# Fiber cones of rational normal scrolls are Cohen–Macaulay

Kuei-Nuan Lin and Yi-Huang Shen  Department of Mathematics, The Penn State
University, McKeesport, PA, 15132, USA<EMAIL_ADDRESS>CAS Wu Wen-Tsun Key
Laboratory of Mathematics, School of Mathematical Sciences, University of
Science and Technology of China, Hefei, Anhui, 230026, P.R. China
<EMAIL_ADDRESS>
###### Abstract.

In this short paper, we show that the fiber cones of rational normal scrolls
are Cohen–Macaulay. As an application, we compute their Castelnuovo–Mumford
regularities and $\bm{a}$-invariants, as well as the reduction number of the
defining ideals of the rational normal scrolls. We also characterize the
Gorensteinness of the fiber cone.

2020 Mathematics Subject Classification. Primary 13C40, 13A30, 13P10;
Secondary 14M12

Keyword: Rational normal scroll, Rees algebra, Fiber cone, Regularity,
Cohen–Macaulay, Singularity

##  1\. Introduction

In this short note, we show that the fiber cones of rational normal scrolls
are Cohen–Macaulay. Let $R$ be a standard graded polynomial ring over some
field ${\mathbb{K}}$ and $I$ be an ideal of $R$ minimally generated by some
forms $f_{1},\ldots,f_{s}$ of the same degree. Those forms define a rational
map $\phi$ whose image is a variety $X$. The bi-homogeneous coordinate ring of
the graph of $\phi$ is the Rees algebra of the ideal $I$, and the homogeneous
coordinate ring of the image is the fiber cone (special fiber ring) of this
ideal. Here, we are mostly interested in these two types of blow-up algebras
when the ideal $I$ describes the rational normal scroll.

Rational normal scrolls are typical determinantal varieties, central in the
study of algebraic varieties. The study of determinantal varieties has
attracted earnest attentions of algebraic geometers and commutative
algebraists, partly due to the beautiful structures involved and the
interesting applications to the applied mathematics and statistics; see, for
instance, [BV], [MR3836659], [arXiv:2005.02909] and [arXiv:2003.14232], to
name but a few.

The study of Cohen–Macaulay property of the Rees algebra and the related fiber
cone is an active research area. Most of the work is focusing on giving
certain assumptions on the ground ring and the ideal; see for example,
[MR1723128], [MR0563540], [MR1128652], and [MR2431031]. However, not much is
known about the Cohen–Macaulayness of these blow-up algebras of the
determinantal varieties. The first well-known case is probably the maximal
minors of a generic matrix where the fiber cone is the Grassmannian variety.
Hochster [MR314833] in 1973 showed that the Grassmannian is Cohen–Macaulay. In
1983, Eisenbud and Huneke [MR0696134] showed that the posets defining the Rees
algebra and fiber cone are wonderful, and obtained the Cohen–Macaulayness of
those algebras. Recently, the first author of this paper and her coauthors
proved the Cohen–Macaulayness of Rees algebras and its fibers cones of closed
determinantal facet ideals in [ALL] and sparse determinantal ideals in
[CDFGLPS].

All the work mentioned above came from generic matrices and their
specializations. Blow-up algebras of rational normal scrolls were first
studied by Conca, Herzog and Valla [Sagbi] in 1996. And they proved that the
Rees algebra and the fiber cone of a balanced rational normal scroll are
Cohen–Macaulay normal domains. The difficulty of determining the
Cohen–Macaulayness of those blow-up algebras is partly due to the lack of
necessary information regarding the defining ideals of those algebras.
Fortunately, Sammartano solved this implicitization problem in full generality
for the rational normal scrolls in [MR4068250] recently. And this opened the
door of understanding those algebras for us.

We need to be more precise now. In the following, let $n_{1},\dots,n_{d}$ be a
sequence of positive integers and $c=\sum_{i=1}^{d}n_{i}$. For each
$i=1,\dots,d$, let $\mathcal{C}_{i}\subseteq{\mathbb{P}}^{c+d-1}$ be a
rational normal curve of degree $n_{i}$ with complementary linear spans and
let $\varphi_{i}:{\mathbb{P}}^{1}\to\mathcal{C}_{i}$ be the corresponding
isomorphism. Whence, the related _rational normal scroll_ is simply

$\mathcal{S}_{n_{1},\dots,n_{d}}\coloneqq\bigcup_{p\in{\mathbb{P}}^{1}}\overline{\\{\varphi_{1}(p),\varphi_{2}(p),\dots,\varphi_{d}(p)\\}}\subseteq{\mathbb{P}}^{c+d-1}.$

It is well-known ([MR1416564]) that $\mathcal{S}_{n_{1},\dots,n_{d}}$ is
uniquely determined by the sequence $n_{1},\dots,n_{d}$ up to projective
equivalence. And in suitable coordinates, the ideal $I_{n_{1},\dots,n_{d}}$ of
$\mathcal{S}_{n_{1},\dots,n_{d}}$ is generated by the maximal minors of the
matrix

$\mathbf{X}\coloneqq\scalebox{0.9}{$\pNiceArray{cccc|cccc|c|cccc}x_{1,0}&x_{1,1}\cdots
x_{1,n_{1}-1}x_{2,0}x_{2,1}\cdots x_{2,n_{2}-1}\cdots x_{d,0}x_{d,1}\cdots
x_{d,n_{d}-1}\\\ x_{1,1}x_{1,2}\cdots x_{1,n_{1}}x_{2,1}x_{2,2}\cdots
x_{2,n_{2}}\cdots x_{d,1}x_{d,2}\cdots x_{d,n_{d}}.$}$ (1)

In the following, we will write $R={\mathbb{K}}[x_{i,j}\mid 0\leq j\leq
n_{i}-1]$ for the ground ring, and $I=I_{n_{1},\dots,n_{d}}=I_{2}(\mathbf{X})$
for this ideal. Recall that the _Rees algebra_ of $I$ is
$\mathcal{R}(I):=\oplus_{i\geq 0}I^{i}t^{t}\subseteq R[t]$, and the related
_fiber cone_ will be
$\mathcal{F}(I)=\mathcal{R}(I)\otimes_{R}\mathbb{K}\cong\mathbb{K}[I]\subseteq
R$, where $t$ is a new variable. As mentioned above, Sammartano gave the
implicit equations of the defining ideals of these blow-up algebras in
[MR4068250]. Meanwhile, in the _balanced case_ , i.e., when $|n_{i}-n_{j}|\leq
1$ for all $i,j$, Conca, Herzog, and Valla showed in [Sagbi] that both
$\mathcal{R}(I)$ and $\mathcal{F}(I)$ are Cohen–Macaulay normal domains.

In this paper, we will _drop_ the balanced assumption. We can still obtain the
Cohen–Macaulayness of the fiber cone $\mathcal{F}(I)$ in 2.1. By this, we
generalize the corresponding result of Conca, Herzog, and Valla. As a quick
application, we compute the Castelnuovo–Mumford regularity and
$\bm{a}$-invariant of $\mathcal{F}(I)$ in 3.4. And the reduction number of the
ideal $I$ is provided as well. As the second application, we characterize the
Gorensteinness of the fiber cone.

##  2\. Cohen–Macaulayness of the fiber cones of rational normal scrolls

This section is devoted to prove that the fiber cone associated with the
rational normal scroll is Cohen–Macaulay. To achieve that, we associate the
initial ideal of the defining ideal of the fiber cone with its Stanley–Reisner
complex $\Delta$. We will show that the ideal of the Alexander dual of this
complex has linear quotients. Consequently, the original simplicial complex is
shellable and Cohen–Macaulay. Hence, the fiber cone is Cohen–Macaulay as well.

