# On the minimal generating sets
of the Eulerian ideal

Jorge Neves  University of Coimbra, Department of Mathematics, CMUC, 3000-143
Coimbra, Portugal<EMAIL_ADDRESS>and  Gonçalo Varejão  University of
Coimbra, Department of Mathematics, CMUC, 3000-143 Coimbra, Portugal
<EMAIL_ADDRESS>
###### Abstract.

We study the minimal homogeneous generating sets of the Eulerian ideal
associated with a simple graph and its maximal generating degree. We show that
the Eulerian ideal is a lattice ideal and use this to give a characterization
of binomials belonging to a minimal homogeneous generating set. In this way,
we obtain an explicit minimal homogeneous generating set. We find an upper
bound for the maximal generating degree in terms of the graph. This invariant
is half the number of edges of a largest Eulerian subgraph of even cardinality
without even-chords. We show that for bipartite graphs this invariant is the
maximal generating degree. In particular, we prove that if the graph is
bipartite, the Eulerian ideal is generated in degree $2$ if and only if the
graph is chordal. Furthermore, we show that the maximal generating degree is
also $2$ when the graph is a complete graph.

Partially supported by the Centre for Mathematics of the University of Coimbra
(funded by the Portuguese Government through FCT/MCTES, DOI
10.54499/UIDB/00324/2020). Partially supported by the Centre for Mathematics
of the University of Coimbra - UIDB/00324/2020, funded by the Portuguese
Government through FCT/MCTES and by the Portuguese Republic / MCTES, the ESF,
and the POR Centro, through FCT – Fundação para a Ciência e a Tecnologia,
I.P., DOI 10.54499/2021.05420.BD. The authors use Macaulay2 [4] in the
computations of examples.

_Data availability_. Data sharing not applicable to this article as no
datasets were generated or analyzed during the current study.

##  1\. Introduction

Currently, one of the most popular research areas in commutative algebra is
the study of properties of ideals, of the ring of polynomials over a field,
associated with graphs. This originated with the famous paper [14] and has
motivated the study of several other ideals associated with graphs. Among
these, the Eulerian ideal is a recent addition and the focus of this paper. It
was defined in [8], by the first author, together with Vaz Pinto and
Villarreal, inspired by the vanishing ideal parameterized by a graph,
introduced in [12]. The Eulerian ideal is a binomial ideal of a polynomial
ring whose variables are indexed by the edges of a simple graph. It is known
that this ideal has a generating set that contains a set of binomials
canonically associated with the Eulerian subgraphs of the graph — see
Subsection 2.1 — and that, moreover, this ideal contains the toric ideal
associated with the graph.

Several properties of Eulerian ideals have been studied. In [8], the quotient
of the polynomial ring by it was shown to be a Cohen-Macaulay graded ring of
dimension $1$; in [7], a Gröbner basis for the ideal was given and a
combinatorial characterization of its Castelnuovo-Mumford regularity and
degree was derived; in [10], the ideal and the above results were generalized
for $k$-uniform hypergraphs and the Hilbert function of the quotient was
characterized by a combinatorial formula and computed in some cases; finally,
in [9], a monomial basis of the socle of the quotient and a description of the
socle degrees was found.

Through all of these results the graph invariant that appears in common, and
that translates the properties of the Eulerian ideal, is the notion of
$T$-join. Given a subset, $T$, of vertices of the graph, a $T$-join is a
subset of edges, $J$, such that, in the subgraph with edge set $J$, $T$ is the
set of odd degree vertices. We will focus on $T$-joins with cardinality of a
fixed parity. For more on $T$-joins, see [6] and Subsection 2.2, below.

This work has two main goals; to characterize minimal binomial generating sets
of the Eulerian ideal and to study its maximal generating degree. In the case
of the toric ideal of a graph, an explicit description of the minimal
generating sets, in terms of the closed even walks of the graph, was obtained
in [13, Theorem 4.13]. For the Eulerian ideal, we have found that a minimal
generating set consists of binomials that identify $T$-joins that are not
equivalent under a certain equivalence relation — see Definition 3.2 and
Theorem 3.3. The equivalence relation is an adaptation of the notion of Markov
basis, as introduced by Diaconis and Sturmfels in [3] and used in [2, Theorem
4.12] to obtain a description of all minimal generating sets of binomials of a
lattice ideal. Since, as we show in Proposition 2.7, the Eulerian ideal is a
lattice ideal, in Theorem 3.5, we use this result to obtain an explicit
minimal generating set for the Eulerian ideal.

Regarding the maximal generating degree, we start by showing that this
invariant is $2$ for complete graphs — see Theorem 4.5. Then, in Theorem 4.6,
by showing that the binomials associated with even cycles with an even-chord
or with edge-disjoint unions of two odd cycles with at most one vertex in
common that possess an even-chord are not needed in a minimal generating set
of the Eulerian ideal, we obtain an upper bound for the maximal generating
degree. This result is analogous to the results [11, Lemmas 3.1, 3.2 and
Theorem 1.2] for the toric ideal. As an application, in Corollary 4.7, we show
that if the graph is bipartite, then the Eulerian ideal is generated in degree
$2$ if and only if the graph is chordal.

This paper is organized as follows. In Section 2, we define the ideal, the
graph theoretical notions that we need, exhibit the lattice associated with
the Eulerian ideal and recall two background results: the characterization of
its binomials and the description of a Gröbner basis. In Section 3, we
characterize when a binomial in the Eulerian ideal is in the ideal generated
by binomials of lower degree and we describe a minimal generating set. In
Section 4, we give an upper bound for the maximal generating degree of the
Eulerian ideal and show that this invariant is $2$ for bipartite chordal
graphs and complete graphs.

##  2\. Preliminaries

Let $G=(V_{G},E_{G})$ be a simple graph. Let us assume that
$V_{G}=\\{1,\dots,n\\}$ and that $s=|E_{G}|>0$. Let $\mathbb{K}$ denote a
field and let us consider the polynomial rings $\mathbb{K}[x_{1},\dots,x_{n}]$
and $\mathbb{K}[E_{G}]=\mathbb{K}[t_{e}:e\in E_{G}]$. Throughout, we use the
multi-index notation for monomials in a polynomial ring, so that, for example,
if $\alpha\in\mathbb{N}^{E_{G}}$,

$\textstyle\mathbf{t}^{\alpha}=\prod\limits_{e\in E_{G}}t_{e}^{\alpha(e)}.$

Also, if $J\subseteq E_{G}$, we denote by $\mathbf{t}_{J}$ the product of all
variables $t_{e}$, with $e\in J$.

###  2.1. The ideal

Let $\varphi$ be the ring homomorphism
$\mathbb{K}[E_{G}]\to\mathbb{K}[x_{1},\dots,x_{n}]$ that maps $t_{e}\mapsto
x_{i}x_{j}$, for every $e=\\{i,j\\}$ in $E_{G}$.

