# An illustrated view of differential operators of a reduced quotient of an
affine semigroup ring

Christine Berkesch  Department of Mathematics
University of Minnesota
Minneapolis, MN<EMAIL_ADDRESS>,  C-Y. Jean Chan  Department of Mathematics
Central Michigan University
Mt. Pleasant, MI<EMAIL_ADDRESS>,  Patricia Klein  Department of
Mathematics
University of Minnesota
Minneapolis, MN<EMAIL_ADDRESS>,  Laura Felicia Matusevich  Department of
Mathematics
Texas A & M
College Station, TX<EMAIL_ADDRESS>,  Janet Page  Department of
Mathematics
University of Michigan
Ann Arbor, MI<EMAIL_ADDRESS>and  Janet Vassilev  Department of Mathematics
and Statistics
University of New Mexico
Albuquerque, NM<EMAIL_ADDRESS>
###### Abstract.

Through examples, we illustrate how to compute differential operators on a
quotient of an affine semigroup ring by a radical monomial ideal, when working
over an algebraically closed field of characteristic 0.

CB was partially supported by NSF DMS 2001101. LFM was partially supported by
the Simons Foundation.

## Introduction

In this paper, we provide illustrative examples and visualizations of some
differential operators on the quotient of an affine semigroup ring by a
radical monomial ideal. These examples motivate our work in [BCK+21]. For a
finitely-generated algebra $R$ over a field, let $D(R)$ denote the ring of
differential operators of $R$ and use $\ast$ to denote an action of a
differential operator. A foundational result in this area relates the ring of
differential operators of an arbitrary quotient of a polynomial ring by an
ideal $J$ to differential operators on that polynomial ring and the idealizer
of $J$.

######  Proposition 1.1.

[SS88, Propostion 1.6] Let $S$ be the coordinate ring of a nonsingular affine
variety over an algebraically closed field of characteristic $0$, and let $J$
be an $S$-ideal. Then there is an isomorphism

(1.1) $D\left(\frac{S}{J}\right)\cong\frac{\mathbb{I}(J)}{JD(S)},$

where $\mathbb{I}(J):=\\{\delta\in D(R)\mid\delta\ast J\subset J\\}$, which is
called the _idealizer_ of $J$.

The phrasing of [SS88, Proposition 1.6] given above differs from the original
by making use of the equivalence of the conditions, with notation as above, of
$\delta JD(S)\subset JD(S)$ and $\delta\ast J\subset J$ [ST01, Lemma 2.3.1].
Traves [Tra99] uses Proposition 1.1 to give concrete descriptions of rings of
differential operators of Stanley–Reisner rings, and Saito and Traves [ST01]
use the same to compute rings of differential operators of affine semigroup
rings. For a non-regular affine semigroup ring $R_{A}$ over an algebraically
closed field of characteristic $0$, Proposition 1.1 fails, even for a radical
monomial ideal $J$ in $R_{A}$. However, the differential operators on $R_{A}$
that induce maps on $R_{A}/J$ are precisely those in $\mathbb{I}(J)$. Further,
there is an embedding of rings

$\displaystyle\frac{\mathbb{I}(J)}{D(R_{A},J)}\hookrightarrow
D\left(\frac{R_{A}}{J}\right),$

so the description of $D(R_{A})$ by Saito and Traves can still be used to
compute this subring of $D(R_{A}/J)$.

One primary goal of this article is to visually illustrate the computation of
$\mathbb{I}(J)/D(R_{A},J)$ when $R_{A}$ is an affine semigroup ring and $J$ is
a radical monomial ideal in $J$; in [BCK+21], we provide an explicit formula
for this computation. In particular, the pictures we provide explain how to
compute differential operators of the form

$D(I,J):=\\{\delta\in D(R)\mid\delta\ast I\subseteq J\\},$

where $I$ and $J$ are subsets of the ring $R$, and $\ast$ denotes an action by
a differential operator. These sets were originally instrumental in both the
work of Smith and Stafford [SS88] and of Musson [Mus89, Section 1], and they
are essential building blocks of our computations, as well.

Our second major goal in this paper is to compare $JD(R_{A})$ and $D(R_{A},J)$
for an affine semigroup ring $R_{A}$ and radical monomial ideal $J$ of
$R_{A}$. Towards this end, the first set of examples we consider consists of
quotients of the coordinate rings of rational normal curves. These quotients
are all isomorphic to $\mathbb{C}[x,y]/\langle xy\rangle$, which is handled in
[Tra99]. From the standpoint of comparing $JD(R_{A})$ and $D(R_{A},J)$, what
we will see is that $JD(R_{A})=D(R_{A},J)$ for the rational normal curves of
degrees $1$ and $2$ but fail to coincide in all degrees larger than $2$. For
rational normal curves of degree at most $2$, $J$ is principally generated and
so is $JD(R_{A})$. In this case, it is straightforward to see (by definition)
that $D(R_{A},J)$ is also principal and is isomorphic to $JD(R_{A})$. However,
for degree $n>2$, $J$ has $n-1$ generators. We will see that this number of
generators greatly impacts $\mathbb{I}(J)/JD(R_{A})$ but not
$\mathbb{I}(J)/D(R_{A},J)$.

In [BCK+21], we consider cases where

(1.2) $JD(R_{A})=D(R_{A},J),$

for the particular ideal $J$ generated by all monomials corresponding to the
interior of the semigroup, i.e., $J=\omega_{R_{A}}$. When $R_{A}$ is
Gorenstein and normal, (1.2) holds since $\omega_{R_{A}}$ is principal. We
show that the converse is also true; that is, (1.2) is equivalent to $R_{A}$
being a Gorenstein ring.

To provide intuition beyond the two-dimensional case, we also offer a three-
dimensional normal affine semigroup ring $R_{A}$ for which we compute
$\mathbb{I}(J)/D(R_{A},J)$ for two choices of $J$. Then, we return to the two-
dimensional setting to consider a scored but not normal example, as well as a
non-scored example, computing differential operators for quotients of both
rings.

### Outline

Section 2 fixes notation to be used throughout the article and describes the
main result of [ST01]. Sections 3, 4, and 5 describe
$\mathbb{I}(J)/D(R_{A},J)$ for rational normal cones of degrees 2, 3, and
higher, respectively. Section 6 considers a three-dimensional normal affine
semigroup ring modulo two different choices of radical monomial ideal $J$, and
Section 7 computes $\mathbb{I}(J)/D(R_{A},J)$ for two different non-normal
two-dimensional affine semigroup rings, where $J=\omega_{R_{A}}$ is the
radical monomial ideal corresponding to the interior of the semigroup
$\mathbb{N}A$.

### Acknowledgements

We are grateful to the organizers, Karen Smith, Sandra Spiroff, Irena Swanson,
and Emily Witt, of _Workshop: Women in Commutative Algebra (WICA)_ , at which
this work began. Additionally, we are thankful to BIRS for their hospitality
in hosting this meeting. We also owe thanks to Jack Jeffries, who pointed us
to an example that improved our work. We truly appreciate the comments of the
referee which added extra insight into our work and greatly improved the
paper.

##  2\. Background and Notation

In this section, we fix notation and conventions to be assumed throughout the
article. Although the results we discuss in this paper hold over any
algebraically closed field of characteristic 0, we will use in this
illustrated view the field of complex numbers for the sake of simplicity.
Having fixed notation, we will then state and discuss [ST01, Theorem 3.2.2].

######  Definition 2.1.

Let $A$ be a $k\times\ell$ matrix with entries in $\mathbb{Z}$. Let
$\mathbb{N}A$ denote the semigroup in $\mathbb{Z}^{k}$ that is generated by
the columns of $A$. The _affine semigroup ring_ determined by $A$ is

$R_{A}=\mathbb{C}[\mathbb{N}A]=\bigoplus_{{\bm{a}}\in\mathbb{N}A}\mathbb{C}\cdot{\bm{t}}^{\bm{a}},$

where ${\bm{t}}^{\bm{a}}=t_{1}^{a_{1}}t_{2}^{a_{2}}\cdots t_{k}^{a_{k}}$ for
$\bm{a}=(a_{1},a_{2},\ldots,a_{k})\in\mathbb{N}A$. Throughout this article, we
assume that the group generated by the columns of $A$ is the full ambient
lattice, so $\mathbb{Z}A=\mathbb{Z}^{k}$, and also that the real positive cone
over $A$, $\mathbb{R}_{\geq 0}A$, is pointed, meaning that it is strongly
convex.

######  Definition 2.2.

A semigroup $\mathbb{N}A$ is _normal_ if $\mathbb{N}A=\mathbb{R}_{\geq
0}A\cap\mathbb{Z}A$. A semigroup $\mathbb{N}A$ is _scored_ if the difference
$(\mathbb{R}_{\geq 0}A\cap\mathbb{Z}A)\setminus\mathbb{N}A$ consists of a
finite union of hyperplane sections of $\mathbb{R}_{\geq 0}A\cap\mathbb{Z}A$
that are all parallel to facets of the cone $\mathbb{R}_{\geq 0}A$. An affine
semigroup ring $\mathbb{C}[\mathbb{N}A]$ is said to be _normal_ , or _scored_
, if $\mathbb{N}A$ is normal, or scored, respectively. Note that normal
semigroups are scored.

When we write a facet $\sigma$ of $\mathbb{R}_{\geq 0}A$ (or $A$ or
$\mathbb{N}A$), we will always mean the integer points in the corresponding
facet of $\mathbb{R}_{\geq 0}A$. When $A$ is normal, this is the same as the
semigroup generated by the columns of $A$ that lie in the corresponding facet
of $\mathbb{R}_{\geq 0}A$.

Throughout, we use $\langle-\rangle$ to indicate ideals in the commutative
rings $R_{A}$ or the polynomial ring
$\mathbb{C}[\theta]=\mathbb{C}[\theta_{1},\ldots,\theta_{k}]$. It will be
clear from the context if the ideals are in $R_{A}$ or in
$\mathbb{C}[\theta]$.

Notice that the $\mathbb{Z}^{k}$-graded prime ideals in $\mathbb{R}_{A}$ are
in one-to-one correspondence with the faces of $A$ (or $\mathbb{R}_{\geq
0}A$), as a face $\tau$ of $A$ corresponds to the multigraded prime
$R_{A}$-ideal

$P_{\tau}:=\left\langle\bm{t}^{\bm{d}}\,\big{|}\,\bm{d}\in\mathbb{N}A\setminus\tau\right\rangle.$

In this paper, we compute rings of the form $\mathbb{I}(J)/D(R_{A},J)$, where
$J$ is a radical monomial ideal in $R_{A}$; as such, $J$ is always as
intersection of primes of the form $P_{\tau}$, for various faces $\tau$ of
$A$.

We are mainly following the description of Saito and Traves [ST01], although
Musson and Van den Bergh describe the ring of differential operators of a
toric ring $\mathbb{C}[\mathbb{N}A]$ first in [Mus87], [Mus94] and [MVdB98]
when viewed as a subring of the ring of differential operators of the Laurent
polynomials, i.e., $D(\mathbb{C}[\mathbb{Z}^{k}])=\mathbb{C}\\{t_{1}^{\pm
1},\ldots,t_{k}^{\pm 1},\partial_{1},\ldots,\partial_{k}\\}/\sim$, where
$\partial_{i}$ denotes the differential operator $\frac{\partial}{\partial
t_{i}}$ and $\sim$ denotes the usual relations on the free associative algebra
$\mathbb{C}\\{t_{1}^{\pm 1},\ldots,t_{k}^{\pm
1},\partial_{1},\ldots,\partial_{k}\\}$ that describe the behavior of
differential operators. To explain the differential operators of
$\mathbb{C}[\mathbb{N}A]$ as presented by Saito and Traves, note first that
$D(\mathbb{C}[\mathbb{Z}^{k}])$ carries a $\mathbb{Z}^{k}$-grading, where
$e_{i}$ denotes the $i$-th column of the identity matrix $I$ and
$\deg(t_{i})=e_{i}=-\deg(\partial_{i})$. Note also that if $\bm{a}_{i}$ is a
column of $A$, then $\deg(\bm{t}^{\bm{a}_{i}})=\bm{a}_{i}$. Set
$\theta_{j}=t_{j}\partial_{j}$ for $1\leq j\leq k$, and set

$\Omega({\bm{d}}):=\\{\bm{a}\in\mathbb{N}A\mid\bm{a}+{\bm{d}}\notin\mathbb{N}A\\}=\mathbb{N}A\setminus(-{\bm{d}}+\mathbb{N}A).$

We note also that for any $\bm{t}^{\bm{m}}\in R_{A}$ and
$f(\theta)\in\mathbb{C}[\theta]$,
$f(\theta)\ast\bm{t}^{\bm{m}}=f(\bm{m})\bm{t}^{\bm{m}}$. The _idealizer of
$\Omega({\bm{d}})$_ is defined to be

$\mathbb{I}(\Omega({\bm{d}})):=\langle
f(\theta)\in\mathbb{C}[\theta]=\mathbb{C}[\theta_{1},\theta_{2}\ldots,\theta_{k}]\mid
f(\bm{a})=0\text{ for all }\bm{a}\in\Omega({\bm{d}})\rangle,$

which is viewed as an ideal in the ring $\mathbb{C}[\theta]$, where
$\theta_{i}=t_{i}\partial_{i}\in D(\mathbb{C}[\mathbb{Z}^{d}])$ is of degree
$\bm{0}$. We will soon see that $\mathbb{I}(\Omega({\bm{d}}))$ consists of
$f(\theta)$ such that $\bm{t}^{\bm{d}}f(\theta)\in D(R_{A})$.

To compute $\mathbb{I}(\Omega({\bm{d}}))$ for a normal semigroup ring,
consider a facet $\sigma$ of $A$, recalling that by this we mean all lattice
points on the corresponding facet of the cone $\mathbb{R}_{\geq 0}A$. The
primitive integral support function $h_{\sigma}$ is a uniquely determined
linear form on $\mathbb{R}^{k}$ such that:

  1. (1)

$h_{\sigma}(\mathbb{R}_{\geq 0}A)\geq 0$,

  2. (2)

$h_{\sigma}(\mathbb{R}\sigma)=0$, and

  3. (3)

$h_{\sigma}(\mathbb{Z}^{k})=\mathbb{Z}.$

We write $\sigma_{1},\sigma_{2},\ldots,\sigma_{m}$ for the facets of $A$, so
that for the remainder of the paper, we set

$h_{j}:=h_{j}(\theta)=h_{\sigma_{j}}(\theta).$

For a non-negative integer $n$, we will use the following notation to denote
this descending factorial:

$\left(\alpha,n\right)!:=\prod\limits_{i=0}^{n}(\alpha-i)=\alpha(\alpha-1)\cdots(\alpha-n),$

where $\alpha$ is a function, which could be constant or already evaluated.
For example,

$\left(h_{j},h_{j}({-\bm{d}})-1)\right)!=\prod\limits_{i=0}^{h_{j}({-\bm{d}})-1}(h_{j}-i)$

will be a common expression throughout this article, as it is a factor in the
generator of the idealizer $\mathbb{I}(\Omega(\bm{d}))$. To streamline our
presentation, we make the convention that $\left(\alpha,n\right)!=1$ for all
$n<0$.

######  Theorem 2.3.

[ST01, Theorem 3.2.2] If $R_{A}$ is a pointed, normal affine semigroup ring
with $\mathbb{Z}A=\mathbb{Z}^{k}$, then

$\displaystyle D(R_{A})$
$\displaystyle=\bigoplus\limits_{{\bm{d}}\in\mathbb{Z}^{d}}\bm{t}^{\bm{d}}\cdot\mathbb{I}(\Omega({\bm{d}})),$
(2.1)
$\displaystyle=\bigoplus\limits_{{\bm{d}}\in\mathbb{Z}^{d}}\bm{t}^{\bm{d}}\cdot\left\langle\prod\limits_{h_{i}({\bm{d}})<0}\left(h_{i},h_{i}(-{\bm{d}})-1\right)!\right\rangle,$

where the outer product runs over those $j=1,2,\dots,m$ with
$h_{j}({\bm{d}})<0$.

We now turn to an example that illustrates Theorem 2.3, as well as our
notation.

######  Example 2.4.

Let $R=\mathbb{C}[s,st,st^{2},st^{3}]$, where $\deg(s)=1=\deg(t)$. Let
$\partial_{1}$ and $\partial_{2}$ denote the partial derivatives with respect
to $s$ and $t$, and let $\theta_{1}=s\partial_{1}$ and
$\theta_{2}=t\partial_{2}$. By Theorem 2.3, $D(R)_{\bm{d}}$, for some
$\bm{d}=(d_{1},d_{2})\in\mathbb{Z}^{2}$, is generated by an element in the
form of $s^{d_{1}}t^{d_{2}}\cdot f(\theta_{1},\theta_{2})$, where
$f(\theta_{1},\theta_{2})$ is a polynomial in $\theta_{1}$ and $\theta_{2}$.

To understand the philosophy from [ST01], fix
$\bm{d}=(d_{1},d_{2})\in\mathbb{Z}^{2}$, and consider a differential operator
on $R=k[s,st,st^{2},st^{3}]$ of the form
$\delta=s^{d_{1}}t^{d_{2}}f(\theta_{1},\theta_{2})$. By definition,
$\delta\ast(s^{m_{1}}t^{m_{2}})\in R$ for any monomial $s^{m_{1}}t^{m_{2}}\in
R$. We can check that

$\delta\ast s^{m_{1}}t^{m_{2}}=f(\bm{m})s^{d_{1}+m_{1}}t^{d_{2}+m_{2}}.$

In particular, if $s^{d_{1}+m_{1}}t^{d_{2}+m_{2}}\notin R$ then we must have
that $\delta\ast s^{m_{1}}t^{m_{2}}=0$ so that $f(\bm{m})=0$. In summary,
$\delta\ast(s^{m_{1}}t^{m_{2}})$ may be nonzero if and only if
$s^{d_{1}+m_{1}}t^{d_{2}+m_{2}}\in R$.

These multidegrees $\bm{m}=(m_{1},m_{2})$ where $\delta\ast
s^{m_{1}}t^{m_{2}}$ vanishes are exactly the points in $\Omega({\bm{d}})$. We
illustrate examples of $\Omega({\bm{d}})$ for specific $\bm{d}$ in Figure 2.1
by the integral points that are on the dotted lines in the first quadrant,
including those on the positive horizontal axis.

$\bm{d}$$-\bm{d}$ |  | $\bm{d}$$-\bm{d}$
---|---|---
Figure 2.1. Half-lines of vanishing for various $\bm{d}$; here,
$\bm{d}=(-1,2)$ and $(-2,-1)$.

To explicitly compute the polynomial $f=f(\theta_{1},\theta_{2})$ for a fixed
$\bm{d}\in\mathbb{Z}^{2}$, we will use the primitive integral support
functions for the two facets of $A$, which are $h_{1}=\theta_{2}$ and
$h_{2}=3\theta_{1}-\theta_{2}$. Both of these must divide $f$, along with all
of the linear forms representing the dotted lines in Figure 2.1. Specifically,
$\mathbb{I}(\Omega(-1,2))$ is generated by

$f(\theta_{1},\theta_{2})=h_{2}(h_{2}-1)(h_{2}-2)(h_{2}-3)(h_{2}-4)=(h_{2},4)!,$

and $\mathbb{I}(\Omega(-2,-1))$ is generated by

$f(\theta_{1},\theta_{2})=h_{1}h_{2}(h_{2}-1)(h_{2}-2)(h_{2}-3)(h_{2}-4)=(h_{1},0)!(h_{2},4)!.$

For $\bm{d}=(-1,2)$, we have $h_{2}(-\bm{d})=5$ while $h_{1}(-\bm{d})=-2<0$,
so there are no linear forms involving $h_{1}$ in this generator. Similarly
for $\bm{d}=(-2,-1)$, $h_{1}(-\bm{d})=1$, $h_{2}(-\bm{d})=5$. In particular,
$(h_{1},h_{1}(-\bm{d})-1)!=(h_{1},0)!=h_{1}$ and
$(h_{2},h_{2}(-\bm{d})-1)!=(h_{2},4)!=(h_{2}-4)(h_{2}-3)(h_{2}-2)(h_{2}-1)h_{2}$.