To establish the expected linear-quotients property, we have to resort to the
binary-tree description of the facets of the simplicial complex $\Delta$ in
[MR4068250]. To be more accurate, instead of just using the lexicographic
order on the minimal monomial generators of the Alexander dual ideal, we need
to group these generators first using the leaves of the binary trees; see 2.4.
The subsequent proof is also quite involved, since we have to analyze the
binary trees in more detail and pin down those linear generators of the
successive colon ideals.

Recall that a rational normal scroll is uniquely determined by a sequence of
positive integers ${\bm{n}}=(n_{1},...,n_{d})$ with $n_{1}\leq\cdots\leq
n_{d}$ by [MR1416564]. Using the terminology stated in the above introduction,
we are ready to present the main result of this paper.

######  Theorem 2.1.

Let $I_{n_{1},\dots,n_{d}}$ be the defining ideal of the rational normal
scroll $\mathcal{S}_{n_{1},\dots,n_{d}}\subseteq{\mathbb{P}}^{c+d-1}$ for
$c=n_{1}+\cdots+n_{d}$. Then the fiber cone
$\mathcal{F}(I_{n_{1},\dots,n_{d}})$ is Cohen–Macaulay.

We need some preparations before laying out its proof. Let us start by
recalling the indispensable constructions in [MR4068250] for the fiber cone
$\mathcal{F}(I)$ where $I=I_{n_{1},\dots,n_{d}}$. First of all, a matrix
$\mathbf{M}=\mathbf{M}_{n_{1},\dots,n_{d}}$ was introduced in [MR4068250,
Section 2] as follows. One starts with the first column of the $i$-th
catalecticant block $\mathbf{X}_{i}$ of the original matrix
$\mathbf{X}=(\mathbf{X}_{1},\mathbf{X}_{2},\dots,\mathbf{X}_{d})$ for each $i$
increasingly in $i$, then the second column, and so on until one has used all
columns except the last one for each block; when a block runs out of columns
one simply skips it. The first $c-d$ columns of $\mathbf{M}$ form the
submatrix

$\setcounter{MaxMatrixCols}{12}\begin{pmatrix}x_{i(2),0}&x_{i(2)+1,0}&\cdots&x_{d,0}&x_{i(3),1}&x_{i(3)+1,1}&\cdots&x_{d,1}&x_{i(4),2}&\cdots&\cdots&x_{d,n_{d}-2}\\\
x_{i(2),1}&x_{i(2)+1,1}&\cdots&x_{d,0}&x_{i(3),2}&x_{i(3)+1,2}&\cdots&x_{d,1}&x_{i(4),3}&\cdots&\cdots&x_{d,n_{d}-1}\end{pmatrix},$

where the $i(l)$ denotes the least integer $i$ such that $n_{i}\geq l$. The
last $d$ columns of $\mathbf{M}$ are of the last column of the $i$-th block of
$\mathbf{X}$ for each $i$, but ordered decreasingly in $i$:

$\begin{pmatrix}x_{d,n_{d}-1}&x_{d-1,n_{d-1}-1}&\cdots&x_{2,n_{2}-1}&x_{1,n_{1}-1}\\\
x_{d,n_{d}}&x_{d-1,n_{d}-1}&\cdots&x_{2,n_{2}}&x_{1,n_{1}}\end{pmatrix}.$

Interested readers are invited to go through the examples in [MR4068250,
Example 2.1].

Let

$\pi:{\mathbb{K}}[T_{\alpha,\beta}\mid 1\leq\alpha<\beta\leq
c]\twoheadrightarrow\mathcal{F}(I)$

be the ring epimorphism determined by
$T_{\alpha,\beta}\mapsto\det(\mathbf{M}_{\alpha,\beta})$, where
$\mathbf{M}_{\alpha,\beta}$ is the $2\times 2$ submatrix of $\mathbf{M}$ using
the $\alpha$-th and $\beta$-th columns. Then
$P\coloneqq\operatorname{ker}(\pi)$ is called _the defining ideal_ of the
fiber cone $\mathcal{F}(I)$. Under suitable monomial ordering introduced in
[MR4068250], the initial ideal $\operatorname{in}(P)$ is squarefree and
quadratic. One can then consider its associated Stanley–Reisner complex
$\Delta$, which will be called the _initial complex_ of the fiber cone
$\mathcal{F}(I)$ following [MR4068250].

###### Proof of 2.1.

When $c<d+4$, the fiber cone is the coordinate ring of the Grassmann variety
$\mathbb{G}(1,c-1)\subset P^{\binom{c}{2}-1}$; see [MR4068250, Remark 3.15].
Thus, the fiber cone is Cohen–Macaulay by [MR314833].

Now, it remains to consider the case when $c\geq d+4$. Notice that
$\mathcal{F}(I)$ being Cohen–Macaulay is equivalent to saying that the
defining ideal $P$ has this property. By [MR2724673, Corollary 3.3.5], or more
strongly, by [arXiv1805.11923, Corollary 2.7], it suffices to show that
$\operatorname{in}(P)$ is Cohen–Macaulay. But this in turn is equivalent to
showing that $\operatorname{in}(P)^{\vee}$, the Alexander dual, has a linear
resolution by [MR2724673, Theorem 8.1.9]. Because of this, we will show in 2.6
that $\operatorname{in}(P)^{\vee}$ has linear quotients. This is sufficient
for our purpose by [MR2724673, Proposition 8.2.1]. ∎

In the following, we shall assume that $c\geq d+4$. Before we really start
proving the linear quotients property, we still need some preparations. Under
the natural identification $T_{\alpha,\beta}\leftrightarrow(\alpha,\beta)$,
the vertex set of the aforementioned simplicial complex $\Delta$ is

$V=\Set{(\alpha,\beta)}{1\leq\alpha<\beta\leq c}.$

Following [MR4068250], the vertices of $\Delta$ will also be described as open
intervals in the real line ${\mathbb{R}}$ with integral endpoints. And we will
use the familiar notions of length, intersection, and containment of
intervals.

As usual, if $n$ is a positive integer, $[n]$ will be the set
$\\{1,2,\dots,n\\}$. Then, with respect to the above matrix $\mathbf{M}$, for
each $\alpha\in[c-d-2]$ and each $i\in[d]$, let $\gamma_{\alpha,i}$ be the
least index $\gamma\geq\alpha+2$ such that the $\gamma$-th column of
$\mathbf{M}$ involves variables from the block $\mathbf{X}_{i}$. Then, there
exists some $\ell_{\alpha}$ with $2\leq\ell_{\alpha}\leq d+1$ such that

$\Set{\gamma_{\alpha,1},\dots,\gamma_{\alpha,d}}=\Set{\alpha+2,\dots,\alpha+\ell_{\alpha}}\cup\Set{c-d+\ell_{\alpha},\dots,c};$
(2)

c.f. [MR4068250, Definition 3.6]. The key observation that we shall apply is
the following result.

######  Lemma 2.2 ([MR4068250, Proposition 3.8]).

Suppose that $c\geq d+4$. A subset $F\subseteq V$ is a facet of $\Delta$ if
and only if the Hasse diagram $\mathcal{T}_{F}$ of the poset $(F,\subseteq)$
satisfies the following conditions:

  1. i

$\mathcal{T}_{F}$ is a rooted binary tree with the root $(1,c)$;

  2. ii

there exists some $\alpha\in[c-d-2]$ such that the leaves of $\mathcal{T}_{F}$
are the intervals

$\Set{(\beta,\beta+1)}{\beta\in\\{\alpha,\ldots,\alpha+\ell_{\alpha}\\}\cup\\{c-d+\ell_{\alpha}-1,\ldots,c-1\\}};$
(3)

  3. iii

if $\mathfrak{I}\in F$ is a node of $\mathcal{T}_{F}$ with only one child
$\mathfrak{I}_{1}$, then
$\operatorname{length}(\mathfrak{I})=\operatorname{length}(\mathfrak{I}_{1})+1$
and the unique unitary interval in $\mathfrak{I}\setminus\mathfrak{I}_{1}$
does not belong to $F$;

  4. iv

if $\mathfrak{I}\in F$ is a node of $\mathcal{T}_{F}$ with two children
$\mathfrak{I}_{1},\mathfrak{I}_{2}$, then
$\mathfrak{I}_{1}\cap\mathfrak{I}_{2}=\varnothing$ and
$\operatorname{length}(\mathfrak{I}_{1})+\operatorname{length}(\mathfrak{I}_{2})=\operatorname{length}(\mathfrak{I})$.