######  Definition 2.1.

The Eulerian ideal of $G$ is the ideal of $\mathbb{K}[E_{G}]$ defined by

$I(G)=\varphi^{-1}(x_{i}^{2}-x_{j}^{2}:i,j\in V_{G}).$

Given two edges of $G$, $e=\\{i,j\\}$ and $\ell=\\{u,v\\}$, it is easy to see
that the binomial $t_{e}^{2}-t_{\ell}^{2}$ is in $I(G)$,

$\textstyle\varphi(t_{e}^{2}-t_{\ell}^{2})=x_{i}^{2}x_{j}^{2}-x_{u}^{2}x_{v}^{2}=x_{i}^{2}(x_{j}^{2}-x_{u}^{2})+x_{u}^{2}(x_{i}^{2}-x_{v}^{2}).$

Therefore the Eulerian ideal is trivial if and only $G$ has only one edge.
From now on let us assume that $|E_{G}|\geq 2$.

To describe a generating set of $I(G)$, let us fix some terminology. For a set
of edges $J\subseteq E_{G}$ and a vertex $i\in V_{G}$, we define the degree of
$i$ in $J$ as

$\textstyle\deg_{J}(i)=\sum\limits_{e\in J}|e\cap\\{i\\}|.$

We say that a subset of edges $C\subseteq E_{G}$ is Eulerian if $\deg_{C}(i)$
is even, for every $i\in V_{G}$. Given an Eulerian subset $C$ of even
cardinality and a partition $C=J\operatorname{\mathbin{\dot{\cup}}}K$ into
sets of same cardinality, we form a homogeneous binomial in
$\mathbb{K}[E_{G}]$,

$\textstyle\mathbf{t}_{J}-\mathbf{t}_{K}.$

Binomials obtained in this way are called Eulerian binomials associated with
$C$. Let us denote by $\mathscr{E}$ the set of all Eulerian binomials,
associated with all such $C$ and choices of partitions. We recall from [7,
Theorem 3.3] the following result.

######  Theorem 2.2.

The set $\\{t_{e}^{2}-t_{\ell}^{2}:e,\ell\in E_{G}\\}\cup\mathscr{E}$ is a
Gröbner basis for $I(G)$, with respect to the graded reverse lexicographic
order on $\mathbb{K}[E_{G}]$, associated with any choice of ordering of
$E_{G}$.

###  2.2. Joins

Given $T\subseteq V_{G}$, a $T$-join is a subset of edges, $J\subseteq E_{G}$,
such that

$T=\\{v\in V_{G}:\deg_{J}(v)\text{ is odd}\\}.$

If $p\in\mathbb{Z}_{2}$, we say that $J$ is a $(T,p)$-join whenever
$|J|+2\mathbb{Z}=p$. Throughout, we abbreviate $0+2\mathbb{Z}$ and
$1+2\mathbb{Z}$ by $0$ and $1$, respectively. It is clear that, given any
subset of edges $J\subseteq E_{G}$, there is a unique pair $(T,p)$ such that
$J$ is a $(T,p)$-join. For example, an Eulerian subset of even cardinality is
an $(\emptyset,0)$-join. However, given a pair $(T,p)$, there may or may not
exist a $(T,p)$-join. For one, by [6, Proposition 12.7], if a $(T,p)$-join
exists, then the intersection of $T$ with the vertex set of each connected
component of the graph must have even cardinality.

######  Example 2.3.

Let us consider the graph in Figure 1.

$\scriptstyle 1$$\scriptstyle 2$$\scriptstyle 3$$\scriptstyle 4$$\scriptstyle
5$$\scriptstyle 6$ Figure 1. A hexagon with an even-chord.

Set $T=\\{2,5\\}$. Then $\\{\\{2,5\\}\\}$ is a $(T,1)$-join, as are
$\\{\\{1,2\\},\\{1,6\\},\\{5,6\\}\\}$ and
$\\{\\{2,3\\},\\{3,4\\},\\{4,5\\}\\}.$ However, this graph has no
$(T,0)$-joins, for $T=\\{2,5\\}$.

Throughout, let us denote by $\Delta$ the symmetric difference of two sets. By
an elementary argument (see, for example, [6, Proposition 12.6]), one can show
that if $J$ is a $T$-join and $J^{\prime}$ is a $T^{\prime}$-join then
$J\Delta J^{\prime}$ is a $(T\Delta T^{\prime})$-join. It follows that if $J$
is a $(T,p)$-join and $J^{\prime}$ is a $(T^{\prime},p^{\prime})$-join then
$J\Delta J^{\prime}$ is a $(T\Delta T^{\prime},p+p^{\prime})$-join. It is then
easy to show that, when $G$ is bipartite, either there exists a $(T,0)$-join
or there exists a $(T,1)$-join and, when $G$ is non-bipartite, there exist
both a $(T,0)$-join and a $(T,1)$-join (see [7, Lemma 4.6]).

The relation between $(T,p)$-joins and $I(G)$ may be expressed as follows.

######  Proposition 2.4.

For any monomials $\mathbf{t}^{\alpha}=\mathbf{t}^{2\mu}\mathbf{t}_{J}$ and
$\mathbf{t}^{\beta}=\mathbf{t}^{2\nu}\mathbf{t}_{K}$, of $\mathbb{K}[E_{G}]$,
let $(T_{1},p_{1})$ and $(T_{2},p_{2})$ be such that $J$ is a
$(T_{1},p_{1})$-join and $K$ is a $(T_{2},p_{2})$-join. Then
$\mathbf{t}^{\alpha}-\mathbf{t}^{\beta}\in I(G)$ if and only if
$(T_{1},p_{1})=(T_{2},p_{2})$ and $\mathbf{t}^{\alpha}-\mathbf{t}^{\beta}$ is
homogeneous.

###### Proof.

See the proof of [10, Proposition 3]. ∎

By [7, Theorem 4.3], the reduction of a monomial by the Gröbner basis
identifies a unique $(T,p)$-join of minimal cardinality. Moreover,
$(T,p)$-joins of minimum cardinality are parity joins in the sense of [7,
Definition 4.11]. These notions are used to express the regularity of $I(G)$ —
[7, Theorem 4.13] — and also to identify a set of standard monomials for the
quotient of the polynomial ring by $I(G)$ — [7, Theorem 4.9] and [9,
Proposition 2.7].

######  Definition 2.5.

Let us denote by $\mathscr{T}_{G}$ the set of all pairs $(T,p)$ for which
there exists a $(T,p)$-join. Given a pair $(T,p)\in\mathscr{T}_{G}$, we denote
by $\mathcal{J}^{*}(T,p)$ the set of all $(T,p)$-joins with minimum
cardinality.

######  Proposition 2.6.

Given $(T,p)\in\mathscr{T}_{G}$ and $J$ a $(T,p)$-join, the following are
equivalent:

  1. i)

$J\in\mathcal{J}^{*}(T,p)$;

  2. ii)

for every even cardinality Eulerian subset $C\subseteq E_{G}$,

$\textstyle|J\Delta C|\geq|J|\iff|J\cap C|\leq\frac{|C|}{2}.$

###### Proof.

If $J$ is a $(T,p)$-join and $C$ is an even cardinality Eulerian subset then
$J\Delta C$ is also a $(T,p)$-join. Hence $(i)\Rightarrow(ii)$. If $K$ is
another $(T,p)$-join then $J\Delta K$ is an even cardinality Eulerian set and
hence $(ii)\Rightarrow(i)$. ∎

###  2.3. The lattice property

Fix $e_{1},\dots,e_{s}$ an ordering of the edge set $E_{G}$. The incidence
matrix of $G$ is the $n\times s$ matrix $B$, where each entry $B_{ij}$ equals
$1$ if the vertex $i$ is in the edge $e_{j}$, and $0$ otherwise. To identify
the lattice associated with $I(G)$, let us use the induced identification of
$\mathbb{N}^{E_{G}}$ with $\mathbb{N}^{s}$.