As shown in Example 2.4, $\mathbb{I}(\Omega({\bm{d}}))$ is an ideal in
$\mathbb{C}[\theta]$, and any polynomial $f(\theta)\in\mathbb{C}[\theta]$ has
multidegree $\bm{0}$; for any monomial ${\bm{t}}^{\bm{m}}\in R_{A}$,
$f(\theta)\ast{\bm{t}}^{\bm{m}}=f(\bm{m})\bm{t}^{\bm{m}}$, which belongs to
$R_{A}$. So to determine if the $\bm{d}$-th graded piece of $D(R_{A})$ applied
to a monomial lands in $R_{A}$ or an $R_{A}$-ideal $J$, it is enough to test
the membership of any monomial $\bm{t}^{\bm{d}+\bm{m}}$ and then adjust the
$\theta$-portion of the differential operator appropriately. Membership
failure for $R_{A}$ only occurs if $\bm{m}$ lies in $\mathbb{N}A$ but outside
of the cone $-\bm{d}+\mathbb{R}_{\geq 0}A$, whereas for $J\neq R_{A}$, failure
is more likely to occur. Since the monomials whose exponents lie on a face of
$-\bm{d}+\mathbb{R}_{\geq 0}A$ will lie on the corresponding face $\tau$ of
$A$, if an associated prime of $J$ is a prime ideal associated to a face that
contains $\tau$, then membership failure will also occur for $\bm{m}$ on this
face. In either case, there are only finitely many potential linear forms to
be determined, and they are of the form
$h_{\sigma}(\theta)-h_{\sigma}(-{\bm{d}})$ where $h_{\sigma}$ is the primary
support function of a facet $\sigma$.

In [Eri98, Tra99, Tri97], Eriksson, Traves, and Tripp separately computed the
ring of differential operators of a Stanley–Reisner ring over an arbitrary
field, i.e., the quotient of any polynomial ring over a field by a squarefree
monomial ideal. We include here the ring of differential operators of an
ordinary double point $R=\mathbb{C}[x,y]/\langle xy\rangle$ using the
viewpoint presented by the above authors, as it exhibits behavior akin to our
computations in this article. In [Mus89], Musson also considered the
differential operators on an ordinary double point.

######  Example 2.5.

The ring of differential operators on $\mathbb{C}[x,y]$ is the Weyl algebra
$W=\mathbb{C}\\{x,y,\partial_{x},\partial_{y}\\}$, which is the free
associative algebra generated by $x,y,\partial_{x},\partial_{y}$, subject to
the relations:

$\\{xy-
yx,\partial_{x}\partial_{y}-\partial_{y}\partial_{x},\partial_{x}y-y\partial_{x},\partial_{y}x-x\partial_{y},\partial_{x}x-x\partial_{x}+1,\partial_{y}y-y\partial_{y}+1\\}.$

The Weyl algebra $W$ is a graded ring with
$\deg(x)=\deg(y)=1=-\deg(\partial_{x})=-\deg(\partial_{y})$.

For the ordinary double point ring $R=\mathbb{C}[x,y]/\langle xy\rangle$,
Traves shows in [Tra99, Theorem 3.5] that the idealizer of $\langle xy\rangle$
in $W$ is also graded and generated by

$\\{1,x^{m},y^{n},x^{m}y^{n},x^{m}\partial_{x}^{i},y^{n}\partial_{y}^{j},x^{m}y^{n}\partial_{x}^{i}\partial_{y}^{j}\mid
i\geq m>0,j\geq n>0\\}.$

Notice in particular that $x^{m}\partial_{x}^{m}$ and $y^{n}\partial_{y}^{n}$
both have degree 0. Setting $\theta_{x}=x\partial_{x}$ and
$\theta_{y}=y\partial_{y}$, from the Weyl algebra relations, it follows that

$\partial_{x}^{i}=x^{-i}\cdot\prod\limits_{\ell=0}^{i-1}(\theta_{x}-\ell)=x^{-i}\left(\theta_{x},i-1\right)!\quad\text{and}\quad\partial_{y}^{j}=y^{-j}\cdot\prod\limits_{\ell=0}^{j-1}(\theta_{y}-\ell)=\left(\theta_{y},j-1\right)!.$

Hence $x^{m}\partial_{x}^{i}=x^{m-i}\cdot\left(\theta_{x},i-1\right)!$ and
$y^{n}\partial_{y}^{j}=y^{n-j}\cdot\left(\theta_{y},j-1\right)!$, and
$x^{m}y^{n}\partial_{x}^{i}\partial_{y}^{j}$ has multidegree $(m-i,n-j)$ in
$W$. In fact,

$W=\bigoplus\limits_{(m,n)\in\mathbb{Z}^{2}}x^{m}y^{n}\cdot\langle\left(\theta_{x},m-1\right)!\left(\theta_{y},n-1\right)!\rangle,$

which is a presentation of the Weyl algebra using the Saito–Traves approach
from Theorem 2.3. From this viewpoint, Traves showed that

$\mathbb{I}(\langle
xy\rangle)_{(m,n)}=\begin{cases}x^{m}y^{n}\cdot\mathbb{C}[\theta_{x},\theta_{y}]&\text{
if }m,n\geq 0,\\\
x^{m}y^{n}\cdot\langle\left(\theta_{x},-m\right)!\rangle&\text{ if }m<0,n\geq
0,\\\ x^{m}y^{n}\cdot\langle\left(\theta_{y},-n\right)!\rangle&\text{ if
}n<0,m\geq 0,\\\
x^{m}y^{n}\cdot\langle\left(\theta_{x},-m\right)!\left(\theta_{y},-n\right)!\rangle&\text{
if }m,n<0.\end{cases}$

Further, $\langle xy\rangle W$ can be expressed as a multigraded $W$-ideal
that is contained in $\mathbb{I}(\langle xy\rangle)$ as follows:

$\langle xy\rangle
W_{(m,n)}=\begin{cases}x^{m}y^{n}\cdot\mathbb{C}[\theta_{x},\theta_{y}]&\text{
if }m,n>0,\\\ x^{m}y^{n}\cdot\langle\left(\theta_{x},-m\right)!\rangle&\text{
if }m\leq 0,n>0,\\\
x^{m}y^{n}\cdot\langle\left(\theta_{y},-n\right)!\rangle&\text{ if }n\leq
0,m>0,\\\
x^{m}y^{n}\cdot\langle\left(\theta_{x},-m\right)!\left(\theta_{y},-n\right)!\rangle&\text{
if }m,n\leq 0.\\\ \end{cases}$

Now, applying Proposition 1.1 yields a computation for the ring of
differential operators for the ordinary double point
$R=\mathbb{C}[x,y]/\langle xy\rangle$:

(2.2) $D\left(\frac{\mathbb{C}[x,y]}{\langle
xy\rangle}\right)_{(m,n)}=\begin{cases}0&\text{ if }mn\neq 0,\\\
\displaystyle\frac{\mathbb{C}[\theta_{x},\theta_{y}]}{\langle\theta_{x}\theta_{y}\rangle}&\text{
if }m=n=0,\\\
x^{m}\cdot\displaystyle\frac{\langle\left(\theta_{x},-m\right)!\rangle}{\langle\left(\theta_{x},-m\right)!\theta_{y}\rangle}&\text{
if }m\neq 0,n=0,\\\
y^{n}\cdot\displaystyle\frac{\langle\left(\theta_{y},-n\right)!\rangle}{\langle\left(\theta_{y},-m\right)!\theta_{x}\rangle}&\text{
if }n\neq 0,m=0.\\\ \end{cases}$

We next fix some notation to be used in the remainder of this paper. Whenever
the dimension of the semigroup ring under consideration is $k=2$, instead of
using $t_{1},t_{2}$ as our variables, we will use $s,t$. Further, when
considering subsets of $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$ that contain the
lattice points that describe a set of monomials in our semigroup, such as
lines or planes, we will describe them with the variables $x,y,$ and $z$, for
example, the line $y=2x-1$ in $\mathbb{R}^{2}$ or the plane $x-z=2$ in
$\mathbb{R}^{3}$.

Consider the matrix

$A_{n}=\begin{bmatrix}1&1&1&\cdots&1\\\ 0&1&2&\cdots&n\\\ \end{bmatrix}.$

We call $R_{A_{n}}=\mathbb{C}[\mathbb{N}A_{n}]=\mathbb{C}[s,st,st^{2},\ldots
st^{n}]$ the ring of the rational normal curve of degree $n$, since it is the
coordinate ring of the affine cone of the projective rational normal curve.
This ring will be the subject of the next three sections. The two facets of
$A_{n}$ are

$\displaystyle\sigma_{1}=\mathbb{N}\begin{bmatrix}1\\\
0\end{bmatrix}=\\{(x,y)\in\mathbb{N}^{2}\mid x\geq
0,y=0\\}\quad\text{and}\quad\sigma_{2}=\mathbb{N}\begin{bmatrix}1\\\
n\end{bmatrix}=\\{(x,y)\in\mathbb{N}^{2}\mid x\geq 0,y=nx\\}.$

The prime ideal associated to $\sigma_{1}$ is $P_{1}=\langle
st,st^{2},\ldots,st^{n}\rangle$, and the prime ideal associated to
$\sigma_{2}$ is $P_{2}=\langle s,st,\ldots,st^{n-1}\rangle$. We will consider
the radical ideal

$J=P_{1}\cap P_{2}=\begin{cases}\langle
st,st^{2},\ldots,st^{n-1}\rangle&\text{if }n>1,\\\ \langle
s^{2}t\rangle&\text{if }n=1.\end{cases}$

Observe that $R_{A_{n}}/J\cong\mathbb{C}[x,y]/\langle xy\rangle$ for all
$n>0$. At the end of Section 5, we revisit Example 2.5 and present a
$\mathbb{C}$-algebra isomorphism between $D(\mathbb{C}[x,y]/\langle
xy\rangle)$ and $D(R_{A_{n}}/J)$ and compare it to our calculations for
$\mathbb{I}(J)/D(R_{A_{n}},J)$.

##  3\. Differential operators on the rational normal curve of degree 2

In this section, we compute the idealizer, $\mathbb{I}(I)$, along with the
subset of differential operators $D(R_{A_{2}},I)$ for the ideals
$I=P_{1}=\langle st,st^{2}\rangle$ and $I=J=\langle st\rangle$ over the ring
of the rational normal curve of degree $2$,
$R_{A_{2}}=\mathbb{C}[s,st,st^{2}]$. To aid our computations we include
illustrations of the lattice representing the multidegrees in the plane broken
down into four chambers where the operators will be determined by similar
expressions. The facets of $A_{2}$ are

$\displaystyle\sigma_{1}=\\{(x,y)\in\mathbb{N}^{2}\mid x\geq
0,y=0\\}\quad\text{and}\quad\sigma_{2}=\\{(x,y)\in\mathbb{N}^{2}\mid x,y\geq
0,y=2x\\},$

which have primitive integral support functions

$h_{1}=\theta_{2}\quad\text{and}\quad h_{2}=2\theta_{1}-\theta_{2}.$

Figure 3.1. Chambers of $D(R_{A_{2}})$

Figure 3.1 illustrates the integer lattice, divided into four _chambers_ that
are colored as follows:

  * C1:

The red multidegrees correspond to monomials in $J$, the gray multidegrees
correspond to monomials in $P_{1}\setminus J$ and the blue multidegrees
correspond to monomials in $R_{A_{2}}\setminus P_{1}$,

  * C2:

The yellow multidegrees are the $\bm{d}$ with $h_{1}({\bm{d}})\geq 0$ and
$h_{2}({\bm{d}})<0$,

  * C3:

The violet multidegrees are the $\bm{d}$ with both $h_{1}({\bm{d}})<0$ and
$h_{2}({\bm{d}})<0$, and

  * C4:

The green multidegrees are the $\bm{d}$ with $h_{1}({\bm{d}})<0$ and
$h_{2}({\bm{d}})\geq 0$.

Still following the convention $\left(h,n\right)!=1$ if $n<0$, by Theorem 2.3,
the graded pieces of $D(R_{A_{2}})$ are

$\hskip
144.54ptD(R_{A_{2}})_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\left\langle\left(h_{1},h_{1}({-\bm{d}})-1\right)!\left(h_{2},h_{2}({-\bm{d}})-1\right)!\right\rangle.$

Broken down by chambers, this amounts to:

$D(R_{A_{2}})_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{if
}\bm{d}\in\textbf{C1},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-2d_{1}+d_{2}-1\right)!\rangle&\text{if
}\bm{d}\in\textbf{C2},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}-1\right)!\left(h_{2},-2d_{1}+d_{2}-1\right)!\rangle&\text{if
}\bm{d}\in\textbf{C3},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}-1\right)!\rangle&\text{if
}\bm{d}\in\textbf{C4}.\\\ \end{cases}$

######  Example 3.1.

We first compute the graded pieces of the sets of differential operators
$\mathbb{I}(P_{1})$ and $D(R_{A_{2}},P_{1})$. Later, since $\mathbb{C}[x]\cong
R_{A_{2}}/P_{1}$, we will exhibit a $\mathbb{C}$-algebra isomorphism between
$D(\mathbb{C}[x])$ and $D(R_{A_{2}}/P_{1})$.

To begin, recall that

$\mathbb{I}(P_{1})=\\{\delta\in D(R_{A_{2}})\mid\delta*P_{1}\subseteq
P_{1}\\}\quad\text{and}\quad D(R_{A_{2}},P_{1})=\\{\delta\in
D(R_{A_{2}})\mid\delta*R\subseteq P_{1}\\}.$

Now if $\bm{d}\in\textbf{C1}$ and $s^{m_{1}}t^{m_{2}}\in P_{1}$ or if
$\bm{d}\in\textbf{C1}\setminus\sigma_{1}$ and $s^{m_{1}}t^{m_{2}}\in
R_{A_{2}}$, then for any $g(\theta)\in\mathbb{C}[\theta]$,

$s^{d_{1}}t^{d_{2}}\cdot
g(\theta)*s^{m_{1}}t^{m_{2}}=g(\bm{m})s^{d_{1}+m_{1}}t^{d_{2}+m_{2}}\in
P_{1},\text{ so}$ $\mathbb{I}(P_{1})_{\bm{d}}=D(R_{A_{2}})_{\bm{d}}\text{ for
all }\bm{d}\in\textbf{C1}\quad\text{and}\quad
D(R_{A_{2}},P_{1})_{\bm{d}}=D(R_{A_{2}})_{\bm{d}}\text{ for all
}\bm{d}\in\textbf{C1}\setminus\sigma_{1}.$

A Lines parallel
to $\sigma_{2}$.

B Lines parallel
to $\sigma_{1}$.

Figure 3.2. Lines parallel to the facets.

With the aid of Figure 3.2, we will explain how to determine
$\mathbb{I}(P_{1})$ and $D(R_{A_{2}},P_{1})$ in the other chambers. The red
lattice points in Figure 3.2, indicate the monomials in $P_{1}$.

First, note that if $\bm{d}\in\textbf{C2}$ and $s^{m_{1}}t^{m_{2}}\in P_{1}$
or $\bm{d}\in\textbf{C2}\setminus(-\sigma_{1})$ and $s^{m_{1}}t^{m_{2}}\in
R_{A_{2}}$, and $\bm{m}$ lies on one of the lines $y=2x-r$ shown in Figure 2A,
then $s^{d_{1}}t^{d_{2}}\left(h_{2},-2d_{1}+d_{2}-1\right)!$, the generator of
$D(R_{A_{2}})_{\bm{d}}$, applied to such a monomial will either be a constant
times a monomial represented by a red lattice point on one of the lines
$y=2x-j$ for $0\leq j\leq r$ or, when
$\left(h_{2}(\bm{m}),-2d_{1}+d_{2}-1\right)!=0$, it will be $0$. Hence,

$\mathbb{I}(P_{1})_{\bm{d}}=D(R_{A_{2}})_{\bm{d}}\text{ for
}\bm{d}\in\textbf{C2}\quad\text{and}\quad
D(R,P_{1})_{\bm{d}}=D(R_{A_{2}})_{\bm{d}}\text{ for
}\bm{d}\in\textbf{C2}\setminus(-\sigma_{1}).$

Now if $\bm{d}\in\textbf{C4}$ and $s^{m_{1}}t^{m_{2}}\in P_{1}$ or
$\bm{d}\in\textbf{C4}\cup\sigma_{1}$ and $s^{m_{1}}t^{m_{2}}\in R_{A_{2}}$ and
$\bm{m}$ lies on one of the lines $y=r$ shown in Figure 2B, then
$s^{d_{1}}t^{d_{2}}\left(h_{1},-d_{2}-1\right)!$, the generator of
$D(R_{A_{2}})_{\bm{d}}$, applied to such a monomial will either be a constant
times a monomial represented by a red lattice point on one of the lines $y=j$
for $0\leq j\leq r$ or, when $\left(h_{1}(\bm{m}),-2d_{1}+d_{2}-1\right)!=0$,
it will be $0$. However, the monomials represented by the lattice points on
$y=0$ do not lie in $P_{1}$. The monomials represented by the red lattice
points on the line $y=-d_{2}$ are precisely the monomials whose image is a
term represented by a lattice point on $\sigma_{1}$. Hence, we need to further
right-multiply any operator in $D(R_{A_{2}})_{\bm{d}}$ by $\theta_{2}+d_{2}$,
so that

$\displaystyle\mathbb{I}(P_{1})_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle\text{
for }\bm{d}\in\textbf{C4}\text{ \ and \
}D(R_{A_{2}},P_{1})=s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle\text{
for }\bm{d}\in\textbf{C4}\cup\sigma_{1}.$

Using similar reasoning, we can easily see that

$\displaystyle\mathbb{I}(P_{1})_{\bm{d}}$
$\displaystyle=s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-2d_{1}+d_{2}-1\right)!\rangle\text{
for }\bm{d}\in\textbf{C3}\text{ and}$ $\displaystyle
D(R_{A_{2}},P_{1})_{\bm{d}}$
$\displaystyle=s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-2d_{1}+d_{2}-1\right)!\rangle\text{
for }\bm{d}\in\textbf{C3}\cup(-\sigma_{1}).$

Putting these all together, the graded pieces of $\mathbb{I}(P_{1})$ and
$D(R_{A_{2}},P_{1})$ are as follows:

$\mathbb{I}(P_{1})_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{if
}{\bm{d}}\in\textbf{C1}=\mathbb{N}A_{2},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-2d_{1}+d_{2}-1\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C2},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-2d_{1}+d_{2}-1\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C3},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C4},\end{cases}$
$D(R_{A_{2}},P_{1})_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{if
}s^{d_{1}}t^{d_{2}}\in P_{1},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-2d_{1}+d_{2}-1\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C2}\setminus(-\sigma_{1}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-2d_{1}+d_{2}-1\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C3}\cup(-\sigma_{1}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C4}\cup\sigma_{1}.\end{cases}$

