# On the Rees algebras of $t$-path ideals of cycles

Oleksandra Gasanova, Jürgen Herzog, Jiawen Shan

###### Abstract.

In this paper we explore certain properties of the Rees algebra of
$I_{t}(C_{n})$, the $t$-path ideal of an $n$-cycle. Our main focus is on the
cases when such ideals are of fiber type.

##  1\. Introduction

Let $G$ be a finite simple graph with the vertex set
$V=\\{x_{1},\ldots,x_{n}\\}$ and let $1\leq t\leq n$ be an integer. A sequence
$\\{x_{i_{1}},x_{i_{2}},\ldots,x_{i_{t}}\\}$ is called a $t$-path of $G$ if
$\\{x_{i_{1}},x_{i_{2}}\\},\\{x_{i_{2}},x_{i_{3}}\\},\ldots,\\{x_{i_{t-1}},x_{i_{t}}\\}$
are distinct edges of $G$. The $t$-path ideal $I_{t}(G)$ associated to $G$ is
the squarefree monomial ideal

$I_{t}(G)=(x_{i_{1}}\cdots
x_{i_{t}}\mid\\{x_{i_{1}},\ldots,x_{i_{t}}\\}\textrm{~{}is~{}a}~{}t\textrm{-path~{}of}~{}G)$

in the polynomial ring $S=K[x_{1},\ldots,x_{n}]$.

Note that if we set $t=2$, we recover the definition of the edge ideal,
defined and studied by Villareal in [11]. Here it was also proven that for a
connected graph $G$ the ideal $I_{2}(G)$ is of linear type if and only if $G$
is a tree or has a unique cycle of odd length. The notion of a path ideal as
defined above was first introduced by Conca and De Negri in [3] in 1999 as a
generalisation of the notion of an edge ideal. The authors also showed that if
$G$ is a tree, then $I_{t}(G)$ is of linear type and $R(I_{t}(G))$ is normal
and Cohen-Macaulay. In [1] Brumatti and da Silva focused on the case when
$G=C_{n}$ is a cycle, and discussed in which cases the ideal $I_{t}(C_{n})$ is
of linear type.

The next best possible situation after linear type is so-called _fiber type_.
Villarreal proved in [12, Theorem 8.2.1] that $I_{2}(G)$ is of fiber type. In
[6] it was shown that all polymatroidal ideals are of fiber type.

In this paper we are mainly focusing on path ideals of cycles $C_{n}$, and
study which $I_{t}(C_{n})$ are of fiber type, but not of linear type. The
paper is organized as follows. In Section 2 we give the necessary background:
we recall the definitions of the Rees algebra, the symmetric algebra and the
fiber cone of an ideal, and explain the notions of linear type and fiber type.
In Section 3 we discuss the dimension and defining relations of the fiber cone
$F(I_{t}(C_{n}))$. In Section 4 we compute a Gröbner basis of the defining
ideal of $R(I_{n-2}(C_{n}))$. It allows us to conclude that $I_{n-2}(C_{n})$
is of fiber type, and to compute the Hilbert series and Cohen-Macaulay type of
$R(I_{n-2}(C_{n}))$. In Section 5 we compute a Gröbner basis of the defining
ideal of $R(I_{\frac{n}{2}}(C_{n}))$ when $n$ is even, and conclude that
$I_{\frac{n}{2}}(C_{n})$ is also of fiber type. Section 6 contains a summary
or results regarding linear and fiber type of $I_{t}(C_{n})$, and a few open
questions.

##  2\. Preliminaries and background

In this section we provide some algebraic background and notation which will
be used throughout this paper.

Let $I=(u_{1},\ldots,u_{m})$ be an equigenerated graded ideal of the
polynomial ring $S=K[x_{1},\ldots,x_{n}]$ over the field $K$. Let $s$ be a
variable over $S$. We denote by $R(I)=S[u_{1}s,\ldots,u_{m}s]$ the Rees
algebra of $I$. Then $R(I)$ is bigraded with $\deg x_{i}=(1,0)$ for
$i=1,\ldots,n$ $\deg u_{j}s=(0,1)$ for $j=1,\ldots,m$.

Let $T=S[y_{1},\ldots,y_{m}]$ be the polynomial ring over $S$ in the variables
$y_{1},\ldots,y_{m}$. We define a bigrading on $T$ by setting $\deg
x_{i}=(1,0)$ for $i=1,\ldots,n$ and $\deg y_{j}=(0,1)$ for $j=1,\ldots,m$.
Then the natural surjective bigraded $K$-algebra homomorphism

$\varphi:T\longrightarrow R(I)$

with $\varphi(x_{i})=x_{i}$ for $i=1,\ldots,n$ and $\varphi(y_{j})=u_{j}s$ for
$j=1,\ldots,m$ is bigraded.

Let $J=\operatorname{Ker}(\varphi)$ be the defining ideal of $R(I)$. One says
that $I$ satisfies the $x$-condition if there exists a monomial order $<$ on
$T$ such that for any minimal monomial generator $u$ of
$\operatorname{in}_{<}(J)$ we have $\deg_{x_{i}}u\leq 1$ for all $i$. The
$x$-condition guarantees that all powers of $I$ have linear resolution (see
[5, Corollary 10.1.7]). Moreover, if $<$ satisfies some extra properties, then
all powers of $I$ have linear quotients (see [5, Theorem 10.1.9]).

Let $\alpha=(u_{ij})_{m\times r}$ be a relation matrix of $I$. Then for
$j=1,\ldots,r$, the polynomials
$\ell_{j}=\sum_{i=1}^{m}u_{ij}y_{i}\in\operatorname{Ker}(\varphi)$. Let
$L=(\ell_{1},\ldots,\ell_{r})$. Then $T/L$ is isomorphic to the symmetric
algebra $S(I)$. Since $L\subseteq J$, there exists a surjective bigraded
$K$-algebra homomorphism $\psi:S(I)\longrightarrow R(I)$.

The ideal $I$ is called of _linear type_ , if $\psi$ is an isomorphism, that
is, if $\operatorname{Ker}(\varphi)=L$.

In a sense the next best case after linear type is fiber type, which is
defined as follows. Let ${\mathfrak{m}}=(x_{1},\ldots,x_{n})$ be the graded
maximal ideal of $S$. The $K$-algebra $F(I)=R(I)/{\mathfrak{m}}R(I)$ is called
the _fiber cone_ of $I$. If $I$ is equigenerated, then $F(I)\cong
K[u_{1},\ldots,u_{m}]$.

The map $\varphi:T\longrightarrow R(I)$ induces a surjective $K$-algebra
homomorphism $\varphi^{\prime}:T/{\mathfrak{m}}T\longrightarrow
R(I)/{\mathfrak{m}}R(I).$ Since $T/{\mathfrak{m}}T\cong
K[y_{1},\ldots,y_{m}]$, the map $\varphi^{\prime}$ may be identified with

$K[y_{1},\ldots,y_{m}]\longrightarrow K[u_{1},\ldots,u_{m}],\quad y_{i}\mapsto
u_{i}.$

The elements in $H=\operatorname{Ker}(\varphi^{\prime})$ are called _the fiber
relations_ of $I$. They are polynomials in $R=K[y_{1},\ldots,y_{m}]$.

Obviously, $HT\subseteq J$ and therefore, $L+HT\subseteq J$. The ideal is
called of _fiber type_ , if $J=L+HT$. Note that $\text{linear
type}\Rightarrow\text{fiber type}\Rightarrow\text{general case},$ and none of
the implications can be reversed.

In this paper we are mainly focusing on path ideals of cycles $C_{n}$, and
study which $I_{t}(C_{n})$ are of fiber type, but not of linear type.
Throughout the paper, we consider $1<t<n-1$, since the cases $t=0$ and $t=n$
are trivial, and $I_{t}(C_{n})$ are of linear type when $t=1$ and $t=n-1$. We
have $m=n$ since the number of $t$-paths equals the number of vertices, and
the homomorphism $\varphi$ defined above will map $y_{j}$ to $x_{j}\cdots
x_{j+t-1}s$. Since we are working with an $n$-cycle, the indices of all the
variables (both $x_{i}$ and $y_{j}$) will be taken modulo $n$. In particular,
we will interchangeably use $x_{0}$ and $x_{n}$ ($y_{0}$ and $y_{n}$) to mean
the same thing.

##  3\. Dimension and relations of $F(I_{t}(C_{n}))$

In this section we focus on some properties of the fiber cone of the path
ideals of cycles.

######  Theorem 3.1.

Let $d=\gcd(n,t)$. Then $\dim F(I_{t}(C_{n}))=n-d+1$. In particular,
$F(I_{t}(C_{n}))$ is a polynomial ring if and only if $\gcd(n,t)=1$.

###### Proof.

It is known that the dimension of a toric ring equals the rank of the matrix
whose column vectors are the exponent vectors of the monomial generators of
the algebra. In our case,

$A=\begin{pmatrix}c_{0}&c_{n-1}&\ldots&c_{2}&c_{1}\\\
c_{1}&c_{0}&\ldots&c_{3}&c_{2}\\\ \vdots&\vdots&~{}~{}&\vdots&\vdots&\\\
c_{n-2}&c_{n-3}&\ldots&c_{0}&c_{n-1}\\\
c_{n-1}&c_{n-2}&\ldots&c_{1}&c_{0}\end{pmatrix},$

where $c_{0}=c_{1}=\cdots=c_{t-1}=1$ and $c_{t}=c_{t+1}=\cdots=c_{n-1}=0$.
This is a so-called circulant matrix, and by [8] its rank equals
$n-\deg\gcd(f(x),x^{n}-1)$, where $f(x)=c_{0}+c_{1}x+\cdots+c_{n-1}x^{n-1}$.
In our case,

$f(x)=1+x+\cdots+x^{t-1}=\prod_{\begin{subarray}{c}\xi^{t}=1\\\
\xi\not=1\end{subarray}}(x-\xi)$

and

$x^{n}-1=\prod_{{\xi^{n}=1}}(x-\xi).$

Therefore, our goal is to show that these polynomials have exactly $d-1$ roots
in common. Equivalently, our goal is to show that $x^{t}-1$ and $x^{n}-1$ have
exactly $d$ roots in common. On the one hand, every $d$th root of $1$ is also
a $n$th and a $t$th root of $1$, so we have at least $d$ common roots. On the
other hand, if $\xi$ is both a $n$th and a $t$th root of $1$, it has to be a
$d$th root of $1$. Indeed, $d=k_{1}n+k_{2}t$ for some integers $k_{1},k_{2}$,
which then implies $\xi^{d}=\xi^{k_{1}n+k_{2}t}=\xi^{k_{1}n}\xi^{k_{2}t}=1$.
Therefore, we can not have more than $d$ common roots. We conclude that the
two polynomials have exactly $d$ common roots, as desired. ∎

######  Theorem 3.2.

Consider the $K$-algebra homomorphism
$\varphi:R=K[y_{1},\ldots,y_{n}]\rightarrow F(I_{t}(C_{n}))$ given by
$\varphi(y_{i})=x_{i}x_{i+1}\cdots x_{i+t-1}$. Let $d=\gcd(n,t)$ and let

$m_{i}=\prod_{\begin{subarray}{c}1\leq j\leq n\\\ j\equiv
i\operatorname{mod}d\end{subarray}}y_{j}$

for $i=1,\ldots,d$. Then the defining ideal $H$ of $F(I_{t}(C_{n}))$ coincides
with the ideal $H^{\prime}=(m_{1}-m_{d},m_{2}-m_{d},\ldots,m_{d-1}-m_{d})$.

###### Proof.

We may assume that $d>1$, otherwise the theorem follows directly from 3.1. Let
$n=n^{\prime}d$ and $t=t^{\prime}d$. Note that $\varphi(m_{i})=(x_{1}\cdots
x_{n})^{t^{\prime}}$ for all $i$. We will prove this for $i=1$, the proof for
the other $i$ is analogous. Let

$v_{1}=x_{1}\cdots x_{d},v_{2}=x_{d+1}\cdots
x_{2d},\ldots,v_{n^{\prime}}=x_{n-d+1}\cdots x_{n}.$

In other words, $v_{j}$ encodes a $d$-path ending at $dj$ for all
$j=1,\ldots,n^{\prime}$. Then we have

$\displaystyle\varphi(y_{1})$ $\displaystyle=v_{1}v_{2}\cdots
v_{t^{\prime}-1}v_{t^{\prime}},$ $\displaystyle\varphi(y_{d+1})$
$\displaystyle=v_{2}v_{3}\cdots v_{t^{\prime}}v_{t^{\prime}+1},$
$\displaystyle\vdots$ $\displaystyle\varphi(y_{n-2d+1})$
$\displaystyle=v_{n^{\prime}-1}v_{n^{\prime}}\cdots
v_{t^{\prime}-3}v_{t^{\prime}-2},$ $\displaystyle\varphi(y_{n-d+1})$
$\displaystyle=v_{n^{\prime}}v_{1}\cdots v_{t^{\prime}-2}v_{t^{\prime}-1}.$

Therefore, $\varphi(m_{1})$ is the product of all of these expressions. We can
see that the product of each column equals $v_{1}v_{2}\ldots
v_{n^{\prime}}=x_{1}\cdots x_{n}$, and there are $t^{\prime}$ columns in
total, which implies $\varphi(m_{1})=(x_{1}\cdots x_{n})^{t^{\prime}}$. This
shows that $m_{i}-m_{d}\in H$ for $i=1,\ldots,d-1$. Therefore,
$H^{\prime}\subseteq H$. 3.1 implies that $\operatorname{height}H=d-1$,
therefore, it is enough to show that $H^{\prime}$ is a prime ideal of height
$d-1$. The initial terms of the set
$\mathcal{G}=\\{m_{1}-m_{d},m_{2}-m_{d},\ldots,m_{d-1}-m_{d}\\}$ with respect
to the lexicographical order $<$ are the monomials $m_{1},\ldots,m_{d-1}$.
Since these monomials are pairwise coprime, it follows that $\mathcal{G}$ is a
Gröbner basis of $H^{\prime}$, and hence
$\operatorname{in}_{<}(H^{\prime})=(m_{1},\ldots,m_{d-1})$. Therefore,
$\operatorname{height}H^{\prime}=\operatorname{height}\operatorname{in}_{<}(H^{\prime})=d-1$.
From this Gröbner basis we also see that all variables dividing $m_{d}$ are
non-zerodivisors modulo $\operatorname{in}_{<}(H^{\prime})$, and therefore
modulo $H^{\prime}$. Since $H^{\prime}=(m_{i}-m_{j}\mid 1\leq i<j\leq n)$, we
obtain by symmetry that all $y_{i}$ are non-zerodivisors modulo $H^{\prime}$.