######  Proposition 2.7.

$I(G)$ is the lattice ideal associated with the lattice

$\textstyle\mathscr{L}=\\{\theta\in\mathbb{Z}^{s}:B\theta\in(2\mathbb{Z})^{n}\text{
and }\sum_{e\in E_{G}}\theta(e)=0\\}.$

###### Proof.

Recall that, given $\mathscr{L}\subseteq\mathbb{Z}^{s}$ a lattice, the lattice
ideal associated with $\mathscr{L}$, which we denote by $I_{\mathscr{L}}$, is
the ideal generated by $\mathbf{t}^{\alpha}-\mathbf{t}^{\beta}$, where
$\alpha,\beta\in\mathbb{N}^{s}$ and $\alpha-\beta\in\mathscr{L}$. Given
$\alpha,\beta\in\mathbb{N}^{E_{G}}\cong\mathbb{N}^{s}$, let
$\mu,\nu\in\mathbb{N}^{E_{G}}$ and $J,K\subseteq E_{G}$ be such that
$\mathbf{t}^{\alpha}=\mathbf{t}^{2\mu}\mathbf{t}_{J}$ and
$\mathbf{t}^{\beta}=\mathbf{t}^{2\nu}\mathbf{t}_{K}$. By Proposition 2.4, the
Eulerian ideal is generated by the set of binomials
$\mathbf{t}^{\alpha}-\mathbf{t}^{\beta}$ such that

(1)
$\textstyle\deg(\mathbf{t}^{\alpha})=\deg(\mathbf{t}^{\beta})\iff\sum\limits_{e\in
E_{G}}\bigl{(}\alpha(e)-\beta(e)\bigr{)}=0$

and such that $J$ and $K$ are $(T,p)$-joins, for some
$(T,p)\in\mathscr{T}_{G}$. Let us show that this set coincides with the
generating set of $I_{\mathscr{L}}$. We have

$\textstyle(B\alpha)_{i}=\sum\limits_{j=1}^{s}B_{ij}\cdot\alpha(e_{j})=\sum\limits_{j:\
i\in e_{j}}\alpha(e_{j})=\sum\limits_{j:\ i\in e_{j}}|\\{e_{j}\\}\cap
J|+2\mu(e_{j}).$

Since, for each $i\in V_{G}$, the sum $\sum_{j:\ i\in e_{j}}|\\{e_{j}\\}\cap
J|$ counts the number of edges in $J$ that contain $i$, which is equal to
$\deg_{J}(i)$, we get $(B\alpha)_{i}\equiv_{2}\deg_{J}(i)$. Hence

(2)
$\textstyle(B(\alpha-\beta))_{i}=(B\alpha)_{i}-(B\beta)_{i}\equiv_{2}\deg_{J}(i)-\deg_{K}(i).$

Now, if $\mathbf{t}^{\alpha}=\mathbf{t}^{2\mu}\mathbf{t}_{J}$ and
$\mathbf{t}^{\beta}=\mathbf{t}^{2\nu}\mathbf{t}_{K}$ are such that (1) is
satisfied and $J$ and $K$ are $(T,p)$-joins then, by (2),
$\bigl{(}B(\alpha-\beta)\bigr{)}_{i}$ is even, for every $i\in V_{G}$, and
hence $\alpha-\beta\in\mathscr{L}$. Conversely, if
$\alpha-\beta\in\mathscr{L}$ then, if $J$ and $K$ are obtained as above, by
(1) they have cardinalities of the same parity and moreover, using (2), we
deduce that there exists $(T,p)\in\mathscr{T}_{G}$ such that both $J$ and $K$
are $(T,p)$-joins. ∎

##  3\. A minimal generating set

Consider $G$ as in Figure 1. This graph contains three Eulerian subsets of
edges, two of which are squares and one of which is a hexagon. Accordingly,
$I(G)$ is generated by $\\{t_{e}^{2}-t_{\ell}^{2}:e,\ell\in E_{G}\\}$ together
with the set of Eulerian binomials associated with these Eulerian subsets.
However, the generating set of binomials obtained is not minimal. Using
Macaulay2, one can check that the degree $3$ Eulerian binomials associated
with the hexagon,

$\begin{array}[]{cc}t_{23}t_{34}t_{45}-t_{12}t_{16}t_{56},&t_{16}t_{34}t_{45}-t_{12}t_{23}t_{56},\\\
t_{12}t_{34}t_{45}-t_{16}t_{23}t_{56},&t_{16}t_{23}t_{45}-t_{12}t_{34}t_{56},\\\
t_{12}t_{23}t_{45}-t_{16}t_{34}t_{56},&t_{12}t_{16}t_{45}-t_{23}t_{34}t_{56},\\\
t_{16}t_{23}t_{34}-t_{12}t_{45}t_{56},&t_{12}t_{23}t_{34}-t_{16}t_{45}t_{56},\\\
t_{12}t_{16}t_{34}-t_{23}t_{45}t_{56},&t_{12}t_{16}t_{23}-t_{34}t_{45}t_{56},\end{array}$

are not needed.

######  Definition 3.1 ([3, page 367]).

Let $\mathscr{L}\subseteq\mathbb{Z}^{s}$ be a lattice. A Markov basis of
$\mathscr{L}$ is a finite set $M\subseteq\mathscr{L}$ such that, for every
$\alpha,\beta\in\mathbb{N}^{s}$ such that $\alpha-\beta\in\mathscr{L}$, there
exists a finite sequence, $\gamma_{0},\dots,\gamma_{r}$, of elements in
$\mathbb{N}^{s}$, where $\gamma_{0}=\alpha$, $\gamma_{r}=\beta$ and either
$\gamma_{i}-\gamma_{i+1}\in M$ or $\gamma_{i+1}-\gamma_{i}\in M$, for
$i=0,\dots,r-1$.

A Markov basis of $\mathscr{L}$ yields a generating set of binomials of the
associated lattice ideal and vice-versa, see [5, Lemma A.1]. Definition 3.1
motivates an equivalence relation on $\mathcal{J}^{*}(T,p)$. From now on, to
help in the translation, let us fix an ordering of $E_{G}$,
$e_{1},\dots,e_{s}$, and work with the identification
$\mathbb{N}^{E_{G}}\cong\mathbb{N}^{s}$. Suppose that we have two minimum
cardinality $(T,p)$-joins, $J$ and $K$, for a fixed $(T,p)\in\mathscr{T}_{G}$.
Denote $\mathbf{t}^{\alpha}=\mathbf{t}_{J},\mathbf{t}^{\beta}=\mathbf{t}_{K}$
and let $M$ be a Markov basis for the lattice $\mathscr{L}$ associated with
$I(G)$. Then there is a sequence of monomials,
$\mathbf{t}^{\gamma_{0}},\dots,\mathbf{t}^{\gamma_{r}}$, in
$\mathbb{K}[E_{G}]$ such that $\gamma_{0}=\alpha$, $\gamma_{r}=\beta$ and
$\gamma_{i}-\gamma_{i+1}\in\mathscr{L}$, for all $i=0,\dots,r-1$. It follows
that, for all $i$,