Now taking the quotient, we obtain:

$\left(\displaystyle\frac{\mathbb{I}(P_{1})}{D(R_{A_{2}},P_{1})}\right)_{\bm{d}}=\begin{cases}0&\text{
if }{\bm{d}}\notin\mathbb{Z}\sigma_{1},\\\
s^{d_{1}}\cdot\dfrac{\mathbb{C}[\theta]}{\langle h_{1}\rangle}&\text{ if
}{\bm{d}}\in\sigma_{1},\\\
s^{d_{1}}\cdot\dfrac{\langle\left(h_{2},-2d_{1}-1\right)!\rangle}{\langle
h_{1}\left(h_{2},-2d_{1}-1\right)!\rangle}&\text{ if
}{\bm{d}}\in(-\sigma_{1}\setminus{\bf 0}).\end{cases}$

Viewing $D(\mathbb{C}[x])$ as a $\mathbb{Z}$-graded algebra over
$\mathbb{C}[\theta_{x}]$, we note that

$D(\mathbb{C}[x])=\sum\limits_{d=1}^{\infty}\partial_{x}^{d}\cdot\mathbb{C}[\theta_{x}]\oplus\sum\limits_{d=0}^{\infty}x^{d}\mathbb{C}[\theta_{x}]=\sum\limits_{d\in\mathbb{Z}}x^{d}\left(\theta_{x},-d-1\right)!\cdot\mathbb{C}[\theta_{x}].$

The map

$\phi\colon\sum\limits_{d\in\mathbb{Z}}x^{d}\cdot\left(\theta_{x},-d-1\right)!\cdot\mathbb{C}[\theta_{x}]\rightarrow\sum\limits_{d\in\mathbb{Z}}s^{d}\cdot\dfrac{\langle\left(h_{2},-2d-1\right)!\rangle}{\langle
h_{1}\left(h_{2},-2d-1\right)!\rangle},$

which is defined on the generators by

$\phi(x^{d}\left(\theta_{x},-d-1\right)!)=s^{d}\left(h_{2},-2d-1\right)!+\langle
h_{1}\left(h_{2},-2d-1\right)!\rangle,$

although a $\mathbb{C}$-vector space isomorphism, does not produce a ring
isomorphism. The graded pieces in negative degree are generated by polynomials
in $\theta$ whose degrees are twice as large as large as the degrees in
$\theta_{x}$ given in the Weyl algebra. In fact,

$D(R_{A_{2}}/P_{1})=\begin{cases}0&\text{ if
}{\bm{d}}\notin\mathbb{Z}\sigma_{1},\\\
s^{d_{1}}\cdot\dfrac{\mathbb{C}[\theta]}{\left\langle\dfrac{h_{1}}{2}\right\rangle}&\text{
if }{\bm{d}}\in\sigma_{1},\\\
s^{d_{1}}\cdot\dfrac{\left\langle\left(\dfrac{h_{2}}{2},-d_{1}-1\right)!\right\rangle}{\left\langle\dfrac{h_{1}}{2}\left(\dfrac{h_{2}}{2},-d_{1}-1\right)!\right\rangle}&\text{
if }{\bm{d}}\in(-\sigma_{1}\setminus{\bf 0}).\end{cases}$

Therefore, we can produce an isomorphism of graded rings $\psi\colon
D(\mathbb{C}[x])\rightarrow D(R_{A_{2}}/P_{1})$ defined for any
$m\in\mathbb{N}$ by

$\psi(x^{m})=s^{m},\quad\text{and}\quad\psi(\partial_{x}^{m})=s^{-m}\cdot\left(\frac{h_{2}}{2},m-1\right)!+\left\langle\frac{h_{1}}{2}\left(\frac{h_{2}}{2},m-1\right)!\right\rangle.$

######  Example 3.2.

We will now compute the graded pieces of the sets of differential operators
$\mathbb{I}(J)$ and $D(R_{A_{2}},J)$, as well as the graded pieces of
$JD(R_{A_{2}})$. Recall that

$\mathbb{I}(J)=\\{\delta\in D(R_{A_{2}})\mid\delta*J\subseteq
J\\}\quad\text{and}\quad D(R_{A_{2}},J)=\\{\delta\in
D(R_{A_{2}})\mid\delta*R\subseteq J\\}.$

Figure 3.3. Vanishing
for $\bm{d}=(-1,0)$

We will soon give a general formula for the graded pieces of $\mathbb{I}(J)$
and $D(R_{A_{2}},J)$; however, for illustrative purposes, first consider the
graded piece of $D(R_{A_{2}})$ at $(-1,0)$:
$s^{-1}\cdot\langle\left(h_{2},1\right)!\rangle.$ Applying
$s^{-1}\cdot\left(h_{2},1\right)!$ to a monomial whose exponent lies in the
two parallel half-lines $\sigma_{2}$ and $y=2x-1$ in $\mathbb{N}A$ will yield
$0$, which certainly lives inside $J$. However, when we let
$s^{-1}\cdot\left(h_{2},1\right)!$ act on a monomial whose exponent is a
member of the half-lines $y=2x-2$ or $y=0$ lying inside $\mathbb{N}A_{2}$, we
obtain an integer multiple of a monomial whose exponent lies in one of the
facets of $A_{2}$, $\sigma_{2}$ or $\sigma_{1}$ respectively, and these are
not in $J$. The remaining monomials in $J$ yield another element of $J$ when
they are acted upon by any operator in
$s^{-1}\cdot\langle\left(h_{2},1\right)!\rangle$ .

In Figure 3.3, the two light blue lines indicate the two half-lines
representing the multidegrees of monomials in $R_{A_{2}}$ that, after
application of an element in $D(R_{A_{2}})_{(-1,0)}$, fails to yield an
element in $J$. To correct for this lack of membership in $J$ for the
monomials on $y=2x-2$, every element of $D(R_{A_{2}})_{(-1,0)}$ should be
multiplied by $(h_{2}-2)$; applying $s^{-1}\cdot\left(h_{2},2\right)!$ to
these monomials yields $0$. The application of
$s^{-1}\cdot\left(h_{2},2\right)!$ to the remaining monomials in $J$ will
output a term inside $J$. Thus,

$\mathbb{I}(J)_{(-1,0)}=s^{-1}\cdot\langle\left(h_{2},2\right)!\rangle.$

Similarly, for every operator $\delta\in D(R_{A_{2}})_{(-1,0)}$,
$\delta(h_{2}-2)h_{1}\ast s^{m_{1}}t^{m_{2}}=0$ for every $\bm{m}$ on $y=2x-2$
or in $\sigma_{1}$. Also, $\delta(h_{2}-2)h_{1}\ast s^{m_{1}}t^{m_{2}}\in J$
for all other $\bm{m}$ in $\mathbb{N}A$, so

$D(R_{A_{2}},J)_{(-1,0)}=s^{-1}\cdot\langle
h_{1}\left(h_{2},2\right)!\rangle.$

In fact, using a similar argument applied to any $\bm{d}\in\textbf{C2}$ in the
case of $\mathbb{I}(J)_{\bm{d}}$ and
$\bm{d}\in\textbf{C2}\setminus(-\sigma_{1})$ for $D(R_{A_{2}},J)_{\bm{d}}$,
applying an operator in $D(R_{A_{2}})_{\bm{d}}$ to a monomial with exponent on
the half-lines $y=2x-j$ for $0\leq j<h_{2}({-\bm{d}})$ will give $0$; whereas,
these operators applied to a monomial with multidegrees on
$y=2x+h_{2}({\bm{d}})$ yields a constant multiple of a monomial with exponent
in $\sigma_{2}$. Hence, right-multiplying
$s^{d_{1}}t^{d_{2}}\cdot\left(h_{2},h_{2}({-\bm{d}})-1\right)!$ by
$h_{2}+h_{2}({\bm{d}})$ produces an operator that, when applied to monomials
with multidegrees on the lines $y=2x+h_{2}({\bm{d}})$, becomes $0$, and

$\displaystyle\mathbb{I}(J)_{\bm{d}}$
$\displaystyle=s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-2d_{1}+d_{2}\right)!\rangle\quad\text{for
all $\bm{d}\in\textbf{C2}$},\quad\text{and}$ $\displaystyle
D(R_{A_{2}},J)_{\bm{d}}$
$\displaystyle=s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-2d_{1}+d_{2}\right)!\rangle\quad\text{for
all $\bm{d}\in\textbf{C2}\setminus(-\sigma_{1})$}.$

We will discuss the multidegrees $\bm{d}\in\textbf{C2}\cap(-\sigma_{1})$
momentarily, when we turn to C3.

Figure 3.4. Vanishing
for $\bm{d}=(-1,-2)$

Determining the graded piece of multidegree $\bm{d}$ for both $\mathbb{I}(J)$
in C4 and $D(R_{A_{2}},J)$ in $\textbf{C4}\setminus(-\sigma_{2})$ is quite
similar to the arguments we used to determine $\mathbb{I}(J)_{\bm{d}}$ for
$\bm{d}$ in C2 and $D(R_{A_{2}},J)_{\bm{d}}$ for $\bm{d}$ in
$\textbf{C2}\setminus(\sigma_{1})$, respectively. We will briefly describe
$\mathbb{I}(J)_{(-1,-2)}$ and $D(R_{A_{2}},J)_{(-1,-2)}$ with the aid of
Figure 3.4 and then immediately describe the general case. Recall that
$D(R_{A_{2}})_{(-1,-2)}=s^{-1}t^{-2}\cdot\langle\left(h_{1},1\right)!\rangle$.
Similar to the argument for degree $\bm{d}=(-1,0)$ above, the monomials
corresponding to the multidegrees which lie on the two light blue lines ($y=2$
and $\sigma_{2}$) in Figure 3.4 are the only exponents of monomials that fail
to land inside $J$ after the application of
$s^{-1}t^{-2}\cdot\left(h_{1},1\right)!$.

To correct this deficiency, right-multiply by $(h_{1}-2)$ for
$\mathbb{I}(J)_{(-1,-2)}$ and $(h_{1}-2)h_{2}$ for $D(R_{A_{2}},J)_{(-1,-2)}$.
Notice that applying the operator $s^{-1}t^{-2}\cdot\left(h_{1},2\right)!$ to
a monomial corresponding to $\bm{d}\in\mathbb{N}A_{2}$ along the half-lines
$y=2$ or the operator $s^{-1}t^{-2}\cdot\left(h_{1},2\right)!h_{2}$ to a
monomial corresponding to $\bm{d}\in\mathbb{N}A_{2}$ along $y=2$ or $y=2x$ now
yields $0$, and no problems are created for the remaining monomials in $J$ or
$R_{A_{2}}$, respectively. Thus,

$\mathbb{I}(J)_{(-1,-2)}=s^{-1}t^{-2}\cdot\langle\left(h_{1},2\right)!\rangle\quad\text{and}\quad
D(R_{A_{2}},J)_{(-1,-2)}=s^{-1}t^{-2}\cdot\langle\left(h_{1},2\right)!h_{2}\rangle.$

In fact,

$\displaystyle\mathbb{I}(J)_{\bm{d}}$
$\displaystyle=s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle\quad\text{for
all }{\bm{d}}\in\textbf{C4},\quad\text{and}$ $\displaystyle
D(R_{A_{2}},J)_{\bm{d}}$
$\displaystyle=s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle\quad\text{for
all }{\bm{d}}\in\textbf{C4}\setminus(-\sigma_{2}).$

We will return to $D(R_{A_{2}},J)_{\bm{d}}$ for
$\bm{d}\in\textbf{C4}\cap(-\sigma_{2})$ when we turn to C3.

Figure 3.5. Vanishing
for $\bm{d}=(-1,-1)$

Determining $\mathbb{I}(J)_{\bm{d}}$ for $\bm{d}\in\textbf{C3}$ or
$D(R_{A_{2}},J)_{\bm{d}}$ for
$\bm{d}\in\textbf{C3}\cup(-\sigma_{1})\cup(-\sigma_{2})$ again is akin to the
arguments we gave above for $D(R_{A_{2}},J)_{(-1,0)}$ and
$D(R_{A_{2}},J)_{(-1,-2)}$. With the aid of Figure 3.5, we will briefly
describe

$\hskip 144.54pt\mathbb{I}(J)_{(-1,-1)}\quad\text{and}\quad
D(R_{A_{2}},J)_{(-1,-1)}.$

This explanation can easily be extended from $\bm{d}=(-1,-1)$ to all
multidegrees $\bm{d}$ in C3 (or in
$\textbf{C3}\cup(-\sigma_{1})\cup(-\sigma_{2})$ in the case of
$D(R_{A_{2}},J)$). If the operator $s^{-1}t^{-1}\cdot h_{1}h_{2}$, which is
the generator for $D(R_{A_{2}})_{(-1,-1)}$, is applied to any of the monomials
corresponding to multidegree $\bm{d}\in\mathbb{N}A_{2}$ along the two light
blue half-lines in Figure 3.5 (the portion of $y=1$ or $y=2x-1$ in C1), we
obtain an integer multiple of a monomial with exponent in the facets
$\sigma_{1}$ or $\sigma_{2}$, respectively, which is not in $J$, and no
problems are created for the remaining monomials in $J$ (or $R_{A_{2}}$ for
$D(R_{A_{2}},J)$).

Hence, right-multiplying by $(h_{1}-1)(h_{2}-1)$ yields a new operator
$s^{-1}t^{-1}\cdot\left(h_{1},1\right)!\left(h_{2},1\right)!$ that will send
to $0$ all monomials with multidegrees $\bm{d}\in\mathbb{N}A_{2}$ along the
half-lines $y=2x-1$ and $y=1$. No problems are created for the remaining
monomials in $R_{A_{2}}$ and we obtain

$\mathbb{I}(J)_{(-1,-1)}=D(R_{A_{2}},J)_{(-1,-1)}=s^{-1}t^{-1}\cdot\langle\left(h_{1},1\right)!\left(h_{2},1\right)!\rangle.$

In fact,

$\displaystyle\mathbb{I}(J)_{\bm{d}}=D(R_{A_{2}},J)_{\bm{d}}$
$\displaystyle=s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2})\right)!\left(h_{2},-2d_{1}+d_{2}\right)!\rangle\quad\text{
for all }\bm{d}\in\textbf{C3},$

and

$\displaystyle D(R_{A_{2}},J)_{\bm{d}}$
$\displaystyle=s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2})\right)!\left(h_{2},-2d_{1}+d_{2}\right)!\rangle\quad\text{
for all }\bm{d}\in(-\sigma_{1})\cup(-\sigma_{2}).$

Hence, the graded pieces of $\mathbb{I}(J)$ and $D(R_{A_{2}},J)$ are as
follows:

$\mathbb{I}(J)_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{if
}{\bm{d}}\in\textbf{C1}=\mathbb{N}A_{2},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-2d_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C2},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-2d_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C3},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C4},\end{cases}$
$D(R_{A_{2}},J)_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{if
}s^{d_{1}}t^{d_{2}}\in J,\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-2d_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in(\textbf{C2}\setminus(-\sigma_{1}))\cup(\sigma_{2}\setminus\\{\bm{0}\\}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-2d_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C3}\cup(-\sigma_{1})\cup(-\sigma_{2}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in(\textbf{C4}\setminus(-\sigma_{2}))\cup(\sigma_{1}\setminus\\{\bm{0}\\}).\end{cases}$

Now taking the quotient, we obtain:

$\left(\displaystyle\frac{\mathbb{I}(J)}{D(R_{A_{2}},J)}\right)_{\bm{d}}=\begin{cases}0&\text{
if }{\bm{d}}\notin(\mathbb{Z}\sigma_{1}\cup\mathbb{Z}\sigma_{2}),\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\mathbb{C}[\theta]}{\langle
h_{1}h_{2}\rangle}&\text{ if }{\bm{d}=\bm{0}},\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\mathbb{C}[\theta]}{\langle h_{1}\rangle}&\text{
if }{\bm{d}}\in\sigma_{1}\setminus\\{\bm{0}\\},\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\mathbb{C}[\theta]}{\langle h_{2}\rangle}&\text{
if }{\bm{d}}\in\sigma_{2}\setminus\\{\bm{0}\\},\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\langle\left(h_{2},-2d_{1}\right)!\rangle}{\langle
h_{1}\left(h_{2},-2d_{1}\right)!\rangle}&\text{ if
}{\bm{d}}\in(-\sigma_{1}),\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\langle\left(h_{1},-d_{2}\right)!\rangle}{\langle
h_{2}\left(h_{1},-d_{2}\right)!\rangle}&\text{ if
}{\bm{d}}\in(-\sigma_{2}).\end{cases}$

As both $J$ and $D(R_{A_{2}})$ are graded, we can similarly determine
$JD(R_{A_{2}})$. Our goal for the remainder of the section is to compute the
graded pieces of $JD(R_{A_{2}})$, in order to observe that, in this case,
$D(R_{A_{2}},J)=JD(R_{A_{2}})$.

To begin, note that for all $s^{d_{1}}t^{d_{2}}\in\mathbb{Z}^{2}$,

$D(R_{A_{2}})_{(d_{1}-1,d_{2}-1)}=s^{d_{1}-1}t^{d_{2}-1}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-2d_{1}+d_{2}\right)!\rangle.$

For $s^{m_{1}}t^{m_{2}}\in J$ the graded piece at the multidegree
$(d_{1}-m_{1},d_{2}-m_{2})$ will be

$s^{d_{1}-m_{1}}t^{d_{2}-m_{2}}\cdot\langle\left(h_{1},-(d_{2}-m_{2})-1\right)!\left(h_{2},-2(d_{1}-m_{1})+d_{2}-m_{2}-1\right)!\rangle.$

Note that since $m_{1}$ and $m_{2}$ are both positive,

$s^{m_{1}}t^{m_{2}}s^{d_{1}-m_{1}}t^{d_{2}-m_{2}}=s^{d_{1}}t^{d_{2}},$

and

$\langle\left(h_{1},-(d_{2}-c_{2})-1\right)!\left(h_{2},-2(d_{1}-c_{1})+d_{2}-c_{2}-1\right)!\rangle\subseteq\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-2d_{1}+d_{2}\right)!\rangle.$

Thus, to determine $(JD(R_{A_{2}}))_{\bm{d}}$ for any
$\bm{d}\in\mathbb{Z}^{2}$, it is enough to (left) multiply
$D(R_{A_{2}})_{(d_{1}-1,d_{2}-1)}$ by $st$. Hence, from our previous
computation, it follows that

$(JD(R_{A_{2}}))_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{
if }s^{d_{1}}t^{d_{2}}\in J,\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-2d_{1}+d_{2}\right)!\rangle&\text{
if
}{\bm{d}}\in(\textbf{C2}\setminus(-\sigma_{1}))\cup(\sigma_{1}\setminus\\{\bm{0}\\}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-2d_{1}+d_{2}\right)!\rangle&\text{
if }{\bm{d}}\in\textbf{C3}\cup(-\sigma_{1})\cup(-\sigma_{2}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle&\text{ if
}{\bm{d}}\in(\textbf{C4}\setminus(-\sigma_{2}))\cup(\sigma_{1}\setminus\\{\bm{0}\\}).\end{cases}$

Combining all cases together, it follows that for an arbitrary
$\bm{d}\in\mathbb{Z}^{2}$, there is an equality

$(JD(R_{A_{2}}))_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\left\langle\left(h_{1},h_{1}(-{\bm{d}})\right)!\left(h_{2},h_{2}(-{\bm{d}})\right)!\right\rangle=D(R_{A_{2}},J)_{\bm{d}}.$

##  4\. Differential operators on the rational normal curve of degree 3

In this section, we determine some subsets of the ring of differential
operators for the ring of the rational normal curve of degree $3$ determined
by its interior ideal $J$. We contrast these computations with the degree $2$
case that was determined in Section 3. Although the description follows the
same reasoning as the degree $2$ setting, we ultimately get quite different
behavior for the operators that end up in $JD(R_{A_{3}})$.

