ for $1\leq i<j\leq\ell$. Therefore $I(D)$
is unmixed. ∎

######  Definition 4.3.

A weighted oriented graph $D$ has the minimal-strong property if each strong
vertex cover is a minimal vertex cover.

######  Remark 4.4.

Using Proposition 2.5, we have that $D$ has the minimal-strong property if and
only if $L_{3}(C)=\emptyset$ for each strong vertex cover $C$ of $D$.

######  Definition 4.5.

$D^{\prime}$ is a c-minor of $D$ if there is a stable set $S$ of $D$, such
that $D^{\prime}=D\setminus N_{G}[S]$.

######  Lemma 4.6.

If $D$ has the minimal-strong property, then $D^{\prime}=D\setminus N_{G}[x]$
has the minimal-strong property, for each $x\in V$.

###### Proof.

We take a strong vertex cover $C^{\prime}$ of $D^{\prime}=D\setminus N_{G}[x]$
where $x\in V$. Thus, $C=C^{\prime}\cup N_{D}(x)$ is a vertex cover of $D$. If
$y^{\prime}\in L_{3}(C^{\prime})$, then by Proposition 2.4,
$N_{D^{\prime}}(y^{\prime})\subseteq C^{\prime}$. Consequently,
$N_{D}(y^{\prime})\subseteq C^{\prime}\cup N_{D}(x)=C$ implying $y^{\prime}\in
L_{3}(C)$. Hence, $L_{3}(C^{\prime})\subseteq L_{3}(C)$. Now, we take $y\in
L_{3}(C)$, then $N_{D}(y)\subseteq C$. This implies $y\notin N_{D}(x)$, since
$x\notin C$. Then, $y\in C^{\prime}$ and $N_{D^{\prime}}(y)\cup(N_{D}(y)\cap
N_{D}(x))=N_{D}(y)\subseteq C=C^{\prime}\cup N_{D}(x)$. So,
$N_{D^{\prime}}(y)\subseteq C^{\prime}$ implies $y\in L_{3}(C^{\prime})$.
Therefore $L_{3}(C)=L_{3}(C^{\prime})$.

Now, if $y\in L_{3}(C)=L_{3}(C^{\prime})$, then there is $z\in
C^{\prime}\setminus L_{1}(C^{\prime})$ with $w(z)\neq 1$, such that $(z,y)\in
E(D^{\prime})$. If $z\in L_{1}(C)$, then there exist $z^{\prime}\notin C$ such
that $(z,z^{\prime})\in E(D)$. Since $z^{\prime}\notin C$, we have
$z^{\prime}\notin C^{\prime}$, then $z\in L_{1}(C^{\prime})$. A contradiction,
consequently $z\notin L_{1}(C)$. Hence, $C$ is strong. This implies
$L_{3}(C)=\emptyset$, since $D$ has the minimal-strong property. Thus,
$L_{3}(C^{\prime})=L_{3}(C)=\emptyset$. Therefore $D^{\prime}$ has the
minimal-strong property. ∎

######  Proposition 4.7.

If $D$ is unmixed and $x\in V$, then $D^{\prime}=D\setminus N_{G}[x]$ is
unmixed.

###### Proof.

By Theorem 4.2, $G$ is unmixed and $D$ has the minimal-strong property. Hence,
by [6], $G^{\prime}=G\setminus N_{G}[x]$ is unmixed. Also, by Lemma 4.6 we
have that $D^{\prime}$ has the minimal-strong property. Therefore, by Theorem
4.2, $D^{\prime}$ is unmixed. ∎

######  Theorem 4.8.

If $D$ is unmixed, then a c-minor of $D$ is unmixed.

###### Proof.

If $D^{\prime}$ is a c-minor of $D$, then there is a stable
$S=\\{a_{1},\ldots,a_{s}\\}$ such that $D^{\prime}=D\setminus N_{G}[S]$. Since
$S$ is stable, $D^{\prime}=(\cdots((D\setminus N_{G}[a_{1}])\setminus
N_{G}[a_{2}])\setminus\cdots)\setminus N_{G}[a_{s}]$. Hence, by induction and
Proposition 4.7, $D^{\prime}$ is unmixed. ∎

######  Proposition 4.9.

If $V(D)$ is a strong vertex cover of $D$, then $I(D)$ is mixed.

###### Proof.

By Proposition 2.4 $V(D)$ is not minimal, since $L_{3}(V(D))=V(D)$. Therefore,
by Theorem 4.2, $I(D)$ is mixed. ∎

######  Remark 4.10.

If $V=V^{+}$, then $I(D)$ is mixed.

###### Proof.

If $x_{i}\in V$, then by Remark 3.3 $N^{-}_{D}(x_{i})\neq\emptyset$, since
$V=V^{+}$. Thus, there is $x_{j}\in V$ such that $(x_{j},x_{i})\in E(D)$.
Also, $w(x_{j})\neq 1$ and $x_{j}\in V=L_{3}(V)$. So, $V$ is a strong vertex
cover. Hence, by Proposition 4.9, $I(D)$ is mixed. ∎

In the following three results we assume that $D_{1},\ldots,D_{r}$ are the
connected components of $D$. Furthermore $G_{i}$ is the underlying graph of
$D_{i}$.