Let $y=y_{d}\cdots y_{n}$. Then $y$ is a non-zerodivisor modulo $H^{\prime}$.
Therefore, the natural map $R/H^{\prime}\rightarrow R_{y}/H^{\prime}R_{y}$ is
injective. We will show that $R_{y}/H^{\prime}R_{y}$ is a domain. This will
then imply that $R/H^{\prime}$ is a domain, as desired. We know that
$R_{y}=K[y_{1},\ldots,y_{d-1},y_{d}^{\pm 1},\ldots,y_{n}^{\pm 1}]$ and that
$H^{\prime}R_{y}$ is generated by all the $m_{i}-m_{d}=y_{i}y_{d+i}\cdots
y_{n-d+i}-y_{d}y_{2d}\cdots y_{n}$, $i=1,\ldots,d-1$. Equivalently, we can say
that $H^{\prime}R_{y}$ is generated by all the $y_{i}-(y_{d}y_{2d}\cdots
y_{n})\cdot(y_{d+i}\cdots y_{n-d+i})^{-1}$, $i=1,\ldots,d-1$. This implies
that $R_{y}/H^{\prime}R_{y}\cong K[y_{d}^{\pm 1},y_{d+1}^{\pm
1},\ldots,y_{n}^{\pm 1}]$, which finishes the proof. ∎

##  4\. The structure of the Rees ring of $I_{n-2}(C_{n})$

Let $L$ be the defining ideal of the symmetric algebra $S(I_{n-2}(C_{n}))$. By
[1, Proposition 2.5] we know that $L=(f_{1},\ldots,f_{n-1},g_{1})$, where
$f_{j}={x_{j-2}y_{j}}-x_{j}y_{j+1}$ for $j=1,\ldots,n-1$ and
$g_{1}=x_{0}y_{1}-x_{n-2}y_{0}$. Note that if we extend the definition of
$f_{j}$ to $j=n$, we will obtain $g_{1}=-f_{n}$. However, for reasons that
will become clear later, it is not convenient to make such an extension for
Gröbner bases computations. For all other purposes we may say
$L=(f_{1},\ldots,f_{n})$. As a preparation for a computation of a Gröbner
basis for $L$ (and then later for $J$, the defining ideal of
$R(I_{n-2}(C_{n}))$), we extend the set of its generators.

######  Proposition 4.1.

$L=(f_{1},\ldots,f_{n-1},g_{1},\ldots,g_{\lfloor\frac{n}{2}\rfloor}),$ where

$g_{k}={x_{2k-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\ i\text{
odd}\end{subarray}}y_{i}}-x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{
even}\end{subarray}}y_{i}~{}~{}for~{}~{}k=1,\ldots,\lfloor\frac{n}{2}\rfloor.$

###### Proof.

We will show by induction that $g_{k}\in L=(f_{1},\ldots,f_{n-1},g_{1})$ for
all $k=1,\ldots,\lfloor\frac{n}{2}\rfloor$. The base case holds by definition.
Now, modulo $L$ we have:

$\displaystyle x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\ i\text{
even}\end{subarray}}y_{i}=y_{2k-2}\left(x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k-2)\\\
i\text{
even}\end{subarray}}y_{i}\right)=y_{2k-2}\left(x_{2k-4}\prod_{\begin{subarray}{c}i\in[0,2k-2)\\\
i\text{ odd}\end{subarray}}y_{i}\right)$ $\displaystyle=$ $\displaystyle
x_{2k-2}y_{2k-1}\prod_{\begin{subarray}{c}i\in[0,2k-2)\\\ i\text{
odd}\end{subarray}}y_{i}=x_{2k-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{ odd}\end{subarray}}y_{i}.$

Here the second equality follows from the induction hypothesis and the third
equality comes from $x_{2k-4}y_{2k-2}-x_{2k-2}y_{2k-1}\in L$. ∎

######  Theorem 4.2.

The generators of $L$ given in 4.1 form a Gröbner basis of $L$ with respect to
any monomial order $<$ on $T$ which is the product order of $<_{1}$ and
$<_{2}$, where $<_{1}$ is the reverse lexicographic order on $R$ induced by
$y_{1}>y_{2}>\ldots>y_{n-1}>y_{0}$, and $<_{2}$ is an arbitrary monomial order
on $S$.

The proof of this fact can be found in the appendix. There, and further in
this paper, the leading terms of binomials are underlined for convenience
whenever a Gröbner basis or an initial ideal is being computed.

Recall that by 3.2 we have $H=0$ if $n$ is odd, and if $n$ is even we have
$H=(h)$, where

$h=\prod_{\begin{subarray}{c}i\in[0,n)\\\ i\text{
odd}\end{subarray}}y_{i}-\prod_{\begin{subarray}{c}i\in[0,n)\\\ i\text{
even}\end{subarray}}y_{i}.$

We will now describe a Gröbner basis of $L+HT$, which, if $n$ is even, will
later turn out to be equal to $J$, the defining ideal of $R(I_{n-2}(C_{n}))$.

######  Corollary 4.3.

Let $n$ be even. Then the polynomials
$\\{f_{1},\ldots,f_{n-1},g_{1},\ldots,g_{\frac{n}{2}},h\\}$ form a Gröbner
basis for $L+HT$ with respect to any monomial order given in 4.2.

The proof of this fact can be found in the appendix.

We are now ready to describe the defining ideal $J$ of $R(I_{n-2}(C_{n}))$.

######  Theorem 4.4.

Let as before $L=(f_{1},\ldots,f_{n})$, where
$f_{j}=x_{n-2+j}y_{j}-x_{j}y_{j+1}$ for $j=1,\ldots,n$. As before, let
$H=(h)$, where
$h=\prod_{i=0}^{\frac{n}{2}-1}y_{2i}-\prod_{i=0}^{\frac{n}{2}-1}y_{2i+1}$.
Then $J=L$ if $n$ is odd, and $J=L+HT$ if $n$ is even.

###### Proof.

The proof is known when $n$ is odd, see [1, Theorem 3.4]. So we only have to
deal with the case when $n$ is even. We set
$J^{\prime}=(f_{1},\ldots,f_{n},h)$. Then obviously $J^{\prime}\subseteq J$.

We have $\dim R(I_{n-2}(C_{n}))=\dim S+1=n+1$. Thus $J^{\prime}=J$, once we
have shown that $J^{\prime}$ is a prime ideal and that $\dim
T/J^{\prime}=n+1$.

It can be seen from the Gröbner basis of $J^{\prime}$ given in 4.3 that
$y_{0}$ is a non-zerodivisor modulo $J^{\prime}$. By rotational symmetry it
follows that all $y_{i}$ are non-zerodivisors modulo $J^{\prime}$. This
implies that $T/J^{\prime}\rightarrow T_{y}/J^{\prime}T_{y}$ is injective,
where $y=y_{0}y_{1}\cdots y_{n-1}$, and
$J^{\prime}T_{y}=(\\{x_{i}-x_{i-2}w_{i}\mid i=1,\ldots,n-1\\},h)$ with
$w_{i}=y_{i}y_{i+1}^{-1}$. In particular, $x_{1}-x_{n-1}w_{1}$ and
$x_{3}-x_{1}w_{3}$ belong to $J^{\prime}T_{y}$. It follows that
$x_{3}-x_{1}w_{3}+w_{3}(x_{1}-x_{n-1}w_{1})=x_{3}-x_{n-1}w_{1}w_{3}\in
J^{\prime}T_{y}$, and we may replace $x_{3}-x_{1}w_{3}$ by
$x_{3}-x_{n-1}w_{1}w_{3}$ in the set of generators of $J^{\prime}T_{y}$, since
$x_{3}-x_{1}w_{3}$ can also be written as a linear combination of
$x_{1}-x_{n-1}w_{1}$ and $x_{3}-x_{n-1}w_{1}w_{3}$. By applying similar
substitutions, we may replace $x_{i}-x_{i-2}w_{i}$ by
$x_{i}-x_{n-1}\prod_{j=0}^{\frac{i-1}{2}}w_{2j+1}$, if $i$ is odd, and we may
replace it by $x_{i}-x_{0}\prod_{j=0}^{\frac{i-2}{2}}w_{2j+2}$, if $i$ is
even. In particular, $x_{n-1}-x_{n-3}w_{n-1}$ can be replaced by
$x_{n-1}-x_{n-1}\prod_{j=0}^{\frac{n-2}{2}}w_{2j+1}$, which after
multiplication with the unit $\prod_{j=0}^{\frac{n}{2}-1}y_{2j}$, can be
replaced by $x_{n-1}h$. Similarly, $x_{n}-x_{n-2}w_{n}$ can be replaced by
$x_{n}h$.

We conclude that
$T_{y}/J^{\prime}T_{y}\cong(R[x_{0},x_{n-1}]/(h)R[x_{0},x_{n-1}])_{y}$. It
follows that $T_{y}/J^{\prime}T_{y}$ is a domain, since $h$ is an irreducible
polynomial in the polynomial ring $R[x_{0},x_{n-1}]$. This proves that
$J^{\prime}$ is a prime ideal.

Finally, notice that

$\displaystyle\dim T/J^{\prime}=\dim
T_{y}/J^{\prime}T_{y}=\dim(R[x_{0},x_{n-1}]/(h)R)_{y}$ $\displaystyle=\dim
R[x_{0},x_{n-1}]/(h)R[x_{0},x_{n-1}]=n+1,$

as desired.

∎

The preceding theorem together with 4.3 yields the following structural
results about the Rees ring of $I_{n-2}(C_{n})$.

######  Corollary 4.5.

Let $n\geq 3$ and let $I=I_{n-2}(C_{n})$. Then the following holds:

  1. (a)

$R(I)$ is a normal Cohen–Macaulay domain.

  2. (b)

$I$ is of linear type if $n$ is odd, and $I$ is of fiber type if $n$ is even.

  3. (c)

$I^{s}$ has linear quotients for all $s\geq 1.$

###### Proof.

(a) Let $J$ as before be the defining ideal of $R(I)$. By 4.2 and 4.3, $J$ has
a squarefree initial ideal. By [10, Proposition 13.15] this implies that
$R(I)$ is a normal. Then, since $R(I)$ is a toric ring, we obtain that $R(I)$
is a Cohen–Macaulay domain, see [7, Theorem 1].

(b) This can be seen from the generators of $J$ given in 4.4.

(c) It is shown in 4.2 and 4.3 that $J$ satisfies the $x$-condition. Any of
the orders given in 4.2 satisfies the assumptions of [5, Theorem 10.1.9]. This
yields the desired conclusion. ∎

In the following we give an interesting interpretation of the fiber relation
$h$ of the ideal $I=I_{n-2}(C_{n})$ when $n$ is even. With the notation as in
4.4 we set $L=(f_{1},\ldots,f_{n})$. Then the symmetric algebra $S(I)$ of $I$
has the presentation $T/L$. Exchanging the role of the $x_{i}$ and the
$y_{j}$, we may view $T/L$ also as the symmetric algebra of an $R$-module $E$,
which is called the Jacobian dual of $I$. The relation matrix $A$ of $E$ is
defined by the equation ${\mathbf{f}}=A{\mathbf{x}}$, where
${\mathbf{f}}=[f_{1},\ldots,f_{n}]^{T}$ and
${\mathbf{x}}=[x_{1},\ldots,x_{n}]^{T}$. It can be seen that $A$ is a skew-
symmetric matrix and that the Pfaffian of $A$ is just the fiber relation $h$.

We are ready to compute the Hilbert series and Cohen-Macaulay type of
$R(I_{n-2}(C_{n}))$.

######  Theorem 4.6.

Let $I=I_{n-2}(C_{n})$. Then:

  1. _(a)_

if $n=2s+1$, we have

$\operatorname{HS}(R(I))=\frac{\sum_{k=0}^{s-1}{(1+z)^{2k+1}z^{s-1-k}}+z^{s}}{(1-z)^{2s+2}},$

  2. _(b)_

if $n=2s$, we have

$\operatorname{HS}(R(I))=\frac{\sum_{k=0}^{s-1}{(1+z)^{2k}z^{s-1-k}}}{(1-z)^{2s+1}}$

###### Proof.

As before, let $f_{1}=\underline{x_{n-1}y_{1}}-x_{1}y_{2}$,
$f_{2}=\underline{x_{0}y_{2}}-x_{2}y_{3}$, $\ldots,$
$f_{n-1}=\underline{x_{n-3}y_{n-1}}-x_{n-1}y_{0}$ and let

$g_{k}=\underline{x_{2k-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\ i\text{
odd}\end{subarray}}y_{i}}-x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{ even}\end{subarray}}y_{i}$

for $k=1,\ldots,\lfloor\frac{n}{2}\rfloor$. Recall from 4.2 and 4.1 that the
polynomials $f_{1},\ldots,f_{n-1},g_{1},\ldots,g_{\frac{n}{2}}$ form a Gröbner
basis of $L=(f_{1},\ldots,f_{n-1},g_{1})$. If $n=2s+1$, we have $J=L$ by [1,
Theorem 3.4]. Also recall from 4.4 that if $n$ is even, we have $J=L+hT$,
where $h=\underline{\prod_{\begin{subarray}{c}i\in[0,n)\\\ i\text{
odd}\end{subarray}}y_{i}}-\prod_{\begin{subarray}{c}i\in[0,n)\\\ i\text{
even}\end{subarray}}y_{i}$, and
$f_{1},\ldots,f_{n-1},g_{1},\ldots,g_{\frac{n}{2}},h$ form a Gröbner basis of
$J$.