In this case we have $|F|=c+d$.

In the following, when we say the node $(\alpha^{\prime},\beta^{\prime})$ is a
_left child_ of the node $(\alpha^{\prime\prime},\beta^{\prime\prime})$ in the
binary tree $\mathcal{T}_{F}$, we mean it is a child and
$\alpha^{\prime}=\alpha^{\prime\prime}$. The notions of _right child_ , _left
sibling_ and _right sibling_ can be similarly defined.

We will call the set in (3) as the _leaves set_ of $F$ and denote it by
$\mathfrak{L}_{\alpha}$.

Notice that it is uniquely determined by its smallest point $\alpha\in[c-d-2]$
for the given matrix $\mathbf{M}$. The following easy fact will be needed
later.

######  Lemma 2.3.

For each $\alpha\in[c-d-3]$, the cardinalities
$|\mathfrak{L}_{\alpha}\setminus\mathfrak{L}_{\alpha+1}|=|\mathfrak{L}_{\alpha+1}\setminus\mathfrak{L}_{\alpha}|=1$.

###### Proof.

Recall that for each such $\alpha$ and each $i\in[d]$, the integer
$\gamma_{\alpha,i}$ is the least index $\gamma\geq\alpha+2$ such that the
$\gamma$-th column of $\mathbf{M}$ involves variables from the block
$\mathbf{X}_{i}$. Furthermore, the union in (2) is actually disjoint.

Therefore, $\gamma_{\alpha+1,i}\geq\gamma_{\alpha,i}$. And when
$\gamma_{\alpha,i}\geq(\alpha+1)+2$, then
$\gamma_{\alpha+1,i}=\gamma_{\alpha,i}$. Now, suppose that
$\alpha+2=\gamma_{\alpha,i_{0}}$. Then for $i\neq i_{0}$, we have
$\gamma_{\alpha,i}=\gamma_{\alpha+1,i}$. Whence,
$\mathfrak{L}_{\alpha+1}\setminus\mathfrak{L}_{\alpha}=\\{(\gamma_{\alpha+1,i_{0}},\gamma_{\alpha+1,i_{0}}+1)\\}$
and
$\mathfrak{L}_{\alpha}\setminus\mathfrak{L}_{\alpha+1}=\\{(\alpha,\alpha+1)\\}$.
∎

It is time to introduce the expected linear ordering. Let
$S={\mathbb{K}}[T_{v}\mid v\in V]$ be the base ring. Endow it with the
lexicographic order $>_{{\operatorname{lex}}}$ with respect to the linear
order on the variables $T_{v}$’s such that

$T_{v_{1}}>T_{v_{2}}\quad\Leftrightarrow\quad\text{the leftmost nonzero
component of $v_{1}-v_{2}$ is negative};$

here, by abuse of notation, we also think of $V$ as a set of integral points
on ${\mathbb{Z}}_{+}^{2}$. Recall that the facet $F$ of the initial complex
$\Delta$ is bijectively related to the monomial

$\widehat{{\bm{T}}^{F}}\coloneqq\prod_{v\in V\setminus F}T_{v}$

in the minimal monomial generating set $G((\operatorname{in}(P))^{\vee})$ by
[MR2724673, Lemma 1.5.3]. Therefore, to impose a linear order on
$G((\operatorname{in}(P))^{\vee})$ amounts to giving a linear order of the
facets. For later reference, facets with the common leaves set given by (3)
will be grouped into the family _$\mathcal{G}_{\alpha}$_.

######  Notation 2.4.

The expected linear order on the facets, denoted by $\succ$, will be as
follows.

  1. a

If two facets $F\in\mathcal{G}_{\alpha}$ and
$F^{\prime}\in\mathcal{G}_{\alpha^{\prime}}$ respectively with
$\alpha>\alpha^{\prime}$, then $F\succ F^{\prime}$.

  2. b

Next, consider the facets $F$ and $F^{\prime}$ within the same group
$\mathcal{G}_{\alpha}$. Then $F\succ F^{\prime}$ if and only if
$\widehat{{\bm{T}}^{F}}>_{{\operatorname{lex}}}\widehat{{\bm{T}}^{F^{\prime}}}$.

######  Example 2.5.

The first facet of the group $\mathcal{G}_{\alpha}$ with respect to our
ordering $\succ$ is the one with vertices

$\displaystyle(1,c),(2,c),\dots,(c-2,c)$

in addition to the leaves given in (3).

For example, let ${\bm{n}}=(2,2,4,4)$. Then the matrix

$\mathbf{M}=\setcounter{MaxMatrixCols}{12}\begin{pmatrix}x_{1,0}&x_{2,0}&x_{3,0}&x_{4,0}&x_{3,1}&x_{4,1}&x_{3,2}&x_{4,2}&x_{4,3}&x_{3,3}&x_{2,1}&x_{1,1}\\\
x_{1,1}&x_{2,1}&x_{3,1}&x_{4,1}&x_{3,2}&x_{4,2}&x_{3,3}&x_{4,3}&x_{4,4}&x_{3,4}&x_{2,2}&x_{1,2}\end{pmatrix}.$

If we choose $\alpha=2$, then the leaves set is

$\Set{(2,3),(3,4),(4,5),(5,6),(10,11),(11,12)};$

see also [MR4068250, Example 3.9]. The collection of open intervals in the
Figure 1 is the first facet with respect to the given $\alpha=2$ in the
corresponding complex $\Delta$.

123456789101112 Figure 1. The first facet in $\mathcal{G}_{\alpha}$ when
$\alpha=2$ and ${\bm{n}}=(2,2,4,4)$

And the Hasse diagram of the poset as considered in 2.2 is pictured in Figure
2.

(1,12)(2,3)(2,12)(3,4)(11,12)(3,12)(4,5)(4,12)(5,6)(5,12)(6,12)(7,12)(8,12)(9,12)(10,11)(10,12)

Figure 2. Binary tree of the first facet in $\mathcal{G}_{\alpha}$ when
$\alpha=2$ and ${\bm{n}}=(2,2,4,4)$

Now, we are ready to show that the Alexander dual ideal
$(\operatorname{in}(P))^{\vee}$ has linear quotients.

######  Proposition 2.6.

When $c\geq d+4$, the ideal $(\operatorname{in}(P))^{\vee}$ has linear
quotients with respect to the given ordering $\succ$.

###### Proof.

Let $F$ be a facet of $\Delta$ and suppose that $F\in\mathcal{G}_{\alpha}$. We
want to show that the minimal monomial generating set _$G_{F}$_ $\coloneqq
G(I_{F})$ of the colon ideal

$I_{F}\coloneqq\braket{\widehat{{\bm{T}}^{F^{\prime}}}\mid F^{\prime}\succ
F}:\widehat{{\bm{T}}^{F}}$

is linear. Notice that

$\braket{\widehat{{\bm{T}}^{F^{\prime}}}}:\widehat{{\bm{T}}^{F}}={\bm{T}}^{F\setminus
F^{\prime}}\coloneqq\prod_{v\in F\setminus F^{\prime}}T_{v}.$

Suppose that $F$ is the first facet within $\mathcal{G}_{\alpha}$ with respect
to the designated linear order $\succ$. If $\alpha=c-d-2$, then $F$ is
actually the first facet of $\Delta$ and there is nothing to show in this
case. Otherwise, $\alpha<c-d-2$. Let $F^{\prime}$ be the first facet of
$\mathcal{G}_{\alpha+1}$. It is clear that $F^{\prime}\succ F$ and

$F\setminus
F^{\prime}=\mathfrak{L}_{\alpha}\setminus\mathfrak{L}_{\alpha+1}=\\{({\alpha},{\alpha}+1)\\}$

by 2.3 and 2.5. Notice that for any facet $F^{\prime\prime}$ preceding $F$,
$({\alpha},{\alpha}+1)\in F\setminus F^{\prime\prime}$. Whence, the colon
ideal $I_{F}$ is principal and linear.