######  Lemma 4.11.

Let $C$ be a vertex cover of $D$, then
$L_{1}(C)=\bigcup_{i=1}^{r}L_{1}(C_{i})$ and
$L_{3}(C)=\bigcup_{i=1}^{r}L_{3}(C_{i})$, where $C_{i}=C\cap V(D_{i})$.

###### Proof.

We take $x\in C$, then $x\in C_{j}$ for some $1\leq j\leq r$. Thus,
$N_{D}(x)=N_{D_{j}}(x)$. In particular $N_{D}^{+}(x)=N_{D_{j}}^{+}(x)$, so
$C\cap N_{D}^{+}(x)=C_{j}\cap N_{D_{j}}^{+}(x)$. Hence,
$L_{1}(C)=\bigcup_{i=1}^{r}L_{1}(C_{i})$. On the other hand,

$x\in L_{3}(C)\Leftrightarrow N_{D}(x)\subseteq C\Leftrightarrow
N_{D_{j}}(x)\subseteq C_{j}\Leftrightarrow x\in L_{3}(C_{j})$.

Therefore, $L_{3}(C)=\bigcup_{i=1}^{r}L_{3}(C_{i})$. ∎

######  Lemma 4.12.

Let $C$ be a vertex cover of $D$, then $C$ is strong if and only if each
$C_{i}=C\cap V(D_{i})$ is strong with $i\in\\{1,\ldots,r\\}$.

###### Proof.

$\Rightarrow)$ We take $x\in L_{3}(C_{j})$. By Lemma 4.11, $x\in L_{3}(C)$ and
there is $z\in N_{D}^{-}(x)\cap V^{+}$ with $z\in C\setminus L_{1}(C)$, since
$C$ is strong. So, $z\in N_{D_{j}}^{-}(x)$ and $z\in V(D_{j})$, since $x\in
D_{j}$. Consequently, by Lemma 4.11, $z\in C_{j}\setminus L_{1}(C_{j})$.
Therefore $C_{j}$ is strong.

$\Leftarrow)$ We take $x\in L_{3}(C)$, then $x\in C_{i}$ for some $1\leq i\leq
r$. Then, by Lemma 4.11, $x\in L_{3}(C_{i})$. Thus, there is $a\in
N_{D_{i}}^{-}(x)$ such that $w(a)\neq 1$ and $a\in C_{i}\setminus
L_{1}(C_{i})$, since $C_{i}$ is strong. Hence, by Lemma 4.11, $a\in C\setminus
L_{1}(C)$. Therefore $C$ is strong.

∎

######  Corollary 4.13.

$I(D)$ is unmixed if and only if $I(D_{i})$ is unmixed for each $1\leq i\leq
r$.

###### Proof.

$\Rightarrow)$ By Theorem 4.8, since $D_{i}$ is a c-minor of $D$.

$\Leftarrow)$ By Theorem 4.2, $G_{i}$ is unmixed thus $G$ is unmixed. Now, if
$C$ is a strong vertex cover, then by Lemma 4.11, $C_{i}=C\cap V(D_{i})$ is a
strong vertex cover. Consequently, $L_{3}(C_{i})=\emptyset$, since $I(D_{i})$
is unmixed. Hence, by Lemma 4.11,
$L_{3}(C)=\bigcup_{i=1}^{r}L_{3}(C_{i})=\emptyset$. Therefore, by Theorem 4.2,
$I(D)$ is unmixed. ∎

######  Definition 4.14.

Let $G$ be a simple graph whose vertex set is $V(G)=\\{x_{1},\ldots,x_{n}\\}$
and edge set $E(G)$. A whisker of $G$ is a graph $H$ whose vertex set is
$V(H)=V(G)\cup\\{y_{1},\ldots,y_{n}\\}$ and whose edge set is
$E(H)=E(G)\cup\\{\\{x_{1},y_{1}\\},\ldots,\\{x_{n},y_{n}\\}\\}$.

######  Definition 4.15.

Let $D$ and $H$ be weighted oriented graphs. $H$ is a weighted oriented
whisker of $D$ if $D\subseteq H$ and the underlying graph of $H$ is a whisker
of the underlying graph of $D$.

######  Theorem 4.16.

Let $H$ a weighted oriented whisker of $D$, where
$V(D)=\\{x_{1},\ldots,x_{n}\\}$ and $V(H)=V(D)\cup\\{y_{1},\ldots,y_{n}\\}$,
then the following conditions are equivalents:

  1. 1)

$I(H)$ is unmixed.

  2. 2)

If $(x_{i},y_{i})\in E(H)$ for some $1\leq i\leq n$, then $w(x_{i})=1$.

###### Proof.

$2)\Rightarrow 1)$ We take a strong vertex cover $C$ of $H$. Suppose
$x_{j},y_{j}\in C$, then $y_{j}\in L_{3}(C)$, since
$N_{D}(y_{j})=\\{x_{j}\\}\subseteq C$. Consequently, $(x_{j},y_{j})\in E(G)$
and $w(x_{j})\neq 1$, since $C$ is strong. This is a contradiction by
condition $2)$. This implies, $|C\cap\\{x_{i},y_{i}\\}|=1$ for each $1\leq
i\leq n$. So, $|C|=n$. Therefore, by Theorem 4.2, $I(H)$ is unmixed.