The ring of the rational normal curve of degree $3$ is
$R_{A_{3}}=\mathbb{C}[s,st,st^{2},st^{3}]$, and we will compute
$\mathbb{I}(J)/D(R_{A_{3}},J)$, where $J=\langle st,st^{2}\rangle$ is a
radical ideal. The facets of $A_{3}$ are

$\displaystyle\sigma_{1}=\\{(x,y)\in\mathbb{N}^{2}\mid x\geq
0,y=0\\}\quad\text{and}\quad\sigma_{2}=\\{(x,y)\in\mathbb{N}^{2}\mid x,y\geq
0,y=3x\\},$

which have primitive integral support functions

$h_{1}=\theta_{2}\quad\text{and}\quad h_{2}=3\theta_{1}-\theta_{2}.$

Figure 4.1 illustrates the integer lattice, divided into four chambers that
are colored as follows:

Figure 4.1. Chambers of $D(R_{A_{3}})$

  * C1:

The red multidegrees correspond to monomials in $J$, and the blue multidegrees
correspond to monomials in $R_{A_{2}}\setminus J$,

  * C2:

The yellow multidegrees are the $\bm{d}$ with $h_{1}({\bm{d}})\geq 0$ and
$h_{2}({\bm{d}})<0$,

  * C3:

The violet multidegrees are the $\bm{d}$ with both $h_{1}({\bm{d}})<0$ and
$h_{2}({\bm{d}})<0$, and

  * C4:

The green multidegrees are the $\bm{d}$ with $h_{1}({\bm{d}})<0$ and
$h_{2}({\bm{d}})\geq 0$.

Still following the convention $\left(h,n\right)!=1$ if $n<0$, by Theorem 2.3,
the graded pieces of $D(R_{A_{3}})$ are

$\hskip
144.54ptD(R_{A_{3}})_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\left\langle\left(h_{1},h_{1}({-\bm{d}})-1\right)!\left(h_{2},h_{2}({-\bm{d}})-1\right)!\right\rangle.$

Broken down by chambers, this amounts to:

$\hskip
144.54ptD(R_{A_{3}})_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{if
}\bm{d}\in\textbf{C1},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-3d_{1}+d_{2}-1\right)!\rangle&\text{if
}\bm{d}\in\textbf{C2},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}-1\right)!\left(h_{2},-3d_{1}+d_{2}-1\right)!\rangle&\text{if
}\bm{d}\in\textbf{C3},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}-1\right)!\rangle&\text{if
}\bm{d}\in\textbf{C4}.\\\ \end{cases}$

Determining the graded pieces of $\mathbb{I}(J)$ and $D(R_{A_{3}},J)$ in
$D(R_{A_{3}})$ is very similar to our computations in Section 3. Here we
include some visualizations for
$\bm{d}\in\\{(-1,0),(-1,-3),(-1,-1),(-1,-2)\\}$ in Figure 4.2 to aid our in
our description of how to obtain $\mathbb{I}(J)_{\bm{d}}$ and
$D(R_{A_{3}},J)_{\bm{d}}$. Then, we will list the expressions for
$\mathbb{I}(J)_{\bm{d}}$ and $D(R_{A_{3}},J)_{\bm{d}}$ by chamber, as we did
in Section 3.

A $\bm{d}=(-1,0)$.

B $\bm{d}=(-1,-3)$.

C $\bm{d}=(-1,-1)$.

D $\bm{d}=(-1,-2)$.

Figure 4.2. Vanishing at various $\bm{d}$

For each $\bm{d}$ in the illustrations in Figure 4.2, when we apply elements
of $D(R_{A_{3}})_{\bm{d}}$ to any of the monomials represented by lattice
points along the light blue lines, we obtain an integer multiple of a monomial
whose exponent lies on a facet. As in Section 3, these lines determine the
linear multiples $h_{i}-h_{i}(\bm{d})$ we must append to
$\mathbb{I}(\Omega(\bm{d}))$ to obtain either $\mathbb{I}(J)_{\bm{d}}$ or
$D(R_{A_{3}},J)$. Note that the light blue lines
$h_{i}(\bm{x})-h_{i}(\bm{d})=0$ that have contain a red dot determine the
$h_{i}-h_{i}(\bm{d})$ we right-multiply by to obtain $\mathbb{I}(J)_{\bm{d}}$
and that the light blue lines $h_{i}(\bm{x})-h_{i}(\bm{d})=0$ that have non-
empty intersection with the the entire cone determine the
$h_{i}-h_{i}(\bm{d})$ we right-multiply by to obtain
$D(R_{A_{3}},J)_{\bm{d}}$. Below we summarize the graded pieces of
$\mathbb{I}(J)$ and $D(R_{A_{3}},J)$, as we did in Section 3:

$\mathbb{I}(J)_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{if
}{\bm{d}}\in\mathbb{N}A_{3},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C4},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-3d_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C2},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-3d_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C3},\end{cases}$ (4.1) $\displaystyle
D(R_{A_{3}},J)_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{if
}s^{d_{1}}t^{d_{2}}\in J,\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in(\textbf{C4}\setminus(-\sigma_{2}))\cup(\sigma_{1}\setminus\\{\bm{0}\\}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-3d_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in(\textbf{C2}\setminus-\sigma_{1})\cup(\sigma_{2}\setminus\\{\bm{0}\\}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-3d_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C3}\cup(-\sigma_{1})\cup(-\sigma_{2}).\end{cases}$

Taking the quotient, we obtain:

$\left(\displaystyle\frac{\mathbb{I}(J)}{D(R_{A_{3}},J)}\right)_{\bm{d}}=\begin{cases}0&\text{if
}{\bm{d}}\notin(\mathbb{Z}\sigma_{1}\cup\mathbb{Z}\sigma_{2}),\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\mathbb{C}[\theta]}{\langle
h_{1}h_{2}\rangle}&\text{if }{\bm{d}=\bm{0}},\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\mathbb{C}[\theta]}{\langle
h_{1}\rangle}&\text{if }{\bm{d}}\in(\sigma_{1}\setminus\\{\bm{0}\\}),\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\mathbb{C}[\theta]}{\langle
h_{2}\rangle}&\text{if }{\bm{d}}\in(\sigma_{2}\setminus\\{\bm{0}\\}),\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\langle\left(h_{2},-3d_{1}\right)!\rangle}{\langle
h_{1}\left(h_{2},-3d_{1}\right)!\rangle}&\text{if
}{\bm{d}}\in(-\sigma_{1}\setminus\\{\bm{0}\\}),\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\langle\left(h_{1},-d_{2}\right)!\rangle}{\langle
h_{2}\left(h_{1},-d_{2}\right)!\rangle}&\text{if
}{\bm{d}}\in(-\sigma_{2}\setminus\\{\bm{0}\\}).\end{cases}$

In the remainder of the section, we compute the graded pieces of
$JD(R_{A_{3}})$ to show that they are not equal to the graded pieces of
$D(R_{A_{3}},J)$. This means that Proposition 1.1 does not hold for this ring.

A $(JD(R_{A_{3}}))_{\bm{d}}$ for $\bm{t}^{\bm{d}}\in J$.

B $(JD(R_{A_{3}}))_{\bm{d}}$ for $\bm{d}\in\textbf{C2}$.

C $(JD(R_{A_{3}}))_{\bm{d}}$ for $\bm{d}\in\textbf{C4}$.

D $(JD(R_{A_{3}}))_{\bm{d}}$ for $\bm{d}\in\textbf{C3}$.

Figure 4.3. Visualizing $(JD(R_{A_{3}}))_{\bm{d}}$.

To begin the computation, if $s^{m_{1}}t^{m_{2}}\in J$, then

$\displaystyle D(R_{A_{3}})_{(d_{1}-m_{1},d_{2}-m_{2})}$
$\displaystyle=s^{d_{1}-m_{1}}t^{d_{2}-m_{2}}\cdot\langle\left(h_{1},-(d_{2}-m_{2})-1\right)!\left(h_{2},-3(d_{1}-m_{1})+d_{2}-m_{2}-1\right)!\rangle.$

Further, since $m_{1},m_{2}\geq 1$,
$s^{m_{1}}t^{m_{2}}s^{d_{1}-m_{1}}t^{d_{2}-m_{2}}=s^{d_{1}}t^{d_{2}}$, and
there is a containment

$\displaystyle\langle\left(h_{1},-(d_{2}-m_{2})-1\right)!\left(h_{2},-3(d_{1}-m_{1})+d_{2}-m_{2}-1\right)!\rangle$
$\displaystyle\qquad\qquad\subseteq\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-3d_{1}+d_{2}+1\right)!,\left(h_{1},-d_{2}+1\right)!\left(h_{2},-3d_{1}+d_{2}\right)!\rangle.$

Thus, to determine the graded piece of $(JD(R_{A_{3}}))_{\bm{d}}$ for any
$\bm{d}\in\mathbb{Z}^{2}$, it is enough to consider $D(R_{A_{3}})$ in
multidegrees $(d_{1}-1,d_{2}-1)$ and $(d_{1}-1,d_{2}-2)$, where

$\displaystyle D(R_{A_{3}})_{(d_{1}-1,d_{2}-1)}$
$\displaystyle=s^{d_{1}-1}t^{d_{2}-1}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-3d_{1}+d_{2}+1\right)!\rangle\quad\text{and}$
$\displaystyle D(R_{A_{3}})_{(d_{1}-1,d_{2}-2)}$
$\displaystyle=s^{d_{1}-1}t^{d_{2}-2}\cdot\langle\left(h_{1},-d_{2}+1\right)!\left(h_{2},-3d_{1}+d_{2}\right)!\rangle.$

We will now break down this computation by chambers from Figure 4.1, leaving
off some half-lines along the way and addressing them as special cases later
on. For C1, Figure 3A helps to visualize that if $s^{d_{1}}t^{d_{2}}\in J$,
then at least one of the two multidegrees $(d_{1}-1,d_{2}-1)$ or
$(d_{1}-1,d_{2}-2)$ lives in $\mathbb{N}A$. Since
$D(R_{A_{3}})_{\bm{m}}=s^{m_{1}}t^{m_{2}}\cdot\mathbb{C}[\theta]$ for all
$\bm{m}\in\mathbb{N}A$,
$(JD(R_{A_{3}}))_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]$ for
$s^{d_{1}}t^{d_{2}}\in J$. We will consider the blue multidegrees in C1 with
their neighbors in C2 and C4.

Now consider the graded pieces of $(JD(R_{A_{3}}))_{\bm{d}}$ for $\bm{d}$ in
$\sigma_{2}$ or C2, excluding the two half-lines in that chamber given by
$-\sigma_{1}$ and $y=1$. Since we excluded the $x$-axis and $y=1$, both
$(d_{1}-1,d_{2}-1)$ and $(d_{1}-1,d_{2}-2)$ lie in C2 for all $\bm{d}$ under
consideration, see Figure 3B. Since

$\displaystyle D(R_{A_{3}})_{(d_{1}-1,d_{2}-1)}$
$\displaystyle=s^{d_{1}-1}t^{d_{2}-1}\cdot\langle\left(h_{2},-3d_{1}+d_{2}+1\right)!\rangle\quad\text{and}$
$\displaystyle D(R_{A_{3}})_{(d_{1}-1,d_{2}-2)}$
$\displaystyle=s^{d_{1}-1}t^{d_{2}-2}\cdot\langle\left(h_{2},-3d_{1}+d_{2}\right)!\rangle,$

and there is a containment
$\left\langle\left(h_{2},-3d_{1}+d_{2}+1\right)!\right\rangle\subseteq\left\langle\left(h_{2},-3d_{1}+d_{2}\right)!;\right\rangle$,
for such $\bm{d}$,

$(JD(R_{A_{3}}))_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\left\langle\left(h_{2},-3d_{1}+d_{2}\right)!\right\rangle.$

Now consider $(JD(R_{A_{3}}))_{\bm{d}}$ for $\bm{d}$ in $\sigma_{1}$ or in C4,
excluding those multidegrees that lie on $-\sigma_{2}$ or the half-line
$y=3x-1$. Since we excluded the multidegrees on $-\sigma_{2}$ and $y=3x-1$,
both $(d_{1}-1,d_{2}-1)$ and $(d_{1}-1,d_{2}-2)$ also lie in C4, as seen in
Figure 3C. Since

$\displaystyle D(R_{A_{3}})_{(d_{1}-1,d_{2}-1)}$
$\displaystyle=s^{d_{1}-1}t^{d_{2}-1}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle\quad\text{and}\quad$
$\displaystyle D(R_{A_{3}})_{(d_{1}-1,d_{2}-2)}$
$\displaystyle=s^{d_{1}-1}t^{d_{2}-2}\cdot\langle\left(h_{1},-d_{2}+1\right)!\rangle,$
and there is a containment
$\displaystyle\langle\left(h_{1},-d_{2}+1\right)!\rangle$
$\displaystyle\subseteq\langle\left(h_{1},-d_{2}\right)!\rangle,$ it follows
that for such $\bm{d}$, $\displaystyle(JD(R_{A_{3}}))_{\bm{d}}$
$\displaystyle=s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle.$

For $(JD(R_{A_{3}}))_{\bm{d}}$ for $\bm{d}$ in C3, $-\sigma_{1}$, or
$-\sigma_{2}$, see Figure 3D,

$\displaystyle D(R_{A_{3}})_{(d_{1}-1,d_{2}-1)}$
$\displaystyle=s^{d_{1}-1}t^{d_{2}-1}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-3d_{1}+d_{2}+1\right)!\rangle,$
$\displaystyle D(R_{A_{3}})_{(d_{1}-1,d_{2}-2)}$
$\displaystyle=s^{d_{1}-1}t^{d_{2}-2}\cdot\langle\left(h_{1},-d_{2}+1\right)!\left(h_{2},-3d_{1}+d_{2}\right)!\rangle.$

Since the ideals

$\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-3d_{1}+d_{2}+1\right)!\rangle\quad\text{and}\quad\langle\left(h_{1},-d_{2}+1\right)!\left(h_{2},-3d_{1}+d_{2}\right)!\rangle$

are incomparable, it follows that for such $\bm{d}$,

$\displaystyle(JD$
$\displaystyle(R_{A_{3}}))_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-3d_{1}+d_{2}+1\right)!,\left(h_{1},-d_{2}+1\right)!\left(h_{2},-3d_{1}+d_{2}\right)!\rangle.$

Figure 4.4. ($JD(R_{A_{3}}))_{\bm{d}}$ for $\bm{d}$ on exceptional lines in C2
and C4

The multidegrees $\bm{d}$ that we have not yet considered for
$(JD(R_{A_{3}}))_{\bm{d}}$ are those in C2 along $y=1$ and in C4 along
$y=3x-1$. For such ${\bm{d}}$, one of $(d_{1}-1,d_{2}-1)$ or
$(d_{1}-1,d_{2}-2)$ belongs to C3, see Figure 4.4. First, for $\bm{d}$ in C2
along the line $y=1$, $(d_{1}-1,-1)\in\textbf{C3}$, and

$\hskip 144.54ptD(R_{A_{3}})_{(d_{1}-1,-1)}=s^{d_{1}-1}t^{-1}\cdot\langle
h_{1}\left(h_{2},-3d_{1}+1\right)!\rangle.$

Combining the fact that
$D(R_{A_{3}})_{(d_{1}-1,0)}=s^{d_{1}-1}\cdot\langle\left(h_{2},-3d_{1}+2\right)!\rangle$,
we compute that

$\hskip
144.54pt(JD(R_{A_{3}}))_{\bm{d}}=s^{d_{1}}t\cdot\langle\left(h_{2},-3d_{1}+2\right)!,h_{1}\left(h_{2},-3d_{1}+1\right)!\rangle.$

Second, for $\bm{d}\in\textbf{C4}$ along the line $y=3x-1$,

$\hskip
144.54ptD(R_{A_{3}})_{(d_{1}-1,3d_{1}-2)}=s^{d_{1}-1}t^{3d_{1}-2}\cdot\langle\left(h_{1},-3d_{1}+1\right)!h_{2}\rangle.$

Connecting with the fact that

$\hskip
144.54ptD(R_{A_{3}})_{(d_{1}-1,3d_{1}-3)}=s^{d_{1}-1}t^{3d_{1}-3}\cdot\langle\left(h_{1},-3d_{1}+2\right)!\rangle,$

we determine that

$\hskip
144.54pt(JD(R_{A_{3}}))_{\bm{d}}=s^{d_{1}}t^{3d_{1}-1}\cdot\langle\left(h_{1},-3d_{1}+1\right)!h_{2},\left(h_{1},-3d_{1}+2\right)!\rangle.$

Having now computed $JD(R_{A_{3}})$ in all multidegrees, we have that

$(JD(R_{A_{3}}))_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{if
}s^{d_{1}}t^{d_{2}}\in J,\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-3d_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\left(\textbf{C2}\setminus(\sigma_{1}\cup\\{y=1\\})\right)\cup(\sigma_{2}\setminus\\{\bm{0}\\}),\\\
s^{d_{1}}t^{d_{2}}\cdot\\\ \ \ \
\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-3d_{1}+d_{2}+1\right)!,&\\\ \ \
\quad\left(h_{1},-d_{2}+1\right)!\left(h_{2},-3d_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C3}\cup(-\sigma_{1}\cup-\sigma_{2})\cup(\mathbb{N}A_{3}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\left(\textbf{C4}\setminus(\sigma_{2}\cup\\{y=3x-1\\})\right)\cup(\sigma_{1}\setminus\\{\bm{0}\\}).\end{cases}$

Now compare this with $D(R_{A_{3}},J)$ from (4.1); it becomes clear that
$(JD(R_{A_{3}}))_{\bm{d}}\neq D(R_{A_{3}},J)_{\bm{d}}$ whenever $\bm{d}$
belongs to any of the following: $-\sigma_{1}$, $-\sigma_{2}$, C3, C2 along
the half-line $\\{y=1\\}$, or C4 along the half-line $\\{y=3x-1\\}$. In
particular, $\mathbb{I}(J)/D(R_{A_{3}},J)\neq\mathbb{I}(J)/JD(R_{A_{3}})$, in
contrast to Proposition 1.1.

##  5\. Differential operators on a rational normal curve

The techniques used in Sections 3 and 4 can also be applied to compute
$\mathbb{I}(J)_{\bm{d}}$, $D(R_{A_{n}},J)_{\bm{d}}$,
$(JD(R_{A_{n}}))_{\bm{d}}$ and
$\left(\mathbb{I}(J)/D(R_{A_{n}},J)\right)_{\bm{d}}$, where the radical ideal
$J=\langle st,st^{2},\ldots,st^{n-1}\rangle$ is again the intersection of the
primes defined by the facets of $A_{n}$. We denote the facets of $A_{n}$ by

$\displaystyle\sigma_{1}=\\{(x,y)\in\mathbb{N}^{2}\mid x\geq
0,y=0\\}\quad\text{and}\quad\sigma_{2}=\\{(x,y)\in\mathbb{N}^{2}\mid x,y\geq
0,y=3x\\},$

which have primitive integral support functions

$h_{1}=\theta_{2}\quad\text{and}\quad h_{2}=3\theta_{1}-\theta_{2}.$

Again, we divide $\mathbb{Z}^{2}$ into chambers, analogous to those used in
Sections 3 and 4, so

  * C1

$=\\{\bm{d}\in\mathbb{Z}^{2}\mid\bm{d}\in\mathbb{N}A\\}$,

  * C2

$=\\{\bm{d}\in\mathbb{Z}^{2}\mid h_{1}({\bm{d}})\geq 0,h_{2}({\bm{d}})<0\\}$,

  * C3

$=\\{\bm{d}\in\mathbb{Z}^{2}\mid h_{1}({\bm{d}})<0,h_{2}({\bm{d}})<0\\}$, and

  * C4

$=\\{\bm{d}\in\mathbb{Z}^{2}\mid h_{1}({\bm{d}})<0,h_{2}({\bm{d}})\geq 0\\}$.