Let $K_{n}$ denote the initial ideal of $J$ with respect to the reverse
lexicographic order. Then if $n=2s+1$ we have

$K_{2s+1}=(x_{2s}y_{1},x_{0}y_{2},\ldots,x_{2s-2}y_{2s},x_{0}y_{1},x_{2}y_{1}y_{3},\ldots,\\\
x_{2s-4}y_{1}y_{3}\cdots y_{2s-3},x_{2s-2}y_{1}y_{3}\cdots y_{2s-1}).$

Define an auxiliary ideal

$K^{\prime}_{2s+1}=K_{2s+1}+(y_{1}y_{3}\cdots
y_{2s-1})=(x_{2s}y_{1},x_{0}y_{2},\ldots,x_{2s-2}y_{2s},x_{0}y_{1},x_{2}y_{1}y_{3},\ldots,\\\
x_{2s-4}y_{1}y_{3}\cdots y_{2s-3},y_{1}y_{3}\cdots y_{2s-1}).$

Now, if $n=2s$, the corresponding initial ideal is

$K_{2s}=(x_{2s-1}y_{1},x_{0}y_{2},\ldots,x_{2s-3}y_{2s-1},x_{0}y_{1},x_{2}y_{1}y_{3},\ldots,\\\
x_{2s-4}y_{1}y_{3}\cdots y_{2s-3},y_{1}y_{3}\cdots y_{2s-1}).$

Note that our Gröbner basis is not reduced and the leading term of $h$ divides
the leading term of $g_{s}$. From this point on, the proof will be divided
into two steps:

Step 1: We will show that $(a)$ is equivalent to

$\operatorname{HS}(T/K^{\prime}_{2s+1})=\frac{\sum_{k=0}^{s-1}{(1+z)^{2k+1}z^{s-1-k}}}{(1-z)^{2s+2}}.$

Indeed, we have $K_{2s+1}+(y_{1}y_{3}\cdots y_{2s-1})=K^{\prime}_{2s+1}$ and
$K_{2s+1}:(y_{1}y_{3}\cdots
y_{2s-1})=(x_{2s},x_{1},x_{3},\ldots,x_{2s-3},x_{0},x_{2},\ldots,x_{2s-2})=(x_{2s},x_{0},x_{1},\ldots,x_{2s-2})$.
Therefore,

$\displaystyle\operatorname{HS}(T/K_{2s+1})$
$\displaystyle=\operatorname{HS}(T/(K_{2s+1}+(y_{1}y_{3}\cdots
y_{2s-1})))+z^{s}\operatorname{HS}(T/(K_{2s+1}:(y_{1}y_{3}\cdots y_{2s-1})))$
$\displaystyle=\operatorname{HS}(T/K^{\prime}_{2s+1})+\frac{z^{s}}{{(1-z)}^{2s+2}}.$

Step 2: We will prove the theorem by induction on $n$. There will be two
induction steps, one transitioning from $n$ even to $n$ odd, and the other one
transitioning from $n$ odd to $n$ even.

(1) Let us assume we know the theorem is true for all $n\leq 2s$. We want to
prove it for $n=2s+1$.

We are working inside the ring $T=K[x_{0},\ldots,x_{2s},y_{0},\ldots,y_{2s}]$.
The factor ring $T/K_{2s}T$ has the Hilbert series given by (b), divided by
$(1-z)^{2}$ (since we have $2$ extra variables $x_{2s}$ and $y_{2s}$). Let
$\tilde{K}_{2s}\subset T$ be the ideal obtained from $K_{2s}T$ by switching
$x_{2s-1}\leftrightarrow x_{2s}$. Clearly, $T/\tilde{K}_{2s}$ has the same
Hilbert series as $T/K_{2s}T$ (which we know by the induction hypothesis).
Now,

$\tilde{K}_{2s}+(x_{2s-2}y_{2s})=(x_{2s}y_{1},x_{0}y_{2},\ldots,x_{2s-3}y_{2s-1},x_{2s-2}y_{2s}x_{0}y_{1},x_{2}y_{1}y_{3},\\\
\ldots,x_{2s-4}y_{1}y_{3}\cdots y_{2s-3},y_{1}y_{3}\cdots
y_{2s-1})=K^{\prime}_{2s+1}$

and $\tilde{K}_{2s}:(x_{2s-2}y_{2s})=\tilde{K}_{2s}$. Therefore

$\operatorname{HS}(T/\tilde{K}_{2s})=\operatorname{HS}(T/K^{\prime}_{2s+1})+z^{2}\operatorname{HS}(T/\tilde{K}_{2s})$

and hence

$\displaystyle\operatorname{HS}(T/K^{\prime}_{2s+1})$
$\displaystyle=(1-z^{2})\operatorname{HS}(T/\tilde{K}_{2s}t)=(1-z^{2})\frac{\sum_{k=0}^{s-1}{(1+z)^{2k}z^{s-1-k}}}{(1-z)^{2s+3}}$
$\displaystyle=\frac{\sum_{k=0}^{s-1}{(1+z)^{2k+1}z^{s-1-k}}}{(1-z)^{2s+2}},$

which by Step 1 is equivalent to having the desired Hilbert series for
$T/K_{2s+1}$.

(2) Let us assume we know the statement is true for all $n\leq 2s+1$, we want
to prove it for $n=2s+2$.

Now we are working inside the ring
$T=K[x_{0},\ldots,x_{2s+1},y_{0},\ldots,y_{2s+1}]$. The factor ring
$T/K_{2s+1}T$ has the Hilbert series given by (a), divided by $(1-z)^{2}$
(since we have extra variables $x_{2s+1}$ and $y_{2s+1}$). Let
$\tilde{K}_{2s+2}\subset T$ be the ideal obtained from $K_{2s+2}$ by switching
$x_{2s}\leftrightarrow x_{2s+1}$. Clearly, $T/\tilde{K}_{2s+2}$ has the same
Hilbert series as $T/K_{2s+2}$ (which we do not know yet, but will compute).
Now,

$\tilde{K}_{2s+2}+(y_{2s+1})=(x_{2s}y_{1},x_{0}y_{2},\ldots,x_{2s-3}y_{2s-1},x_{2s-2}y_{2s},y_{2s+1},\\\
x_{0}y_{1},x_{2}y_{1}y_{3},\ldots,x_{2s-4}y_{1}y_{3}\cdots
y_{2s-3},y_{1}y_{3}\cdots y_{2s-1})=K_{2s+1}+(y_{2s+1}).$

and

$\tilde{K}_{2s+2}:(y_{2s+1})=(x_{2s}y_{1},x_{0}y_{2},\ldots,x_{2s-3}y_{2s-1},x_{2s-2}y_{2s},x_{2s-1},\\\
x_{0}y_{1},x_{2}y_{1}y_{3},\ldots,x_{2s-4}y_{1}y_{3}\cdots
y_{2s-3},y_{1}y_{3}\cdots y_{2s-1})=K^{\prime}_{2s+1}+(x_{2s-1}).$

By the induction hypothesis we know the Hilbert series of $T/K_{2s+1}T$ and
hence by Step 1 the Hilbert series of $T/K^{\prime}_{2s+1}T$. Therefore, we
also know the Hilbert series of $T/(K^{\prime}_{2s+1}+(x_{2s-1}))T$ since
$x_{2s-1}$ is a non-zerodivisor modulo $K^{\prime}_{2s+1}T$. Then we obtain

$\displaystyle\operatorname{HS}(T/K_{2s+2})=\operatorname{HS}(T/\tilde{K}_{2s+2})$
$\displaystyle=\operatorname{HS}(T/(K_{2s+1}+(y_{2s+1}))T)+z\operatorname{HS}(T/(K^{\prime}_{2s+1}+(x_{2s-1}))T)$
$\displaystyle=\frac{\sum_{k=0}^{s-1}{(1+z)^{2k+1}z^{s-1-k}}+z^{s}}{(1-z)^{2s+3}}+\frac{z\sum_{k=0}^{s-1}{(1+z)^{2k+1}z^{s-1-k}}}{(1-z)^{2s+3}}$
$\displaystyle=\frac{z^{s}+(1+z)\sum_{k=0}^{s-1}{(1+z)^{2k+1}z^{s-1-k}}}{(1-z)^{2s+3}}=\frac{z^{s}+\sum_{k=0}^{s-1}{(1+z)^{2(k+1)}z^{s-(k+1)}}}{(1-z)^{2s+3}}$
$\displaystyle=\frac{z^{s}+\sum_{k=1}^{s}{(1+z)^{2k}z^{s-k}}}{(1-z)^{2s+3}}=\frac{\sum_{k=0}^{s}{(1+z)^{2k}z^{s-k}}}{(1-z)^{2s+3}}.$

∎

Let $I=I_{n-2}(C_{n})$. Since $R(I)$ is a domain and its $h$-vector is
symmetric, it follows from 4.6 and [9, Theorem 4.4] that $R(I)$ is Gorenstein.
4.6 also implies that $R(I)$ is not Gorenstein, if $n$ is odd. We actually
have the following

######  Theorem 4.7.

Let $n>1$ be an odd integer and let $I=I_{n-2}(C_{n})$. Then the Cohen-
Macaulay type of $R(I)$ is $2$.

###### Proof.

The Rees algebra $R(I)$ modulo the image of the sequence
${\mathbf{s}}=x_{1}-y_{3},x_{2}-y_{4},\ldots,x_{i-2}-y_{i},\ldots,x_{n-2}-y_{n},x_{n}$
in $R(I)$ is isomorphic to $K[x_{1},\ldots,x_{n-1}]/\bar{J}$, where

$\bar{J}=(x_{1}^{2}-x_{2}x_{3},x_{2}^{2}-x_{3}x_{4},\ldots,x_{n-3}^{2}-x_{n-2}x_{n-1},x_{n-2}^{2},x_{n-1}^{2},x_{1}x_{2}).$

We denote by $<$ the lexicographical order induced by
$x_{1}>x_{2}>\cdots>x_{n}$. Then $\operatorname{in}_{<}(\bar{J})$ contains
$x_{1}^{2},x_{2}^{2},\ldots,x_{n}^{2}$. This implies that $\dim
R(I)/({\mathbf{s}})=0$. By 4.5(a) the ring $R(I)$ is Cohen-Macaulay. Since
$\dim R(I)$ coincides with the length of the sequence ${\mathbf{s}}$, it
follows therefore from [2, Theorem 2.1.2(c)] that ${\mathbf{s}}$ is an
$R(I)$-sequence. From this we deduce that $R(I)$ and
$\bar{R}=R(I)/({\mathbf{s}})$ have the same Cohen-Macaulay type, see [5,
Proposition A.6.2]. Thus is suffices to show that the Cohen-Macaulay type of
$\bar{R}$ is $2$.

Let
$A=(x_{1}^{2}-x_{2}x_{3},\ldots,x_{n-3}^{2}-x_{n-2}x_{n-1},x_{n-2}^{2},x_{n-1}^{2})\subset
K[x_{1},\ldots,x_{n-1}]$. Then $\bar{J}=A+(x_{1}x_{2})$, and $A$ is generated
by a regular sequence since
$\operatorname{in}_{<}(A)=(x_{1}^{2},\ldots,x_{n-1}^{2})$ is generated by a
regular sequence. Given this, it is well-known that the canonical module
$\omega_{\bar{R}}$ is isomorphic to $(A:(x_{1}x_{2}))/A$. For the convenience
of the reader we show why this is the case.

The ring $\bar{S}=K[x_{1},\ldots,x_{n-1}]/A$ is a complete intersection and
hence Gorenstein. In particular, up to a shift in the grading, we have that
$\omega_{\bar{S}}\cong\bar{S}$. Then it follows from [2, Theorem 3.3.7] that,
up to a shift,

$\omega_{\bar{R}}\cong\operatorname{Hom}_{\bar{S}}(\bar{R},\bar{S})\cong
0:_{\bar{S}}(x_{1}x_{2})\bar{S}.$

The last equation results from the fact that
$\bar{R}=\bar{S}/(x_{1}x_{2})\bar{S}$.

By [2, Proposition 3.3.11], the minimal number of generators of
$\omega_{\bar{R}}$ is the Cohen-Macaulay type of $\bar{R}$. Thus it remains to
be shown that $(A:(x_{1}x_{2}))/A$ is generated by $2$ elements.

In fact, we will show that $A:(x_{1}x_{2})=A+(x_{1},x_{2}x_{4}\ldots
x_{n-3}x_{n-1})$. In other words, we want to determine for which polynomials
$p$ we have $px_{1}x_{2}\in A$. The inclusion $A+(x_{1},x_{2}x_{4}\ldots
x_{n-3}x_{n-1})\subseteq A:(x_{1}x_{2})$ is obvious, so we will focus on the
other one. Since $\operatorname{in}_{<}(A)=(x_{1}^{2},\ldots,x_{n-1}^{2})$,
the squarefree monomials of $K[x_{1},\ldots,x_{n-1}]$ form a $K$-basis of
$K[x_{1},\ldots,x_{n-1}]/A$. One way of proving that an element belongs to $A$
is to prove that it reduces to $0$ by the division algorithm with respect to a
Gröbner basis of $A$. In this case the reduction algorithm can be described as
substituting $x_{1}^{2}\rightarrow x_{2}x_{3},\ldots,x_{n-3}^{2}\rightarrow
x_{n-2}x_{n-1},x_{n-2}^{2}\rightarrow 0,x_{n-1}^{2}\rightarrow 0$. Let
$\bar{p}$ denote the remainder of $p$ with respect to this Gröbner basis.
Since $p\equiv\bar{p}\operatorname{mod}A$, we may assume without loss of
generality that $p=\bar{p}$. In other words, we may assume that all terms of
$p$ are squarefree. Now we need to apply the reduction algorithm to
$x_{1}x_{2}p$. In the first step of the proof we will show by induction the
following claim: for every pair of squarefree monomials $m_{1}\not=m_{2}$ such
that $\overline{x_{1}x_{2}m_{1}}=\overline{x_{1}x_{2}m_{2}}$ we have
$\overline{x_{1}x_{2}m_{1}}=\overline{x_{1}x_{2}m_{2}}=0$. This will imply
that each monomial in the support of $x_{1}x_{2}p$ has to reduce to $0$
individually, which will reduce our problem to the case where $p$ is a
monomial. In the second step of the proof we will show by induction the
following claim: if $m$ is a squarefree monomial such that
$\overline{x_{1}x_{2}m}=0$, then $m\in A+(x_{1},x_{2}x_{4}\ldots
x_{n-3}x_{n-1})$. Note that if $n=3$, we have $A=(x_{1}^{2},x_{2}^{2})$ and
$A:(x_{1}x_{2})=A+(x_{1},x_{2})=(x_{1},x_{2})$. Both claims hold trivially in
this case, and this settles the base case for inductive proofs of both claims.