Suppose instead that $F$ is not the first facet in $\mathcal{G}_{\alpha}$.
This situation is more involved. We will break the long proof into three
parts, according to the following plans.

  1. (1)

Firstly, we describe the expected linear generating set $G_{F}$.

  2. (2)

Next, we show that the variables described above really show up as some colons
of monomials, and they are the only variables possessing this property.

  3. (3)

Finally, we show that we don’t have minimal monomial generators in $G_{F}$ of
higher degrees.

After showing the above three parts, it is clear then that the ideal
$(\operatorname{in}(P))^{\vee}$ has linear quotients. At this point, one may
suggest only proving the last step. But to successfully give the proof, we
have to clear higher-degree generators by the linear ones.

###  2.1. Linear generators of the colon ideal $I_{F}$

We separate the vertices in $F$ into collections for each fixed facet
$F\in\mathcal{G}_{\alpha}$. For $b\in[c-1]$, let

$F_{b}\coloneqq\Set{v\in F\mid v=(b,*)}.$ (4)

If the cardinality _$k_{b}$_ $\coloneqq|F_{b}|\geq 1$, we may assume that

$F_{b}=\Set{(b,i_{b,1}),(b,i_{b,2}),\dots,(b,i_{b,k_{b}})}$

with $i_{b,1}<i_{b,2}<\cdots<i_{b,k_{b}}$. It is clear that $(b,i_{b,j+1})$ is
the parent node of $(b,i_{b,j})$ when $1\leq j<k_{b}$. Now, assume that
$k_{b}\geq 1$, and we describe whether $T_{b,i_{b,j}}\in G_{F}$ for
$j\in[k_{b}]$.

  1. a

If $j=k_{b}$, then $T_{b,i_{b,k_{b}}}\notin G_{F}$.

  2. b

Suppose that $2\leq j<k_{b}$. If

    1. i

the node at $(b,i_{b,j})$ has a right sibling in $\mathcal{T}_{F}$, or

    2. ii

the node at $(b,i_{b,j})$ has two children in $\mathcal{T}_{F}$,

then $T_{b,i_{b,j}}\in G_{F}$. Otherwise, $T_{b,i_{b,j}}\notin G_{F}$.

  3. c

Suppose that $j=1<k_{b}$.

    1. i

Suppose that $(b,i_{b,1})$ is a leaf in $\mathcal{T}_{F}$, namely that
$i_{b,1}=b+1$.

      1. 1

If $b\neq\alpha$, then $T_{b,i_{b,1}}\notin G_{F}$.

      2. 2

Suppose that $b=\alpha$, i.e., $(b,i_{b,1})$ is the leftmost leaf in
$\mathcal{T}_{F}$.

         * •

If $b=c-d-2$, then $\mathcal{G}_{\alpha}$ is the first group to be considered.
Whence, we expect $T_{b,i_{b,1}}\notin G_{F}$.

         * •

Otherwise, we expect $T_{b,i_{b,1}}=T_{\alpha,\alpha+1}\in G_{F}$ and
$\alpha\neq c-d+2$.

    2. ii

Otherwise, $i_{b,1}>b+1$. Whence, we always have $T_{b,i_{b,1}}\in G_{F}$.

For later reference, we denote the set of the expected linear generators of
$F$ with respect to $b$ described above as $LG_{F,b}$. Then in the next two
subsections, we will see that the minimal generating set $G_{F}$ is precisely
the set

$LG_{F}\coloneqq\bigcup_{b\in[c-1]}LG_{F,b}.$

######  Example 2.7.

Consider the case when ${\bm{n}}=(n_{1},n_{2},n_{3})=(2,4,5)$. It is not
difficult to see that

$\displaystyle F=\\{(1,2),\,(1,3),\,$
$\displaystyle(1,4),\,(1,6),\,(1,7),\,(1,8),\,(1,9),\,(1,11),\,$
$\displaystyle(2,3),\,(3,4),\,(4,5),\,(4,6),\,(9,11),\,(10,11)\\}$

is a facet of the associated complex $\Delta$. Consider $b=1$. The subset
$F_{b}$ consists of the vertices written on the first line. And our criterion
above says that the corresponding linear generators for $b=1$ are

$LG_{F,1}=\\{T_{1,2},T_{1,3},T_{1,4},T_{1,6},T_{1,9}\\}\subseteq LG_{F}.$

Indeed, it is not difficult to see that equality holds in this special case.

###  2.2. Why the vertex of $F$ shows or not shows up in $G_{F}$?

Before we proceed, notice first the following consequences of 2.2.

######  Observations 2.8.

Let $F$ be a facet of the aforementioned initial complex $\Delta$. Recall that
it is convenient to consider the vertex $(\alpha,\beta)$ in $\Delta$ as an
open interval in the real line ${\mathbb{R}}$ with integral endpoints.

  1. i

Vertices of $F$ are _non-crossing_ , i.e., if we consider the vertices
$\mathfrak{I}_{1},\mathfrak{I}_{2}$ of $F$ as open intervals, then
$\mathfrak{I}_{1}\cap\mathfrak{I}_{2}=\varnothing$, or
$\mathfrak{I}_{1}\subseteq\mathfrak{I}_{2}$, or
$\mathfrak{I}_{2}\subseteq\mathfrak{I}_{1}$.

  2. ii

If $(\alpha,\beta)\in F$ is not the root and has no sibling, then its parent
node is either $(\alpha-1,\beta)$ or $(\alpha,\beta+1)$. Correspondingly,
$(\alpha-1,\alpha)$ or $(\beta,\beta+1)$ is not a leaf of $\mathcal{T}_{F}$.

  3. iii

If as an open interval, $(\alpha,\beta)$ contains no leaf of
$\mathcal{T}_{F}$, then the vertex $(\alpha,\beta)$ does not belong to $F$.

  4. iv

If both $(\alpha,\beta)$ and $(\alpha,\beta^{\prime})$ belong to $F$ with
$\beta<\beta^{\prime}$ such that $(\beta,\beta^{\prime})$ contains no leaf of
$\mathcal{T}_{F}$, then $(\alpha,\beta+j)\in F$ for all $j$ with
$0<j<\beta^{\prime}-\beta$. To see this, notice that for each such fixed $j$,
$F$ must have at least one vertex of the form $(\beta+j,\gamma)$ or
$(\alpha^{\prime},\beta+j)$. In the first subcase, $\gamma\leq\beta^{\prime}$
is impossible by the previous observation. But when $\gamma>\beta^{\prime}$,
this is also impossible, since the two open intervals $(\beta+j,\gamma)$ and
$(\alpha,\beta^{\prime})$ cross. As for the second subcase, by the previous
observation, it is clear that $\alpha^{\prime}<\beta$. If
$\alpha<\alpha^{\prime}<\beta$, then the two open intervals
$(\alpha^{\prime},\beta+j)$ and $(\alpha,\beta)$ cross. And if
$\alpha^{\prime}<\alpha$, then the two open intervals
$(\alpha^{\prime},\beta+j)$ and $(\alpha,\beta^{\prime})$ cross. Therefore,
$\alpha^{\prime}=\alpha$ and $(\alpha,\beta+j)$ is a vertex of $F$, as
expected.