These computations yield the following formulas:

$\mathbb{I}(J)_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{if
}{\bm{d}}\in\mathbb{N}A_{n},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C4},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-nd_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C2},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-nd_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C3},\end{cases}$
$D(R_{A_{n}},J)_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{if
}s^{d_{1}}t^{d_{2}}\in J,\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in(\textbf{C4}\setminus(-\sigma_{2}))\cup(\sigma_{1}\setminus\\{\bm{0}\\}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-nd_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in(\textbf{C2}\setminus(-\sigma_{1}))\cup(\sigma_{2}\setminus\\{\bm{0}\\}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},-d_{2}\right)!\left(h_{2},-nd_{1}+d_{2}\right)!\rangle&\text{if
}{\bm{d}}\in\textbf{C3}\cup(-\sigma_{1}\cup-\sigma_{2}),\\\ \end{cases}$
$(JD(R_{A_{n}}))_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{if
}s^{d_{1}}t^{d_{2}}\in J,\\\
s^{d_{1}}t^{d_{2}}\cdot\left\langle\left(h_{1},-d_{2}\right)!\right\rangle&\text{if
}{\bm{d}}\in\textbf{C4}\setminus\left(\displaystyle\bigcup_{j=0}^{n-2}\\{y=nx-j\\}\right),\\\
s^{d_{1}}t^{d_{2}}\cdot\left\langle\left(h_{2},-nx+y\right)!\right\rangle&\text{if
}{\bm{d}}\in\textbf{C2}\setminus\left(\displaystyle\bigcup_{j=0}^{n-2}\\{y=j\\}\right),\\\
s^{d_{1}}t^{d_{2}}\cdot\left\langle\left(h_{1},-d_{2}+j\right)!\,\cdot\right.&\\\
\ \
\left.\left(h_{2},-nd_{1}+d_{2}+n-2-j\right)!\right\rangle_{j=0}^{n-2}&\text{if
}{\bm{d}}\text{ in
}\textbf{C3}\cup\left[\textbf{C2}\cap\left(\displaystyle\bigcup_{j=0}^{n-2}\\{y=j\\}\right)\right]\cup\\\
&\qquad\qquad\qquad\ \ \
\left[\textbf{C4}\cap\left(\displaystyle\bigcup_{j=0}^{n-2}\\{y=nx-j\\}\right)\right].\end{cases}$

Recall from Section 4 that $JD(R_{A_{3}})\neq D(R_{A_{3}},J)$; we see that
this is true for all rings of rational normal curves
$R_{A_{n}}=\mathbb{C}[s,st,\dots,st^{n}]$ with $n\geq 3$. Specifically,
comparing $D(R_{A_{n}},J)_{\bm{d}}$ and $(JD(R_{A_{n}}))_{\bm{d}}$ for various
$\bm{d}\in\mathbb{Z}^{2}$, the graded pieces differ whenever $\bm{d}$ belongs
to $\sigma_{1}$, $\sigma_{2}$, C3, the half-lines in C2 inside

$\\{(x,y)\in\mathbb{R}^{2}\mid x<0,y=i,0\leq i\leq n-2\\},$

or the half-lines in C4 inside

$\\{(x,y)\in\mathbb{R}^{2}\mid x<0,y=nx-i,0\leq i\leq n-2\\}.$

Figure 5.1 has these multidegrees $\bm{d}$ in blue for the case $n=7$.

Figure 5.1.  $(JD(R_{A_{7}}))_{\bm{d}}$ and $D(R_{A_{7}},J)_{\bm{d}}$ differ
at blue $\bm{d}$

As with the rational normal cones in degrees $2$ and $3$ from Sections 3 and
4, for $\bm{d}\in\mathbb{Z}^{2}$,

(5.1)
$\left(\dfrac{\mathbb{I}(J)}{D(R_{A_{n}},J)}\right)_{\bm{d}}=\begin{cases}0&\text{if
}{\bm{d}}\in\mathbb{Z}^{2}\setminus(\mathbb{Z}\sigma_{1}\cup\mathbb{Z}\sigma_{2}),\\\
\dfrac{\mathbb{C}[\theta]}{\langle h_{1}h_{2}\rangle}&\text{if
}{\bm{d}=\bm{0}},\\\
s^{d_{1}}\cdot\dfrac{\langle\left(h_{2},-nd_{1}\right)!\rangle}{\langle
h_{1}\left(h_{2},-nd_{1}\right)!\rangle}&\text{if
}{\bm{d}}\in\mathbb{Z}\sigma_{1}\setminus\\{\bm{0}\\},\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\langle\left(h_{1},-d_{2}\right)!\rangle}{\langle
h_{2}\left(h_{1},-d_{2}\right)!\rangle}&\text{if
}{\bm{d}}\in\mathbb{Z}\sigma_{2}\setminus\\{\bm{0}\\}.\end{cases}\hskip
108.405pt$

Example 2.5 considered the ordinary double point $\mathbb{C}[x,y]/\langle
xy\rangle$, which is isomorphic to $R_{A_{n}}/J$ for all $n\geq 1$. Comparing
$D(\mathbb{C}[x,y]/\langle xy\rangle))_{\bm{d}}$ from (2.2) and
$\left(\mathbb{I}(J)/D(R_{A_{n}},J)\right)_{\bm{d}}$ from (5.1), we see that
there is an isomorphism between the graded components, viewed as
$\mathbb{C}$-vector spaces, given by $\varphi\colon
D\left({\mathbb{C}[x,y]}/{\langle xy\rangle}\right)\rightarrow
D\left({R_{A_{n}}}/{J}\right)$ with, for $m\in\mathbb{Z}$,

$\displaystyle\varphi\left(\frac{\mathbb{C}[\theta_{x},\theta_{y}]}{\langle\theta_{x}\theta_{y}\rangle}\right)$
$\displaystyle=\displaystyle\frac{\mathbb{C}[\theta]}{\langle
h_{1}h_{2}\rangle},\hskip 108.405pt$
$\displaystyle\varphi\left(x^{m}\displaystyle\frac{\langle\left(\theta_{x},-m\right)!\rangle}{\langle\left(\theta_{x},-m\right)!\theta_{y}\rangle}\right)$
$\displaystyle=s^{m}\cdot\displaystyle\frac{\langle\left(h_{2},-nm\right)!\rangle}{\langle\left(h_{2},-nm\right)!h_{1}\rangle},\quad\text{and}\hskip
108.405pt$
$\displaystyle\varphi\left(y^{m}\cdot\displaystyle\frac{\langle\left(\theta_{y},-m\right)!\rangle}{\langle\left(\theta_{y},-m\right)!\theta_{x}\rangle}\right)$
$\displaystyle=s^{m}t^{nm}\cdot\displaystyle\frac{\langle\left(h_{1},-nm\right)!\rangle}{\langle\left(h_{1},-nm\right)!h_{2}\rangle}.\hskip
108.405pt$

However, this isomorphism is not an isomorphism of rings. Again as in Example
3.1, the degree of the polynomial generator in $\theta_{1}$ and $\theta_{2}$
in multidegree $\bm{d}$ is $n$ times the degree of the polynomial generator of
in $\theta_{x}$ and $\theta_{y}$. In fact, noting that along
$\mathbb{Z}\sigma_{2}$, $d_{2}=nd_{1}$, we have

(5.2) $D\left(R_{A_{n}}/J\right)_{\bm{d}}=\begin{cases}0&\text{if
}{\bm{d}}\in\mathbb{Z}^{2}\setminus(\mathbb{Z}\sigma_{1}\cup\mathbb{Z}\sigma_{2}),\\\
\dfrac{\mathbb{C}[\theta]}{\left\langle\dfrac{h_{1}}{n}\dfrac{h_{2}}{n}\right\rangle}&\text{if
}{\bm{d}=\bm{0}},\\\
s^{d_{1}}\cdot\dfrac{\left\langle\left(\dfrac{h_{2}}{n},-d_{1}\right)!\right\rangle}{\left\langle\dfrac{h_{1}}{n}\left(\dfrac{h_{2}}{n},-d_{1}\right)!\right\rangle}&\text{if
}{\bm{d}}\in\mathbb{Z}\sigma_{1}\setminus\\{\bm{0}\\},\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\left\langle\left(\dfrac{h_{1}}{n},-d_{1}\right)!\right\rangle}{\left\langle\dfrac{h_{2}}{n}\left(\dfrac{h_{1}}{n},-d_{1}\right)!\right\rangle}&\text{if
}{\bm{d}}\in\mathbb{Z}\sigma_{2}\setminus\\{\bm{0}\\}.\end{cases}$

Now comparing (2.2) and (5.2), we see there is an isomorphism of the ring of
differential operators between the two rings given by, for each
$m\in\mathbb{Z}$,

$\psi(\left(\theta_{x},-m\right)!+\langle\left(\theta_{x},-m\right)!\theta_{y}\rangle)=s^{-m}\cdot\left(\frac{h_{2}}{n},-m\right)!+\left\langle\left(\frac{h_{2}}{n},-m\right)!h_{1}\right\rangle,$
$\psi(\left(\theta_{y},-m\right)!+\langle\left(\theta_{y},-m\right)!\theta_{x}\rangle)=s^{-m}t^{-nm}\cdot\left(\frac{h_{1}}{n},-m\right)!+\left\langle\left(\frac{h_{1}}{n},-m\right)!h_{2}\right\rangle,$
$\psi(x^{m})=s^{m},\quad\text{and}\quad\psi(y^{m})=s^{m}t^{nm}.$

##  6\. Higher dimensional examples

Thus far, we have considered differential operators determined by radical
ideals in two-dimensional semigroup rings. In this section, we turn to looking
at some of the differential operators in the three-dimensional semigroup ring
given by

$A=\begin{bmatrix}1&1&1&1\\\ 0&1&0&1\\\ 0&0&1&1\\\ \end{bmatrix}$

and describe some subsets of differential operators of $R_{A}$ given by two
different radical ideals $J$. For both choices of $J$, we will compute the
graded pieces $\mathbb{I}(J)_{\bm{d}}$, $D(R_{A},J)_{\bm{d}}$ and
$\left(\mathbb{I}(J)/D(R_{A},J)\right)_{\bm{d}}$. For the given matrix $A$,

$R_{A}=\mathbb{C}[\mathbb{N}A]=\mathbb{C}[t_{1},t_{1}t_{2},t_{1}t_{3},t_{1}t_{2}t_{3}].$

The facets of the cone $\mathbb{R}_{\geq 0}A$ are

$\sigma_{1}=\mathbb{N}\\{e_{1},e_{1}+e_{2}\\},\ \ \
\sigma_{2}=\mathbb{N}\\{e_{1},e_{1}+e_{3}\\},\ \ \
\sigma_{3}=\mathbb{N}\\{e_{1}+e_{3},e_{1}+e_{2}+e_{3}\\},\ \ \
\sigma_{4}=\mathbb{N}\\{e_{1}+e_{2},e_{1}+e_{2}+e_{3}\\},$

and the corresponding primitive integral support functions are

$h_{1}=h_{1}(\theta)=\theta_{3},\quad h_{2}=h_{2}(\theta)=\theta_{2},\quad
h_{3}=h_{3}(\theta)=\theta_{1}-\theta_{3},\quad
h_{4}=h_{4}(\theta)=\theta_{1}-\theta_{2}.$

The prime ideals associated to the facets are

$P_{\sigma_{1}}=\langle t_{1}t_{3},t_{1}t_{2}t_{3}\rangle,\quad
P_{\sigma_{2}}=\langle t_{1}t_{2},t_{1}t_{2}t_{3}\rangle,\quad
P_{\sigma_{3}}=\langle t_{1},t_{1}t_{2}\rangle,\quad P_{\sigma_{4}}=\langle
t_{1},t_{1}t_{3}\rangle,$

and the prime ideals associated to the rays (or $1$-dimensional faces) of the
cone are

$\displaystyle P_{\sigma_{1}\cap\sigma_{2}}=\langle
t_{1}t_{2},t_{1}t_{3},t_{1}t_{2}t_{3}\rangle,\qquad$ $\displaystyle
P_{\sigma_{2}\cap\sigma_{3}}=\langle t_{1},t_{1}t_{2},t_{1}t_{2}t_{3}\rangle,$
$\displaystyle P_{\sigma_{3}\cap\sigma_{4}}=\langle
t_{1},t_{1}t_{2},t_{1}t_{3}\rangle,$ $\displaystyle
P_{\sigma_{1}\cap\sigma_{4}}=\langle t_{1},t_{1}t_{3},t_{1}t_{2}t_{3}\rangle.$

Figure 6.1. Chambers for
$\mathbb{C}[t_{1},t_{1}t_{2},t_{1}t_{3},t_{1}t_{2}t_{3}]$

In contrast with the two dimensional cases considered in Sections 3, 4, and 5,
the increased dimension of the semigroup $\mathbb{N}A$ allows for many more
choices for radical monomial ideal $J$ in $R_{A}$. We will compute
$\mathbb{I}(J)/D(R_{A},J)$ for two such choices,

$\displaystyle\hskip 173.44756ptJ=P_{\sigma_{1}}\cap P_{\sigma_{2}}\cap
P_{\sigma_{3}}\cap P_{\sigma_{4}}=\langle
t_{1}^{2}t_{2}t_{3}\rangle\quad\text{and}$ $\displaystyle\hskip
173.44756ptJ=P_{\sigma_{1}}\cap P_{\sigma_{2}}\cap
P_{\sigma_{3}\cap\sigma_{4}}=\langle
t_{1}^{2}t_{2}t_{3},t_{1}^{2}t_{2}^{2}t_{3},t_{1}^{2}t_{2}t_{3}^{2}\rangle.$

For both choices of $J$, to compute the graded pieces of $\mathbb{I}(J)$, we
will divide up $\mathbb{Z}^{3}$ into 14 chambers, depending on various
combinations of signs of $h_{i}({\bm{d}})$. Table 6.1 describes the 14
chambers in a list, while Figure 6.1 illustrates them, with chamber C1 (given
by $\mathbb{N}A$) in the top right with some of the lattice points of
$\mathbb{N}A$ shown. Note the $x$-axis is the vertical axis in this picture.

We must modify the chambers slightly to compute the graded pieces of
$D(R_{A},J)$, as we now consider which differential operators, when applied to
an element in $R_{A}$, yield an element in $J$. Although all of the operators
in the graded pieces of $\mathbb{I}(J)$ lying on the facets send $J$ into $J$,
it is not necessarily the case that these operators applied to an element of
$R_{A}$ will output an element in $J$. So we will switch the lattice points on
the facets that correspond to monomials that do not lie in $J$ to lie in the
adjacent chambers by interchanging $\geq$ with $>$. We will describe these new
_regions_ within the examples.

######  Example 6.1.

First consider the ideal

$J=P_{\sigma_{1}}\cap P_{\sigma_{2}}\cap P_{\sigma_{3}}\cap
P_{\sigma_{4}}=\langle t_{1}^{2}t_{2}t_{3}\rangle\quad\text{in}\quad
R_{A}=\mathbb{C}[t_{1},t_{1}t_{2},t_{1}t_{3},t_{1}t_{2}t_{3}].$

Figure 6.2. Cone of $R_{A}$ with ideal $J=\langle t_{1}^{2}t_{2}t_{3}\rangle$

$\begin{array}[]{@{\extracolsep{\fill}}
|c|l|l|}\hline\cr\text{Chamber}&\text{Halfspace inequalities}&\text{Lattice
point inequalities}\\\ \hline\cr\textbf{C1}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}}),h_{3}({\bm{d}}),h_{4}({\bm{d}})\geq
0\\}&\mathbb{N}A\\\ \hline\cr\textbf{C2}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{3}({\bm{d}})\geq
0,h_{2}({\bm{d}})>0,h_{4}({\bm{d}})<0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{2}>d_{1}\geq d_{3}\geq 0\\}\\\
\hline\cr\textbf{C3}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{2}({\bm{d}}),h_{4}({\bm{d}})\geq
0,h_{1}({\bm{d}})>0,h_{3}({\bm{d}})<0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{3}>d_{1}\geq d_{2}\geq 0\\}\\\
\hline\cr\textbf{C4}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{3}({\bm{d}})\geq
0,h_{4}({\bm{d}})>0,h_{2}({\bm{d}})<0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{1}\geq d_{3}\geq 0>d_{2}\\}\\\
\hline\cr\textbf{C5}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{2}({\bm{d}}),h_{4}({\bm{d}})\geq
0,h_{3}({\bm{d}})>0,h_{1}({\bm{d}})<0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{1}\geq d_{2}\geq 0>d_{3}\\}\\\
\hline\cr\textbf{C6}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}})\geq
0,h_{3}({\bm{d}}),h_{4}({\bm{d}})<0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{3}\geq 0,d_{2}\geq 0,d_{2}>d_{1},d_{3}>d_{1}\\}\\\
\hline\cr\textbf{C7}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{2}({\bm{d}}),h_{3}({\bm{d}})\geq
0,h_{1}({\bm{d}}),h_{4}({\bm{d}})<0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{2}\geq 0>d_{3},d_{2}>d_{1}\geq d_{3}\\}\\\
\hline\cr\textbf{C8}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{4}({\bm{d}})\geq
0,h_{2}({\bm{d}}),h_{3}({\bm{d}})<0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{3}\geq 0>d_{2},d_{3}>d_{1}\geq d_{2}\\}\\\
\hline\cr\textbf{C9}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{3}({\bm{d}}),h_{4}({\bm{d}})\geq
0,h_{1}({\bm{d}}),h_{2}({\bm{d}})<0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid
0>d_{2},0>d_{3},d_{1}\geq d_{2},d_{1}\geq d_{3}\\}\\\
\hline\cr\textbf{C10}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{3}({\bm{d}}),h_{4}({\bm{d}})<0,h_{2}({\bm{d}})\geq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{2}\geq 0>d_{3}>d_{1}\\}\\\
\hline\cr\textbf{C11}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{2}({\bm{d}}),h_{3}({\bm{d}}),h_{4}({\bm{d}})<0,h_{1}({\bm{d}})\geq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{3}\geq 0>d_{2}>d_{1}\\}\\\
\hline\cr\textbf{C12}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}}),h_{3}({\bm{d}})<0,h_{4}({\bm{d}})\geq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid 0>d_{3}>d_{1}\geq d_{2}\\}\\\
\hline\cr\textbf{C13}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}}),h_{4}({\bm{d}})<0,h_{3}({\bm{d}})\geq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid 0>d_{2}>d_{1}\geq d_{3}\\}\\\
\hline\cr\textbf{C14}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}}),h_{3}({\bm{d}}),h_{4}({\bm{d}})<0\\}&-\text{Int}(\mathbb{N}A)\\\
\hline\cr\end{array}$

Table 6.1. The chambers used to compute $\mathbb{I}(J)$ in Examples 6.1 and
6.2.