To prove the first claim by induction, assume that the exists a pair of
monomials $m_{1}=x_{1}^{\alpha_{1}}\cdots x_{n-1}^{\alpha_{n-1}}$ and
$m_{2}=x_{1}^{\beta_{1}}\cdots x_{n-1}^{\beta_{n-1}}$ with each $\alpha_{i}$
and $\beta_{i}$ either $0$ or $1$ such that $m_{1}\not=m_{2}$ and
$\overline{x_{1}x_{2}m_{1}}=\overline{x_{1}x_{2}m_{2}}\not=0$. First of all
note that none of the $m_{i}$ is divisible by $x_{1}$, in other words,
$\boxed{\alpha_{1}=\beta_{1}=0}$. Indeed, if $\alpha_{1}=1$, then
$x_{1}x_{2}m_{1}$ is divisible by $x_{1}^{2}x_{2}\in A$ because $x_{1}\in
A:(x_{1}x_{2})$, and therefore $x_{1}x_{2}m_{1}$ reduces to $0$. The same
holds if $\beta_{1}=1$.

Now notice that $\alpha_{2}=\beta_{2}$. Indeed, if, say, $\alpha_{2}=0$ and
$\beta_{2}=1$, then $\overline{x_{1}x_{2}m_{1}}=x_{1}x_{2}m_{1}$ (since
$x_{1}x_{2}m_{1}$ is squarefree), while
$x_{1}x_{2}m_{2}=x_{1}x_{2}^{2}x_{3}^{\beta_{3}}\cdots
x_{n-1}^{\beta_{n-1}}\rightarrow x_{1}x_{3}x_{4}x_{3}^{\beta_{3}}\cdots
x_{n-1}^{\beta_{n-1}}$. From this first reduction step it is already clear
that $x_{2}$ does not divide $\overline{x_{1}x_{2}m_{2}}$, while $x_{2}$ does
divide $\overline{x_{1}x_{2}m_{1}}$, which is a contradiction. If
$\alpha_{2}=\beta_{2}=0$, we have $\overline{x_{1}x_{2}m_{1}}=x_{1}x_{2}m_{1}$
and $\overline{x_{1}x_{2}m_{2}}=x_{1}x_{2}m_{2}$, which then implies
$m_{1}=m_{2}$, which is a contradiction. So we will assume
$\boxed{\alpha_{2}=\beta_{2}=1}$. Let $m_{1}=x_{2}m_{1}^{\prime}$ and
$m_{2}=x_{2}m_{2}^{\prime}$, where $m_{1}^{\prime}=x_{3}^{\alpha_{3}}\cdots
x_{n-1}^{\alpha_{n-1}}$ and $m_{2}^{\prime}=x_{3}^{\beta_{3}}\cdots
x_{n-1}^{\beta_{n-1}}$. Then
$x_{1}x_{2}m_{1}=x_{1}x_{2}^{2}m_{1}^{\prime}\rightarrow
x_{1}x_{3}x_{4}m_{1}^{\prime}$ and similarly
$x_{1}x_{2}m_{2}=x_{1}x_{2}^{2}m_{2}^{\prime}\rightarrow
x_{1}x_{3}x_{4}m_{2}^{\prime}$. Clearly, $x_{1}$ will not be involved in any
further reductions. Therefore, it is enough to consider
$x_{3}x_{4}m_{1}^{\prime}$ and $x_{3}x_{4}m_{2}^{\prime}$. Note that
$m_{1}^{\prime}$ and $m_{2}^{\prime}$ are squarefree monomials in
$K[x_{3},x_{4},\ldots,x_{n-1}]$, and the only reduction rules we can use to
reduce them are $x_{3}^{2}\rightarrow x_{4}x_{5},\ldots,x_{n-3}^{2}\rightarrow
x_{n-2}x_{n-1},x_{n-1}^{2}\rightarrow 0,x_{n}\rightarrow 0$. This means we are
doing a reduction of $x_{3}x_{4}m_{1}^{\prime}$ and $x_{3}x_{4}m_{2}^{\prime}$
modulo
$A^{\prime}=(x_{3}^{2}-x_{4}x_{5},\ldots,x_{n-3}^{2}-x_{n-2}x_{n-1},x_{n-1}^{2},x_{n})$
inside $K[x_{3},x_{4},\ldots,x_{n-1}]$. By the induction hypothesis we know
that
$\overline{x_{3}x_{4}m_{1}^{\prime}}=\overline{x_{3}x_{4}m_{2}^{\prime}}\not=0$
implies $m_{1}^{\prime}=m_{2}^{\prime}$ therefore, $m_{1}=m_{2}$, which is a
contradiction.

Now it remains to prove the second claim. Let $m=x_{1}^{\alpha_{1}}\cdots
x_{n-1}^{\alpha_{n-1}}$ be a squarefree monomial such that
$\overline{x_{1}x_{2}m}=0$. Assume $m\not\in A+(x_{1},x_{2}x_{4}\ldots
x_{n-3}x_{n-1})$ . We first note that $\boxed{\alpha_{1}=0}$, otherwise we
immediately get a contradiction. If $\alpha_{2}=0$, then
$\overline{x_{1}x_{2}m_{1}}=x_{1}x_{2}m_{1}\not=0$. Therefore,
$\boxed{\alpha_{2}=1}$. Let $m=x_{2}m^{\prime}$, where
$m^{\prime}=x_{3}^{\alpha_{3}}\cdots x_{n-1}^{\alpha_{n-1}}$. Then
$x_{1}x_{2}m=x_{1}x_{2}^{2}m^{\prime}\rightarrow x_{1}x_{3}x_{4}m^{\prime}$.
Clearly, $x_{1}$ will not be involved in any further reductions, so it
suffices to consider $x_{3}x_{4}m^{\prime}$. Note that $m^{\prime}$ is a
squarefree monomial in $K[x_{3},x_{4},\ldots,x_{n-1}]$, and the only reduction
rules we can use to reduce $x_{3}x_{4}m^{\prime}$ are $x_{3}^{2}\rightarrow
x_{4}x_{5},\ldots,x_{n-3}^{2}\rightarrow x_{n-2}x_{n-1},x_{n-1}^{2}\rightarrow
0,x_{n}\rightarrow 0$. In other words, we are doing a reduction of
$x_{3}x_{4}m^{\prime}$ modulo
$A^{\prime}=(x_{3}^{2}-x_{4}x_{5},\ldots,x_{n-3}^{2}-x_{n-2}x_{n-1},x_{n-1}^{2},x_{n})$
inside $K[x_{3},x_{4},\ldots,x_{n-1}]$. By the induction hypothesis we know
that the squarefree monomials modulo $A^{\prime}$ which reduce to $0$ are
either multiples of $x_{3}$ or multiples of $x_{4}x_{6}\cdots x_{n-1}$. Assume
$m^{\prime}$ is a multiple of $x_{3}$ modulo $A^{\prime}$ (therefore, also
modulo $A$). Then, since $m=x_{2}m^{\prime}$, it follows that $m$ is a
multiple of $x_{2}x_{3}$, and hence a multiple of $x_{1}^{2}$ modulo $A$,
which is a contradiction. If $m^{\prime}$ is a multiple of $x_{4}\cdots
x_{n-1}$, then modulo $A$ we have that $m$ a multiple of $x_{2}x_{4}\cdots
x_{n-1}$, which is a contradiction. ∎

##  5\. The structure of the Rees ring of $I_{\frac{n}{2}}(C_{n})$

The methods and the results in this section will be similar to those in
Section 4. For the rest of this section, let $n\geq 4$ be an even integer. Let
$L$ be the defining ideal of the symmetric algebra
$S(I_{\frac{n}{2}}(C_{n}))$. By [1, Proposition 2.5] we know that
$L=(f_{1},\ldots,f_{n-1},g_{1})$, where
$f_{j}={x_{\frac{n}{2}+j}y_{j}}-x_{j}y_{j+1}$ for $j=1,\ldots,n-1$ and
$g_{1}=x_{0}y_{1}-x_{\frac{n}{2}}y_{0}$. Note that if we extend the definition
of $f_{j}$ to $j=n$, we will obtain $g_{1}=-f_{n}$. As a preparation for a
computation of a Gröbner basis for $J$ (which is the defining ideal of
$R(I_{\frac{n}{2}}(C_{n}))$), we first extend the set of generators of $L$.

######  Proposition 5.1.

$L=(f_{1},\ldots,f_{n-1},g_{1},\ldots,g_{\frac{n}{2}-1}),$ where

$g_{k}={y_{k}\prod_{i=0}^{k-1}x_{i}}-y_{0}\prod_{i=\frac{n}{2}}^{k-1+\frac{n}{2}}x_{i}\quad\textrm{for}\quad
k=1,\ldots,\frac{n}{2}-1.$

###### Proof.

We will show by induction that $g_{k}\in(f_{1},\ldots,f_{n-1},g_{1})$ for all
$k=1,\ldots,\frac{n}{2}-1$. The base case holds by definition. Now, modulo $L$
we have:

$\displaystyle y_{k}\prod_{i=0}^{k-1}x_{i}$
$\displaystyle=(x_{k-1}y_{k})\prod_{i=0}^{k-2}x_{i}\overset{f_{k-1}}{=}x_{\frac{n}{2}+k-1}y_{k-1}\prod_{i=0}^{k-2}x_{i}=x_{\frac{n}{2}+k-1}(x_{k-2}y_{k-1})\prod_{i=0}^{k-3}x_{i}$
$\displaystyle\overset{f_{k-2}}{=}(x_{\frac{n}{2}+k-1}x_{\frac{n}{2}+k-2})y_{k-2}\prod_{i=0}^{k-3}x_{i}=\cdots\overset{f_{1}}{=}\prod_{i=1+\frac{n}{2}}^{k-1+\frac{n}{2}}x_{i}(x_{0}y_{1})\overset{g_{1}}{=}y_{0}\prod_{i=\frac{n}{2}}^{k-1+\frac{n}{2}}x_{i}.$

∎

Recall that by 3.2 we have $H=(h_{1},\ldots,h_{\frac{n}{2}-1})$, where

$h_{l}={y_{l}y_{l+\frac{n}{2}}}-y_{0}y_{\frac{n}{2}}\quad\textrm{for}\quad
l=1,\ldots,\frac{n}{2}-1.$

We will now describe a Gröbner basis of $L+HT$, which will later turn out to
be equal to $J$, the defining ideal of $R(I_{\frac{n}{2}}(C_{n}))$.

######  Theorem 5.2.

Let $<$ be the reverse lexicographic order on $T$ induced by
$x_{1}>\ldots>x_{n-1}>x_{0}>y_{1}>\ldots>y_{n-1}>y_{0}$. Then the polynomials

$f_{1},\ldots,f_{n-1},g_{1},\ldots,g_{\frac{n}{2}-1},h_{1},\ldots,h_{\frac{n}{2}-1}$

form a Gröbner basis for $L+HT$.

The proof of this fact can be found in the Appendix.

We are now ready to describe the defining ideal $J$ of
$R(I_{\frac{n}{2}}(C_{n}))$.

######  Theorem 5.3.

Let as before $L=(f_{1},\ldots,f_{n})$, where
$f_{j}=x_{\frac{n}{2}+j}y_{j}-x_{j}y_{j+1}$ for $j=1,\ldots,n$. As before, let
$H=(h_{1},\ldots,h_{\frac{n}{2}-1})$, where
$h_{l}={y_{l}y_{l+\frac{n}{2}}}-y_{0}y_{\frac{n}{2}}\quad\textrm{for}\quad
l=1,\ldots,\frac{n}{2}-1.$ Then $J=L+HT$.

###### Proof.

Let $J^{\prime}=(f_{0},\ldots,f_{n-1},h_{1},\ldots,h_{\frac{n}{2}-1})$. We
apply the same strategy as in the proof of 4.4 in order to show that
$J^{\prime}=J$. It follows from the description of the Gröbner basis of
$J^{\prime}$, given in 5.2, that $x_{\frac{n}{2}}$ is a nonzero-divisor modulo
$\operatorname{in}(J^{\prime})$. Then $x_{\frac{n}{2}}$ is a nonzero-divisor
modulo $J^{\prime}$, as well. By symmetry it follows that all $x_{i}$ are
nonzero-divisors modulo $J^{\prime}$. Consequently, $x=\prod_{i=1}^{n}x_{i}$
is a nonzero-divisor modulo $J^{\prime}$. Therefore, the natural homomorphism
$T/J^{\prime}\rightarrow(T/J^{\prime})_{x}$ is injective. Thus $J^{\prime}$
will be a prime ideal, if $J^{\prime}T_{x}$ is a prime ideal. The relations
$f_{1},\ldots,f_{n-1}$ show that $y_{i}=w_{i}y_{i+1}$ modulo $J^{\prime}T_{x}$
for $i=0,\ldots,n-1$, where $w_{i}$ are some monomials in $x_{i}^{\pm 1}$.
Thus, for each $1\leq i\leq n-1$ we get
$y_{i}=w_{i}y_{i+1}=w_{i}w_{i+1}y_{i+2}$. Proceeding in this way, we see that
for all $i=1,\ldots,n-1$ we have $y_{i}=v_{i}y_{0}$ for some unit $v_{i}$,
where each $v_{i}$ is a product of certain $x_{k}^{\pm 1}$ and $v_{n}=1$.