(3) $\mathbf{t}^{\gamma_{i}}-\mathbf{t}^{\gamma_{i+1}}\in I(G).$

When $i=0$, we get that $\mathbf{t}_{J}-\mathbf{t}^{\gamma_{1}}\in I(G)$. By
Proposition 2.4, since $J$ has minimum cardinality, $\mathbf{t}^{\gamma_{1}}$
is then a squarefree monomial $\mathbf{t}_{L^{1}}$, for some minimum
cardinality $(T,p)$-join $L^{1}.$ By induction on $i$, using Proposition 2.4
and (3), we find a sequence $L^{0},L^{1},\dots,L^{r}\in\mathcal{J}^{*}(T,p)$,
where $L^{0}=J$, $L^{r}=K$ and $\mathbf{t}^{\gamma_{i}}=\mathbf{t}_{L^{i}}$,
for $i=0,\dots,r$.

######  Definition 3.2.

Let $J,K\in\mathcal{J}^{*}(T,p)$ be minimum cardinality $(T,p)$-joins. If
there exist $L^{0},L^{1},\dots,L^{r}\in\mathcal{J}^{*}(T,p)$ such that
$J=L^{0}$, $K=L^{r}$ and $L^{i}\cap L^{i+1}\neq\emptyset,$ for all $0\leq
i\leq r-1$, then we say that $J$ is equivalent to $K$ and write $J\sim K$.

Clearly, for each $(T,p)\in\mathscr{T}_{G}$, the binary relation on
$\mathcal{J}^{*}(T,p)$, defined above, is an equivalence relation. The next
result characterizes the binomials of $I(G)$ that belong to some minimal
homogeneous generating set.

######  Theorem 3.3.

A degree $d$ homogeneous binomial in $I(G)$ is not in the ideal generated by
the elements of $I(G)$ of degree less than $d$ if and only if it is either of
the form $t_{e}^{2}-t_{\ell}^{2}$, for some $e,\ell\in E_{G}$, or of the form
$\mathbf{t}_{J}-\mathbf{t}_{K}$, where $J$ and $K$ are nonequivalent minimum
cardinality $(T,p)$-joins, for a given $(T,p)\in\mathscr{T}_{G}.$

###### Proof.

To prove the implication “$\Rightarrow$”, take a nonzero binomial

$f=\mathbf{t}^{2\mu}\mathbf{t}_{J}-\mathbf{t}^{2\nu}\mathbf{t}_{K}\in I(G)$

of degree $d$ and consider $(T,p)\in\mathscr{T}_{G}$ such that $J$ and $K$ are
$(T,p)$-joins. Let us suppose that $f$ is not of the forms given above and
prove that then $f$ is a $\mathbb{K}[E_{G}]$-linear combination of binomials,
in $I(G)$, having degree less than $d$. Without loss of generality, we may
assume that the monomials of $f$ are coprime. Then $J\cup K$ is a disjoint
union and is an even cardinality Eulerian set. If $d=2$, either
$J\operatorname{\mathbin{\dot{\cup}}}K=\emptyset$ and
$f=t_{e}^{2}-t_{\ell}^{2}$, for some $e,\ell\in E_{G}$, or
$J\operatorname{\mathbin{\dot{\cup}}}K$ forms a square. In the latter,
$\mu=0=\nu$ and both $J$ and $K$ satisfy Proposition 2.6, thus are minimum
cardinality $(T,p)$-joins. Furthermore $J$ and $K$ are nonequivalent, because
otherwise we could find another minimum cardinality $(T,p)$-join, $N$,
intersecting $J$ and yielding an even cardinality Eulerian set, $J\Delta N$,
of cardinality $2$, which is impossible. We suppose then that $d\geq 3$ and
let $|J|\leq|K|$ and $\deg(\mathbf{t}^{2\mu})\geq\deg(\mathbf{t}^{2\nu})$. If
$K=\emptyset,$ $f=\mathbf{t}^{2\mu}-\mathbf{t}^{2\nu}$ is in the ideal
generated by the differences of squares of variables. If $K\neq\emptyset$ and
$\mu\neq 0$, taking $\ell\in K$ and an edge $e\in E_{G}$ such that $t_{e}^{2}$
divides $\mathbf{t}^{2\mu}$, $f$ can be written as a combination of binomials,
in $I(G)$, of degree less than $d$,

$\textstyle\mathbf{t}^{2\mu}\mathbf{t}_{J}-\mathbf{t}^{2\nu}\mathbf{t}_{K}=\frac{\mathbf{t}^{2\mu}}{t_{e}^{2}}\mathbf{t}_{J}\bigl{(}t_{e}^{2}-t_{\ell}^{2}\bigr{)}+t_{\ell}\bigl{(}\frac{\mathbf{t}^{2\mu}}{t_{e}^{2}}\mathbf{t}_{J\operatorname{\mathbin{\dot{\cup}}}\ell}-\mathbf{t}^{2\nu}\mathbf{t}_{K\setminus\ell}\bigr{)}.$

We are left with considering $K\neq\emptyset,\mu=0$ and
$f=\mathbf{t}_{J}-\mathbf{t}_{K}$, yielding two last cases.

If $|J|=|K|$ is not the minimum cardinality of a $(T,p)$-join. By Proposition
2.6, there is an even cardinality Eulerian set $D\subseteq E_{G}$ such that

(4) $\textstyle|J\cap D|>\frac{|D|}{2}.$

We can then choose $A\subseteq J\cap D$ such that $|A|=\frac{|D|}{2}$. It
follows that

(5) $f=\mathbf{t}_{J\setminus A}(\mathbf{t}_{A}-\mathbf{t}_{D\setminus
A})+\mathbf{t}_{J\setminus A}\mathbf{t}_{D\setminus A}-\mathbf{t}_{K}.$

Let us write $\mathbf{t}_{J\setminus A}\mathbf{t}_{D\setminus
A}=\mathbf{t}^{2\xi}\mathbf{t}_{L}$, for some $\xi\in\mathbb{N}^{E_{G}}$ and
$L\subseteq E_{G}$. Since $f$ and $\mathbf{t}_{A}-\mathbf{t}_{D\setminus A}$
belong to $I(G)$, (5) implies that the binomial
$\mathbf{t}^{2\xi}\mathbf{t}_{L}-\mathbf{t}_{K}$ is in $I(G)$. By (4), we
deduce that $\xi\neq 0$ and so, by the previous argument,
$\mathbf{t}^{2\xi}\mathbf{t}_{L}-\mathbf{t}_{K}$ is in the ideal generated by
the binomials, in $I(G)$, with degrees less than $d$.