  5. v

Suppose that $(\gamma,\gamma+1)\in V$ is not a leaf of $\mathcal{T}_{F}$.
Consider the smallest interval (node) in $\mathcal{T}_{F}$ that contains
$(\gamma,\gamma+1)$. Then by 2.2 and the minimality, this node is either
$(\gamma^{\prime},\gamma+1)$ or $(\gamma,\gamma^{\prime})$ for some
$\gamma^{\prime}$. And correspondingly it has a unique child node
$(\gamma^{\prime},\gamma)$ or $(\gamma+1,\gamma^{\prime})$.

Another fact is that the variable $T_{v}\in G_{F}$ if and only if $F\setminus
F^{\prime}=\\{v\\}$ for some $F^{\prime}\succ F$. And for this pair $F$ and
$F^{\prime}$, if $F\setminus F^{\prime}=\\{(i,j)\\}$ while
$F^{\prime}\setminus F=\\{(i^{\prime},j^{\prime})\\}$, then

either $i^{\prime}>i$ or $i^{\prime}=i$ and $j^{\prime}>j$.

Now, we are ready to justify the linear generating set $LG_{F}$ presented in
the previous subsection 2.1. Correspondingly, we also have three cases.

  1. a

If $j=k_{b}$, we show that $T_{b,i_{b,k_{b}}}\notin G_{F}$. We don’t have to
worry about the case when $b=1$, since $(1,c)$ is the common root. Suppose now
for contradiction that there is some $F^{\prime}$ with $F^{\prime}\succ F$ and
$F\setminus F^{\prime}=\\{(b,i_{b,k_{b}})\\}$. Thus, we have two subcases.

Suppose first that $F\in\mathcal{G}_{\alpha}$ and
$F^{\prime}\in\mathcal{G}_{\alpha^{\prime}}$ with $\alpha<\alpha^{\prime}$.
Since necessarily
$(\alpha,\alpha+1)\in\mathfrak{L}_{\alpha}\setminus\mathfrak{L}_{\alpha^{\prime}}\subseteq
F\setminus F^{\prime}$ while $|F\setminus F^{\prime}|=1$, it follows that
$b=\alpha$ and $i_{b,k_{b}}=\alpha+1$. Since the leftmost leaf of
$\mathcal{T}_{F}$ is $(\alpha,\alpha+1)$ and $|F_{\alpha}|=1$, $F$ must
contain the vertex $(\alpha-1,\alpha+1)$ by item ii of 2.8. Hence,
$F^{\prime}$ must also contain this vertex. But the leftmost leaf of
$\mathcal{T}_{F^{\prime}}$ is $(\alpha^{\prime},\alpha^{\prime}+1)$ with
$\alpha^{\prime}>\alpha$, contradicting the item iii of 2.8.

Suppose now instead that $F$ and $F^{\prime}$ have the same leaves set, i.e.,
$\alpha=\alpha^{\prime}$. Then $(b,i_{b,k_{b}})$ is not a leaf of
$\mathcal{T}_{F}$ and must have some child. The parent node of
$(b,i_{b,k_{b}})$ in $\mathcal{T}_{F}$ must have the form
$(b^{\prime},i_{b,k_{b}})$ for some $b^{\prime}<b$. Notice that the unique
vertex in $F^{\prime}\setminus F$ that the vertex $(b,i_{b,k_{b}})$ is
switched to cannot take the form $(b,b^{\prime\prime})$ with
$b^{\prime\prime}>i_{b,k_{b}}$, since the interval $(b,b^{\prime\prime})$ will
cross the interval $(b^{\prime},i_{b,k_{b}})$. Meanwhile, this unique vertex
cannot take the form $(b^{\prime\prime},b^{\prime\prime\prime})$ with
$b^{\prime\prime}>b$, since $(b^{\prime},i_{b,k_{b}})$ will not have a legal
branch in $\mathcal{T}_{F^{\prime}}$; see iii and iv of 2.2. Therefore, we
have a contradiction to the fact stated at the end of 2.8.

Thus, $T_{b,i_{b,k_{b}}}\notin G_{F}$.

  2. b

Suppose that $2\leq j<k_{b}$.

    1. i

Suppose that the node $(b,i_{b,j})$ has a sibling in $\mathcal{T}_{F}$. It is
clear that the parent node of $(b,i_{b,j})$ is $(b,i_{b,j+1})$ and
$(b,i_{b,j})$ is the left child in the binary tree. Whence, the sibling of
$(b,i_{b,j})$ is $(i_{b,j},i_{b,j+1})$. Now, let

$F^{\prime}\coloneqq
F\cup\\{(i_{b,j-1},i_{b,j+1})\\}\setminus\\{(b,i_{b,j})\\}.$

Obviously $F^{\prime}\succ F$. And it is easy to verify that $F^{\prime}$ is a
legal facet and the node $(b,i_{b,j+1})$ has two children $(b,i_{b,j-1})$ and
$(i_{b,j-1},i_{b,j+1})$ in $\mathcal{T}_{F^{\prime}}$. This implies that
$T_{b,i_{b,j}}\in G_{F}$.

    2. ii

Suppose that the node $(b,i_{b,j})$ has two children in $\mathcal{T}_{F}$.
They have to be $(b,i_{b,j-1})$ and $(i_{b,j-1},i_{b,j})$. We similarly define

$F^{\prime}\coloneqq
F\cup\\{(i_{b,j-1},i_{b,j+1})\\}\setminus\\{(b,i_{b,j})\\},$

and see that $T_{b,i_{b,j}}\in G_{F}$.

    3. iii

Suppose that the node $(b,i_{b,j})$ has only one child and has no right
sibling. Whence, $i_{b,j\pm 1}=i_{b,j}\pm 1$ respectively and
$(i_{b,j-1},i_{b,j+1})$ contains no leaf in $\mathcal{T}_{F}$ by 2.8 ii.
Suppose for contradiction that we have some facet $F^{\prime}$ such that
$F^{\prime}\succ F$ and $F\setminus F^{\prime}=\\{(b,i_{b,j})\\}$. Since
$(b,i_{b,j})$ is not a leaf, $F$ and $F^{\prime}$ have identical leaves set.
Therefore $(i_{b,j-1},i_{b,j+1})$ contains no leaf in
$\mathcal{T}_{F^{\prime}}$ as well. The node $(b,i_{b,j-1})$ must be a
grandchildren of $(b,i_{b,j+1})$ in $\mathcal{T}_{F^{\prime}}$ by 2.8 iv and
2.2 iii and iv. Meanwhile, $(b,i_{b,j})$ is the only possible option to
connect $(b,i_{b,j-1})$ with $(b,i_{b,j+1})$. This makes $F=F^{\prime}$, a
contradiction. Therefore, $T_{b,i_{b,j}}\notin G_{F}$.

  3. c

Suppose that $j=1<k_{b}$.

    1. i

We first consider the case when $(b,i_{b,1})$ is a leaf in $\mathcal{T}_{F}$,
namely that $i_{b,1}=b+1$. Suppose that $T_{b,i_{b,1}}\in G_{T}$. Then we have
$F\in\mathcal{G}_{\alpha}$ and $F^{\prime}\in\mathcal{G}_{\alpha^{\prime}}$
for some $\alpha<\alpha^{\prime}$ such that $F\setminus
F^{\prime}=\\{(b,b+1)\\}$. It is clear then that $\alpha<\alpha^{\prime}\leq
c-d-2$. And again, since
$(\alpha,\alpha+1)\in\mathfrak{L}_{\alpha}\setminus\mathfrak{L}_{\alpha^{\prime}}\subseteq
F\setminus F^{\prime}$ while $|F\setminus F^{\prime}|=1$, it follows that
$b=\alpha$.