It follows that

$T_{x}/J^{\prime}T_{x}\cong K[y_{n},x_{1}^{\pm 1},\ldots,x_{n}^{\pm
1}]/(x_{\frac{n}{2}}y_{n}-x_{n}v_{1}y_{n},\\\
\\{v_{l}v_{\frac{n}{2}+l}y_{n}^{2}-v_{n}v_{\frac{n}{2}}y_{n}^{2}:l=1,\ldots,\frac{n}{2}-1\\}).$

Since $J^{\prime}\subseteq J=\ker(\phi)$ sending $y_{n}\mapsto
x_{n}x_{1}\cdots x_{\frac{n}{2}-1}s$, we conclude that these binomials vanish
under this substitution. Since they are homogeneous in $y_{n}$, it follows
that the coefficients of $y_{n}$ in the first binomial must be equal, and the
coefficients of $y_{n}^{2}$ in the remaining binomials must also be equal.

This implies that $T_{x}/J^{\prime}T_{x}\cong K[y_{n},x_{1}^{\pm
1},\ldots,x_{n}^{\pm 1}]$, from which we conclude that $J^{\prime}$ is a prime
ideal of height $n+1$. Since $J^{\prime}\subseteq J$ and since
$\operatorname{height}J=n+1$, we have $J^{\prime}=J$, as desired.

∎

######  Corollary 5.4.

Let $n\geq 4$ and let $I=I_{\frac{n}{2}}(C_{n})$. Then the following holds:

  1. (a)

$R(I)$ is a normal Cohen–Macaulay domain.

  2. (b)

$I$ is of fiber type.

###### Proof.

(a) Let $J$ as before be the defining ideal of $R(I)$. By 5.2, $J$ has a
squarefree initial ideal. By [10, Proposition 13.15] this implies that $R(I)$
is a normal. Then, since $R(I)$ is a toric ring, we obtain that $R(I)$ is a
Cohen–Macaulay domain, see [7, Theorem 1].

(b) This can be seen from the generators of $J$ given in 5.3.

∎

##  6\. Summary of results and open problems

In this section we discuss some observations based on computations in
Macaulay2. We get the following table: the entry in the $t$th column and the
$n$th row (where rows start at $n=3$) is $L$, $F$ or $\times$, which means
that $I_{t}(C_{n})$ is of linear type, fiber type or neither, respectively.

$13$$12$$11$$10$$9$$8$$7$$6$$5$$4$$3$$2$$1$$t$$n$${\color[rgb]{1,1,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,0}\pgfsys@color@cmyk@stroke{0}{0}{1}{0}\pgfsys@color@cmyk@fill{0}{0}{1}{0}t=1}$${\color[rgb]{1,1,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,0}\pgfsys@color@cmyk@stroke{0}{0}{1}{0}\pgfsys@color@cmyk@fill{0}{0}{1}{0}t=n-1}$${\color[rgb]{1,.75,.75}\definecolor[named]{pgfstrokecolor}{rgb}{1,.75,.75}t=2}$${\color[rgb]{1,.75,.75}\definecolor[named]{pgfstrokecolor}{rgb}{1,.75,.75}t=n-2}$${\color[rgb]{0,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,1}\pgfsys@color@cmyk@stroke{1}{0}{0}{0}\pgfsys@color@cmyk@fill{1}{0}{0}{0}t=\lfloor\frac{n}{2}\rfloor}$$C_{3}$$C_{4}$$C_{5}$$C_{6}$$C_{7}$$C_{8}$$C_{9}$$C_{10}$$C_{11}$$C_{12}$$C_{13}$$C_{14}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$$\times$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$$\times$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$$\times$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$$\times$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$$\times$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$$\times$$\times$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$$\times$$\times$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$$\times$$\times$$\times$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$$\times$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$$\times$$\times$$\times$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$$\times$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$$\times$$\times$$\times$$\times$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$$\times$$\times$$\times$$\times$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$$\times$$\times$$\times$$\times$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}F}$${\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}L}$
Figure 1. When $I_{t}(C_{n})$ is of linear type or fiber type or neither?

The following results are either known or proven in this paper:

  * •

$I_{1}(C_{n})$ is of linear type for all $n$ since it is generated by a
regular sequence;

  * •

$I_{n-1}(C_{n})$ is of linear type for all $n$ by [1, Theorem 3.2];

  * •

$I_{2}(C_{n})$ is of linear type if $n$ is odd by [11, Theorem 3.4], and of
fiber type (but not linear type) if $n$ is even by [12, Theorem 8.2.1];

  * •

$I_{n-2}(C_{n})$ is of linear type if $n$ is odd by [1, Theorem 3.4], and of
fiber type (but not linear type) if $n$ is even by 4.5;

  * •

$I_{\lfloor\frac{n}{2}\rfloor}(C_{n})$ is of linear type if $n$ is odd by [1,
Corollary 3.6] and of fiber type (but not linear type) if $n$ is even by 5.4;

  * •

There are some other partial results on linear type given in [1].

The table above leads us to the following:

######  Conjecture 6.1.

_(1)_ Suppose $2\leq t\leq\frac{n}{2}$. Then $I_{t}(C_{n})$ is of fiber type,
but not of linear type, if and only if $t\mid n$.

_(2)_ Suppose $\frac{n}{2}<t<n-2$. Then $I_{t}(C_{n})$ is of fiber type if and
only if $n=9$ and $t=6$.

## Acknowledgement

This project was started when the third author was visiting the first and the
second author at the University of Duisburg-Essen. The third author thanks the
local research group for their hospitality, and would like to acknowledge
International Exchange and Overseas Study Scholarship of Soochow University
which allowed her to perform this trip.

The first author was supported by a fellowship from the Wenner-Gren
Foundations (grant WGF2022-0052).

## Appendix

_4.2._ Recall that
$L=(f_{1},\ldots,f_{n-1},g_{1},\ldots,g_{\lfloor\frac{n}{2}\rfloor}),$ where

$f_{j}=\underline{x_{j-2}y_{j}}-x_{j}y_{j+1},\quad
g_{k}=\underline{{x_{2k-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\ i\text{
odd}\end{subarray}}y_{i}}}-x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{ even}\end{subarray}}y_{i}.$

These generators form a Gröbner basis of $L$ with respect to any monomial
order $<$ on $T$ which is the product order of $<_{1}$ and $<_{2}$, where
$<_{1}$ is the reverse lexicographic order on $R$ induced by
$y_{1}>y_{2}>\ldots>y_{n-1}>y_{0}$, and $<_{2}$ is an arbitrary monomial order
on $S$.

###### Proof of 4.2.

The proof will be performed in three steps.

Step 1: _The $S-$pairs $S(g_{k_{1}},g_{k_{2}})$ all reduce to $0$._ Let $1\leq
k-s<k\leq\lfloor\frac{n}{2}\rfloor$. We would like to show that
$S(g_{k},g_{k-s})$ reduces to $0$. We have:

$\displaystyle S(g_{k},g_{k-s})$
$\displaystyle=S\left(\underline{x_{2k-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{
odd}\end{subarray}}y_{i}}-x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{
even}\end{subarray}}y_{i},\quad\underline{x_{2k-2s-2}\prod_{\begin{subarray}{c}i\in[0,2k-2s)\\\
i\text{
odd}\end{subarray}}y_{i}}-x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k-2s)\\\
i\text{ even}\end{subarray}}y_{i}\right)$
$\displaystyle=g_{k}x_{2k-2s-2}-g_{k-s}x_{2k-2}\prod_{\begin{subarray}{c}i\in[2k-2s,2k)\\\
i\text{ odd}\end{subarray}}y_{i}$
$\displaystyle=x_{n-2}x_{2k-2}\prod_{\begin{subarray}{c}i\in[2k-2s,2k)\\\
i\text{ odd}\end{subarray}}y_{i}\prod_{\begin{subarray}{c}i\in[0,2k-2s)\\\
i\text{
even}\end{subarray}}y_{i}-x_{n-2}x_{2k-2s-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{ even}\end{subarray}}y_{i}$
$\displaystyle=x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k-2s)\\\ i\text{
even}\end{subarray}}y_{i}\left(x_{2k-2}\prod_{\begin{subarray}{c}i\in[2k-2s,2k)\\\
i\text{
odd}\end{subarray}}y_{i}-x_{2k-2s-2}\prod_{\begin{subarray}{c}i\in[2k-2s,2k)\\\
i\text{ even}\end{subarray}}y_{i}\right).$

We will prove by induction on $s$ that the binomial in parentheses reduces to
$0$. Since $s\geq 1$, the interval $[2k-2s,2k)$ is nonempty and we take out
$y_{2k-2s+1}$ and $y_{2k-2s}$ and rewrite the binomial in parentheses as
follows:

$\displaystyle
x_{2k-2}y_{2k-2s+1}\prod_{\begin{subarray}{c}i\in[2k-2s+2,2k)\\\ i\text{
odd}\end{subarray}}y_{i}-\underline{x_{2k-2s-2}y_{2k-2s}\prod_{\begin{subarray}{c}i\in[2k-2s+2,2k)\\\
i\text{ even}\end{subarray}}y_{i}}.$

Since $k-s\geq 1$, $2k\leq n$ and $s\geq 1$, we have $2\leq 2k-2s\leq n-2$,
and thus $2k-2s$ lies inside the range where $f_{i}$’s are defined. Therefore,
one can say that $x_{2k-2s-2}y_{2k-2s}$ is the leading term of
$f_{2k-2s}=\underline{x_{2k-2s-2}y_{2k-2s}}-x_{2k-2s}y_{2k-2s+1}$. Therefore,
reducing the leading term of our binomial using $f_{2k-2s}$, we obtain:

$\displaystyle
x_{2k-2}y_{2k-2s+1}\prod_{\begin{subarray}{c}i\in[2k-2s+2,2k)\\\ i\text{
odd}\end{subarray}}y_{i}-x_{2k-2s}y_{2k-2s+1}\prod_{\begin{subarray}{c}i\in[2k-2s+2,2k)\\\
i\text{ even}\end{subarray}}y_{i}$ $\displaystyle=$ $\displaystyle
y_{2k-2s+1}\left(x_{2k-2}\prod_{\begin{subarray}{c}i\in[2k-2(s-1),2k)\\\
i\text{
odd}\end{subarray}}y_{i}-x_{2k-2(s-1)-2}\prod_{\begin{subarray}{c}i\in[2k-2(s-1),2k)\\\
i\text{ even}\end{subarray}}y_{i}\right),$

and the binomial in parentheses is exactly the one we wanted to reduce, with
$s\rightarrow s-1$. Note that if $s=1$ it equals $0$, which settles the base
case.

Step 2: _The $S-$pairs $S(f_{j},g_{k})$ all reduce to $0$._ Recall that

$f_{j}=\underline{x_{j-2}y_{j}}-x_{j}y_{j+1},\quad
g_{k}=\underline{x_{2k-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\ i\text{
odd}\end{subarray}}y_{i}}-x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{ even}\end{subarray}}y_{i}.$

We need to consider only the cases when the leading terms of these binomials
are not coprime. This can happen in the following two cases:

Case 1: The two leading terms are not coprime because of a common $x$\-
factor, in other words, if $j=2k$. Note that this case is impossible if $2k=n$
since $j$ only takes values until $n-1$. Then we have

$\displaystyle
S(f_{2k},g_{k})=S\left(\underline{x_{2k-2}y_{2k}}-x_{2k}y_{2k+1},\quad\underline{x_{2k-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{
odd}\end{subarray}}y_{i}}-x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{ even}\end{subarray}}y_{i}\right)$
$\displaystyle=f_{2k}\prod_{\begin{subarray}{c}i\in[0,2k)\\\ i\text{
odd}\end{subarray}}y_{i}-g_{k}y_{2k}=x_{n-2}y_{2k}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{
even}\end{subarray}}y_{i}-x_{2k}y_{2k+1}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{ odd}\end{subarray}}y_{i}.$

We recall that the indices of variables should be interpreted modulo $n$,
while the intervals should be interpreted in the usual way. For example, if
$n=2k+1$, then $y_{2k+1}$ should be interpreted as $y_{0}$ and therefore in
this case we avoid writing for instance
$y_{2k+1}\prod_{\begin{subarray}{c}i\in[0,2k)\\\ i\text{
odd}\end{subarray}}y_{i}=\prod_{\begin{subarray}{c}i\in[0,2k+2)\\\ i\text{
odd}\end{subarray}}y_{i}$ in order to avoid the confusion of $y_{0}$ being
present in the product of $y_{i}$ with odd indices. That’s why all the
intervals we use in products have to be subsets of $[0,n)$.

In order to further reduce this binomial, we need to consider two final
subcases:

Subcase 1.a: $2k=n-1$. In this case our binomial becomes

$x_{n-2}y_{n-1}\prod_{\begin{subarray}{c}i\in[0,n-1)\\\ i\text{
even}\end{subarray}}y_{i}-\underline{x_{n-1}y_{0}\prod_{\begin{subarray}{c}i\in[0,n-1)\\\
i\text{ odd}\end{subarray}}y_{i}}.$

We will show that the leading term reduces to the other term:

$\displaystyle x_{n-1}y_{0}\prod_{\begin{subarray}{c}i\in[0,n-1)\\\ i\text{
odd}\end{subarray}}y_{i}=\prod_{\begin{subarray}{c}i\in[0,2)\\\ i\text{
even}\end{subarray}}y_{i}\cdot(x_{n-1}y_{1})\cdot\prod_{\begin{subarray}{c}i\in[2,n-1)\\\
i\text{ odd}\end{subarray}}y_{i}\xrightarrow{f_{1}}$
$\displaystyle\prod_{\begin{subarray}{c}i\in[0,2)\\\ i\text{
even}\end{subarray}}y_{i}\cdot
x_{1}y_{2}\cdot\prod_{\begin{subarray}{c}i\in[2,n-1)\\\ i\text{
odd}\end{subarray}}y_{i}=\prod_{\begin{subarray}{c}i\in[0,4)\\\ i\text{
even}\end{subarray}}y_{i}\cdot(x_{1}y_{3})\cdot\prod_{\begin{subarray}{c}i\in[4,n-1)\\\
i\text{ odd}\end{subarray}}y_{i}\xrightarrow{f_{3}}$
$\displaystyle\prod_{\begin{subarray}{c}i\in[0,4)\\\ i\text{
even}\end{subarray}}y_{i}\cdot
x_{3}y_{4}\cdot\prod_{\begin{subarray}{c}i\in[4,n-1)\\\ i\text{
odd}\end{subarray}}y_{i}=\prod_{\begin{subarray}{c}i\in[0,6)\\\ i\text{
even}\end{subarray}}y_{i}\cdot(x_{3}y_{5})\cdot\prod_{\begin{subarray}{c}i\in[6,n-1)\\\
i\text{ odd}\end{subarray}}y_{i}\xrightarrow{\cdots}$

We see that right after the reduction by $f_{2s+1}$ the monomial is

$\prod_{\begin{subarray}{c}i\in[0,2s+2)\\\ i\text{
even}\end{subarray}}y_{i}\cdot(x_{2s+1}y_{2s+2})\cdot\prod_{\begin{subarray}{c}i\in[2s+2,n-1)\\\
i\text{ odd}\end{subarray}}y_{i}.$

We set $s=k-1$ and obtain that after the reduction by
$f_{2s+1}=f_{2k-1}=f_{n-2}$ we get

$\prod_{\begin{subarray}{c}i\in[0,n-1)\\\ i\text{
even}\end{subarray}}y_{i}\cdot(x_{n-2}y_{n-1})\cdot\prod_{\begin{subarray}{c}i\in[n-1,n-1)\\\
i\text{ odd}\end{subarray}}y_{i},$

which is precisely what we need.