In the last case, $J$ and $K$ are equivalent minimum cardinality
$(T,p)$-joins. So there exist $L^{0},L^{1},\dots,L^{r}\in\mathcal{J}^{*}(T,p)$
such that $J=L^{0}$, $K=L^{r}$ and $L^{i}\cap L^{i+1}\neq\emptyset,$ for all
$0\leq i\leq r-1$. Hence

$\textstyle
f=\mathbf{t}_{J}-\mathbf{t}_{K}=\sum\limits_{i=0}^{r-1}\mathbf{t}_{L^{i}\cap
L^{i+1}}(\mathbf{t}_{L^{i}\setminus L^{i+1}}-\mathbf{t}_{L^{i+1}\setminus
L^{i}}).$

Let us prove the implication ”$\Leftarrow$”. Take a binomial $f$ of the form
$t_{e}^{2}-t_{\ell}^{2}$, where $e,\ell\in E_{G}$, or of the form
$\mathbf{t}_{J}-\mathbf{t}_{K}$, where $J$ and $K$ are nonequivalent minimum
cardinality $(T,p)$-joins for a given $(T,p)\in\mathscr{T}_{G}$. We must prove
that $f$ is not a $\mathbb{K}[E_{G}]$-linear combination of binomials, in
$I(G)$, of lesser degree. If $\deg(f)=2$, this is clear. So, assume
$\deg(f)\geq 3$ and $f=\mathbf{t}_{J}-\mathbf{t}_{K}$ with $J$ and $K$ as
described. We proceed by contradiction, assuming that $f$ is in the ideal
generated in degree lesser than $\deg(f)$. Then we can write

(6) $\textstyle
f=\sum\limits_{i=1}^{r}c_{i}(\mathbf{t}^{\alpha_{i}}-\mathbf{t}^{\beta_{i}}),$

where, for all $i=1,\dots,r$,
$\mathbf{t}^{\alpha_{i}}-\mathbf{t}^{\beta_{i}}\in I(G)$,
$\gcd(\mathbf{t}^{\alpha_{i}},\mathbf{t}^{\beta_{i}})\neq 1$ and
$c_{i}\in\mathbb{K}$. The monomials $\mathbf{t}_{J}$ and $\mathbf{t}_{K}$
appear in (6), respectively, as linear combinations
$\sum_{i}c_{i}\mathbf{t}_{J}$ and $\sum_{i}c_{i}\mathbf{t}_{K}$. This means
that the remaining monomials on the right side of (6) cancel out. In
particular, all the monomials that are not of the form $\mathbf{t}_{A}$, for
some $A\in\mathcal{J}^{*}(T,p)$, cancel out and we may exclude them from (6).
However, by Proposition 2.4, the monomial $\mathbf{t}^{\alpha_{i}}$ is not of
this form if and only if $\mathbf{t}^{\beta_{i}}$ is neither. So, after
excluding these monomials, we still obtain $f$ as a linear combination of
binomials. Therefore, without loss of generality, we assume that, for all
$i=1,\dots,r,$ there are $J^{i},K^{i}\in\mathcal{J}^{*}(T,p)$ such that
$\mathbf{t}^{\alpha_{i}}=\mathbf{t}_{J^{i}}$ and
$\mathbf{t}^{\beta_{i}}=\mathbf{t}_{K^{i}}$. Then (6) becomes

(7) $\textstyle
f=\sum\limits_{i=1}^{r}c_{i}(\mathbf{t}_{J^{i}}-\mathbf{t}_{K^{i}}).$

Finally, consider the ring homomorphism
$\psi\colon\mathbb{K}[E_{G}]\to\mathbb{K}$ that maps $t_{e}\mapsto 1$, for
every $e$ belonging to some $N\in\mathcal{J}^{*}(T,p)$ equivalent to $J$, and
$t_{e}\mapsto 0$, for every other edge $e\in E_{G}$. Since, for $i=0,\dots,r$,
$J^{i}\sim K^{i}$, either $J^{i}\sim K^{i}\sim J$ or $J^{i}\nsim J$ and
$K^{i}\nsim J$. In the first case,
$\psi(\mathbf{t}_{J^{i}}-\mathbf{t}_{K^{i}})=1-1=0$. In the second,
$\psi(\mathbf{t}_{J^{i}}-\mathbf{t}_{K^{i}})=0-0=0$. This means that the right
side of (7) is sent to zero by $\psi$. On the other hand, since we are
assuming that $J\nsim K$,

$\psi(f)=\psi(\mathbf{t}_{J}-\mathbf{t}_{K})=1-0=1,$

which is a contradiction. ∎

######  Example 3.4.

Let $G$ be the graph of Figure 1 and let us show that the Eulerian binomials
associated with the hexagon are not needed in a generating set. Consider
$\mathbf{t}_{J}-\mathbf{t}_{K}$ one such Eulerian binomial. When $J$ is
contained in one of the squares, we see that $(T,p)=(\\{2,5\\},1)$ and a
minimum cardinality $(T,p)$-join is thus the edge $\\{2,5\\}$. Hence $J$ is
not a minimum cardinality $(\\{2,5\\},1)$-join and, by Theorem 3.3,
$\mathbf{t}_{J}-\mathbf{t}_{K}$ is a linear combination of binomials of lesser
degree. The example is:

$\begin{array}[]{l}t_{12}t_{16}t_{56}-t_{23}t_{34}t_{45}\\\
=t_{12}(t_{16}t_{56}-t_{12}t_{25})+t_{25}(t_{12}^{2}-t_{23}^{2})+t_{23}(t_{23}t_{25}-t_{34}t_{45}).\end{array}$

We may therefore assume that $J$ intersects a square of $G$, say $Q\subseteq
E_{G}$, in exactly $2$ edges. Then, as both squares of $G$ contain $3$ edges
of the hexagon, a third edge of $Q$ belongs to $K$ (and not to $J$). Let us
show that $J\sim K$, which by Theorem 3.3, proves our claim. Consider $J\Delta
Q\subseteq E_{G}$. Then

$|J\Delta Q|=|J|+|Q|-2|J\cap Q|=3+4-4=3$

and thus $J\Delta Q$ is another minimum cardinality $(T,p)$-join. Clearly
$J\cap(J\Delta Q)\neq\emptyset$ and $K\cap(J\Delta Q)\neq\emptyset$, so $J\sim
K$. An example of this is

$\begin{array}[]{l}t_{12}t_{34}t_{56}-t_{23}t_{45}t_{16}\\\
=t_{34}(t_{12}t_{56}-t_{16}t_{25})+t_{16}(t_{34}t_{25}-t_{23}t_{45}).\end{array}$

We now give an explicit description of a minimal binomial generating set of
$I(G)$. In the remainder of this article, let us denote by
$\mathscr{T}_{G}^{*}$ the set of all $(T,p)\in\mathscr{T}_{G}$ for which there
are at least two nonequivalent minimum cardinality $(T,p)$-joins.

######  Theorem 3.5.

Let $G$ be a graph. Choose $\ell\in E_{G}$. For each
$(T,p)\in\mathscr{T}_{G}^{*}$, choose
$\mathscr{S}(T,p)\subseteq\mathcal{J}^{*}(T,p)$ a set of representatives of
the equivalence classes of $\sim$ in $\mathcal{J}^{*}(T,p)$ and choose
$K(T,p)\in\mathscr{S}(T,p)$. The set

(8) $\textstyle\\{t_{e}^{2}-t_{\ell}^{2}:e\in
E_{G}\setminus\ell\\}\cup\bigcup\limits_{(T,p)\in\mathscr{T}_{G}^{*}}\\{\mathbf{t}_{J}-\mathbf{t}_{K(T,p)}:J\in\mathscr{S}(T,p)\setminus
K(T,p)\\}$

is a minimal homogeneous generating set of $I(G)$.