In Figure 6.2, the multidegrees in the gray cone whose vertex lies at
$(2,1,1)$ correspond to the monomials that lie in $J$, and the monomials in
the outer cone lie in $R_{A}\setminus J$. Note also that the view of the cone
that we see in Figure 6.2 is from the side of the $xz$-plane.

Since none of the points on the facets of the cone have monomials that lie in
$J$, the 14 regions that we consider in determining $D(R_{A},J)$ are listed in
Table 6.2.

If ${\bm{d}}$ is in C1 (which are the lattice points corresponding to points
in the semigroup) when considering the idealizer or R1 (which are the lattice
points corresponding to the monomials in $J$) when considering $D(R_{A},J)$,
then $D(R_{A})_{\bm{d}}$ is generated by multiplication by
${\bm{t}^{\bm{d}}}$. Multiplying any element of $J$ by ${\bm{t}^{\bm{d}}}$
remains in $J$ making the graded pieces of
$\mathbb{I}(J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\mathbb{C}[\theta]$ and
$D(R_{A},J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\mathbb{C}[\theta]$.

A ${\bm{d}}=(1,2,1)\in\textbf{C2}$ and $\\{x-y+h_{4}({\bm{d}})=0\\}$

B ${\bm{d}}=(-1,0,0)\in\textbf{C6}$, $\\{x-y+h_{4}({\bm{d}})=0\\}$, and
$\\{x-z+h_{4}({\bm{d}})=0\\}$

C ${\bm{d}}=(-2,0,-1)\in\textbf{C10}$, $\\{z+h_{1}({\bm{d}})=0\\}$,
$\\{x-z+h_{3}({\bm{d}})=0\\}$, and $\\{x-y+h_{4}({\bm{d}})=0\\}$

Figure 6.3. Vanishing planes

$\begin{array}[]{@{\extracolsep{\fill}}
|c|l|l|}\hline\cr\text{Region}&\text{Halfspace inequalities}&\text{Lattice
point inequalities}\\\ \hline\cr\textbf{R1}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}}),h_{3}({\bm{d}}),h_{4}({\bm{d}})>0\\}&\text{Int}(\mathbb{N}A)\\\
\hline\cr\textbf{R2}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}}),h_{3}({\bm{d}})>0,h_{4}({\bm{d}})\leq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{2}\geq d_{1}>d_{3}>0\\}\\\
\hline\cr\textbf{R3}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}}),h_{4}({\bm{d}})>0,h_{3}({\bm{d}})\leq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{3}\geq d_{1}>d_{2}>0\\}\\\
\hline\cr\textbf{R4}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{3}({\bm{d}}),h_{4}({\bm{d}})>0,h_{2}({\bm{d}})\leq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{1}>d_{3}>0\geq d_{2}\\}\\\
\hline\cr\textbf{R5}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{2}({\bm{d}}),h_{3}({\bm{d}}),h_{4}({\bm{d}})>0,h_{1}({\bm{d}})\leq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{1}>d_{2}>0\geq d_{3}\\}\\\
\hline\cr\textbf{R6}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}})>0,h_{3}({\bm{d}}),h_{4}({\bm{d}})\leq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{2}>0,d_{3}>0,d_{2}\geq d_{1},d_{3}\geq
d_{1}\\}\\\ \hline\cr\textbf{R7}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{2}({\bm{d}}),h_{3}({\bm{d}})>0,h_{1}({\bm{d}}),h_{4}({\bm{d}})\leq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{2}>0\geq d_{3},d_{2}\geq
d_{1}>d_{3}\\}\\\ \hline\cr\textbf{R8}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{4}({\bm{d}})>0,h_{2}({\bm{d}}),h_{3}({\bm{d}})\leq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{3}>0\geq d_{2},d_{3}\geq
d_{1}>d_{2}\\}\\\ \hline\cr\textbf{R9}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{3}({\bm{d}}),h_{4}({\bm{d}})>0,h_{1}({\bm{d}}),h_{2}({\bm{d}})\leq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{1}>d_{2},d_{1}>d_{3},0\geq d_{2},0\geq
d_{3}\\}\\\ \hline\cr\textbf{R10}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{3}({\bm{d}})\leq
0,h_{4}({\bm{d}})<0,h_{2}({\bm{d}})>0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{2}>0\geq d_{3}\geq d_{1}\\}\\\
\hline\cr\textbf{R11}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{2}({\bm{d}}),h_{4}({\bm{d}})\leq
0,h_{3}({\bm{d}})<0,h_{1}({\bm{d}})>0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{3}>0\geq d_{2}\geq d_{1}\\}\\\
\hline\cr\textbf{R12}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{3}({\bm{d}})\leq
0,h_{2}({\bm{d}})<0,h_{4}({\bm{d}})>0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid 0\geq
d_{3}\geq d_{1}>d_{2}\\}\\\
\hline\cr\textbf{R13}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{2}({\bm{d}}),h_{4}({\bm{d}})\leq
0,h_{1}({\bm{d}})<0,h_{3}({\bm{d}})>0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid 0\geq
d_{2}\geq d_{1}>d_{3}\\}\\\
\hline\cr\textbf{R14}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}}),h_{3}({\bm{d}}),h_{4}({\bm{d}})\leq
0\\}&-\mathbb{N}A\\\ \hline\cr\end{array}$

Table 6.2. Regions to compute $D(R_{A},J)$ in Example 6.1.

Since the multidegrees in the chambers and regions corresponding to
$\textbf{C2}-\textbf{C5}$ and $\textbf{R2}-\textbf{R5}$ will all need
operators adjusted only for monomials in $J$ whose exponents are parallel to a
single facet in $A$, we will only describe the process to determine the graded
pieces of $\mathbb{I}(J)$ and $D(R_{A},J)$ for chamber C2 and region R2,
respectively, as precisely the same type of argument holds for the other
chambers and regions. If ${\bm{d}}$ is in C2 in the case of the idealizer or
in R2 in the case of $D(R_{A},J)$, consider the lattice points corresponding
to the monomials in $J$ on the plane $x-y+h_{4}({\bm{d}})=0$. When an element
in $\mathbb{I}(\Omega({\bm{d}}))$ is applied to such a monomial in $J$, the
result is a constant times a monomial with exponent on the facet $\sigma_{4}$,
which is not in $J$. Hence we need to multiply $\mathbb{I}(\Omega({\bm{d}}))$
by $h_{4}+h_{4}({\bm{d}})$, so for ${\bm{d}}$ in C2,

$\mathbb{I}(J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\langle\left(h_{4},h_{4}({-\bm{d}})\right)!\rangle$

and for ${\bm{d}}$ in R2,

$D(R_{A},J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\langle\left(h_{4},h_{4}(-{\bm{d}})\right)!\rangle.$

Figure 3A illustrates the plane in C2 that determines the linear form that we
multiply by $\mathbb{I}(\Omega(\bm{d}))$ to obtain $\mathbb{I}(J)_{\bm{d}}$.

Since the multidegrees in the chambers and regions corresponding to
$\textbf{C6}-\textbf{C9}$ and $\textbf{R6}-\textbf{R6}$ will all need
operators adjusted only for monomials in $J$ whose exponents are parallel to
two facets of $A$, we will only describe the process to determine the graded
pieces of $\mathbb{I}(J)$ and $D(R_{A},J)$ for chamber C6 and region R6,
respectively, as precisely the same type of argument holds for the other
chambers and regions. If ${\bm{d}}$ is in C6 in the case of the idealizer or
R6 in the case of $D(R_{A},J)$, consider the lattice points corresponding to
the monomials in $J$ on the planes $x-y+h_{4}({\bm{d}})=0$ and
$x-z+h_{3}({\bm{d}})=0$. When an element in $\mathbb{I}(\Omega({\bm{d}}))$ is
applied to such monomial in $J$, the result is a constant times a monomial
with exponent on $x-y=0$ or $x-z=0$, which is not in $J$. Hence we need to
multiply $\mathbb{I}(\Omega({\bm{d}}))$ by
$(h_{3}+h_{3}({\bm{d}}))(h_{4}+h_{4}({\bm{d}}))$, so for ${\bm{d}}$ in C6,

$\mathbb{I}(J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\langle\left(h_{3},h_{3}({-\bm{d}})\right)!\left(h_{4},h_{4}({-\bm{d}})\right)!\rangle,$

and for ${\bm{d}}$ in R6,

$D(R_{A},J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\langle\left(h_{3},h_{3}(-{\bm{d}})\right)!\left(h_{4},h_{4}({-\bm{d}})\right)!\rangle.$

Figure 3B illustrates the two planes in C6 that determine the linear forms
that we multiply $\mathbb{I}(\Omega(\bm{d}))$ by to obtain
$\mathbb{I}(J)_{\bm{d}}$.

Since the multidegrees in the chambers and regions corresponding to
$\textbf{C10}-\textbf{C13}$ and $\textbf{R10}-\textbf{R13}$ will all need
operators adjusted only for monomials in $J$ whose exponents are parallel to
three facets of $A$, we will only describe the process to determine the graded
piece of $\mathbb{I}(J)$ and $D(R_{A},J)$ for chamber C10 and region R10,
respectively, as precisely the same type of argument holds for the other
chambers and regions. If ${\bm{d}}$ is in C10 in the case of the idealizer or
R10 in the case of $D(R_{A},J)$, consider the lattice points corresponding to
the monomials in $J$ on the planes $z+h_{1}({\bm{d}})=0$,
$x-z+h_{3}({\bm{d}})=0$ and $x-y+h_{4}({\bm{d}})=0$. When an element of
$\mathbb{I}(\Omega({\bm{d}}))$ is applied to such a monomial in $J$, the
result is a constant times a monomial whose exponent is in $\sigma_{1}$,
$\sigma_{3}$, or $\sigma_{4}$, which is not in $J$. Hence we need to multiply
$\mathbb{I}(\Omega({\bm{d}}))$ by
$(h_{1}+h_{1}({\bm{d}}))(h_{3}+h_{3}({\bm{d}}))(h_{4}+h_{4}({\bm{d}}))$, so
for ${\bm{d}}$ in C10,

$\mathbb{I}(J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\langle\left(h_{1},h_{1}({-\bm{d}})\right)!\left(h_{3},h_{3}(-{\bm{d}})\right)!\left(h_{4},h_{4}(-{\bm{d}})\right)!\rangle,$

and for ${\bm{d}}$ in R10,

$D(R_{A},J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\langle\left(h_{1},h_{1}({-\bm{d}})\right)!\left(h_{3},h_{3}({-\bm{d}})\right)!\left(h_{4},h_{4}(-{\bm{d}})\right)!\rangle.$

Figure 3C illustrates the three planes in C10 that determine the linear forms
that we multiply by $\mathbb{I}(\Omega(\bm{d}))$ to obtain
$\mathbb{I}(J)_{\bm{d}}$.

If ${\bm{d}}$ is in C14 in the case of the idealizer or R14 in the case of
$D(R_{A},J)$, consider the lattice points corresponding to the monomials in
$J$ on the planes $z+h_{1}({\bm{d}})=0$, $y+h_{2}({\bm{d}})=0$,
$x-z+h_{3}({\bm{d}})=0$ and $x-y+h_{4}({\bm{d}})=0$. When an element of
$\mathbb{I}(\Omega({\bm{d}}))$ is applied to such a monomial in $J$, the
result is a constant times a monomial with exponent on one of the facets
$\sigma_{i}$ of $A$, which is not in $J$. Hence we need to multiply
$\mathbb{I}(\Omega({\bm{d}}))$ by
$(h_{1}+h_{1}({\bm{d}}))(h_{2}+h_{2}({\bm{d}}))(h_{3}+h_{3}({\bm{d}}))(h_{4}+h_{4}({\bm{d}}))$,
so for ${\bm{d}}$ in C14,

$\mathbb{I}(J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\langle\left(h_{1},h_{1}({-\bm{d}})\right)!\left(h_{2},h_{2}({-\bm{d}})\right)!\left(h_{3},h_{3}({-\bm{d}})\right)!\left(h_{4},h_{4}({-\bm{d}})\right)!\rangle,$

and for ${\bm{d}}$ in R14,

$D(R_{A},J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\langle\left(h_{1},h_{1}({-\bm{d}})\right)!\left(h_{2},h_{2}({-\bm{d}})\right)!\left(h_{3},h_{3}({-\bm{d}})\right)!\left(h_{4},h_{4}({-\bm{d}})\right)!\rangle.$

Combining the information from all $14$ chambers C1-C14 and regions R1-R14,
the general formula for the graded piece of the idealizer of $J$ at
${\bm{d}}\in\mathbb{Z}^{3}$ is

$\mathbb{I}(J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\left\langle\prod\limits_{h_{i}({\bm{d}})<0}\left(h_{i},h_{i}(-{\bm{d}})\right)!\right\rangle,$

and the general formula for the graded piece of the $D(R_{A},J)$ at
${\bm{d}}\in\mathbb{Z}^{3}$ is

$D(R_{A},J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\left\langle\prod\limits_{h_{i}({\bm{d}})\leq
0}\left(h_{i},h_{i}(-{\bm{d}})\right)!\right\rangle.$

For any $\bm{d}\in\mathbb{Z}^{3}$,

$\left(\dfrac{\mathbb{I}(J)}{D(R_{A},J)}\right)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\displaystyle\frac{\left\langle\prod\limits_{h_{i}({\bm{d}})<0}\left(h_{i},h_{i}(-{\bm{d}})\right)!\right\rangle}{\left\langle\prod\limits_{h_{i}({\bm{d}})\leq
0}\left(h_{i},h_{i}(-{\bm{d}})\right)!\right\rangle}.$

######  Example 6.2.

Now consider the ideal

$J=P_{\sigma_{1}}\cap P_{\sigma_{2}}\cap P_{\sigma_{3}\cap\sigma_{4}}=\langle
t_{1}^{2}t_{2}t_{3},t_{1}^{2}t_{2}^{2}t_{3},t_{1}^{2}t_{2}t_{3}^{2}\rangle$

in $R_{A}=\mathbb{C}[t_{1},t_{1}t_{2},t_{1}t_{3},t_{1}t_{2}t_{3}]$. In Figure
6.4, we give two views of the cone with the lattice points in $J$ shaded in
grey. Note that multidegrees of the monomials in $J$ lie in the interior of
the cone and on the interiors of the two faces $\sigma_{3}$ and $\sigma_{4}$,
so this is the grey portion of the figure.

Figure 6.4. Cone of $R_{A}$ with $J=P_{\sigma_{1}}\cap P_{\sigma_{2}}\cap
P_{\sigma_{3}\cap\sigma_{4}}$

Since the only points on the facets of the cone that have monomials that lie
in $J$ lie in the interiors of $\sigma_{3}$ and $\sigma_{4}$, the 14 _regions_
that we consider in determining $D(R_{A},J)$ appear in Table 6.3.

$\begin{array}[]{@{\extracolsep{\fill}}
|c|l|l|}\hline\cr\text{Region}&\text{Halfspace inequalities}&\text{Lattice
point inequalities}\\\
\hline\cr\textbf{R1}&\begin{array}[]{l}\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}}),h_{3}({\bm{d}})>0,h_{4}({\bm{d}})\geq 0\\}\
\cup\vspace{-6pt}\\\ \qquad\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}}),h_{4}({\bm{d}})>0,h_{3}({\bm{d}})\geq
0\\}\end{array}&\begin{array}[]{l}\\{\bm{d}\in\mathbb{Z}^{3}\mid{\bm{x}}^{\bm{d}}\in
J\\}\end{array}\\\ \hline\cr\textbf{R2}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}})>0,h_{3}({\bm{d}})\geq
0,h_{4}({\bm{d}})<0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{2}>d_{1}\geq
d_{3}>0\\}\\\ \hline\cr\textbf{R3}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}})>0,h_{4}({\bm{d}})\geq
0,h_{3}({\bm{d}})<0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{3}>d_{1}\geq
d_{2}>0\\}\\\
\hline\cr\textbf{R4}&\begin{array}[]{l}\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{3}({\bm{d}}),h_{4}({\bm{d}})>0,h_{2}({\bm{d}})\leq 0\\}\
\vspace{-6pt}\cup\\\ \qquad\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{3}({\bm{d}})=0,h_{1}({\bm{d}}),h_{4}({\bm{d}})>0,h_{2}({\bm{d}})<0\\}\end{array}&\begin{array}[]{l}\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{1}>d_{3}>0\geq d_{2}\\}\ \vspace{-6pt}\cup\\\
\qquad\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{1}=d_{3}>0>d_{2}\\}\end{array}\\\
\hline\cr\textbf{R5}&\begin{array}[]{l}\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{2}({\bm{d}}),h_{3}({\bm{d}}),h_{4}({\bm{d}})>0,h_{1}({\bm{d}})\leq 0\\}\
\vspace{-6pt}\cup\\\ \qquad\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{4}({\bm{d}})=0,h_{2}({\bm{d}}),h_{3}({\bm{d}})>0,h_{1}({\bm{d}})<0\\}\end{array}&\begin{array}[]{l}\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{1}\geq d_{2}>0\geq d_{3}\\}\ \vspace{-6pt}\cup\\\
\qquad\\{\bm{d}\in\mathbb{C}^{3}\mid d_{1}=d_{2}>0>d_{3}\\}\end{array}\\\
\hline\cr\textbf{R6}&\begin{array}[]{l}\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}})>0\text{ and
}h_{3}({\bm{d}}),h_{4}({\bm{d}})<0\\}\ \vspace{-6pt}\cup\\\
\qquad\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}})>0,h_{2}({\bm{d}})>0,h_{3}({\bm{d}})=h_{4}({\bm{d}})=0\\}\end{array}&\begin{array}[]{l}\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{3}>0,d_{2}>0,d_{2}>d_{1},d_{3}>d_{1}\\}\ \vspace{-6pt}\cup\\\
\qquad\qquad\qquad\qquad\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{1}=d_{2}=d_{3}>0\\}\end{array}\\\
\hline\cr\textbf{R7}&\begin{array}[]{l}\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{2}({\bm{d}}),h_{3}({\bm{d}})>0,h_{1}({\bm{d}})\leq 0,h_{4}({\bm{d}})<0\\}\
\vspace{-6pt}\cup\\\ \qquad\qquad\quad\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{2}({\bm{d}})>0,h_{3}({\bm{d}})=0,h_{1}({\bm{d}})<0\\}\end{array}&\begin{array}[]{l}\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{2}>0\geq d_{3},d_{2}>d_{1}>d_{3}\\}\ \vspace{-6pt}\cup\\\
\qquad\qquad\qquad\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{2}>0>d_{1}=d_{3}\\}\end{array}\\\
\hline\cr\textbf{R8}&\begin{array}[]{l}\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{4}({\bm{d}})>0,h_{2}({\bm{d}})\leq 0,h_{3}({\bm{d}})<0\\}\
\vspace{-6pt}\cup\\\ \qquad\qquad\quad\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}})>0,h_{4}({\bm{d}})=0,h_{2}({\bm{d}})<0\\}\end{array}&\begin{array}[]{l}\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{3}>0\geq d_{2},d_{3}>d_{1}>d_{2}\\}\ \vspace{-6pt}\cup\\\
\qquad\qquad\qquad\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{3}>0>d_{1}=d_{2}\\}\end{array}\\\
\hline\cr\textbf{R9}&\begin{array}[]{l}\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{3}({\bm{d}}),h_{4}({\bm{d}})>0,h_{1}({\bm{d}}),h_{2}({\bm{d}})\leq 0\\}\
\vspace{-6pt}\cup\\\ \quad\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}})<0,h_{4}({\bm{d}})>0,h_{3}({\bm{d}})=0\\}\
\vspace{-6pt}\cup\\\ \qquad\quad\\{\bm{d}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}}<0,h_{3}({\bm{d}})>0,h_{4}({\bm{d}})=0\\}\end{array}&\begin{array}[]{l}\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{1}>d_{2},d_{3},0\geq d_{2},d_{3}\\}\ \vspace{-6pt}\cup\\\
\qquad\\{\bm{d}\in\mathbb{Z}^{3}\mid 0>d_{1}=d_{2},0>d_{1}>d_{3}\\}\
\vspace{-6pt}\cup\\\ \qquad\qquad\\{\bm{d}\in\mathbb{Z}^{3}\mid
0>d_{1}=d_{3},0>d_{1}>d_{2}\\}\end{array}\\\
\hline\cr\textbf{R10}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid h_{1}({\bm{d}})\leq
0,h_{3}({\bm{d}}),h_{4}({\bm{d}})<0,h_{2}({\bm{d}})>0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{2}>0\geq d_{3}>d_{1}\\}\\\
\hline\cr\textbf{R11}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid h_{2}({\bm{d}})\leq
0,h_{4}({\bm{d}}),h_{3}({\bm{d}})<0,h_{1}({\bm{d}})>0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{3}>0\geq d_{2}>d_{1}\\}\\\
\hline\cr\textbf{R12}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{3}({\bm{d}})<0,h_{2}({\bm{d}})\leq 0,h_{4}({\bm{d}})\geq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid 0\geq d_{2}>d_{1}\geq d_{3}\\}\\\
\hline\cr\textbf{R13}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{2}({\bm{d}}),h_{4}({\bm{d}})<0,h_{1}({\bm{d}})\leq 0,h_{3}({\bm{d}})\geq
0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid 0\geq d_{3}>d_{1}\geq d_{2}\\}\\\
\hline\cr\textbf{R13}&\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}}),h_{4}({\bm{d}})\leq
0,h_{3}({\bm{d}})>0\\}&\\{\bm{d}\in\mathbb{Z}^{3}\mid 0\geq d_{2}\geq
d_{1}>d_{3}\\}\\\
\hline\cr\textbf{R14}&\begin{array}[]{l}\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}})h_{3}({\bm{d}})\leq 0,h_{4}({\bm{d}})<0\\}\
\vspace{-6pt}\cup\\\ \qquad\\{{\bm{d}}\in\mathbb{Z}^{3}\mid
h_{1}({\bm{d}}),h_{2}({\bm{d}})h_{4}({\bm{d}})\leq
0,h_{3}({\bm{d}})<0\\}\end{array}&\begin{array}[]{l}\\{\bm{d}\in\mathbb{Z}^{3}\mid
d_{1}<d_{2}\leq 0,d_{1}\leq d_{3}\leq 0\\}\ \vspace{-6pt}\cup\\\
\qquad\\{\bm{d}\in\mathbb{Z}^{3}\mid d_{1}\leq d_{2}\leq 0,d_{1}<d_{3}\leq
0\\}\end{array}\\\ \hline\cr\end{array}$