$1)\Rightarrow 2)$ By contradiction suppose $(x_{i},y_{i})\in E(H)$ and
$w(x_{i})\neq 1$ for some $i$. Since $w(x_{i})\neq 1$ and by Remark 3.3, we
have that $x_{i}$ is not a source. Thus, there is $x_{j}\in V(D)$, such that
$(x_{j},x_{i})\in E(H)$. We take the vertex cover $C=\\{V(D)\setminus
x_{j}\\}\cup\\{y_{j},y_{i}\\}$, then by Proposition 2.4,
$L_{3}(C)=\\{y_{i}\\}$. Furthermore $N_{D}(x_{i})\setminus C=\\{x_{j}\\}$ and
$(x_{j},x_{i})\in E(H)$, then $x_{i}\in L_{2}(C)$. Hence C is strong, since
$L_{3}(C)=\\{y_{i}\\}$, $(x_{i},y_{i})\in E(G)$ and $w(x_{i})\neq 1$. A
contradiction by Theorem 4.2, since $I(H)$ is unmixed. ∎

######  Theorem 4.17.

Let $D$ be a bipartite weighted oriented graph, then $I(D)$ is unmixed if and
only if

  1. 1)

$G$ has a perfect matching
$\\{\\{x_{1}^{1},x_{1}^{2}\\},\ldots,\\{x_{s}^{1},x_{s}^{2}\\}\\}$ where
$\\{x_{1}^{1},\ldots,x_{s}^{1}\\}$ and $\\{x_{1}^{2},\ldots,x_{s}^{2}\\}$ are
stable sets. Furthermore if $\\{x_{j}^{1},x_{i}^{2}\\}$,
$\\{x_{i}^{1},x_{k}^{2}\\}\in E(G)$ then $\\{x_{j}^{1},x_{k}^{2}\\}\in E(G)$.

  2. 2)

If $w(x_{j}^{k})\neq 1$ and
$N_{D}^{+}(x_{j}^{k})=\\{x_{i_{1}}^{k^{\prime}},\ldots,x_{i_{r}}^{k^{\prime}}\\}$
where $\\{k,k^{\prime}\\}=\\{1,2\\}$, then $N_{D}(x_{i_{\ell}}^{k})\subseteq
N_{D}^{+}(x_{j}^{k})$ and $N_{D}^{-}(x_{i_{\ell}}^{k})\cap V^{+}=\emptyset$
for each $1\leq\ell\leq r$.

###### Proof.

$\Leftarrow)$ By $1)$ and [1, Theorem 2.5.7], $G$ is unmixed. We take a strong
vertex cover $C$ of $D$. Suppose $L_{3}(C)\neq\emptyset$, thus there exist
$x_{i}^{k}\in L_{3}(C)$. Since $C$ is strong, there is $x_{j}^{k^{\prime}}\in
V^{+}$ such that $(x_{j}^{k^{\prime}},x_{i}^{k})\in E(D)$,
$x_{j}^{k^{\prime}}\in C\setminus L_{1}(C)$ and
$\\{k,k^{\prime}\\}=\\{1,2\\}$. Furthermore
$N_{D}^{+}(x_{j}^{k^{\prime}})\subseteq C$, since $x_{j}^{k^{\prime}}\notin
L_{1}(C)$. Consequently, by $3)$, $N_{D}(x_{i}^{k^{\prime}})\subseteq
N_{D}^{+}(x_{j}^{k^{\prime}})\subseteq C$ and
$N_{D}^{-}(x_{i}^{k^{\prime}})\cap V^{+}=\emptyset$. A contradiction, since
$x_{i}^{k^{\prime}}\in L_{3}(C)$ and $C$ is strong. Hence $L_{3}(C)=\emptyset$
and $D$ has the strong-minimal property. Therefore $I(D)$ is unmixed, by
Theorem 4.2.

$\Rightarrow)$ By Theorem 4.2, $G$ is unmixed. Hence, by [1, Theorem 2.5.7],
$G$ satisfies $1)$.