Subcase 1.b: $2k\leq n-2$. In this case we have $2k+2\leq n$ and so we can use
the interval $[0,2k+2)$ to rewrite our binomial as

$x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k+2)\\\ i\text{
even}\end{subarray}}y_{i}-x_{2k}\prod_{\begin{subarray}{c}i\in[0,2k+2)\\\
i\text{ odd}\end{subarray}}y_{i}.$

Note that this is simply $-g_{k+1}$. Also note that the indices of $g$’s are
only allowed to up until $\lfloor\frac{n}{2}\rfloor$, and so index $k+1$ is
allowed since $2(k+1)=2k+2\leq n$. This is not the case in the Subcase 1.a,
where $g_{k+1}$ merely does not exist.

Case 2: The two leading terms are not coprime because of a common $y$\-
factor, in other words, if $j$ is an odd number belonging to $[0,2k)$. Then we
get

$\displaystyle
S(f_{j},g_{k})=S\left(\underline{x_{j-2}y_{j}}-x_{j}y_{j+1},\quad\underline{x_{2k-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{
odd}\end{subarray}}y_{i}}-x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{ even}\end{subarray}}y_{i}\right)$
$\displaystyle=f_{j}x_{2k-2}\prod_{\begin{subarray}{c}i\in[0,2k)\setminus\\{j\\}\\\
i\text{
odd}\end{subarray}}y_{i}-g_{k}x_{j-2}=x_{j-2}x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\
i\text{
even}\end{subarray}}y_{i}-\underline{x_{2k-2}x_{j}y_{j+1}\prod_{\begin{subarray}{c}i\in[0,2k)\setminus\\{j\\}\\\
i\text{odd}\end{subarray}}y_{i}}$

Note that the underlined term is indeed the leading term: unless $j+1=n$, we
see that $y_{0}$ is present only in the support of the first term (making it
the smaller one). If $j+1=n$, and noting that $j=n-1$ is an odd number
belonging to $[0,2k)$, we conclude that $2k=n$. Then the above binomial
becomes $x_{n-3}x_{n-2}\prod_{\begin{subarray}{c}i\in[0,n)\\\ i\text{
even}\end{subarray}}y_{i}-{x_{n-2}x_{n-1}y_{0}\prod_{\begin{subarray}{c}i\in[0,n-2)\\\
i\text{ odd}\end{subarray}}y_{i}}$. We see that $y_{0}$ is present on both
sides, then $y_{n-1}$ is not present anywhere, and $y_{n-2}$ is only present
in the first term (making it the smaller one). We will show that the leading
term reduces to the other one:

$\displaystyle
x_{2k-2}x_{j}y_{j+1}\prod_{\begin{subarray}{c}i\in[0,2k)\setminus\\{j\\}\\\
i\text{
odd}\end{subarray}}y_{i}=x_{2k-2}\prod_{\begin{subarray}{c}i\in[j+1,j+3)\\\
i\text{ even}\end{subarray}}y_{i}\cdot
x_{j}y_{j+2}\cdot\prod_{\begin{subarray}{c}i\in[0,2k)\setminus\\{j,j+2\\}\\\
i\text{ odd}\end{subarray}}y_{i}$
$\displaystyle\xrightarrow{f_{j+2}}x_{2k-2}\prod_{\begin{subarray}{c}i\in[j+1,j+3)\\\
i\text{ even}\end{subarray}}y_{i}\cdot
x_{j+2}y_{j+3}\cdot\prod_{\begin{subarray}{c}i\in[0,2k)\setminus\\{j,j+2\\}\\\
i\text{ odd}\end{subarray}}y_{i}$
$\displaystyle=x_{2k-2}\prod_{\begin{subarray}{c}i\in[j+1,j+5)\\\ i\text{
even}\end{subarray}}y_{i}\cdot
x_{j+2}y_{j+4}\cdot\prod_{\begin{subarray}{c}i\in[0,2k)\setminus\\{j,j+2,j+4\\}\\\
i\text{ odd}\end{subarray}}y_{i}$
$\displaystyle\xrightarrow{f_{j+4}}x_{2k-2}\prod_{\begin{subarray}{c}i\in[j+1,j+5)\\\
i\text{ even}\end{subarray}}y_{i}\cdot
x_{j+4}y_{j+5}\cdot\prod_{\begin{subarray}{c}i\in[0,2k)\setminus\\{j,j+2,j+4\\}\\\
i\text{ odd}\end{subarray}}y_{i}$
$\displaystyle=x_{2k-2}\prod_{\begin{subarray}{c}i\in[j+1,j+7)\\\ i\text{
even}\end{subarray}}y_{i}\cdot
x_{j+4}y_{j+6}\cdot\prod_{\begin{subarray}{c}i\in[0,2k)\setminus\\{j,j+2,j+4,j+6\\}\\\
i\text{ odd}\end{subarray}}y_{i}\xrightarrow{\cdots}$

We see that right after the reduction by $j+2s$ the monomial is

$x_{2k-2}\prod_{\begin{subarray}{c}i\in[j+1,j+2s+1)\\\ i\text{
even}\end{subarray}}y_{i}\cdot
x_{j+2s}y_{j+2s+1}\cdot\prod_{\begin{subarray}{c}i\in[0,2k)\setminus\\{j,j+2,\ldots,j+2s\\}\\\
i\text{ odd}\end{subarray}}y_{i}.$

We do it as many times as it is needed to remove $2k-1$ from the set of odd
numbers. In other words, if we set $s=\frac{2k-1-j}{2}$, then $j+2s=2k-1$
(which is in the range of the allowed indices for $f$’s) and we get

$x_{2k-2}\prod_{\begin{subarray}{c}i\in[j+1,2k)\\\ i\text{
even}\end{subarray}}y_{i}\cdot
x_{2k-1}y_{2k}\cdot\prod_{\begin{subarray}{c}i\in[0,j-1)\\\ i\text{
odd}\end{subarray}}y_{i}.$

It is not difficult to notice that any monomial of type $x_{i}x_{i+1}y_{i+2}$
(for any $i=0,1,\ldots,n-1$, and indices interpreted modulo $n$) reduces to
$x_{n-2}x_{n-1}y_{0}$. Indeed,

$x_{n-1}x_{0}y_{1}\xrightarrow{f_{1}}x_{0}x_{1}y_{2}\xrightarrow{f_{2}}x_{1}x_{2}y_{3}\xrightarrow{\ldots}\ldots\xrightarrow{f_{n-2}}x_{n-3}x_{n-2}y_{n-1}\xrightarrow{f_{n-1}}x_{n-2}x_{n-1}y_{0}.$

Therefore, in the monomial above we may replace $x_{2k-2}x_{2k-1}y_{2k}$ with
$x_{n-2}x_{n-1}y_{0}$ and get

$\displaystyle x_{n-2}\prod_{\begin{subarray}{c}i\in[j+1,2k)\\\ i\text{
even}\end{subarray}}y_{i}\cdot\left[x_{n-1}y_{0}\cdot\prod_{\begin{subarray}{c}i\in[0,j-1)\\\
i\text{ odd}\end{subarray}}y_{i}\right]$

Note that the monomial in square brackets is almost exactly the same as the
leading term in Subcase 1.a, with the only difference that here the product is
taken over a shorter interval. The reduction process of this monomial will be
exactly the same as before, but with fewer steps. Analogously to Subcase 1.a,
we do reductions with $f_{1},f_{3},\ldots$, and we see that right after the
reduction by $f_{2s+1}$ the monomial in square brackets is

$\prod_{\begin{subarray}{c}i\in[0,2s+2)\\\ i\text{
even}\end{subarray}}y_{i}\cdot(x_{2s+1}y_{2s+2})\cdot\prod_{\begin{subarray}{c}i\in[2s+2,j-1)\\\
i\text{ odd}\end{subarray}}y_{i}.$

if $j=1$, this monomial already equals the one we want to reduce to and we are
done. Otherwise we set $s=\frac{j-3}{2}$ and obtain that after the reduction
by $f_{2s+1}=f_{j-2}$ we get

$\prod_{\begin{subarray}{c}i\in[0,j-1)\\\ i\text{
even}\end{subarray}}y_{i}\cdot(x_{j-2}y_{j-1})\cdot\prod_{\begin{subarray}{c}i\in[j-1,j-1)\\\
i\text{ odd}\end{subarray}}y_{i}=\prod_{\begin{subarray}{c}i\in[0,j+1)\\\
i\text{ even}\end{subarray}}y_{i}\cdot x_{j-2}.$

Now that we have reduced the part in square brackets, the entire monomial then
reduces to

$x_{n-2}\prod_{\begin{subarray}{c}i\in[j+1,2k)\\\ i\text{
even}\end{subarray}}y_{i}\prod_{\begin{subarray}{c}i\in[0,j+1)\\\ i\text{
even}\end{subarray}}y_{i}\cdot
x_{j-2}=x_{j-2}x_{n-2}\prod_{\begin{subarray}{c}i\in[0,2k)\\\ i\text{
even}\end{subarray}}y_{i},$

which is precisely what we need.

Step 3: _The $S-$pairs $S(f_{i},f_{j})$ all reduce to $0$._ This is obvious
since the leading terms of $f_{i}$ and $f_{j}$ are coprime for $i\not=j$.

∎

_4.3._ Let $n$ be even. Then the polynomials
$\\{f_{1},\ldots,f_{n-1},g_{1},\ldots,g_{\frac{n}{2}},h\\}$ form a Gröbner
basis for $L+HT$ with respect to any monomial order given in 4.2. Here, as
before, $H=(h)$, where

$h=\underline{\prod_{\begin{subarray}{c}i\in[0,n)\\\ i\text{
odd}\end{subarray}}y_{i}}-\prod_{\begin{subarray}{c}i\in[0,n)\\\ i\text{
even}\end{subarray}}y_{i}.$

###### Proof of 4.3.

Let $n=2s$. We only need to check that the $S$-pairs involving $h$ reduce to
$0$. Note that $g_{s}=x_{n-2}h$, and so the $S$-pairs involving $h$ will
reduce very similarly to those involving $g_{s}$. We will briefly go through
all the cases, essentially mimicking the proof of 4.2.

Step 1: _The $S-$pairs $S(h,g_{k})$ all reduce to $0$._ If $k=s$, there is
nothing to check, so we will assume $k=s-a$, $a\geq 1$. It is not hard to
verify that $S(g_{s},g_{s-a})=x_{n-2}S(h,g_{s-a})$.

In Step 1 in the proof of 4.2 we computed all $S(g_{k},g_{k-a})$, the result
was divisible by $x_{n-2}$, which we factored out, and the remaining binomial
reduced to $0$. Therefore, with the exact same reduction process
$S(h,g_{s-a})$ will reduce to $0$.

Step 2: _The $S-$pairs $S(f_{j},h)$ all reduce to $0$._ We will mimic the
proof of Step 2 of 4.2. Recall that

$f_{j}=\underline{x_{j-2}y_{j}}-x_{j}y_{j+1},\quad
h=\underline{\prod_{\begin{subarray}{c}i\in[0,n)\\\ i\text{
odd}\end{subarray}}y_{i}}-\prod_{\begin{subarray}{c}i\in[0,n)\\\ i\text{
even}\end{subarray}}y_{i}.$

Again, we need to consider only the cases when the leading terms of these
binomials are not coprime. They cannot have a common $x$-factor, so Case 1 of
Step 2 of 4.2 cannot happen here. If the two leading terms are not coprime, it
has to be because of a common $y$-factor, and this brings us to Case 2 of Step
2 of 4.2. Then, as in the proof of 4.2, $j$ needs to be an odd number, in our
case lying in $[0,n)$. We once again see that
$S(f_{j},g_{s})=x_{n-2}S(f_{j},h)$. Following the proof of Case 2 of Step 2 of
4.2 (with $n=2s$ and $k=s$), we see that $x_{2k-2}$ (which now becomes
$x_{n-2}$) is not involved in any reduction except in the step where
$x_{2k-2}x_{2k-1}y_{2k}$ reduces $x_{n-2}x_{n-1}y_{0}$. In our case, however,
this is actually no reduction since $n=2k$. Therefore, performing the same
chain of reductions we will reduce $S(f_{j},h)$ to $0$. ∎

_5.2._ Let $<$ be the reverse lexicographic order on $T$ induced by
$x_{1}>\ldots>x_{n-1}>x_{0}>y_{1}>\ldots>y_{n-1}>y_{0}$. Then the polynomials

$f_{1},\ldots,f_{n-1},g_{1},\ldots,g_{\frac{n}{2}-1},h_{1},\ldots,h_{\frac{n}{2}-1}$

form a Gröbner basis for $L+HT$. Here
$f_{j}=\underline{x_{\frac{n}{2}+j}y_{j}}-x_{j}y_{j+1}$,

$g_{k}=\underline{y_{k}\prod_{i=0}^{k-1}x_{i}}-y_{0}\prod_{i=\frac{n}{2}}^{k-1+\frac{n}{2}}x_{i},$

and $h_{l}=\underline{y_{l}y_{l+\frac{n}{2}}}-y_{0}y_{\frac{n}{2}}$. As
before, $L=(f_{1},\ldots,f_{n-1},g_{1},\ldots,g_{\frac{n}{2}-1})$ and
$H=(h_{1},\ldots,h_{\frac{n}{2}-1})$.