Table 6.3. Regions for $D(R_{A},J)$ in Example 6.2.

If ${\bm{d}}$ is in $\textbf{C1}=\mathbb{N}A$, when considering the idealizer
or R1, which are the exponents of the monomials in $J$, when considering
$D(R_{A},J)$, then $D(R_{A})_{\bm{d}}$ is generated by ${\bm{t}^{\bm{d}}}$.
The product of any element of $J$ and ${\bm{t}^{\bm{d}}}$ is also an element
of $J$, which means that the graded pieces are
$\mathbb{I}(J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\mathbb{C}[\theta]$ and
$D(R_{A},J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\mathbb{C}[\theta]$.

Since the multidegrees in $\textbf{C2}-\textbf{C5}$ and
$\textbf{R2}-\textbf{R5}$ potentially affect monomials in $J$ whose exponents
are parallel to a single facet of $A$, we will only describe the process to
determine the graded pieces of $\mathbb{I}(J)$ and $D(R_{A},J)$ for chamber C2
and region R2, respectively, as precisely the same type of argument holds for
the other chambers and regions. The reader may refer to Figure 3A to help
visualize the argument below. If ${\bm{d}}$ is in C2 in the case of the
idealizer or R2 in the case of $D(R_{A},J)$, consider exponents of monomials
in $J$ on the plane $x-y+h_{4}({\bm{d}})=0$. When an element of
$\mathbb{I}(\Omega({\bm{d}}))$ is applied to such a monomial in $J$, the
result is a constant times a monomial on the interior of the facet
$\sigma_{4}$, which is in $J$. Hence, for ${\bm{d}}$ in C2,

$\mathbb{I}(J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\langle\left(h_{4},h_{4}({-\bm{d}})-1\right)!\rangle,$

and for ${\bm{d}}$ in R2,

$D(R_{A},J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\langle\left(h_{4},h_{4}({-\bm{d}})-1\right)!\rangle.$

Since the multidegrees in the chambers and regions corresponding to
$\textbf{C6}-\textbf{C9}$ and $\textbf{R6}-\textbf{R9}$ can will all need
operators adjusted only for monomials in $J$ whose exponents are parallel to
two facets of $A$, we will only describe the process to determine the graded
piece of $\mathbb{I}(J)$ and $D(R_{A},J)$ for chamber C6 and region R6,
respectively, as precisely the same type of argument holds for the other
chambers and regions. The reader may refer to Figure 3B to help visualize the
argument below. If ${\bm{d}}$ is in C6 in the case of the idealizer or R6 in
the case of $D(R_{A},J)$, consider the multidegrees of monomials in $J$ on the
planes $x-z+h_{3}({\bm{d}})=0$ or $x-y+h_{4}({\bm{d}})=0$. When an element of
$\mathbb{I}(\Omega({\bm{d}}))$ is applied to such a monomial in $J$, the
result is a constant times a monomial on $\sigma_{3}$ or $\sigma_{4}$. This is
not in $J$ only if this monomial’s exponent is in the intersection of the two
planes. Hence, we need to multiply $\mathbb{I}(\Omega({\bm{d}}))$ by either
$(h_{3}+h_{3}({\bm{d}}))$ or $(h_{4}+h_{4}({\bm{d}}))$, so that for ${\bm{d}}$
in C6,

$\mathbb{I}(J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\left(h_{4},h_{4}({-\bm{d}})-1\right)!\left(h_{3},h_{3}({-\bm{d}})-1\right)!\cdot\langle(h_{3}+h_{3}({\bm{d}})),(h_{4}+h_{4}({\bm{d}}))\rangle,$

and for ${\bm{d}}$ in R6,

$D(R_{A},J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\left(h_{4},h_{4}({-\bm{d}})-1\right)!\left(h_{3},h_{3}({-\bm{d}})-1\right)!\cdot\langle(h_{3}+h_{3}({\bm{d}})),(h_{4}+h_{4}({\bm{d}}))\rangle.$

Since the multidegrees in the chambers and regions corresponding to
$\textbf{C10}-\textbf{C13}$ and $\textbf{R10}-\textbf{R13}$ can will all need
operators adjusted only for monomials in $J$ whose exponents are parallel to a
three facets of $A$, we will only describe the process to determine the graded
piece of $\mathbb{I}(J)$ and $D(R_{A},J)$ for chamber C10 and region R10,
respectively, as precisely the same type of argument holds for the other
chambers and regions. The reader may refer to Figure 3C to help visualize the
argument below. If ${\bm{d}}$ is in C10 in the case of the idealizer or R10 in
the case of $D(R_{A},J)$, consider the multidegrees of monomials in $J$ on the
planes $z+h_{1}({\bm{d}})=0$, $x-z+h_{3}({\bm{d}})=0$, and
$x-y+h_{4}({\bm{d}})=0$. When an element in $\mathbb{I}(\Omega({\bm{d}}))$ is
applied to such a monomial in $J$, the result is a constant times a monomial
on $\sigma_{1}$, which is not in $J$, or a constant times a monomial in
$\sigma_{3}$ or $\sigma_{4}$, which is not in $J$ only if the monomial is in
the intersection of $\sigma_{3}$ and $\sigma_{4}$. Hence, we need to multiply
$\mathbb{I}(\Omega({\bm{d}}))$ by either
$(h_{1}+h_{1}({\bm{d}}))(h_{3}+h_{3}({\bm{d}}))$ or
$(h_{1}+h_{1}({\bm{d}}))(h_{4}+h_{4}({\bm{d}}))$, so that for ${\bm{d}}$ in
C10,

$\mathbb{I}(J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\left(h_{1},h_{1}({-\bm{d}})\right)!\left(h_{3},h_{3}({-\bm{d}})-1\right)!\left(h_{4},h_{4}({-\bm{d}})-1\right)!\cdot\langle(h_{3}+h_{3}({\bm{d}})),(h_{4}+h_{4}({\bm{d}}))\rangle,$

and for ${\bm{d}}$ in R10,

$D(R_{A},J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\left(h_{1},h_{1}({-\bm{d}})\right)!\left(h_{3},h_{3}({-\bm{d}})-1\right)!\left(h_{4},h_{4}({-\bm{d}})-1\right)!\cdot\langle(h_{3}+h_{3}({\bm{d}})),(h_{4}+h_{4}({\bm{d}}))\rangle.$

If ${\bm{d}}$ is in C14 in the case of the idealizer or R14 in the case of
$D(R_{A},J)$, consider the multidegrees of the monomials in $J$ on the planes
$z+h_{1}({\bm{d}})=0$, $y+h_{2}({\bm{d}})=0$, $x-z+h_{3}({\bm{d}})=0$, and
$x-y+h_{4}({\bm{d}})=0$. When an element of $\mathbb{I}(\Omega({\bm{d}}))$ is
applied to such a monomial in $J$, the result is a constant times a monomial
in $\sigma_{1}$ or $\sigma_{2}$, which is not in $J$, or a constant times a
monomial in $\sigma_{3}$ or $\sigma_{4}$, which is not in $J$ if this monomial
lies in both planes parallel to these faces. Hence we need to multiply
$\mathbb{I}(\Omega({\bm{d}}))$ by
$(h_{1}+h_{1}({\bm{d}}))(h_{2}+h_{2}({\bm{d}}))(h_{3}+h_{3}({\bm{d}}))(h_{4}+h_{4}({\bm{d}})-1)$
or
$(h_{1}+h_{1}({\bm{d}}))(h_{2}+h_{2}({\bm{d}}))(h_{3}+h_{3}({\bm{d}})-1)(h_{4}+h_{4}({\bm{d}}))$,
so that for ${\bm{d}}$ in C14,

$\mathbb{I}(J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\prod\limits_{i=1}^{4}\left(h_{i},h_{i}({-\bm{d}})-1\right)!\cdot\left\langle\left(\prod\limits_{i=1}^{3}(h_{i}+h_{i}({\bm{d}}))\right),(h_{4}+h_{4}({\bm{d}}))\left(\prod\limits_{i=1}^{2}(h_{i}+h_{i}({\bm{d}}))\right)\right\rangle$

and for ${\bm{d}}$ in R14,

$D(R_{A},J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\prod\limits_{i=1}^{4}\left(h_{i},h_{i}({-\bm{d}})-1\right)!\cdot\left\langle\left(\prod\limits_{i=1}^{3}(h_{i}+h_{i}({\bm{d}}))\right),(h_{4}+h_{4}({\bm{d}}))\left(\prod\limits_{i=1}^{2}(h_{i}+h_{i}({\bm{d}}))\right)\right\rangle.$

Putting all the information together from all $14$ chambers C1-C14 and regions
R1-R14, the general formula for the graded piece of the idealizer of $J$ at
${\bm{d}}\in\mathbb{Z}^{3}$ is

$\mathbb{I}(J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\prod\limits_{i=1}^{4}\left(h_{i},h_{i}({-\bm{d}})-1\right)!\cdot\left\langle\left(\prod\limits_{\begin{subarray}{c}h_{i}({\bm{d}})<0\\\
\text{for }i\neq
4\end{subarray}}(h_{i}+h_{i}({\bm{d}}))\right),\left(\prod\limits_{\begin{subarray}{c}h_{i}({\bm{d}})<0\\\
\text{for }i\neq 3\end{subarray}}(h_{i}+h_{i}({\bm{d}}))\right)\right\rangle,$

and the general formula for the graded piece of $D(R_{A},J)$ at
${\bm{d}}\in\mathbb{Z}^{3}$ is

$D(R_{A},J)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\prod\limits_{i=1}^{4}\left(h_{i},h_{i}({-\bm{d}})-1\right)!\cdot\left\langle\left(\prod\limits_{\begin{subarray}{c}h_{i}({\bm{d}})\leq
0\\\ \text{for }i\neq
4\end{subarray}}(h_{i}+h_{i}({\bm{d}}))\right),\left(\prod\limits_{\begin{subarray}{c}h_{i}({\bm{d}})\leq
0\\\ \text{for }i\neq
3\end{subarray}}(h_{i}+h_{i}({\bm{d}}))\right)\right\rangle.$

We also have for any $\bm{d}\in\mathbb{Z}^{3}$,

$\left(\dfrac{\mathbb{I}(J)}{D(R_{A},J)}\right)_{\bm{d}}={\bm{t}^{\bm{d}}}\cdot\displaystyle\frac{\
\prod\limits_{i=1}^{4}\left(h_{i},h_{i}({-\bm{d}})-1\right)!\cdot\left\langle\left(\prod\limits_{\begin{subarray}{c}h_{i}({\bm{d}})<0\\\
\text{for }i\neq
4\end{subarray}}(h_{i}+h_{i}({\bm{d}}))\right),\left(\prod\limits_{\begin{subarray}{c}h_{i}({\bm{d}})<0\\\
\text{for }i\neq 3\end{subarray}}(h_{i}+h_{i}({\bm{d}}))\right)\right\rangle\
}{\
\prod\limits_{i=1}^{4}\left(h_{i},h_{i}({-\bm{d}})-1\right)!\cdot\left\langle\left(\prod\limits_{\begin{subarray}{c}h_{i}({\bm{d}})\leq
0\\\ \text{for }i\neq
4\end{subarray}}(h_{i}+h_{i}({\bm{d}}))\right),\left(\prod\limits_{\begin{subarray}{c}h_{i}({\bm{d}})\leq
0\\\ \text{for }i\neq
3\end{subarray}}(h_{i}+h_{i}({\bm{d}}))\right)\right\rangle\ }.$

##  7\. Non-normal examples

Thus far, we have restricted our attention to normal semigroup rings. However,
Saito and Traves determined the ring of differential operators for all
saturated affine semigroup rings in [ST01, Theorem 3.3.1] (see also [ST04,
Theorem 2.1]), and we can use this work to compute $\mathbb{I}(J)$,
$D(R_{A},J)$ and $\mathbb{I}(J)/D(R_{A},J)$ for non-normal $R_{A}$ and a
graded radical $R_{A}$-ideal $J$. We broaden slightly from normal to _scored_
semigroup rings, see Definition 2.2. We include two examples; one is scored
and the other is not. However, the two quotient rings modulo the interior
ideals are isomorphic and we will see that the expressions for
$\mathbb{I}(J)/D(R_{A},J)$ in both rings are isomorphic.

######  Example 7.1.

Consider the matrix $A=\begin{bmatrix}2&3&0\\\ 0&0&1\end{bmatrix}$, which is
associated to the semigroup ring
$R_{A}=\mathbb{C}[\mathbb{N}A]=\mathbb{C}[s^{2},s^{3},t]$. Since $R_{A}$ is
not normal, note that integer points in the faces of $A$ and $\mathbb{N}A$
differ from those in $\mathbb{R}_{\geq 0}A$. We let

$\sigma_{1}=\mathbb{N}\\{e_{2}\\}\quad\text{and}\quad\sigma_{2}=\mathbb{N}\\{e_{1}\\}$

be the integer points in the facets of the cone $\mathbb{R}_{\geq 0}A$. The
primitive integral support functions of $\mathbb{N}A$ are

$h_{1}=h_{1}(\theta)=\theta_{1}\quad\text{and}\quad
h_{2}=h_{2}(\theta)=\theta_{2}.$

The prime ideals associated to the facets of $A$ are

$P_{1}=\langle s^{2},s^{3}\rangle\quad\text{and}\quad P_{2}=\langle t\rangle.$

Set $J=P_{1}\cap P_{2}=\langle s^{2}t,s^{3}t\rangle$. Note that $R_{A}$ is a
Gorenstein ring that is not normal, and, in this case, the ideal $J$, which is
the interior, is not the canonical module of $R_{A}$.

Figure 7.1.
Chambers for $\mathbb{C}[s^{2},s^{3},t]$

We will again use _chambers_ of $\mathbb{Z}^{2}$ to aid in our computations of
the graded pieces of $\mathbb{I}(J)/D(R_{A},J)$; in particular, we show that
$\mathbb{I}(J)/D(R_{A},J)$ is only nonzero in multidegree ${\bm{d}}$ if
${\bm{d}}\in\sigma_{1}\cup\sigma_{2}$, as occurred in the normal cases
computed thus far. In Figure 7.1, the red multidegrees correspond to the
monomials in $J$, and the blue multidegrees are exponents for monomials in the
semigroup ring $R_{A}$ that do not lie in $J$. The monomials at the orange
multidegrees behave somewhat more like the red and blue, in that, for all such
$\bm{d}$, $\mathbb{I}(J)_{\bm{d}}=D(R_{A})_{\bm{d}}$. Thus, together the red,
blue, and orange multidegrees form chamber C1. The yellow multidegrees
constitute C2, the violet points form C3, and both the green and dark green
make C4. The vertical lines $\\{x=-1\\}$ and $\\{x=1\\}$ contain multidegrees
$\bm{d}$ for which extra care is needed to determine $\mathbb{I}(J)_{\bm{d}}$
and $D(R_{A},J)_{\bm{d}}$.

Eriksen showed in [Eri03, Proposition 2.1] that the ring of differential
operators of a numerical semigroup ring $R=\mathbb{C}[t^{\Gamma}]$ is

$D(R)=\bigoplus\limits_{d\in\mathbb{Z}}s^{d_{1}}t^{d_{2}}\cdot\left\langle\prod\limits_{\gamma\in\Omega(d)}(\theta-\gamma)\right\rangle,$

where $\Omega(d)=\\{\gamma\in\Gamma\mid\gamma+d\notin\Gamma\\}$. We encode
these polynomial generators using $G_{\bm{d}}(\theta)$, where

$G_{\bm{d}}(\theta):=\begin{cases}1&\text{ if }d_{1}\in\mathbb{N}\langle
2,3\rangle,\\\ h_{1}&\text{ if }d_{1}=1,\\\ h_{1}-1&\text{if }d_{1}=-1,\\\
\displaystyle\frac{\left(h_{1},1-d_{1}\right)!}{(h_{1}-1)(h_{1}+d_{1})}&\text{
if }d_{1}\leq-2.\end{cases}$

By Theorem 2.3, for any $\bm{d}\in\mathbb{Z}^{2}$,

$D(R_{A})_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\left\langle
G_{\bm{d}}(\theta)\left(h_{2},-d_{2}-1\right)!\right\rangle.$

To determine $\mathbb{I}(J)$ or $D(R_{A},J)$, we must determine the lines of
multidegrees that contain operators that, when applied to monomials in $J$ or
$R$, produce elements in $R\setminus J$. Note that given
$\bm{m}\in\mathbb{N}A$,

(7.1) $\displaystyle
s^{d_{1}}t^{d_{2}}G_{\bm{d}}(\theta)\left(h_{2},-d_{2}-1\right)!\ast
s^{m_{1}}t^{m_{2}}=G_{\bm{d}}(\bm{m})(h_{2}(\bm{m}),-d_{2}-1)!\cdot
s^{d_{1}+m_{1}}t^{d_{2}+m_{2}}.$

If $\bm{d}\in\textbf{C1}$ and $s^{m_{1}}t^{m_{2}}\in J$, then the result of
(7.1) will remain in $J$. Thus, for all $\bm{d}\in\textbf{C1}$,
$D(R_{A},J)_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\langle
G_{\bm{d}}(\theta)\rangle$.