If $w(x_{j}^{k})\neq 1$, then we take
$C=N_{D}^{+}(x_{j}^{k})\cup\\{x_{i}^{k}\mid N_{D}(x_{i}^{k})\not\subseteq
N_{D}^{+}(x_{j}^{k})\\}$ and $k^{\prime}$ such that
$\\{k,k^{\prime}\\}=\\{1,2\\}$. If
$\\{x_{i}^{k},x_{i^{\prime}}^{k^{\prime}}\\}\in E(G)$ and $x_{i}^{k}\notin C$,
then $x_{i^{\prime}}^{k^{\prime}}\in N_{D}(x_{i}^{k})\subseteq
N_{D}^{+}(x_{j}^{k})\subseteq C$. This implies, $C$ is a vertex cover of $D$.
Now, if $x_{i_{1}}^{k}\in L_{3}(C)$, then $N_{D}(x_{i_{1}}^{k})\subseteq C$.
Consequently $N_{D}(x_{i_{1}}^{k})\subseteq N_{D}^{+}(x_{j}^{k})$ implies
$x_{i_{1}}^{k}\notin C$. A contradiction, then $L_{3}(C)\subseteq
N_{D}^{+}(x_{j}^{k})$. Also, $N_{G}^{-}(x_{j}^{k})\neq\emptyset$, since
$w(x_{j}^{k})\neq 1$. Thus $x_{j}^{k}\in L_{2}(C)$, since
$N_{G}^{-}(x_{j}^{k})\cap C=\emptyset$. Hence $C$ is strong, since
$L_{3}(C)\subseteq N_{D}^{+}(x_{j}^{k})$ and $x_{j}^{k}\in V^{+}$. Furthermore
$\\{x_{1}^{\prime},\ldots,x_{s}^{\prime}\\}$ is a minimal vertex cover, then
by Theorem 4.2 $|C|=s$, since $D$ is unmixed. We assume
$N_{D}^{+}(x_{j}^{k})=\\{x_{i_{1}}^{k^{\prime}},\ldots
x_{i_{r}}^{k^{\prime}}\\}$. Since $C$ is minimal, $x_{i_{\ell}}^{k}\notin C$
for each $1\leq\ell\leq r$. So, $N_{D}(x_{i_{\ell}}^{k})\subseteq
N_{D}^{+}(x_{j}^{k})$. Now, suppose $z\in N_{D}^{-}(x_{i_{\ell}}^{k})\cap
V^{+}$, then $z=x_{i_{\ell^{\prime}}}^{k^{\prime}}$ for some
$1\leq\ell^{\prime}\leq r$, since $N_{D}(x_{i_{\ell}}^{k})\subseteq
N_{D}^{+}(x_{j}^{k})$. We take
$C^{\prime}=N_{D}^{+}(x_{j}^{k})\cup\\{x_{i}^{k}\mid
i\notin\\{i_{1},\ldots,i_{r}\\}\\}\cup
N_{D}^{+}(x_{i_{\ell^{\prime}}}^{k^{\prime}})$. Since
$N_{D}(x_{i_{u}}^{k})\subseteq N_{D}^{+}(x_{j}^{k})$ for each $1\leq u\leq r$,
we have that $C^{\prime}$ is a vertex cover. If
$\\{x_{q}^{k},x_{q}^{k^{\prime}}\\}\cap L_{3}(C)=\emptyset$, then
$\\{x_{q}^{k},x_{q}^{k^{\prime}}\\}\subseteq C^{\prime}$, so
$x_{q}^{k^{\prime}}\in N_{D}^{+}(x_{j}^{k})$ implies
$q\in\\{i_{1},\ldots,i_{r}\\}$. Consequently, $x_{q}^{k}\in
N_{D}^{+}(x_{i_{\ell^{\prime}}}^{k^{\prime}})$, since $x_{q}^{k}\in
C^{\prime}$. This implies,
$(x_{j}^{k},x_{q}^{k^{\prime}}),(x_{i_{\ell^{\prime}}}^{k^{\prime}},x_{q}^{k})\in
E(D)$. Moreover, $N_{D}^{+}(x_{i_{\ell^{\prime}}}^{k^{\prime}})\cup
N_{D}^{+}(x_{j}^{k})\subseteq C^{\prime}$, then
$x_{i_{\ell^{\prime}}}^{k^{\prime}}\notin L_{1}(C^{\prime})$ and
$x_{j}^{k}\notin L_{1}(C^{\prime})$. Thus, $C^{\prime}$ is strong, since
$x_{j}^{k},x_{i_{\ell^{\prime}}}^{k^{\prime}}\in V^{+}$. Furthermore, by
Theorem 4.2, $|C^{\prime}|=s$. But $x_{i_{\ell}}^{k^{\prime}}\in
N_{D}^{+}(x_{j}^{k})$ and $x_{i_{\ell}}^{k}\in
N_{D}^{+}(x_{i_{\ell^{\prime}}}^{k^{\prime}})$, hence
$x_{i_{\ell}}^{k^{\prime}},x_{i_{\ell}}^{k}\in C^{\prime}$. A contradiction,
so $N_{D}^{-}(x_{i_{\ell}}^{k})\cap V^{+}=\emptyset$. Therefore $D$ satisfies
$2)$. ∎

######  Lemma 4.18.

If the vertices of $V^{+}$ are sinks, then $D$ has the minimal-strong
property.

###### Proof.

We take a strong vertex cover $C$ of $D$. Hence, if $y\in L_{3}(C)$, then
there is $(z,y)\in E(D)$ with $z\in V^{+}$. Consequently, by hypothesis, $z$
is a sink. A contradiction, since $(z,y)\in E(D)$. Therefore,
$L_{3}(C)=\emptyset$ and $C$ is a minimal vertex cover. ∎

######  Lemma 4.19.

Let $D$ be a weighted oriented graph, where $G\simeq C_{n}$ with $n\geq 6$.
Hence, $D$ has the minimal-strong property if and only if the vertices of
$V^{+}$ are sinks.

###### Proof.

$\Leftarrow)$ By Lemma 4.18.