###### Proof of 5.2..

The proof will be divided into the following steps.

Step 1: _The $S-$pairs $S(f_{j_{1}},f_{j_{2}})$ and $S(h_{l_{1}},h_{l_{2}})$
all reduce to $0$._ This is obvious since the leading terms of $f_{j_{1}}$ and
$f_{j_{2}}$ (resp. $h_{l_{1}}$ and $h_{l_{2}}$) are coprime for
$j_{1}\not=j_{2}$ (resp. $l_{1}\not=l_{2}$).

Step 2: _The $S-$pairs $S(g_{k_{1}},g_{k_{2}})$ all reduce to $0$._ Let $1\leq
k-s<k\leq\frac{n}{2}-1$. We would like to show that $S(g_{k},g_{k-s})$ reduces
to $0$.

$\displaystyle S(g_{k},g_{k-s})$
$\displaystyle=S\left(\underline{y_{k}\prod_{i=0}^{k-1}x_{i}}-y_{0}\prod_{i=\frac{n}{2}}^{k-1+\frac{n}{2}}x_{i},\quad\underline{y_{k-s}\prod_{i=0}^{k-s-1}x_{i}}-y_{0}\prod_{i=\frac{n}{2}}^{k-s-1+\frac{n}{2}}x_{i}\right)$
$\displaystyle=y_{0}y_{k}\prod_{i=k-s}^{k-1}x_{i}\prod_{i=\frac{n}{2}}^{k-s-1+\frac{n}{2}}x_{i}-y_{0}y_{k-s}\prod_{i=\frac{n}{2}}^{k-1+\frac{n}{2}}x_{i}$
$\displaystyle=y_{0}\prod_{i=\frac{n}{2}}^{k-s-1+\frac{n}{2}}x_{i}\left(y_{k}\prod_{i=k-s}^{k-1}x_{i}-\underline{y_{k-s}\prod_{i=k-s+\frac{n}{2}}^{k-1+\frac{n}{2}}x_{i}}\right).$

Therefore, reducing the leading term of the binomial in parentheses using
$f_{k-s},f_{k-s+1},\ldots f_{k-1}$, we obtain:

$\displaystyle
y_{k-s}\prod_{i=k-s+\frac{n}{2}}^{k-1+\frac{n}{2}}x_{i}=\left(y_{k-s}x_{k-s+\frac{n}{2}}\right)\prod_{i=k-s+1+\frac{n}{2}}^{k-1+\frac{n}{2}}x_{i}\xrightarrow{f_{k-s}}x_{k-s}y_{k-s+1}\prod_{i=k-s+1+\frac{n}{2}}^{k-1+\frac{n}{2}}x_{i}$
$\displaystyle=x_{k-s}\left(y_{k-s+1}x_{k-s+\frac{n}{2}+1}\right)\prod_{i=k-s+2+\frac{n}{2}}^{k-1+\frac{n}{2}}x_{i}\xrightarrow{f_{k-s+1}}\ldots\xrightarrow{f_{k-1}}x_{k-s}\cdots
x_{k-1}y_{k}=y_{k}\prod_{i=k-s}^{k-1}x_{i}.$

Alternatively, one can do one reduction step (by $f_{k-s}$) and notice that
the obtained monomial is $x_{k-s}$ times the monomial we are reducing, with
$s\rightarrow s-1$. The rest follows easily by induction. Therefore,
$S(g_{k},g_{k-s})$ reduces to $0$.

Step 3: _The $S-$pairs $S(f_{j},g_{k})$ all reduce to $0$._ Notice that the
leading terms of $f_{j}$ and $h_{l}$ are coprime except
$j=k\in\\{1,\ldots,\frac{n}{2}-1\\}$. In this case we have:

$\displaystyle S(f_{k},g_{k})$
$\displaystyle=S\left(\underline{x_{\frac{n}{2}+k}y_{k}}-x_{k}y_{k+1},\quad\underline{y_{k}\prod_{i=0}^{k-1}x_{i}}-y_{0}\prod_{i=\frac{n}{2}}^{k-1+\frac{n}{2}}x_{i}\right)$
$\displaystyle=\underline{y_{k+1}\prod_{i=0}^{k}x_{i}}-y_{0}\prod_{i=\frac{n}{2}}^{k+\frac{n}{2}}x_{i}.$

Note that this binomial equals $g_{k+1}$, unless $k=\frac{n}{2}-1$. If
$k=\frac{n}{2}-1$, we have:

$\displaystyle y_{\frac{n}{2}}\prod_{i=0}^{\frac{n}{2}-1}x_{i}$
$\displaystyle=x_{0}y_{\frac{n}{2}}\prod_{i=1}^{\frac{n}{2}-1}x_{i}\xrightarrow{f_{\frac{n}{2}}}x_{\frac{n}{2}}y_{\frac{n}{2}+1}\prod_{i=1}^{\frac{n}{2}-1}x_{i}=x_{\frac{n}{2}}(x_{1}y_{\frac{n}{2}+1})\prod_{i=2}^{\frac{n}{2}-1}x_{i}$
$\displaystyle\xrightarrow{f_{\frac{n}{2}+1}}x_{\frac{n}{2}}x_{\frac{n}{2}+1}y_{\frac{n}{2}+2}\prod_{i=2}^{\frac{n}{2}-1}x_{i}=x_{\frac{n}{2}}x_{\frac{n}{2}+1}(x_{2}y_{\frac{n}{2}+2})\prod_{i=3}^{\frac{n}{2}-1}x_{i}$
$\displaystyle\xrightarrow{f_{\frac{n}{2}+2}}\ldots\xrightarrow{f_{n-1}}x_{\frac{n}{2}}x_{\frac{n}{2}+1}\cdots
x_{n-1}y_{0}=y_{0}\prod_{i=\frac{n}{2}}^{n-1}x_{i},$

which is exactly what we need. Therefore, $S(f_{j},g_{k})$ all reduce to $0$.

Step 4: _The $S-$pairs $S(g_{k},h_{l})$ all reduce to $0$._ Notice that the
leading terms of $g_{k}$ and $h_{l}$ are coprime except $k=l$. In this case we
have:

$\displaystyle
S(g_{l},h_{l})=S\left(\underline{y_{l}\prod_{i=0}^{l-1}x_{i}}-y_{0}\prod_{i=\frac{n}{2}}^{l-1+\frac{n}{2}}x_{i},\quad\underline{y_{l}y_{l+\frac{n}{2}}}-y_{0}y_{\frac{n}{2}}\right)=y_{0}\left(\underline{y_{\frac{n}{2}}\prod_{i=0}^{l-1}x_{i}}-y_{l+\frac{n}{2}}\prod_{i=\frac{n}{2}}^{l-1+\frac{n}{2}}x_{i}\right).$

Note that the leading term of the binomial in parentheses is very similar to
the leading term of the $S$-pair in the previous Step 3 (the case
$k=\frac{n}{2}-1$). We do the exact same reduction of this monomial, but with
fewer steps. We use
$f_{\frac{n}{2}},f_{\frac{n}{2}+1},\ldots,f_{\frac{n}{2}+l-1}$ and from Step 3
we see that after these reductions we will obtain $x_{\frac{n}{2}}\cdots
x_{\frac{n}{2}+l-1}y_{\frac{n}{2}+l}=y_{l+\frac{n}{2}}\prod_{i=\frac{n}{2}}^{l-1+\frac{n}{2}}x_{i}$,
which is exactly what we need. We conclude that all $S(g_{k},h_{l})$ reduce to
$0$.

Step 5: _The $S-$pairs $S(f_{j},h_{l})$ all reduce to $0$._ Notice that the
leading terms of $f_{j}$ and $h_{l}$ are coprime except the case
$j=l\in\\{1,\ldots,\frac{n}{2}-1\\}$ and the case
$l\in\\{1,\ldots,\frac{n}{2}-1\\}$, $j=l+\frac{n}{2}$. In the former case we
have:

$\displaystyle S(f_{l},h_{l})$
$\displaystyle=S\left(\underline{x_{\frac{n}{2}+l}y_{l}}-x_{l}y_{l+1},\quad\underline{y_{l}y_{l+\frac{n}{2}}}-y_{0}y_{\frac{n}{2}}\right)=\underline{x_{l}y_{l+1}y_{l+\frac{n}{2}}}-x_{\frac{n}{2}+l}y_{0}y_{\frac{n}{2}}$
$\displaystyle\xrightarrow{f_{l+\frac{n}{2}}}x_{l+\frac{n}{2}}y_{l+1+\frac{n}{2}}y_{l+1}-x_{\frac{n}{2}+l}y_{0}y_{\frac{n}{2}}=x_{\frac{n}{2}+l}h_{l+1}.$

Note that the last equality makes sense unless $l=\frac{n}{2}-1$, in which
case $h_{l+1}$ is not defined. However, in this case the left hand side of
this equality equals $0$. Similarly, in the latter case
$S(f_{l+\frac{n}{2}},h_{l})$ reduces to $x_{l}h_{l+1}$ using $f_{l}$ (and the
same remark about $l=\frac{n}{2}-1$ applies). Therefore, $S(f_{j},h_{l})$ all
reduce to $0$.

∎

## References

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# Improving 2D-3D Dense Correspondences with Diffusion Models for 6D Object
Pose Estimation

Peter Hönig
Automation and Control Institute, TU Wien, Austria
<EMAIL_ADDRESS>
Stefan Thalhammer
Department of Industrial Engineering, UAS Technikum Vienna, Austria
<EMAIL_ADDRESS>
Markus Vincze
Automation and Control Institute, TU Wien, Austria
<EMAIL_ADDRESS>

###### Abstract

Estimating 2D-3D correspondences between RGB images and 3D space is a
fundamental problem in 6D object pose estimation. Recent pose estimators use
dense correspondence maps and Point-to-Point algorithms to estimate object
poses. The accuracy of pose estimation depends heavily on the quality of the
dense correspondence maps and their ability to withstand occlusion, clutter,
and challenging material properties. Currently, dense correspondence maps are
estimated using image-to-image translation models based on GANs, Autoencoders,
or direct regression models. However, recent advancements in image-to-image
translation have led to diffusion models being the superior choice when
evaluated on benchmarking datasets. In this study, we compare image-to-image
translation networks based on GANs and diffusion models for the downstream
task of 6D object pose estimation. Our results demonstrate that the diffusion-
based image-to-image translation model outperforms the GAN, revealing
potential for further improvements in 6D object pose estimation models.

##  1 Introduction

Accurate 6D object pose estimation is essential for a range of perception
tasks, such as autonomous driving, augmented reality, creating digital twins,
or robotic grasping. Numerous pose estimation methods rely on a combination of
RGB and depth [24, 1]. RGB-D sensors, which provide both color and depth data,
are not always available. Depth data is also prone to noise and other
distortions, often caused by reflections of shiny, metallic, and transparent
objects in the scene. To address this issue, pose estimation from RGB images
alone is considered. State-of-the-art methods rely on estimating 2D-3D dense
correspondences [15, 25, 26, 16, 27] between the RGB image and the 3D object
model. Altough these methods excel at inferring object poses with high
visibility, they still face significant challenges posed by clutter,
occlusion, image distortions, and shiny object surfaces [22].

So far, the estimation of 2D-3D correspondences for pose estimation has been
solved by image-to-image translation using direct regression [15, 25, 26],
combinations of Generative Adversarial Networks (GANs) and U-Net architectures
[16], or encoder-decoder Convolutional Neural Networks (CNNs) [27]. The
abovementioned methods estimate dense correspondence maps that contain
normalized XYZ object coordinates [26]. The 6D pose is then estimated by
solving the Point-n-Point (PnP) problem via RANSAC [13] or direct regression
[4, 3] and backpropagation. To estimate accurate poses via PnP, the quality of
the dense correspondence maps, which are estimated by the image-to-image
model, is critical.

Transformer models [7, 12] are superior to traditional CNNs for image-to-image
translation, when evaluated on benchmark tasks. More recently, diffusion
models [19] have been shown to outperform both transformer models and GANs.
However, the benchmarking tasks used to evaluate the performance of image-to-
image translation do not include any tasks related to geometric feature
extraction. Therefore the question arises as to whether the performance
increase of diffusion-based image-to-image translation is directly
transferable to geometric feature extraction tasks.

In this paper, we conduct a comparative study to investigate the potential of
diffusion models for image-to-image translation for estimating 2D-3D dense
correspondences for the downstream task of 6D object pose estimation.

We summarize our contributions as follows:

  * •

We show that diffusion models are superior to GANs for estimating 2D-3D dense
correspondences which leads to a higher accuracy of the subsequent 6D object
pose estimation downstream task.

  * •

Furthermore, our experimental findings demonstrate that diffusion models
outperform GANs in the context of object segmentation. In addition, we show
that diffusion models benefit more significantly from the application of
online data augmentations compared to GANs.

We compare two image-to-image translation networks, namely the GAN-based
Pix2Pix [11] and the Brownian-Bridge Diffusion Model (BBDM) [14] to estimate
dense correspondence maps from RGB images. To solve the subsequent PnP
problem, we use RANSAC E-PnP on the dense correspondences maps. We evaluate on
the challenging Linemod-Occluded (LMO) dataset [2], which contains clutter and
occlusion.

This paper is structured as follows: In Section 2 we give an overview of
current methods for 2D-3D dense correspondence estimation. Our method is
described in Section 3 and the experimental setup is explained in Section 4.
We present our results in Section 5 and conclude in Section 6.

##  2 Related Work

In this section, we summarize the current state-of-the-art regarding types of
2D-3D correspondences and machine learning architectures for 2D-3D dense
correspondences estimation for the downstream task of pose estimation.

In applications without available depth information, 6D object pose estimation
has to be solved from RGB images alone. To solve pose estimation solely from
RGB images, 2D-3D correspondences have to be established. Earlier works create
these 2D-3D correspondences by utilizing (sparse) keypoints [18, 17, 10].
However, more recent works use dense correspondences [27, 16, 15, 25] or
surface embeddings [20]. Dense correspondences are pixel-wise, as opposed to
keypoints, allowing for increased resilience against occlusion or reflections
[22]. uv-coordinates were shown to be effective in [15, 27], but were replaced
with normalized object coordinate spaces (NOCS) [26] recently [25, 16] for
further accuracy improvement.