###### Proof.

Let $S$ denote the set (8). We will show that $S$ satisfies the conditions of
[2, Theorem 4.12], which characterizes every minimal binomial generating set
of a lattice ideal. Given $\alpha\in\mathbb{N}^{s}$, consider

$F_{\alpha}=\\{\mathbf{t}^{\beta}:\mathbf{t}^{\alpha}-\mathbf{t}^{\beta}\in
I(G)\\},$

the fiber of $\mathbf{t}^{\alpha}$ associated with $I(G)$. We define an
equivalence relation on this set as follows. Given
$\mathbf{t}^{\mu},\mathbf{t}^{\nu}\in F_{\alpha}$, set
$\mathbf{t}^{\mu}\equiv\mathbf{t}^{\nu}$ if and only if
$\mathbf{t}^{\mu}-\mathbf{t}^{\nu}$ belongs to the ideal generated by the
elements of $I(G)$ of degree less than $\deg(\mathbf{t}^{\alpha})$. Note that,
since $I(G)$ is homogeneous,

$\deg(\mathbf{t}^{\alpha})=\deg(\mathbf{t}^{\mu})=\deg(\mathbf{t}^{\nu}).$

Define $\mathcal{G}(F_{\alpha})$ to be the graph the vertex set of which is
the set of equivalence classes of $\equiv$ on $F_{\alpha}$ and for which there
is an edge between $[\mathbf{t}^{\mu}]$ and $[\mathbf{t}^{\nu}]$ if and only
if $\mathbf{t}^{\mu}\not\equiv\mathbf{t}^{\nu}$ and there is a binomial
$\mathbf{t}^{\delta}-\mathbf{t}^{\gamma}\in S$ such that
$\mathbf{t}^{\delta}\in[\mathbf{t}^{\mu}]$ and
$\mathbf{t}^{\gamma}\in[\mathbf{t}^{\nu}]$. Since $I(G)$ is a lattice ideal
with a positive associated lattice, according to [2, Theorem 4.12 and Remark
4.13], it suffices to show that when $\mathcal{G}(F_{\alpha})$ is not a single
vertex it is a spanning tree.

Fix $\alpha\in\mathbb{N}^{s}$, and take
$\mathbf{t}^{\beta},\mathbf{t}^{\gamma}\in F_{\alpha}$ such that
$\mathbf{t}^{\beta}\not\equiv\mathbf{t}^{\gamma}$. Then
$\mathbf{t}^{\beta}-\mathbf{t}^{\gamma}$ is a homogeneous binomial in $I(G)$
and, denoting by $d$ its degree, it does not belong to the ideal generated by
the elements of $I(G)$ of degree less than $d$. By Proposition 3.3, there are
two cases to consider. In the first case,
$\mathbf{t}^{\beta}-\mathbf{t}^{\gamma}=t_{e}^{2}-t_{\ell}^{2}$, for some
$e,\ell\in E_{G}$. Now, if $\mathbf{t}^{2\xi}\mathbf{t}_{J}\in F_{\alpha}$,
then $\mathbf{t}^{2\xi}\mathbf{t}_{J}-t_{e}^{2}\in I(G)$ and thus by
Proposition 2.4 the monomial $\mathbf{t}^{2\xi}\mathbf{t}_{J}$ has degree $2$
and $J$ is an $(\emptyset,0)$-join. We deduce that $J=\emptyset$ and
$\mathbf{t}^{2\xi}\mathbf{t}_{J}$ is the square of a variable. We conclude
that $F_{\alpha}=\\{t_{e}^{2}:e\in E_{G}\\}$. Since, using Proposition 3.3,
$t_{e}^{2}\not\equiv t_{\ell}^{2}$, for every $e\neq\ell\in E_{G}$, the vertex
set of $\mathcal{G}(F_{\alpha})$ can be canonically identified with
$F_{\alpha}$ and the edges of $\mathcal{G}(F_{\alpha})$ are of the form
$\\{t_{e}^{2},t_{\ell}^{2}\\}$, with $e\in E_{G}\setminus\ell$. This shows
that $\mathcal{G}(F_{\alpha})$ is a spanning tree.

In the second case,
$\mathbf{t}^{\beta}-\mathbf{t}^{\gamma}=\mathbf{t}_{J}-\mathbf{t}_{K}$, where
$J$ and $K$ are two nonequivalent minimum cardinality $(T,p)$-joins. Now, if
$\mathbf{t}^{2\xi}\mathbf{t}_{L}$ is a monomial in $F_{\alpha}$, then
$\mathbf{t}^{2\xi}\mathbf{t}_{L}-\mathbf{t}_{J}$ is in $I(G)$ and, by
Proposition 2.4, $L$ is a $(T,p)$-join. This implies that $\xi=0$ and hence
$L$ is a minimum cardinality $(T,p)$-join. Therefore
$F_{\alpha}=\\{\mathbf{t}_{L}:L\in\mathcal{J}^{*}(T,p)\\}$. By Proposition
3.3, given $\mathbf{t}_{L},\mathbf{t}_{M}\in F_{\alpha}$ we have

$\mathbf{t}_{L}\equiv\mathbf{t}_{M}\iff L\sim M.$

We conclude that the vertex set of $\mathcal{G}(F_{\alpha})$ is
$\\{[\mathbf{t}_{L}]:L\in\mathscr{S}(T,p)\\}$ and the edge set of
$\mathcal{G}(F_{\alpha})$ consists of the edges
$\\{[\mathbf{t}_{L}],[\mathbf{t}_{K(T,p)}]\\}$, with
$L\in\mathscr{S}(T,p)\setminus K(T,p)$, which again shows that
$\mathcal{G}(F_{\alpha})$ is a spanning tree. ∎

######  Corollary 3.6.

The degrees of the minimal homogeneous generators of $I(G)$ are $2$ and the
cardinalities of minimum cardinality $(T,p)$-joins for
$(T,p)\in\mathscr{T}_{G}^{*}$.

###### Proof.

Use Theorem 3.5. ∎

##  4\. Maximal generating degree

Let us denote the maximal generating degree of $I(G)$ by
$\operatorname{d}(I(G))$. Before we can address the proofs of the main results
of this section, we need to show two preliminary results.

######  Lemma 4.1.

For $(T,p)\in\mathscr{T}_{G}^{*}$, take $J$ and $K$ to be nonequivalent
minimum cardinality $(T,p)$-joins. Then
$J\operatorname{\mathbin{\dot{\cup}}}K$ is either an even cycle or the edge-
disjoint union of two odd cycles, with at most one vertex in common.