$\Rightarrow)$ By contradiction, suppose there is $(z,y)\in E(D)$, with $z\in
V^{+}$. We can assume $G=(x_{1},x_{2},\ldots,x_{n},x_{1})\simeq C_{n}$, with
$x_{2}=y$ and $x_{3}=z$. We take a strong vertex cover $C$ in the following
form: $C=\\{x_{1},x_{3},\ldots,x_{n-1}\\}\cup\\{x_{2}\\}$ if $n$ is even or
$C=\\{x_{1},x_{3},\ldots,x_{n-2}\\}\cup\\{x_{2},x_{n-1}\\}$ if $n$ is odd.
Consequently, if $x\in C$ and $N_{D}(x)\subseteq C$, then $x=x_{2}$. Hence,
$L_{3}(C)=\\{x_{2}\\}$. Furthermore $(x_{3},x_{2})\in E(D)$ with $x_{3}\in
V^{+}$. Thus, $x_{3}$ is not a source, so, $(x_{4},x_{3})\in E(D)$. Then,
$x_{3}\in L_{2}(C)$. This implies $C$ is a strong vertex cover. But
$L_{3}(C)\neq\emptyset$. A contradiction, since $D$ has the minimal-strong
property. ∎

$x_{1}$$x_{2}$$x_{5}$$x_{3}$$x_{4}$$1$$1$$w(x_{5})\neq 1$$1$$w(x_{3})\neq
1$$D_{1}$$x_{1}$$x_{2}$$x_{5}$$x_{3}$$x_{4}$$w(x_{1})\neq 1$$w(x_{2})\neq
1$$1$$1$$1$$D_{2}$$x_{1}$$x_{2}$$x_{5}$$x_{3}$$x_{4}$$1$$1$$w(x_{5})\neq
1$$w(x_{4})\neq 1$$w(x_{3})\neq
1$$D_{3}$$x_{1}$$x_{2}$$x_{5}$$x_{3}$$x_{4}$$1$$w(x_{2})\neq 1$$w(x_{5})\neq
1$$1$$w(x_{3})\neq 1$$D_{4}$

######  Theorem 4.20.

If $G\simeq C_{n}$, then $I(D)$ is unmixed if and only if one of the following
conditions hold:

  1. 1)

$n=3$ and there is $x\in V(D)$ such that $w(x)=1$.

  2. 2)

$n\in\\{4,5,7\\}$ and the vertices of $V^{+}$ are sinks.

  3. 3)

$n=5$, there is $(x,y)\in E(D)$ with $w(x)=w(y)=1$ and $D\not\simeq
D_{1},D\not\simeq D_{2},D\not\simeq D_{3}$.

  4. 4)

$D\simeq D_{4}$.

###### Proof.

$\Rightarrow)$ By Theorem 4.2, $D$ has the minimal- strong property and $G$ is
unmixed. Then, by [1, Exercise 2.4.22], $n\in\\{3,4,5,7\\}$. If $n=3$, then by
Remark 4.10, $D$ satisfies $1)$. If $n=7$, then by Lemma 4.19, $D$ satisfies
$2)$. Now suppose $n=4$ and $D$ does not satisfies $2)$, then we can assume
$x_{1}\in V^{+}$ and $(x_{1},x_{2})\in E(D)$. Consequently, $(x_{4},x_{1})\in
E(G)$, since $w(x_{1})\neq 1$. Furthermore,
$\mathcal{C}=\\{x_{1},x_{2},x_{3}\\}$ is a vertex cover with
$L_{3}(\mathcal{C})=\\{x_{2}\\}$. Thus, $x_{1}\in L_{2}(\mathcal{C})$ and
$(x_{1},x_{2})\in E(D)$ so $\mathcal{C}$ is strong. A contradiction, since
$\mathcal{C}$ is not minimal. This implies $D$ satisfies $2)$. Finally suppose
$n=5$. If $D\simeq D_{1}$, then
$\mathcal{C}_{1}=\\{x_{1},x_{2},x_{3},x_{5}\\}$ is a vertex cover with
$L_{3}(\mathcal{C}_{1})=\\{x_{1},x_{2}\\}$. Also
$(x_{5},x_{1}),(x_{3},x_{2})\in E(D)$ with $x_{5},x_{3}\in V^{+}$.
Consequently, $\mathcal{C}_{1}$ is strong, since $x_{5},x_{3}\in
L_{2}(\mathcal{C}_{1})$. A contradiction, since $\mathcal{C}_{1}$ is not
minimal. If $D\simeq D_{2}$, then
$\mathcal{C}_{2}=\\{x_{1},x_{2},x_{4},x_{5}\\}$ is a vertex cover where
$L_{3}(\mathcal{C}_{2})=\\{x_{1},x_{5}\\}$ and $(x_{2},x_{1}),(x_{1},x_{5})\in
E(D)$ with $x_{2},x_{1}\in V^{+}$. Hence, $\mathcal{C}_{2}$ is strong, since
$x_{2},x_{1}\notin L_{1}(\mathcal{C}_{2})$. A contradiction, since
$\mathcal{C}_{2}$ is not minimal. If $D\simeq D_{3}$,
$\mathcal{C}_{3}=\\{x_{2},x_{3},x_{4},x_{5}\\}$ is a vertex cover where
$L_{3}(\mathcal{C}_{3})=\\{x_{3},x_{4}\\}$ and $(x_{4},x_{3}),(x_{5},x_{4})\in
E(D)$ with $x_{4},x_{5}\in V^{+}$. Thus, $\mathcal{C}_{3}$ is strong, since
$x_{4},x_{5}\notin L_{1}(\mathcal{C}_{3})$. A contradiction, since
$\mathcal{C}_{3}$ is not minimal. Now, since $n=5$ and by $3)$ we can assume
$(x_{2},x_{3})\in E(D)$, $x_{2},x_{3}\in V^{+}$ and there are not two adjacent
vertices with weight $1$. Since $x_{2}\in V^{+}$, $(x_{1},x_{2})\in E(D)$.
Suppose there are not $3$ vertices $z_{1},z_{2},z_{3}$ in $V^{+}$ such that
$(z_{1},z_{2},z_{3})$ is a path in $G$, then $w(x_{4})=w(x_{1})=1$.
Furthermore, $w(x_{5})\neq 1$, since there are not adjacent vertices with
weight $1$. So, $\mathcal{C}_{4}=\\{x_{2},x_{3},x_{4},x_{5}\\}$ is a vertex
cover of $D$, where $L_{3}(\mathcal{C}_{4})=\\{x_{3},x_{4}\\}$. Also
$(x_{2},x_{3})\in E(G)$ with $w(x_{2})\neq 1$. Hence, if $(x_{3},x_{4})\in
E(D)$ or $(x_{5},x_{4})\in E(D)$, then $\mathcal{C}_{4}$ is strong, since
$x_{3},x_{5}\in V^{+}$. But $\mathcal{C}_{4}$ is not minimal. Consequently,
$(x_{4},x_{3}),(x_{4},x_{5})\in E(D)$ and $D\simeq D_{4}$. Now, we can assume
there is a path $(z_{1},z_{2},z_{3})$ in $D$ such that $z_{1},z_{2},z_{3}\in
V^{+}$. Since there are not adjacent vertices with weight $1$, we can suppose
there is $z_{4}\in V^{+}$ such that $\mathcal{L}=(z_{1},z_{2},z_{3},z_{4})$ is
a path. We take $\\{z_{5}\\}=V(D)\setminus V(\mathcal{(}L))$ and we can assume
$(z_{2},z_{3})\in E(D)$. This implies, $(z_{1},z_{2})$, $(z_{5},z_{1})\in
E(D)$, since $z_{1},z_{2}\in V^{+}$. Thus,
$\mathcal{C}_{5}=\\{z_{1},z_{2},z_{3},z_{4}\\}$ is a vertex cover with
$L_{3}(\mathcal{C}_{5})=\\{z_{2},z_{3}\\}$. Then $\mathcal{C}_{5}$ is strong,
since $(z_{1},z_{2}),(z_{2},z_{3})\in E(D)$ with $z_{2}\in
L_{3}(\mathcal{C}_{5})$ and $z_{1}\in L_{2}(\mathcal{C}_{5})$. A
contradiction, since $\mathcal{C}_{5}$ is not minimal.