Next, for ${\bm{d}}\in\textbf{C2}$, when
$\bm{m}\in\mathbb{N}A\cap\\{x=-d_{1}\\}$, then the result of (7.1) is a
constant times a monomial in $\sigma_{1}$, which does not belong to $J$.
Hence, for all $\bm{d}\in\textbf{C2}$,

$\mathbb{I}(J)_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\langle
G_{\bm{d}}(\theta)(h_{1}+d_{1})\rangle.$

Further, if $\bm{d}\in\textbf{C2}\cup(\sigma_{1}\setminus\\{\bm{0}\\})$, then

$D(R_{A},J)_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\langle
G_{\bm{d}}(\theta)(h_{1}+d_{1})\rangle.$

For ${\bm{d}}\in\textbf{C4}$, when $\bm{m}\in\mathbb{N}A\cap\\{y=-d_{2}\\}$,
then the result of (7.1) is a constant times a monomial in $\sigma_{2}$, which
does not belong to $J$. Hence, for all $\bm{d}\in\textbf{C4}$,

$\mathbb{I}(J)_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\langle
G_{\bm{d}}(\theta)\left(h_{2},-d_{2}\right)!\rangle.$

Further, if ${\bm{d}}\in\textbf{C4}\cup(\sigma_{2}\setminus\\{\bm{0}\\})$,
then

$D(R_{A},J)_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},-d_{2}\right)!\rangle.$

Finally, for ${\bm{d}}\in\textbf{C3}$, if
$\bm{m}\in\\{x=-d_{1}\\}\cap\\{\bm{c}\mid s^{c_{1}}t^{c_{1}}\in J\\}$ or if
$\bm{m}\in\\{y=-d_{2}\\}\cap\mathbb{N}A$, respectively, then the result of
(7.1) is either a constant times a monomial in $\sigma_{1}$ or $\sigma_{2}$,
respectively. Hence,

$\mathbb{I}(J)_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\langle
G_{\bm{d}}(\theta)(h_{1}+d_{1})\left(h_{2},-d_{2}\right)!\rangle.$

Finally, if ${\bm{d}}\in\textbf{C3}\cup(-\sigma_{1})\cup(-\sigma_{2})$, then

$D(R_{A},J)_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\langle
G_{\bm{d}}(\theta)(h_{1}+d_{1})\left(h_{2},-d_{2}\right)!\rangle.$

Gathering these computations,

$\mathbb{I}(J)_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\langle
G_{\bm{d}}(\theta)\rangle&\text{ if }{\bm{d}}\in\textbf{C1},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle G_{\bm{d}}(\theta)(h_{1}+d_{1})\rangle&\text{
if }{\bm{d}}\in\textbf{C2},\\\ s^{d_{1}}t^{d_{2}}\cdot\langle
G_{\bm{d}}(\theta)(h_{1}+d_{1})\left(h_{2},-d_{2}\right)!\rangle&\text{ if
}{\bm{d}}\in\textbf{C3},\\\ s^{d_{1}}t^{d_{2}}\cdot\langle
G_{\bm{d}}(\theta)\left(h_{2},-d_{2}\right)!\rangle&\text{ if
}{\bm{d}}\in\textbf{C4},\end{cases}$

and

$D(R_{A},J)_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\langle
G_{\bm{d}}(\theta)\rangle&\text{ if
}\bm{d}\in\textbf{C1}\setminus(\sigma_{1}\cup\sigma_{2}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle G_{\bm{d}}(\theta)(h_{1}+d_{1})\rangle&\text{
if
}{\bm{d}}\in(\textbf{C2}\setminus\sigma_{2})\cup(\sigma_{1}\cap\\{d_{2}>0\\}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle
G_{\bm{d}}(\theta)(h_{1}+d_{1})\left(h_{2},-d_{2}\right)!\rangle&\text{ if
}{\bm{d}}\in\textbf{C3}\cup\\{\bm{0}\\}\cup(-\sigma_{1})\cup(-\sigma_{2}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle
G_{\bm{d}}(\theta)\left(h_{2},-d_{2}\right)!\rangle&\text{ if
}{\bm{d}}\in(\textbf{C4}\setminus\sigma_{1})\cup(\sigma_{2}\cap\\{d_{1}>0\\}).\end{cases}$

Taking the quotient of the above multigraded modules, we then obtain a
description of the graded pieces of $\mathbb{I}(J)/D(R_{A},J)$ as follows:

$\left(\dfrac{\mathbb{I}(J)}{D(R_{A},J)}\right)_{\bm{d}}=\begin{cases}0&\text{
if }{\bm{d}}\notin\mathbb{Z}{\sigma_{1}}\cup\mathbb{Z}{\sigma_{2}},\\\
\displaystyle\frac{\mathbb{C}[\theta]}{\langle h_{1}h_{2}\rangle}&\text{ if
}{\bm{d}}=\bm{0},\\\ s^{d_{1}}t^{d_{2}}\cdot\displaystyle\frac{\langle
G_{\bm{d}}(\theta)\rangle}{\langle G_{\bm{d}}(\theta)h_{2}\rangle},&\text{ if
}d_{1}\neq 0,d_{2}=0,\\\ s^{d_{1}}t^{d_{2}}\cdot\displaystyle\frac{\langle
G_{\bm{d}}(\theta)\rangle}{\langle G_{\bm{d}}(\theta)h_{1}\rangle}&\text{ if
}d_{1}=0,d_{2}>0,\\\ s^{d_{1}}t^{d_{2}}\cdot\displaystyle\frac{\langle
G_{\bm{d}}(\theta)\left(h_{2},-d_{2}\right)!\rangle}{\langle
h_{1}G_{\bm{d}}(\theta)\left(h_{2},-d_{2}\right)!\rangle}&\text{ if
}d_{1}=0,d_{2}<0.\end{cases}$

In this case, $JD(R_{A})=D(R_{A},J)$, so (1.2) happens to hold in this
example. To see this, compute $(JD(R_{A}))_{\bm{d}}$ by looking at
$\mathbb{I}(\Omega(\bm{d}-(2,1))$ and $\mathbb{I}(\Omega(\bm{d}-(3,1))$ for
any $\bm{d}\in\mathbb{Z}^{3}$. For an example of how showing this equality
works, when $(d_{1}-2,d_{2}-1)$ and $(d_{1}-3,d_{2}-1)$ are both in
$\textbf{C2}\setminus\sigma_{2}$, then for $d_{1}<0$,

$G_{(d_{1}-2,d_{2}-1)}(\theta)=\displaystyle\frac{\left(h_{1},3-d_{1}\right)!}{(h_{1}-1)(h_{1}+d_{1}-2)}\quad\text{and}\quad
G_{(d_{1}-3,d_{2}-1)}(\theta)=\displaystyle\frac{\left(h_{1},4-d_{1}\right)!}{(h_{1}-1)(h_{1}+d_{1}-3)}.$

Both of these polynomials are divisible by $G_{\bm{d}}(\theta)(h_{1}+d_{1})$:

$G_{(d_{1}-2,d_{2}-1)}(\theta)=\displaystyle\frac{\left(h_{1},3-d_{1}\right)!}{(h_{1}-1)(h_{1}+d_{1}-2)}=G_{\bm{d}}(\theta)(h_{1}+d_{1})(h_{1}+d_{1}-3)$

and

$G_{(d_{1}-2,d_{2}-1)}(\theta)=\displaystyle\frac{\left(h_{1},4-d_{1}\right)!}{(h_{1}-1)(h_{1}+d_{1}-2)}=G_{\bm{d}}(\theta)(h_{1}+d_{1})(h_{1}+d_{1}-2)(h_{1}+d_{1}-4).$

Further,

$1=(h_{1}+d_{1}-3)^{2}-(h_{1}+d_{1}-2)(h_{1}+d_{1}-4),$

which implies that

$\left\langle G_{\bm{d}}(\theta)(h_{1}+d_{1})\right\rangle=\left\langle
G_{(d_{1}-2,d_{2}-1)}(\theta),G_{(d_{1}-3,d_{2}-1)}(\theta)\right\rangle.$

Hence for $\bm{d}\in\textbf{C2}$, there is an equality
$D(R_{A},J)_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\langle
G_{\bm{d}}(\theta)(h_{1}+d_{1})\rangle=(JD(R_{A}))_{\bm{d}}$.

######  Example 7.2.

Consider the matrix $\widetilde{A}=\begin{bmatrix}1&2&3&0\\\
1&0&0&1\end{bmatrix}$, which is associated to the semigroup ring
$R_{\widetilde{A}}=\mathbb{C}[\mathbb{N}\widetilde{A}]=\mathbb{C}[st,s^{2},s^{3},t]$,
which is not normal, scored, or Gorenstein. As in Example 7.1, we let

$\sigma_{1}=\mathbb{N}\\{e_{2}\\}\quad\text{and}\quad\sigma_{2}=\mathbb{N}\\{e_{1}\\}$

be the integer points in the facets of the cone $\mathbb{R}_{\geq
0}\widetilde{A}$. The primitive integral support functions of
$\mathbb{N}\widetilde{A}$ are

$h_{1}=h_{1}(\theta)=\theta_{1}\quad\text{and}\quad
h_{2}=h_{2}(\theta)=\theta_{2}.$

The prime ideals associated to the facets of $\widetilde{A}$ are

$P_{1}=\langle st,s^{2},s^{3}\rangle\quad\text{and}\quad P_{2}=\langle
st,t\rangle.$

Finally, set $\widetilde{J}=P_{1}\cap P_{2}=\langle st,s^{2}t\rangle$.

Figure 7.2.  Chambers for $\mathbb{C}[st,s^{2},s^{3},t]$

Note that $R_{A}/J$ from Example 7.1 is isomorphic to
$R_{\widetilde{A}}/\widetilde{J}$. Hence, we would also expect that
$\mathbb{I}(J)/D(R_{A},J)\cong\mathbb{I}(J)/D(R_{\widetilde{A}},\widetilde{J})$.
We will determine the graded pieces of $\mathbb{I}(\widetilde{J})$ and
$D\left(R_{\widetilde{A}},\widetilde{J}\right)$, as well as
${\mathbb{I}(\widetilde{J})}/{D(R_{\widetilde{A}},\widetilde{J})}$; then, we
will exhibit an isomorphism between the graded pieces of
${\mathbb{I}(\widetilde{J})}/{D(R_{\widetilde{A}},\widetilde{J})}$ and
$\mathbb{I}(J)/D(R_{A},J)$.

We will again use _chambers_ of $\mathbb{Z}^{2}$ to aid in our computations of
the graded pieces of $\mathbb{I}(J)/D(R_{A},J)$. In Figure 7.2, the red
multidegrees correspond to monomials in $\widetilde{J}$, while the blue ones
(both light and dark) are the remaining $\bm{d}\in\mathbb{Z}^{2}$ for which
$D(R_{\widetilde{A}})_{\bm{d}}=\mathbb{I}(\widetilde{J})_{\bm{d}}$; together,
these red and blue multidegrees form chamber C1. The remaining multidegrees
are split into chambers as in the previous examples: yellow points form C2,
violet points make C3, and green points give C4.

In [ST01, Examples 3.2.7 and 3.3.4], Saito and Traves found that for any
$\bm{d}\in\mathbb{Z}^{2}$,
$D(R_{\widetilde{A}})_{\bm{d}}=s^{d_{1}}t^{d_{2}}\cdot\mathbb{I}(\Omega({\bm{d}}))$,
where

(7.2)
$\displaystyle\mathbb{I}(\Omega({\bm{d}}))=\begin{cases}\left\langle\prod\limits_{i=1}^{2}\left(h_{i},h_{i}({-\bm{d}})-1\right)!\right\rangle&\text{
if }{\bm{d}}\notin e_{1}-\mathbb{N}\widetilde{A},\\\
\prod\limits_{i=1}^{2}\left(h_{i},h_{i}(-{\bm{d}})-1\right)!\cdot&\\\
\qquad\left\langle h_{1}+d_{1}-1,h_{2}+d_{2}\right\rangle&\text{ if
}{\bm{d}}\in e_{1}-\mathbb{N}\widetilde{A}.\end{cases}$

To compute $\mathbb{I}(\widetilde{J})$, we first determine which elements of
$\mathbb{I}(\Omega(d))$ are already in $\mathbb{I}(\widetilde{J})$. To do
this, we first break up (7.2) into cases by chamber:

$\mathbb{I}(\widetilde{J})_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{I}(\Omega({\bm{d}}))&\text{
if }{\bm{d}}\in\textbf{C1},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},h_{1}({-\bm{d}})\right)!\rangle&\text{
if }\bm{d}\in\textbf{C2}\setminus(-\sigma_{2}),\\\
s^{d_{1}}t^{d_{2}}\cdot\left(h_{1},h_{1}({-\bm{d}})\right)!\cdot\langle
h_{1}+d_{1}-1,h_{2}\rangle&\text{ if }\bm{d}\in(-\sigma_{2}),\\\
s^{d_{1}}t^{d_{2}}\cdot\left\langle\left(h_{1},h_{1}({-\bm{d}})\right)!\left(h_{2},h_{2}({-\bm{d}})\right)!\right\rangle&\text{
if }\bm{d}\in\textbf{C3},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},h_{2}({-\bm{d}})\right)!\rangle&\text{
if }\bm{d}\in\textbf{C4}.\end{cases}$

Similarly,

$D(R_{\widetilde{A}},\widetilde{J})_{\bm{d}}=\begin{cases}s^{d_{1}}t^{d_{2}}\cdot\mathbb{C}[\theta]&\text{
if }s^{d_{1}}t^{d_{2}}\in\widetilde{J},\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},h_{1}(-{\bm{d}})\right)!\rangle&\text{
if
}\bm{d}\in(\textbf{C2}\setminus(-\sigma_{2}))\cup(\sigma_{1}\setminus\\{\bm{0}\\}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{2},h_{2}({-\bm{d}})\right)!\rangle&\text{
if }\bm{d}\in\textbf{C4}\setminus(-\sigma_{1}),\\\
s^{d_{1}}t^{d_{2}}\cdot\langle\left(h_{1},h_{1}({-\bm{d}})\right)!\left(h_{2},h_{2}({-\bm{d}})\right)!\rangle&\text{
if }\bm{d}\in\textbf{C3}\cup(-\sigma_{1})\cup(-\sigma_{2}).\end{cases}$

Putting these together, the graded piece of
$\mathbb{I}(\widetilde{J})/D(R_{\widetilde{A}},\widetilde{J})$ at
${\bm{d}}\in\mathbb{Z}^{2}$ is

$\displaystyle\frac{\mathbb{I}(\widetilde{J})}{D(R_{\widetilde{A}},\widetilde{J})}_{\bm{d}}=\begin{cases}0&\text{
if }{\bm{d}}\notin\mathbb{Z}\sigma_{1}\cup\mathbb{Z}\sigma_{2},\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\mathbb{C}[\theta]}{\langle h_{1}\rangle}&\text{
if }\bm{d}\in\sigma_{1}\setminus\\{\bm{0}\\},\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\mathbb{C}[\theta]}{\langle h_{2}\rangle}&\text{
if }\bm{d}\in\sigma_{2}\setminus\\{\bm{0},(1,0)\\},\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\mathbb{C}[\theta]}{\langle\theta_{1}\theta_{2}\rangle}&\text{
if }\bm{d}=\bm{0},\\\ s^{d_{1}}t^{d_{2}}\cdot\dfrac{\langle
h_{1},h_{2}\rangle}{\langle h_{2}\rangle}&\text{ if }{\bm{d}}=(1,0),\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\left(h_{1},h_{1}({-\bm{d}})\right)!\langle
h_{1}+d_{1}-1,h_{2}\rangle}{\langle\left(h_{1},h_{1}({-\bm{d}})\right)!h_{2}\rangle}&\text{
if }\bm{d}\in(-\sigma_{2})\setminus\\{\bm{0}\\},\\\
s^{d_{1}}t^{d_{2}}\cdot\dfrac{\langle\left(h_{2},h_{2}({-\bm{d}})\right)!\rangle}{\langle
h_{1}\left(h_{2},h_{2}({-\bm{d}})\right)!\rangle}&\text{ if
}\bm{d}\in(-\sigma_{1})\setminus\\{\bm{0}\\}.\end{cases}$

Looking at the graded pieces of $\mathbb{I}(J)/D(R_{A},J)$ from Example 7.1
and noting that $G_{\bm{d}}(\theta)=\mathbb{C}[\theta]$ for
$\bm{d}\in(\sigma_{1})\cup(\sigma_{2}\setminus\\{\textbf{0},(1,0)\\}$, it
follows that $\left(\mathbb{I}(J)/D(R_{A},J)\right)_{\bm{d}}$ is identical to
$\left({\mathbb{I}(\widetilde{J})}/{D(R_{\widetilde{A}},\widetilde{J})}\right)_{\bm{d}}$
for all $\bm{d}$ except $\bm{d}=(1,0)$ and
$\bm{d}\in(-\sigma_{2})\setminus\\{\textbf{0}\\}$. So we only need to exhibit
that the graded pieces are isomorphic for
$\bm{d}\in\\{(1,0)\\}\cup(-\sigma_{2})\setminus\\{\textbf{0}\\}$. Note that
for $d_{1}<0$ and $d_{1}\neq-1$,

$\displaystyle\frac{\left\langle\displaystyle\frac{\left(h_{1},1-d_{1}\right)!}{(h_{1}-1)(h_{1}+d_{1})}\right\rangle}{\left\langle\displaystyle\frac{\left(h_{1},1-d_{1}\right)!\,h_{2}}{(h_{1}-1)(h_{1}+d_{1})}\right\rangle}\cong\displaystyle\frac{\langle\left(h_{1},1-d_{1}\right)!\rangle}{\langle\left(h_{1},1-d_{1}\right)!\,h_{2}\rangle}$

by the Third Isomorphism Theorem, and

$\dfrac{\left(h_{1},h_{1}({-\bm{d}})\right)!\langle
h_{1}+d_{1}-1,h_{2}\rangle}{\langle\left(h_{1},h_{1}({-\bm{d}})\right)!h_{2}\rangle}\cong\displaystyle\frac{\langle\left(h_{1},1-d_{1}\right)!\rangle}{\langle\left(h_{1},1-d_{1}\right)!\,h_{2}\rangle}$

by the Second Isomorphism Theorem. A similar argument gives an isomorphism in
the remaining multidegrees. Combining the information on all graded pieces, we
have explicitly shown that

$\dfrac{\mathbb{I}(\widetilde{J})}{D\left(R_{\widetilde{A}},\widetilde{J}\right)}\cong\dfrac{\mathbb{I}(J)}{D\left(R_{A},J\right)}.$

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