###### Proof.

Since $J\operatorname{\mathbin{\dot{\cup}}}K$ is a nonempty Eulerian set, by
[1, Theorem 1], it is the edge-disjoint union of cycles. In particular, it
contains a cycle. Let us consider two cases. In the first case, suppose that
$J\operatorname{\mathbin{\dot{\cup}}}K$ contains an even cycle. Denote it by
$D$ and let us show that $J\operatorname{\mathbin{\dot{\cup}}}K=D$. Since
$|D|=|J\cap D|+|K\cap D|$, by Proposition 2.6, we deduce that $|J|=|J\Delta
D|$. Hence $J\Delta D$ is a minimum cardinality $(T,p)$-join, different from
$J$. If $J\Delta D\neq K$, the sequence $J,J\Delta D,K$ implies that $J$ and
$K$ are equivalent. This is a contradiction. Thus $J\Delta D=K$ and
$J\operatorname{\mathbin{\dot{\cup}}}K=D$. For the second case, let us assume
that $J\operatorname{\mathbin{\dot{\cup}}}K$ does not contain any even cycles.
Then, as $|J\operatorname{\mathbin{\dot{\cup}}}K|$ is even, it contains at
least two edge-disjoint odd cycles, $C_{1},C_{2}\subseteq E_{G}$. Arguing as
in the previous case, we obtain that
$J\operatorname{\mathbin{\dot{\cup}}}K=C_{1}\operatorname{\mathbin{\dot{\cup}}}C_{2}$.
To see that $C_{1}$ and $C_{2}$ have at most one vertex in common, assume
otherwise, that they intersect in two distinct vertices. These vertices
determine, in each $C_{i}$, two paths with lengths of different parity. Out of
these four paths, any two of the same parity form an even cycle contained in
$J\operatorname{\mathbin{\dot{\cup}}}K$. This is a contradiction, so $C_{1}$
and $C_{2}$ have at most one vertex in common. ∎

######  Definition 4.2.

Let $C\subseteq E_{G}$ be a nonempty even cardinality Eulerian set. An edge
$\ell\in E_{G}\setminus C$ is called an even-chord of $C$ if there are even
cardinality Eulerian sets $C_{1},C_{2}$ with $C_{1}\cap C_{2}=\\{\ell\\}$ and
$C=C_{1}\Delta C_{2}$.

######  Remarks 4.3.

$(i)$ A chord of a cycle $C\subseteq E_{G}$ is an edge, of $G$, between two
nonadjacent vertices of $C$. An even-chord of an even cycle is a chord that
separates it into two even cycles. If $G$ is a bipartite graph, any chord of a
cycle separates it into two even cycles and is thus an even-chord. $(ii)$ An
even cardinality Eulerian set with an even-chord need not be an even cycle.
Take the graph of Figure 2.

$\scriptstyle 1$$\scriptstyle 2$$\scriptstyle 3$$\scriptstyle 4$$\scriptstyle
5$$\scriptstyle 6$$\scriptstyle 7$$\scriptstyle 8$$\scriptstyle
9$$\scriptstyle 10$ Figure 2. Even cardinality Eulerian sets with even-chords.

For this graph, an example of an even cardinality Eulerian subset with an
even-chord is the edge-disjoint union of the pentagon on the right with one of
the triangles on the left, say the subset

$\\{\\{6,7\\},\\{7,8\\},\\{8,9\\},\\{9,10\\},\\{6,10\\},\\{1,2\\},\\{1,3\\},\\{2,3\\}\\}.$

The even-chord is the chord of the pentagon, the edge $\\{6,9\\}$, with
$C_{1}$ the square $\\{\\{6,7\\},\\{7,8\\},\\{8,9\\},\\{6,9\\}\\}$ and $C_{2}$
the vertex-disjoint union of two triangles,

$\textstyle\\{\\{1,2\\},\\{1,3\\},\\{2,3\\},\\{6,9\\},\\{6,10\\},\\{9,10\\}\\}.$

Another type of example arises as the union of two triangles with a vertex in
common, say

$\\{\\{1,2\\},\\{1,3\\},\\{2,3\\},\\{1,4\\},\\{1,5\\},\\{4,5\\}\\}.$

It has the edge $\\{2,5\\}$ as an even-chord and, in this case, $C_{1}$ and
$C_{2}$ are the squares $\\{\\{1,3\\},\\{2,3\\},\\{2,5\\},\\{1,5\\}\\}$ and
$\\{\\{1,4\\},\\{1,2\\},\\{2,5\\},\\{4,5\\}\\}$. $(iii)$ If $C$ is the even
cardinality Eulerian subset given by the vertex-disjoint union of two odd
cycles then $C$ has an even-chord if and only if one of the odd cycles has a
chord. To see this, notice that, if $C_{1}$, $C_{2}$ and $\ell\in
E_{G}\setminus C$ satisfy the conditions of Definition 4.2 then
$C\cup\\{\ell\\}$ contains $C_{1}$ and $C_{2}$. However, if $\ell$ is not a
chord of either of the odd cycles of $C$, then $\\{\ell\\}$ does not belong to
any cycle contained in $C\cup\\{\ell\\}$. In particular, $C\cup\\{\ell\\}$
does not contain even cardinality Eulerian subsets, other than $C$ itself,
contradicting that it contains $C_{1}$ and $C_{2}$. Thus $\ell$ must be a
chord of one of the odd cycles of $C$. Conversely, if $\ell$ is a chord of one
of the odd cycles of $C$, it separates that cycle into an even cycle and an
odd cycle, that intersect in $\\{\ell\\}$. Let $C_{1}\subseteq
C\cup\\{\ell\\}$ denote this even cycle and let $C_{2}=C\Delta C_{1}$, which
is the vertex-disjoint union of two odd cycles. Then $C_{1}\cap
C_{2}=\\{\ell\\}$ and $\ell$ is an even-chord of $C$.

######  Lemma 4.4.

Let $J$ and $K$ be disjoint minimum cardinality $(T,p)$-joins, for some
$(T,p)\in\mathscr{T}_{G}$. If $C=J\operatorname{\mathbin{\dot{\cup}}}K$ has an
even-chord, then $J$ and $K$ are equivalent.

###### Proof.

Let $C_{1},C_{2}\subseteq E_{G}$ be even cardinality Eulerian sets such that
$C=C_{1}\Delta C_{2}$ and $|C_{1}\cap C_{2}|=1$. If $|J\cap
C_{i}|\leq\frac{|C_{i}|}{2}-1$, for both $i=1,2$, we would get that

$\textstyle\frac{|C_{1}|}{2}+\frac{|C_{2}|}{2}-1=\frac{|C|}{2}=|J|=|J\cap
C_{1}|+|J\cap C_{2}|\leq\frac{|C_{1}|}{2}+\frac{|C_{2}|}{2}-2,$

which is impossible. Assume then that $|J\cap C_{1}|\geq\frac{|C_{1}|}{2}$. By
Lemma 2.6, it follows that $|J\cap C_{1}|=\frac{|C_{1}|}{2}$ and $J\Delta
C_{1}$ is a minimum cardinality $(T,p)$-join. Let us show that the sequence
$J$, $J\Delta C_{1}$, $K$ gives the equivalence of $J$ and $K$, concluding the
proof. We must check that $J\setminus C_{1}\neq\emptyset$ and $(C_{1}\setminus
J)\cap K\neq\emptyset.$ If $J\setminus C_{1}=\emptyset$, we get

$\textstyle\frac{|C|}{2}=|J|=|J\cap C_{1}|=\frac{|C_{1}|}{2},$

and so $|C_{2}|=2$. This contradicts that nonempty Eulerian sets of even
cardinality must have cardinality at least $4$, thus implying that $J\setminus
C_{1}\neq\emptyset$. Now, to see that $(C_{1}\setminus J)\cap K\neq\emptyset$,
note that

$\textstyle|C_{1}\setminus J|=|C_{1}|-|J\cap C_{1}|=\frac{|C_{1}|}{2}\geq 2.$

So $C_{1}\setminus J$ contains the single edge in $C_{1}\cap C_{2}$ and at
least one other edge of $C$, that must belong to $K$. ∎

######  Theorem 4.5.

If $G$ is a complete graph, $\operatorname{d}(I(G))=2$.