$\Leftarrow)$ If $n\in\\{3,4,5,7\\}$, then by [1, Exercise 2.4.22] $G$ is
unmixed. By Theorem 4.2, we will only prove that $D$ has the minimal-strong
property. If $D$ satisfies $2)$, then by Lemma 4.18, $D$ has the minimal-
strong property. If $D$ satisfies $1)$ and $\mathcal{C}$ is a strong vertex
cover, then by Proposition 2.14, $|\mathcal{C}|\leq 2$. This implies
$\mathcal{C}$ is minimal. Now, suppose $n=5$ and $\mathcal{C}^{\prime}$ is a
strong vertex cover of $D$ with $|\mathcal{C}^{\prime}|\geq 4$. If $D\simeq
D_{4}$, then $x_{2},x_{5}\notin L_{3}(\mathcal{C}^{\prime})$, since
$(N_{D}^{-}(x_{2})\cup N_{D}^{-}(x_{5}))\cap V^{+}=\emptyset$. So
$N_{D}(x_{2})\not\subseteq\mathcal{C}^{\prime}$ and
$N_{D}(x_{5})\not\subseteq\mathcal{C}^{\prime}$. Consequently,
$x_{1}\notin\mathcal{C}^{\prime}$ implies
$\mathcal{C}^{\prime}=\\{x_{2},x_{3},x_{4},x_{5}\\}$. But $x_{4}\in
L_{3}(\mathcal{C}^{\prime})$ and $N_{D}^{-}(x_{4})=\emptyset$. A
contradiction, since $\mathcal{C}^{\prime}$ is strong. Now assume $D$
satisfies $3)$. Suppose there is a path $\mathcal{L}=(x_{1},x_{2},x_{3})$ in
$G$ such that $w(x_{1})=w(x_{2})=w(x_{3})=1$. We can suppose $(x_{4},x_{5})\in
E(D)$ where $V(D)\setminus V(\mathcal{L})=\\{x_{4},x_{5}\\}$. Since
$w(x_{1})=w(x_{3})=1$, $x_{2}\notin L_{3}(\mathcal{C}^{\prime})$. If
$x_{2}\notin\mathcal{C}^{\prime}$, then
$\mathcal{C}^{\prime}=\\{x_{1},x_{3},x_{4},x_{5}\\}$ and $x_{4}\in
L_{3}(\mathcal{C}^{\prime})$. But $N_{D}^{-}(x_{4})=\\{x_{3}\\}$ and
$w(x_{3})=1$. A contradiction, hence $x_{2}\in\mathcal{C}^{\prime}$. We can
assume $x_{3}\notin\mathcal{C}^{\prime}$, since $x_{2}\notin
L_{3}(\mathcal{C}^{\prime})$. This implies
$\mathcal{C}^{\prime}=\\{x_{1},x_{2},x_{4},x_{5}\\}$ and
$L_{3}(\mathcal{C}^{\prime})=\\{x_{1},x_{5}\\}$. Thus, $(x_{5},x_{1})\in
E(D)$, $x_{5},x_{4}\in V^{+}$. Consequently $(x_{3},x_{4})\in E(D)$, since
$x_{4}\in V^{+}$. A contradiction, since $D\not\simeq D_{2}$. Hence, there are
not three consecutive vertices whose weights are $1$. Consequently, since $D$
satisfies $3)$, we can assume $w(x_{1})=w(x_{2})=1$, $w(x_{3})\neq 1$ and
$w(x_{5})\neq 1$. If $w(x_{4})=1$, then $x_{3},x_{5}\notin
L_{3}(\mathcal{C}^{\prime})$ since $N_{D}(x_{3},x_{5})\cap V^{+}=\emptyset$.
This implies $N_{D}(x_{3})\not\subseteq\mathcal{C}^{\prime}$ and
$N_{D}(x_{5})\not\subseteq\mathcal{C}^{\prime}$. Then,
$x_{4}\notin\mathcal{C}^{\prime}$ and
$\mathcal{C}^{\prime}=\\{x_{1},x_{2},x_{3},x_{5}\\}$. Thus,
$(x_{5},x_{1}),(x_{3},x_{2})\in E(D)$, since
$L_{3}(\mathcal{C}^{\prime})=\\{x_{1},x_{2}\\}$. Consequently,
$(x_{4},x_{5}),(x_{4},x_{3})\in E(D)$, since $x_{5},x_{3}\in V^{+}$. A
contradiction, since $D\not\simeq D_{1}$. So, $w(x_{4})\neq 1$ and we can
assume $(x_{5},x_{4})\in E(D)$, since $x_{4}\in V^{+}$. Furthermore
$(x_{1},x_{5})\in E(D)$, since $x_{5}\in V^{+}$. Hence, $(x_{3},x_{4})\in
E(D)$, since $D\not\simeq D_{3}$. Then $(x_{2},x_{3})\in E(D)$, since
$x_{3}\in V^{+}$. This implies $x_{1},x_{2},x_{3},x_{5}\notin
L_{3}(\mathcal{C}^{\prime})$, since $N^{-}_{D}(x_{i})\cap V^{+}=\emptyset$ for
$i\in\\{1,2,3,5\\}$. A contradiction, since $|\mathcal{C}^{\prime}|\geq 4$.
Therefore $D$ has the minimal-strong property. ∎