Conversely, suppose that $b=\alpha<c-d-2$ with $F\in\mathcal{G}_{\alpha}$.
Notice that
$\mathfrak{L}_{\alpha+1}\setminus\mathfrak{L}_{\alpha}=\\{(\beta,\beta+1)\\}$
for some $\beta$ by 2.3, and this $(\beta,\beta+1)\notin F$. Meanwhile, since
the invariant $\ell_{\alpha}$ defined before (2) satisfies $\ell_{\alpha}\geq
2$, both $(\alpha,\alpha+1)$ and $(\alpha+1,\alpha+2)$ are leaves of
$\mathcal{T}_{F}$. Thus, as $k_{\alpha}\geq 2$, the node $(\alpha,\alpha+1)$
has a right sibling $(\alpha+1,i_{\alpha,2})$ by 2.2 iii. It is clear that any
vertex in $\mathcal{T}_{F}$ strictly containing $(\alpha,\alpha+1)$ will
contain its parent node $(\alpha,i_{\alpha,2})$, and consequently will contain
$(\alpha+1,i_{\alpha,2})$. Now, it is not difficult to verify that

$F^{\prime}\coloneqq
F\cup\\{(\beta,\beta+1)\\}\setminus\\{(\alpha,\alpha+1)\\}\in\mathcal{G}_{\alpha+1}$

is a well-defined facet, and therefore, $T_{\alpha,\alpha+1}\in G_{F}$.

    2. ii

Next, we assume that $(b,i_{b,1})$ is not a leaf in $\mathcal{T}_{F}$, namely
that $i_{b,1}>b+1$. Since $j=1$, $(b,b+1)$ is not a leaf and $(b,i_{b,1})$ has
only the right child $(b+1,i_{b,1})$ in $\mathcal{T}_{F}$. We have two
subcases.

       * •

Suppose that $(b,i_{b,1})$ has a right sibling in $\mathcal{T}_{F}$. It has to
be $(i_{b,1},i_{b,2})$.

       * •

Suppose that $(b,i_{b,1})$ does not have a right sibling in $\mathcal{T}_{F}$.
Whence, $i_{b,2}=i_{b,1}+1$ and $(i_{b,1},i_{b,1}+1)$ is not a leaf of
$\mathcal{T}_{F}$ by 2.8 ii.

In either subcase, let

$F^{\prime}\coloneqq F\cup\\{(b+1,i_{b,2})\\}\setminus\\{(b,i_{b,1})\\},$

and we can see as above that $T_{b,i_{b,1}}\in G_{F}$. And this completes our
argument in this subsection.

###  2.3. No minimal monomial generator of higher degrees

In this subsection, we finish the proof by showing that $G_{F}=LG_{F}$.

Suppose for contradiction that $F\in\mathcal{G}_{\alpha}$ and
$F^{\prime}\in\mathcal{G}_{\alpha^{\prime}}$ with $F^{\prime}\succ F$ such
that $\braket{\widehat{{\bm{T}}^{F^{\prime}}}}:\widehat{{\bm{T}}^{F}}$
provides a minimal monomial generator of higher degree in $G_{F}\setminus
LG_{F}$. Whence, none of the linear generators in $LG_{F}$ will ever divide
the principal generator of
$\braket{\widehat{{\bm{T}}^{F^{\prime}}}}:\widehat{{\bm{T}}^{F}}$. If we apply
the identification $T_{i,j}\leftrightarrow(i,j)\in\Delta$, then this is just

$LG_{F}\cap(F\setminus F^{\prime})=\varnothing$, or more simply,
$LG_{F}\subseteq F\cap F^{\prime}$. (5)

If $\alpha\neq\alpha^{\prime}$, let $a\leq\alpha$ be the smallest such that
$(a,\alpha+1)\in F$. We claim that $(a,\alpha+1)\in LG_{F}$. It is clear when
$a=\alpha$ since $\alpha<\alpha^{\prime}\leq c-d-2$. Thus, we will assume
$a<\alpha$. The parent node of $(a,\alpha+1)$ has the form either
$(a^{\prime},\alpha+1)$ for some $a^{\prime}<a$, or
$(a,\alpha^{\prime\prime})$ for some $\alpha^{\prime\prime}>\alpha+1$. The
first case contradicts the minimality of $a$. Thus, we have the second case
and consequently $|F_{a}|\geq 2$.

Whence, $(a,\alpha+1)\in LG_{F}$, still establishing the claim. Meanwhile,
since $\alpha^{\prime}>\alpha$, $(a,\alpha+1)$ does not contain any leaf of
$\mathcal{T}_{F^{\prime}}$. This implies that $(a,\alpha+1)\in F\setminus
F^{\prime}$, contradicting the assumption in (5).

Therefore, we will assume in the following that $\alpha=\alpha^{\prime}$,
i.e., $\mathcal{T}_{F}$ and $\mathcal{T}_{F^{\prime}}$ have a common leaf set.
Now, let $b\in[c-1]$ be the smallest such that $F_{b}\neq F_{b}^{\prime}$.
Here, $F_{b}^{\prime}$ is a subset of $F^{\prime}$, just as $F_{b}$ defined
for $F$ in (4).

  1. a

Suppose that $k_{b}=1$. By the minimality of $b$ and the fact that
$\widehat{{\bm{T}}^{F^{\prime}}}>_{{\operatorname{lex}}}\widehat{{\bm{T}}^{F}}$,
we have $(b,j)\notin F_{b}^{\prime}$ for all $j$ with $b+1\leq j\leq i_{b,1}$.
Therefore, either $(b,i^{\prime})\in F_{b}^{\prime}$ for some
$i^{\prime}>i_{b,1}$ or $F_{b}^{\prime}=\varnothing$ .

In the first subcase, notice that the parent node of $(b,i_{b,1})$ in
$\mathcal{T}_{F}$ must take the form $(b^{\prime},i_{b,1})$ for some
$b^{\prime}<b$. By the minimality of $b$, $(b^{\prime},i_{b,1})\in
F^{\prime}$. But since $(b,i^{\prime})$ will cross $(b^{\prime},i_{b,1})$ when
considered as open intervals, this is impossible.

In the remaining subcase, one can conclude that $F^{\prime}$ contains some
vertices of the form $(b^{\prime},b+1)$ and $(b^{\prime},b)$ such that
$(b,b+1)$ is not a leaf of $\mathcal{T}_{F^{\prime}}$, by 2.8 v. Since
$b^{\prime}<b$, we also have $(b^{\prime},b+1),(b^{\prime},b)\in F$ by the
minimality of $b$. Meanwhile, as $\mathcal{T}_{F}$ and
$\mathcal{T}_{F^{\prime}}$ have a common leaf set, $(b,b+1)$ is not a leaf of
$\mathcal{T}_{F}$ as well. Whence, $i_{b,1}>b+1$ and we have two crossing open
intervals $(b^{\prime},b+1)$ and $(b,i_{b,1})$ for $F$, which is another
contradiction.

  2. b

When $k_{b}=2$, we observe first that $(b,i_{b,1})\in F\cap F^{\prime}$. To
see this, notice that if $(b,i_{b,1})\in F\setminus F^{\prime}$, then
$(b,i_{b,1})$ is not a common leaf. Whence, $(b,i_{b,1})\in LG_{F}$ by our
description in subsection 2.1. Now, we have $(b,i_{b,1})\in
LG_{F}\cap(F\setminus F^{\prime})$, contradicting the assumption in (5).