The core principle of 2D-3D dense correspondences estimation can be defined as
an image-to-image translation task. Given an RGB image $I_{RGB}$, the
objective of the image-to-image translation function $\mathcal{F}$ is to
estimate the 2D-3D dense correspondences image $I_{DC}$, as defined by the
function $I_{DC}=\mathcal{F}(I_{RGB})$. The function $\mathcal{F}$ is
approximated, using a multitude of training pairs. In order to approximate
$\mathcal{F}$, various image-to-image model architectures are used. [27, 15,
25] use direct regression models with encoder-decoder structures. [16] uses a
U-Net architecture with adversarial training. The state of the art in image-
to-image translation has since progressed. VQGAN, a transformer model, has
been shown to outperform GANs and direct regression models [7] regarding
standard benchmarking tasks (e.g. style transfer). The very recently developed
diffusion models [14] show even higher accuracy compared to transformers on
the same benchmark datasets.

So far, research on 6D object pose estimation has focused on finding the ideal
intermediate representation, e.g. uv coordinates [27], NOCS [26] or binary
encoding [20]. Simultaneously, little attention has been paid to the
progression of the state of the art in image-to-image translation. Therefore,
we want to investigate novel image-to-image translation model architectures
and compare them to the established GAN architecture, while building upon
established forms of 2D-3D dense correspondences.

##  3 Method

Figure 1 illustrates the full pose estimation pipeline from RGB image to 6D
object pose. To retrieve a 2D location prior, the region of interest (ROI) is
cropped from the RGB image, with a 2D object detector. The ROI is the input to
the image-to-image translation model. The image-to-image translation model
learns to estimate the 2D-3D dense correspondences from the RGB crops. The 2D
object detector crops the object with a rectangular bounding box. Therefore,
the object is not entirely cropped and background pixels remain. The learning
objective of the image-to-image translation network is therefore twofold. The
networks main objective is to learn to estimate the 2D-3D dense
correspondences. Simultaneously, the network needs to learn implicitly hot to
segment the object from the background. The RANSAC + PnP step estimates the 6D
object pose from the dense correspondences map.

Figure 1: Illustration of 2D-3D Dense Correspondences Estimation using
Diffusion Models. An arbitrary object detector is used to estimate a region of
interest of an object. The diffusion model estimates the normalized object
coordinates map from the RGB input crop. The RANSAC + PnP step solves the
downstream task of 6D object pose estimation.

To generate the training data for the image-to-image translation task, the
object meshes are normalized, to fit into a dimensionless 1x1x1 cube. The
vertices of the object meshes are then colored with RGB values according to
the XYZ positions of the vertices in the normalized object coordinate space
[26]. The normalized and colored meshes are then rendered with the ground
truth translation, rotation, and camera intrinsics. Figure 2 illustrates
example location priors and dense correspondences map renderings in arbitrary
poses with the objects from the Linemod-Occluded dataset. Each object instance
in the training set is re-rendered as a dense correspondences map, resulting
in image pairs for image-to-image translation training.

Figure 2: Illustration of location priors (top row) and rendered dense
correspondences maps (bottom row). Ground truth translation, rotation and
camera intrinsics are used to render the dense correspondences maps as
identical image pair to the RGB location prior.

##  4 Experimental Setup

In this section we explain the image-to-image translation algorithms which are
compared. We also introduce the dataset we use for evaluation, as well as the
evaluation metrics

###  4.1 Image-to-Image Translation Algorithms

We compare two image-to-image translation methods, namely the GAN model
Pix2Pix and the diffusion model BBDM. The location priors and the RANSAC + PnP
step are identical for both methods, only the image-to-image translation
function $I_{DC}=\mathcal{F}(I_{RGB})$ is approximated differently. Both
models are trained under identical conditions. Firstly, models are trained
without any data augmentation, except for resizing the ROI crops to 128x128
pixels, which is the input and output size of both models. In a second
training run, both models are trained with identical data augmentation
parameters, shown in Table 1. For each run, both models are trained for 40
epochs each.

Table 1: Data augmentation. Probabilities, functions, and parameters, as introduced in [25] Probability | Function | Parameters
---|---|---
0.5 | Coarse Dropout | p=0.2, size_percent=0.05
0.5 | Gaussian Blur | 1.2*np.random.rand()
0.5 | Add | (-25, 25), per_channel=0.3
0.3 | Invert | 0.2, per_channel=True
0.5 | Multiply | (0.6, 1.4), per_channel=0.5
0.5 | Multiply | (0.6, 1.4)
0.5 | Linear Contrast | (0.5, 2.2), per_channel=0.3

###  4.2 Dataset

We evaluate the image-to-image translation models on the LMO dataset [2]. It
features 8 household objects sampled in randomized domains [23] and 50 000
synthetically rendered images [5]. These synthetically rendered images are
used for training only. For evaluation, 1214 real-world test images are used.

###  4.3 Location Priors

Two sets of pre-computed location priors are used for object cropping. We use
the YOLOx detection results from the benchmark on object pose estimation (BOP)
challenge 2023 [21] for evaluating the downstream task of pose estimation. To
evaluate object segmentation, ground truth location priors are used.

###  4.4 Evaluation Metrics

We evaluate the quality of the estimated 6D poses, as well as the the 2D-3D
dense correspondences maps and the object segmentation. The 6D object poses
are evaluated using the ADD(-S) score. ADD(-S) refers to the average distance
between model points $m$, for $k_{m}d\geq m$. The error threshold $k_{m}$ is
defined with 10%. The calculation of $m$ is shown in Equation 1.
$\overline{\mathbf{R}}$ and $\overline{\mathbf{T}}$ denote the ground truth
rotation and translation, whereas $\widehat{\mathbf{R}}$ and
$\widehat{\mathbf{T}}$ denote the estimated rotation and translation, while
$\mathbf{x}$ denotes a model point from the model $M$.

$m=\underset{\mathbf{x}\in
M}{\text{avg}}\begin{cases}\left\|(\overline{\mathbf{R}}\mathbf{x}+\overline{\mathbf{T}})-(\widehat{\mathbf{R}}\mathbf{x}+\widehat{\mathbf{T}})\right\|&\text{non
sym.}\\\ \underset{\mathbf{x}_{2}\in
M}{\text{min}}\left\|(\overline{\mathbf{R}}\mathbf{x}+\overline{\mathbf{T}})-(\widehat{\mathbf{R}}\mathbf{x}_{2}+\widehat{\mathbf{T}})\right\|&\text{sym.}\end{cases}$
(1)

To compare the pose estimation results with other methods we rely on the
average recall as calculated for the benchmark on object pose estimation (BOP)
challenge [8]. This AR score is the mean of the average recalls of visible
surface discrepancy (VSD) [8], maximum symmetry-aware surface distance (MSSD)
[6], and maximum symmetry-aware projection distance (MSPD) [9].

To disentangle the image-to-image translation task from the downstream task of
pose estimation, we evaluate the reconstruction loss of the 2D-3D dense
correspondences maps, using the mean squared error (MSE) metric. Equation 2
shows the calculation of mean squared error, where $N$ denotes the number of
pixels in the estimated and ground truth images $\widehat{I}_{i}$ and
$\overline{I}_{i}$.

$\text{MSE}=\frac{1}{N}\sum_{i=1}^{N}(\widehat{I}_{i}-\overline{I}_{i})^{2}$
(2)

The image-to-image task involves the implicit learning of object segmentation,
to seperate the object from the background. Therefore, we also evaluate the
intersection over union (IoU) between the ground truth mask $\overline{M}_{i}$
and estimated masks $\widehat{M}_{i}$, with the number of pixels $N$, as shown
in Equation 3. The mean IoU is a measure of how well the estimated mask and
the ground truth mask overlap.

$\text{IoU}=\frac{\sum_{i=1}^{N}\mathbf{1}_{(\widehat{M}_{i}\land\overline{M}_{i})}}{\sum_{i=1}^{N}\mathbf{1}_{(\widehat{M}_{i}\lor\overline{M}_{i})}}$
(3)

##  5 Results

In this section we show results of the comparative study between Pix2Pix and
BBDM regarding 6D object pose estimation accuracy, reconstruction loss and
object segmentation. We also compare with other 6D object pose estimation
methods, that use 2D-3D dense correspondences and RANSAC + PnP as well.

Table 2 shows the main results of the comparative study between Pix2Pix and
BBDM, by reporting the ADD(-S) score. Overall, the performance of BBDM is
superior to Pix2Pix, with and without augmentation. For the objects ape and
holepuncher, Pix2Pix shows better performance than BBDM, while BBDM is
superior for all remaining objects. The online augmentations influence Pix2Pix
and BBDM differently. While BBDM profits from online augmentation, the
accuracy of Pix2Pix declines. This can be attributed to the different learning
objectives of the diffusion model (BBDM) and the GAN (Pix2Pix).

Table 2: Main results on LMO. ADD(-S) scores for LMO with Pix2Pix and BBDM, without (wo) and with (w) augmentations and YOLOx location priors Object Name | Pix2Pix | BBDM
---|---|---
| wo aug | w aug | wo aug | w aug
ape | 13.10 | 10.90 | 5.10 | 8.00
can | 9.50 | 4.50 | 14.60 | 17.10
cat | 7.00 | 7.60 | 3.50 | 11.70
driller | 23.00 | 11.00 | 11.00 | 19.50
duck | 12.80 | 8.90 | 10.60 | 16.10
eggbox | 3.90 | 3.90 | 16.10 | 20.00
glue | 10.00 | 11.40 | 21.40 | 32.90
holepuncher | 11.00 | 22.00 | 17.50 | 17.50
mean | 11.49 | 10.10 | 12.39 | 17.51

Table 3 shows the results for the mean squared error, representing the
reconstruction loss, and the mean intersection over union scores, describing
the accuracy of the segmentation. BBDM is superior in terms of MSE as well as
mean IoU, indicating that the diffusion model performs reconstruction of the
2D-3D dense correspondences values and segmentation of object with higher
accuracy.

Table 3: Quality of the 2D-3D dense correspondences maps. MSE and mean IoU scores for LMO, without (wo) and with (w) augmentations and ground truth location priors MSE | Mean IoU
---|---
Pix2Pix | BBDM | Pix2Pix | BBDM
wo aug | w aug | wo aug | w aug | wo aug | w aug | wo aug | w aug
0.092 | 0.114 | 0.060 | 0.060 | 0.816 | 0.756 | 0.851 | 0.840

To illustrate the influence of the MSE and mean IoU from Table 3, Figure 3
shows selected examples from the LMO test set. 2D-3D dense correspondences are
estimated by BBDM and Pix2Pix. Absolute errors are calculated between
estimated and ground truth correspondences maps. The three channels from RGB
images are averaged to a single channel error map, shown in the top two rows
of Figure 3. The lower two rows show the influence of the dense
correspondences errors on the 6D object pose. The selected examples in Figure
3 demonstrate how the BBDM and Pix2Pix show errors along the object
boundaries. However, BBDM appears to draw the object boundaries more
accurately and generates smoother object surfaces. Pix2Pix shows more
inaccurate object surfaces and introduces artifacts. This is likely due to
convolution and de-convolution operations, which change the size of the
feature maps, in the GAN architecture of Pix2Pix. BBDMs diffusion principle
preserves feature map sizes throughout the network, maintaining sharpness of
estimated images.

Figure 3: Influence of reconstruction quality on estimated 6D pose. Selected
images from the LMO test data; comparison of pixel-wise error (top two rows),
illustrated with the jet colormap and the influence on the estimated 6D poses
(lower two rows); estimated bounding box in green, ground truth bounding box
in blue.

Table 4 shows the runtimes of the full 6D object pose estimation pipelines and
their subparts, with either Pix2Pix or BBDM for dense correspondences
estimation. Experiments are conducted on a AMD Ryzen 9 5900X CPU and a RTX
3090 GPU. Pix2Pix is $\approx$10 times faster than BBDM, with a mean runtime
of 0.005 seconds. The full pipeline with BBDM takes 0.192 seconds for a single
object instance.

Table 4: Comparison of runtimes. List of runtimes for each step of the 6D object pose estimation pipeline, including YOLOx detection, dense correspondences estimation and RANSAC + PnP step. Step | Pix2Pix | BBDM
---|---|---
Detection (YOLOx) | 0.081 | 0.081
2D-3D Dense Correspondences | 0.005 | 0.061
RANSAC + PnP | 0.050 | 0.050
Sum | 0.136 | 0.192

We compare our results with 6D object pose estimation models, with similar
architecture than ours, namely Pix2Pose [16] and DPOD [27]111Performance
scores of Pix2Pose and DPOD according to the leaderboard of the BOP challenge:
https://bop.felk.cvut.cz/leaderboards/. Both of these methods rely on 2D-3D
dense correspondences as well as a RANSAC + PnP step. As shown in Table 5, our
BBDM implementation outperforms Pix2Pose and DPOD.

Table 5: Comparison with the State of the Art. AR scores for LMO with Pix2Pix and BBDM, without (wo) and with (w) augmentations and YOLOx location priors; comparison with Pix2Pose [11] and DPOD [27] Pix2Pix | BBDM | Pix2Pose | DPOD
---|---|---|---
30.03 | 40.63 | 36.30 | 16.90

##  6 Conclusion

This paper presents a comparative study for 2D-3D dense correspondences
estimation with image-to-image translation networks. We compared the GAN model
Pix2Pix and the diffusion model BBDM under identical training conditions. Our
experiments revealed that the diffusion model is superior to the GAN regarding
the quality of the estimated 2D-3D dense correspondences and the accuracy of
the downstream task of 6D object pose estimation. The diffusion model segments
the object from the background more accurately with sharper boundaries and
estimates smooth object surfaces as compared to the GAN, which struggles with
pattern-like artifacts and object boundaries lacking sharpness. Further
studies will expand this work by implementing a diffusion-based object pose
estimation pipeline with improved PnP step and more resilience toward
symmetries.

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