##  5 Cohen-Macaulay weighted oriented graphs

In this section we study the Cohen-Macaulayness of $I(D)$. In particular we
give a combinatorial characterization of this property when $D$ is a path or
$D$ is complete. Furthermore, we show the Cohen-Macaulay property depends of
the characteristic of $K$.

######  Definition 5.1.

The weighted oriented graph $D$ is Cohen- Macaulay over the field $K$ if the
ring $R/I(D)$ is Cohen-Macaulay.

######  Remark 5.2.

If $G$ is the underlying graph of $D$, then $rad(I(D))=I(G)$.

######  Proposition 5.3.

If $I(D)$ is Cohen-Macaulay, then $I(G)$ is Cohen-Macaulay and $D$ has the
minimal-strong property.

###### Proof.

By Remark 5.2, $I(G)=rad(I(D))$, then by [4, Theorem 2.6], $I(G)$ is Cohen-
Macaulay. Furthermore $I(D)$ is unmixed, since $I(D)$ is Cohen-Macaulay.
Hence, by Theorem 4.2, $D$ has the minimal-strong property. ∎

######  Example 5.4.

In Example 3.13 and Example 3.14 $I(D)$ is mixed. Hence, $I(D)$ is not Cohen-
Macaulay, but $I(G)$ is Cohen-Macaulay.

######  Conjecture 5.5.

$I(D)$ is Cohen- Macaulay if and only if $I(G)$ is Cohen-Macaulay and $D$ has
the minimal-strong property. Equivalently $I(D)$ is Cohen-Macaulay if and only
if $I(D)$ is unmixed and $I(G)$ is Cohen-Macaulay.

######  Proposition 5.6.

Let $D$ be a weighted oriented graph such that $V=\\{x_{1},\ldots,x_{k}\\}$
and whose underlying graph is a path $G=(x_{1},\ldots,x_{k})$. Then the
following conditions are equivalent.

  1. 1)

$R/I(D)$ is Cohen-Macaulay.

  2. 2)

$I(D)$ is unmixed.

  3. 3)

$k=2$ or $k=4$. In the second case, if $(x_{2},x_{1})\in E(D)$ or
$(x_{3},x_{4})\in E(D)$, then $w(x_{2})=1$ or $w(x_{3})=1$ respectively.

###### Proof.

1) $\Rightarrow$ 2) By [1, Corollary 1.5.14].