Therefore, by the minimality of $b$ and the fact that
$\widehat{{\bm{T}}^{F^{\prime}}}>_{{\operatorname{lex}}}\widehat{{\bm{T}}^{F}}$,
we have two subcases: either $(b,i^{\prime})\in F_{b}^{\prime}$ for some
$i^{\prime}>i_{b,2}$ or $F_{b}^{\prime}=\\{(b,i_{b,1})\\}$. One can likewise
exclude the first subcase using the non-crossing argument. Whence, we only
need to check with the $F_{b}^{\prime}=\\{(b,i_{b,1})\\}$ case. In
$\mathcal{T}_{F^{\prime}}$, the parent node of $(b,i_{b,1})$ must take the
form of $(b^{\prime},i_{b,1})$ for some $b^{\prime}<b$. By the minimally of
$b$, $(b^{\prime},i_{b,1})$ belongs to $F\cap F^{\prime}$ and will
consequently be the parent node of $(b,i_{b,1})$ in both $\mathcal{T}_{F}$ and
$\mathcal{T}_{F^{\prime}}$. But the parent node of $(b,i_{b,1})$ in
$\mathcal{T}_{F}$ is $(b,i_{b,2})$, still a contradiction.

  3. c

Assume now that $k_{b}\geq 3$. If $(b,i_{b,j})\in F_{b}^{\prime}$ for all
$1\leq j\leq k_{b}-1$, then we can argue as in the previous case b. Thus, we
have some $j$ with $1\leq j\leq k_{b}-1$ such that $(b,i_{b,j})\notin
F_{b}^{\prime}$. Let $j_{0}$ be the smallest with this property. Notice that

     * •

we always have $(b,i_{b,1})\in F^{\prime}$ as in b, and

     * •

for $j$ with $2\leq j\leq k_{b}-1$, if $(b,i_{b,j})$ has a sibling or two
children, then $(b,i_{b,j})\in LG_{F}\subset F^{\prime}$ by (5).

Thus, $2\leq j_{0}\leq k_{b}-1$ and $(b,i_{b,j_{0}})$ has neither a right
child nor a right sibling in $\mathcal{T}_{F}$.

As a matter of fact, $(b,i_{b,j})$ has no right sibling in $\mathcal{T}_{F}$
for $j_{0}\leq j\leq k_{b}-1$. To see this, suppose for contradiction that
there exists some $j_{1}>j_{0}$ such that $(b,i_{b,j_{1}})$ has a right
sibling. Let $j_{1}$ be the smallest. Therefore, for all $j$ with $j_{0}\leq
j<j_{1}$, $(b,i_{b,j})$ has no right sibling. Notice that $(b,i_{b,j_{0}-1})$
also has no right sibling. Therefore, $(i_{b,j_{0}-1},i_{b,j_{1}})$ contains
no leaf of $\mathcal{T}_{F}$ and $i_{b,j}=i_{b,j_{0}-1}+j-(j_{0}-1)$ for
$j_{0}\leq j\leq j_{1}$ by 2.8 ii. Since
$F,F^{\prime}\in\mathcal{G}_{\alpha}$, it follows that
$(i_{b,j_{0}-1},i_{b,j_{1}})$ contains no leaf of $\mathcal{T}_{F^{\prime}}$.
Meanwhile, as $(b,i_{b,j_{1}})$ has a right sibling in $\mathcal{T}_{F}$,
$(b,i_{b,j_{1}})\in LG_{F}\subset F^{\prime}$ as well. Since we have assumed
that $(b,i_{b,j_{0}-1})\in F^{\prime}$, this forces $(b,i_{b,j})\in
F^{\prime}$ for $j_{0}-1<j\leq j_{1}$ by 2.8 iv, contradicting the choice of
$j_{0}$.

Now, $(b,i_{b,j})$ has no right sibling in $\mathcal{T}_{F}$ for $j_{0}-1\leq
j\leq k_{b}-1$. It follows from 2.8 ii that the interval
$(i_{b,j_{0}-1},i_{b,k_{b}})$ has no leaf of $\mathcal{T}_{F}$, and

$i_{b,j}=i_{b,j_{0}-1}+j-(j_{0}-1)\qquad\text{for}\quad j_{0}-1\leq j\leq
k_{b}.$ (6)

Furthermore, as $(b,i_{b,j_{0}-1})\in F^{\prime}$ while $(b,i_{b,j_{0}})\notin
F^{\prime}$, we have

$(b,i_{b,j})\in F\setminus F^{\prime}\qquad\text{for}\quad j_{0}\leq j\leq
k_{b}$ (7)

by 2.8 iv.

So far, $(b,i_{b,1})\in F\cap F^{\prime}$ and $(b,i_{b,j})\in F_{b}^{\prime}$
for all $j=2,3,\dots,j_{0}-1$ by the minimality of $j_{0}$ . Therefore, we
have

$\Set{(b,i_{b,j})}{1\leq j\leq j_{0}-1}\subseteq F_{b}^{\prime}.$ (8)

If the containment in (8) is strict, then we have some $(b,i^{\prime})\in
F_{b}^{\prime}\setminus F_{b}$. Since $F^{\prime}\succ F$,
$i^{\prime}>i_{b,j_{0}}$. And by (6) with (7), we must have
$i^{\prime}>i_{b,k_{b}}$. If $b=1$, then $(b,i_{b,k_{b}})$ is the common root
$(1,c)$. The existence of such an $i^{\prime}$ is impossible. Therefore, $b>1$
and the parent node of $(b,i_{b,k_{b}})$ in $\mathcal{T}_{F}$ must have the
form $(b^{\prime},i_{b,k_{b}})$ for some $b^{\prime}<b$. By the minimality of
$b$, $(b^{\prime},i_{b,k_{b}})\in F^{\prime}$. Whence, we will arrive at a
contradiction due to crossing. Therefore, the containment in (8) is actually
an equality:

$\Set{(b,i_{b,j})}{1\leq j\leq j_{0}-1}=F_{b}^{\prime}.$

Now, we consider the parent node of $(b,i_{k,j_{0}-1})$ in
$\mathcal{T}_{F^{\prime}}$ and get a similar contradiction due to crossing.
And this completes the proof in this subsection.

In summary, the ideal $(\operatorname{in}(P))^{\vee}$ has linear quotients, as
expected. ∎

######  Corollary 2.9.

When $c\geq d+4$, the initial complex $\Delta$ is shellable.

###### Proof.

It follows from [MR2724673, Proposition 8.2.5] and 2.6. ∎

##  3\. Applications of the Cohen–Macaulay property

In this final section, $I_{n_{1},\dots,n_{d}}$ is still the defining ideal of
the rational normal scroll
$\mathcal{S}_{n_{1},\dots,n_{d}}\subseteq{\mathbb{P}}^{c+d-1}$ for
$c=n_{1}+\cdots+n_{d}$.

As the first application to our main result 2.1, we compute the
Castelnuovo–Mumford regularity of the fiber cone
$\mathcal{F}(I_{n_{1},\dots,n_{d}})$ in 3.4. Its argument depends on the
Cohen–Macaulayness established in section 2 and the regularity result of
balanced cases in [Lin-Shen3]. Again, by the Cohen–Macaulayness, the
regularity of the fiber cone can be related to its ${\bm{a}}$-invariant, as
well as the reduction number of the ideal $I_{n_{1},\dots,n_{d}}$.

It is worth mentioning that, [MR1322960, Section 20.5] provides a concise
introduction to the notion of _Castelnuovo–Mumford regularity_ , as well as
some historical notes. And the _${\bm{a}}$ -invariant_ was introduced by Goto
and Watanabe in [MR494707, Definition 3.1.4]. A nice thing is that, for any
standard graded Cohen–Macaulay algebra $A$ over ${\mathbb{K}}$, one has

${\bm{a}}(A)=\operatorname{reg}(A)-\operatorname{dim}(A),$ (9)

in view of the equivalent definition of regularity in [MR676563, Definitions 1
and 3]. As for the _reduction number_ of an ideal and some related topics, one
may wish to consult [MR2266432, Section 8.2]. Since these three invariants
have been widely studied in the commutative algebra literature, we will not
bother repeating their definitions here.