###### Proof.

By Corollary 3.6, it suffices to show that, if $J$ and $K$ are two
nonequivalent minimum cardinality $(T,p)$-joins, then $|J|=|K|\leq 2$. Assume,
with a view to a contradiction, that $|J|=|K|\geq 3$. By Lemma 4.1, there are
three cases to consider. If $J\operatorname{\mathbin{\dot{\cup}}}K$ is an even
cycle, since $|J\operatorname{\mathbin{\dot{\cup}}}K|\geq 6$ and $G$ is the
complete graph, $J\operatorname{\mathbin{\dot{\cup}}}K$ has an even-chord and,
by Lemma 4.4, $J$ is equivalent to $K$. If
$J\operatorname{\mathbin{\dot{\cup}}}K$ is the edge-disjoint union of two odd
cycles, with exactly one vertex in common, say $i\in V_{G}$, then, for any
vertices $u,v$ of $J\operatorname{\mathbin{\dot{\cup}}}K$, different from $i$,
the edge $\\{u,v\\}$ is an even-chord of $G$. Again, the result follows, by
Lemma 4.4. For the last case, $J\operatorname{\mathbin{\dot{\cup}}}K$ is the
edge-disjoint union of two odd cycles, $C_{1},C_{2}$, with no vertices in
common. Then $J$ must contain at least half of the edges of one of the two odd
cycles, say $C_{1}$. In particular, this means that $J$ contains two adjacent
edges of $C_{1}$, say $\\{u,i\\},\\{i,v\\}\in J$, as in Figure 3.

$\scriptstyle j$$\scriptstyle v$$\scriptstyle i$$\scriptstyle u$ Figure 3.
Vertex-disjoint triangle and pentagon, $J$ in gray.

Fix a vertex $j$ of $C_{2}$ and let $D$ be the square
$\\{\\{u,i\\},\\{i,v\\},\\{v,j\\},\\{j,u\\}\\}$. Then $J\Delta D$ is a minimum
cardinality $(T,p)$-join, equivalent to $J$, and $(J\Delta
D)\operatorname{\mathbin{\dot{\cup}}}K$ is the edge-disjoint union of two odd
cycles with exactly the vertex $j$ in common. By the previous case, $(J\Delta
D)\operatorname{\mathbin{\dot{\cup}}}K$ has an even-chord and $J\sim J\Delta
D\sim K$. ∎

Below we adapt part of the proof of [11, Theorem 1.2].

######  Theorem 4.6.

Let $2d$ be the greatest cardinality of an even cycle without an even-chord,
or an edge-disjoint union of two odd cycles with at most one vertex in common
without an even-chord, of $G$. Then $\operatorname{d}(I(G))\leq d,$ and
equality holds if $G$ is bipartite.

###### Proof.

By Corollary 3.6, $\operatorname{d}(I(G))$ is the greatest cardinality of a
minimum cardinality $(T,p)$-join, among all pairs
$(T,p)\in\mathscr{T}_{G}^{*}$. Then, by Lemmas 4.1 and 4.4,
$2\operatorname{d}(I(G))$ is the cardinality of an even cycle without an even-
chord, or an edge-disjoint union of two odd cycles with at most one vertex in
common without an even-chord, of $G$. Thus $\operatorname{d}(I(G))\leq d.$
Now, if $G$ is bipartite, let $C$ be a chordless even cycle with $|C|=2d$. Set
$T$ as the set of vertices of $C$ and $p=d+2\mathbb{Z}$. Let
$C=J\operatorname{\mathbin{\dot{\cup}}}K$ be the partition of $C$ into two
sets, of cardinality $d$, taking for $J$ every other edge of $C$. Then $J$ and
$K$ are $(T,p)$-joins. Take the binomial $\mathbf{t}_{J}-\mathbf{t}_{K}\in
I(G)$. If $\operatorname{d}(I(G))<d,$ then $\mathbf{t}_{J}-\mathbf{t}_{K}$ is
a $\mathbb{K}[E_{G}]$-linear combination of binomials of lower degree. It
follows that there is an even cycle $D\neq C,$ and a binomial
$\mathbf{t}_{A}-\mathbf{t}_{D\setminus A}\in I(G)$ such that $A\subseteq J$.
Then, since no edges of $J$ are incident, the same applies to $A$ and
consequently to $D\setminus A$. Therefore, as $A\subseteq C$, every vertex of
$D$ is a vertex of $C$ and every edge in $D\setminus C$ is a chord of $C$,
which is a contradiction. We conclude that $\operatorname{d}(I(G))=d.$ ∎

Recall that a bipartite graph is called bipartite chordal if every cycle of
length greater than $4$ has a chord.

######  Corollary 4.7.

If $G$ is bipartite, $\operatorname{d}(I(G))=2$ if and only if $G$ is
bipartite chordal.

###### Proof.

Since $G$ is bipartite, every chord is an even-chord. Let $2d$ be the greatest
cardinality of an even cycle, of $G$, without a chord. By Theorem 4.6,
$\operatorname{d}(I(G))=d$ and $d=2$ if and only if $G$ is chordal bipartite.
∎

If $G$ is the complete graph on $n\geq 6$ vertices, an edge-disjoint union of
two triangles, with no common vertices, does not have an even-chord, as we
mentioned in $(iii)$ of Remarks 4.3. However, by Theorem 4.5,
$\operatorname{d}(I(G))=2$ and hence the inequality in Theorem 4.6 is not
sharp. The following is another example of this.

######  Example 4.8.

Let $G$ be the graph depicted in Figure 4.

$\scriptstyle 1$$\scriptstyle 2$$\scriptstyle 3$$\scriptstyle 4$$\scriptstyle
5$$\scriptstyle 6$ Figure 4. A non-bipartite graph.

Using Macaulay2, one can check that $\operatorname{d}(I(G))=2$. However, the
hexagon

$\\{\\{1,2\\},\\{2,3\\},\\{3,4\\},\\{4,5\\},\\{5,6\\},\\{1,6\\}\\}$

does not have an even-chord, as no edge of $E_{G}\setminus C$ separates $C$ in
two squares.

If $G$ is bipartite, then, by Theorem 4.6 and [11, Lemmas 3.1, 3.2 and Theorem
1.2], the maximal generating degree of $I(G)$ and the maximal generating
degree of the toric ideal of $G$ coincide. This number is half of the length
of a largest chordless even cycle. However, in the non-bipartite case, this
does not occur in general. Using Macaulay2 one can check that for the graph in
Figure 4, the toric ideal of $G$ is generated in degree $2$, but
$\operatorname{d}(I(G))=3$.

$\scriptstyle 1$$\scriptstyle 2$$\scriptstyle 3$$\scriptstyle 4$$\scriptstyle
5$$\scriptstyle 6$ Figure 5. Another non-bipartite graph.

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