2) $\Rightarrow$ 3) By Theorem 4.17, $G$ has a perfect matching, since $D$ is
bipartite. Consequently $k$ is even and
$\\{x_{1},x_{2}\\},\\{x_{3},x_{4}\\},\ldots,\\{x_{k-1},x_{k}\\}$ is a perfect
matching. If $k\geq 6$, then by Theorem 4.17, we have $\\{x_{2},x_{5}\\}\in
E(G)$, since $\\{x_{2},x_{3}\\}$ and $\\{x_{4},x_{5}\\}\in E(G)$. A
contradiction since $\\{x_{2},x_{5}\\}\notin E(G)$. Therefore $k\in\\{2,4\\}$.
Furthermore by Theorem 4.17, $w(x_{2})=1$ or $w(x_{3})=1$ when
$(x_{2},x_{1})\in E(D)$ or $(x_{3},x_{4})\in E(D)$, respectively.

3) $\Rightarrow$ 1) We take $I=I(D)$. If $k=2$, then we can assume
$(x_{1},x_{2})\in E(D)$. So,
$I=(x_{1}x_{2}^{w(x_{2})})=(x_{1})\cap(x_{2}^{w(x_{2})})$. Thus, by Remark
3.12, ${\rm Ass}(I)=\\{(x_{1}),(x_{2})\\}$. This implies, ${\rm ht}(I)=1$ and
$\dim(R/I)=k-1=1$. Also, ${\rm depth}(R/I)\geq 1$, since
$(x_{1},x_{2})\notin{\rm Ass}(I)$. Hence, $R/I$ is Cohen-Macaulay. Now, if
$k=4$, then ${\rm ht}(I)={\rm ht}({\rm rad}(I))={\rm ht}(I(G))=2$.
Consequently, $\dim(R/I)=k-2=2$. Furthermore one of the following sets
$\\{x_{2}-x_{1}^{w(x_{1})},x_{3}-x_{4}^{w(x_{4})}\\}$,
$\\{x_{2}-x_{1}^{w(x_{1})},x_{4}-x_{3}^{w(x_{3})}\\}$,
$\\{x_{1}-x_{2}^{w(x_{2})},x_{4}-x_{3}^{w(x_{3})}\\}$ is a regular sequence of
$R/I$, then ${\rm depth}(R/I)\geq 2$. Therefore, $I$ is Cohen-Macaulay. ∎

######  Theorem 5.7.

If $G$ is a complete graph, then the following conditions are equivalent.

  1. 1)

$I(D)$ is unmixed.

  2. 2)

$I(D)$ is Cohen-Macaulay.

  3. 3)

There are not $D_{1},\ldots,D_{s}$ unicycles orientes subgraphs of $D$ such
that $V(D_{1}),\ldots,V(D_{s})$ is a partition of $V(D)$

###### Proof.

We take $I=I(D)$. Since $I(G)=rad(I)$ and $G$ is complete, ${\rm ht}(I)={\rm
ht}(I(G))=n-1$.

1) $\Rightarrow$ 3) Since ${\rm ht}(I)=n-1$ and $I$ is unmixed,
$(x_{1},\ldots,x_{n})\notin{\rm Ass}(I)$. Thus, by Remark 3.12, $V(D)$ is not
a strong vertex cover of $D$. Therefore, by Proposition 2.14, $D$ satisfies
$3)$.

3) $\Rightarrow$ 2) By Proposition 2.14, $V(D)$ is not a strong vertex cover
of $D$. Consequently, by Remark 3.12, $(x_{1},\ldots,x_{n})\notin{\rm
Ass}(I)$. This implies, ${\rm depth}(R/I)\geq 1$. Furthermore, $\dim(R/I)=1$,
since ${\rm ht}(I)=n-1$. Therefore $I$ is Cohen-Macaulay.

2) $\Rightarrow$ 1) By [1, Corollary 1.5.14].

∎

If $D$ is complete or $D$ is a path, then the Cohen-Macaulay property does not
depend of the field $K$. It is not true in general, see the following example.

######  Example 5.8.

Let $D$ be the following weighted oriented graph:

$x_{3}$$x_{2}$$x_{1}$$x_{9}$$x_{8}$$x_{7}$$x_{6}$$x_{5}$$x_{4}$$x_{11}$$x_{10}$$2$$2$$2$$1$$1$$1$$1$$1$$1$$1$$1$

Hence,

$I(D)=(x_{1}^{2}x_{4},x_{1}^{2}x_{8},x_{1}^{2}x_{5},x_{1}^{2}x_{9},x_{2}^{2}x_{10},x_{2}^{2}x_{5},x_{2}^{2}x_{11},x_{2}^{2}x_{8},x_{2}^{2}x_{6},x_{3}^{2}x_{7},x_{3}^{2}x_{10},x_{3}^{2}x_{6},$

$x_{3}^{2}x_{9},x_{4}x_{8},x_{4}x_{7},x_{4}x_{11},x_{5}x_{10},x_{5}x_{9},x_{5}x_{11},x_{6}x_{8},x_{6}x_{9},x_{6}x_{11},x_{7}x_{10},x_{7}x_{11},x_{9}x_{11})$.

By [5, Example 2.3], $I(G)$ is Cohen- Macaulay when the characteristic of the
field $K$ is zero but it is not Cohen-Macaulay in characteristic $2$.
Consequently, $I(D)$ is not Cohen-Macaulay when the characteristic of $K$ is
$2$. Also, $I(G)$ is unmixed. Furthermore, by Lemma 4.18, $I(D)$ has the
minimal-strong property, then $I(D)$ is unmixed. Using Macaulay$2$ [2] we show
that $I(D)$ is Cohen-Macaulay when the characteristic of $K$ is zero.

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