However, before stating the concrete result and displaying its proof, we still
need to include three important facts here for completeness.

######  Lemma 3.1 ([MR2508056, Proposition 7.43]).

Let $M$ be a graded Cohen–Macaulay module over the polynomial ring
${\mathbb{K}}[x_{1},\dots,x_{n}]$, and $P_{M}(t)$ the numerator Laurent
polynomial of the Hilbert series of $M$. Then
$\operatorname{reg}(M)=\operatorname{deg}(P_{M}(t))$.

######  Lemma 3.2 ([CNPY, Proposition 6.6] or [MR3864202, Proposition 1.2]).

Let $I\subset R={\mathbb{K}}[x_{1},\dots,x_{N}]$ be a homogeneous ideal that
is generated in one degree, say $d$. Assume that the fiber cone
$\mathcal{F}(I)$ is Cohen–Macaulay. Then each minimal reduction of $I$ is
generated by $\operatorname{dim}(\mathcal{F}(I))$ homogeneous polynomials of
degree $d$, and $I$ has the reduction number
$\operatorname{r}(I)=\operatorname{reg}(\mathcal{F}(I))$.

Recall that the rational normal scroll $\mathcal{S}_{n_{1},\dots,n_{d}}$ is
_balanced_ precisely when $|n_{i}-n_{j}|\leq 1$ for all $i,j$. Whence, we can
rearrange the columns of the matrix $\mathbf{X}$ in the equation (1) and
rewrite it as the _Hankel matrix_ ${\bm{H}}_{2,c,d}$ for
$c=n_{1}+\cdots+n_{d}$. And here is the last piece that we need for the
regularity result.

######  Lemma 3.3 ([Lin-Shen3, Theorem 3.1]).

Let $I_{2}({\bm{H}}_{2,c,d})$ be the defining ideal of the rational normal
scroll $\mathcal{S}_{n_{1},\dots,n_{d}}\subseteq{\mathbb{P}}^{c+d-1}$ in the
balanced case. Then, the Castelnuovo–Mumford regularity of the fiber cone
$\mathcal{F}(I_{2}({\bm{H}}_{2,c,d}))$ is

$\operatorname{reg}(\mathcal{F}(I_{2}({\bm{H}}_{2,c,d})))=\begin{cases}\left\lceil(c+d-1)/2\right\rceil,&\text{if
$4+d\leq c$},\\\ c-3,&\text{if $2<c\leq 4+d$}.\end{cases}$

It is time to describe the first result of this section.

######  Theorem 3.4.

Let $I_{n_{1},\dots,n_{d}}$ be the defining ideal of the rational normal
scroll $\mathcal{S}_{n_{1},\dots,n_{d}}\subseteq{\mathbb{P}}^{c+d-1}$ for
$c=n_{1}+\cdots+n_{d}$.

  1. i

The Castelnuovo–Mumford regularity of the fiber cone is given by

$\operatorname{reg}(\mathcal{F}(I_{n_{1},\dots,n_{d}}))=\begin{cases}\left\lceil(c+d-1)/2\right\rceil,&\text{if
$4+d\leq c$},\\\ c-3,&\text{if $2<c<4+d$}.\end{cases}$

  2. ii

The reduction number of $I_{n_{1},\dots,n_{d}}$ is given by

$\operatorname{r}(I_{n_{1},\dots,n_{d}})=\begin{cases}\left\lceil(c+d-1)/2\right\rceil,&\text{if
$4+d\leq c$},\\\ c-3,&\text{if $2<c<4+d$}.\end{cases}$

  3. iii

The ${\bm{a}}$-invariant of the fiber cone is given by

${\bm{a}}(\mathcal{F}(I_{n_{1},\dots,n_{d}}))=\begin{cases}\left\lceil(c+d-1)/2\right\rceil-
c-d,&\text{if $4+d\leq c$},\\\ -c,&\text{if $2<c<4+d$}.\end{cases}$

###### Proof.

Notice that the fiber cone $\mathcal{F}(I_{n_{1},\dots,n_{d}})$ is
Cohen–Macaulay by 2.1. Thus, the item ii follows directly from the item i and
3.2. And similarly, the item iii follows from the item i, Equation (9) and the
fact

$\operatorname{dim}(\mathcal{F}(I_{n_{1},\dots,n_{d}}))=\begin{cases}c+d,&\text{if
$4+d\leq c$,}\\\ 2c-3,&\text{if $2<c<4+d$,}\end{cases}$

by [MR4068250, Proposition 3.1, Corollary 3.10].

As for the item i, just as observed in the proof of [MR4068250, Theorem 3.13],
the Hilbert function of the fiber cone of the rational normal scroll
$\mathcal{S}_{n_{1},\dots,n_{d}}$ depends only on $c$ and $d$ by [MR3341465,
Theorem 3.7]. Whence, the fiber cones of the ideal $I_{n_{1},\dots,n_{d}}$
above and the ideal $I_{2}({\bm{H}}_{2,c,d})$ in [Lin-Shen3, Section 3] have
the same Laurent polynomial. Now, by the Cohen–Macaulayness of
$\mathcal{F}(I_{n_{1},\dots,n_{d}})$ established in 2.1, these fiber cones
have the same Castelnuovo–Mumford regularity by 3.1.

Now, it remains to apply 3.3. ∎

As the second application to our main result 2.1, we characterize the
Gorensteinness of the fiber cone. The corresponding numerical condition in the
balanced case is the following.

######  Lemma 3.5 ([Lin-Shen3, Proposition 2.10]).

Let $I_{2}({\bm{H}}_{2,c,d})$ be the defining ideal of the rational normal
scroll $\mathcal{S}_{n_{1},\dots,n_{d}}\subseteq{\mathbb{P}}^{c+d-1}$ in the
balanced case. Then, the fiber cone $\mathcal{F}(I_{2}({\bm{H}}_{2,c,d}))$ is
Gorenstein if and only if $c\in\Set{2,3,2+d,3+d,4+d}$.

And here is the last result that this paper wants to describe.

######  Theorem 3.6.

Let $I_{n_{1},\dots,n_{d}}$ be the defining ideal of the rational normal
scroll $\mathcal{S}_{n_{1},\dots,n_{d}}\subseteq{\mathbb{P}}^{c+d-1}$ for
$c=n_{1}+\cdots+n_{d}$. Then, the fiber cone
$\mathcal{F}(I_{n_{1},\dots,n_{d}})$ is Gorenstein if and only if
$c\in\Set{2,3,2+d,3+d,4+d}$.

###### Proof.

The fiber cone $\mathcal{F}(I_{n_{1},\dots,n_{d}})$ is Cohen–Macaulay by 2.1.
And since it is isomorphic to the toric ring
${\mathbb{K}}[I_{n_{1},\dots,n_{d}}]$, it is an integral domain. As mentioned
in the proof of 3.4, the fiber cones $\mathcal{F}(I_{n_{1},\dots,n_{d}})$ and
$\mathcal{F}(I_{2}({\bm{H}}_{2,c,d}))$ have identical Hilbert series. Thus, by
both (b) and (c) of [MR1251956, Corollary 4.4.6],
$\mathcal{F}(I_{n_{1},\dots,n_{d}})$ is Gorenstein, if and only
$\mathcal{F}(I_{2}({\bm{H}}_{2,c,d}))$ is so. Now, it suffices to apply 3.5. ∎

######  Acknowledgment.

The authors are grateful to the software system Macaulay2 [M2], for serving as
an excellent source of inspiration. The second author is partially supported
by the “Anhui Initiative in Quantum Information Technologies” (No. AHY150200)
and the “Fundamental Research Funds for the Central Universities”.

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