DIFFERENTIAL CLOSURE OF IDEALS

DIFFERENTIAL CLOSURE OF IDEALS

LILLIAN MCPHERSON, MONROE STEPHENSON, FUXIANG YANG

Abstract. In the following paper the differential closure, inspired by similar closure operations, is explored

in its relation to the radical and DR-modules. Additionally, the differential power is surveyed in respects to

its eventual principality and containment in ordinary powers, particularly in the case of monomial ideals.

Contents

1. Introduction 2

2. Background 2

3. Properties of Differential Closure and Differential Powers 3

3.1. Differential Closure is a Closure Operation 3

3.2. Other Properties 7

4. Differential Powers and Closures of Monomial Ideals 8

4.1. Ideals Generated by Pure Powers of Variables 9

4.2. Monomial Ideals 11

4.3. Containment Problem on Irreducible Monomial Ideals 12

4.4. Containment Problem on Other Monomial Ideals 14

4.5. Differential Powers which are Eventually Principal 16

4.6. Differential Powers which are Eventually Principal in 3 variables 19

5. General Results 20

5.1. Results on More General Ideals 20

5.2. Simple D-Modules 22

6. Open Questions 25

7. Acknowledgments 26

References 26

Date: August 2021.

1

2 LILLIAN MCPHERSON, MONROE STEPHENSON, FUXIANG YANG

1. Introduction

Commutative algebraists are interested in studying closure operations on ideals, primarily as a method toclassify the singularities of a ring. For example, closure operations were used to prove the Briancon-SkodaTheorem, which has direct corollaries describing singularities on hypersurfaces [Hoc07]. Most famouslyHilbert’s Nullstellensatz uses a closure operation to describe the one-to-one correspondence between idealsand varieties. We begin by defining the powers of ideals, which leads us to general closure operations.

Two well-studied notions of powers of an ideal I are the nth ordinary power In, which is the ideal generatedby the products of n elements of I, and the eth Frobenius power, I [pe], which is the ideal generated by peth

powers of elements in I [HH89]. Both of these lead to natural closure operations, the integral closure [HS06]:

I := {r ∈ R | there exists c ∈ R◦ : crn ∈ In for all n� 0}and the tight closure [HH89]:

I∗ := {r ∈ R | there exists c ∈ R◦ : crpe

∈ I [pe] for all e� 0}Inspired by the definitions above, we take another notion of power of an ideal, the differential power anddefine a new closure operation, the differential closure.

Definition 1.1 (Definition 2.2, [DDSG+18]). Let R be a ring and I ⊆ R be an ideal. Define the n-thdifferential power of I to be

I〈n〉 = {r ∈ R | D(r) ∈ I ∀D differential operators of order less than n}

Definition 1.2. Let R be a ring and I ⊆ R be an ideal. Define the differential closure of I to be

Idiff

= {r ∈ R | there exists c ∈ R◦ : crn ∈ I〈n〉 for all n� 0}

The differential power of an ideal coincides with the symbolic power of an ideal for prime ideals inpolynomial rings of characteristic zero [DDSG+18].

Commutative algebraists have previously studied properties of symbolic powers, such as in the containmentresults which inspired sections 4.3 and 4.4 of our paper (see [ELS01]). One major containment result is statedin the following theorem.

Theorem 4.18. Let I = (xβ1

1 , . . . , xβmm ) be an ideal in the ring R = k[x1, . . . , xz] for some characteristic 0

field k. I〈cn〉 ⊆ In where c = max{β1, . . . , βm,m+ 1}.

We also studied the behavior of the differential power I〈n〉 as n gets large.

Theorem 4.26, 4.31. For certain ideals I in a characteristic 0 polynomial ring, there exists a positiveinteger N such that I〈N〉 is principal.

Finally, we studied properties of the differential closure, proving that it is a closure operation. In fact, wehave that the differential closure equals the radical for ideals of simple DR-modules.

Theorem 5.13. Let R be a a regular characteristic 0 ring or positive characteristic p ring, and I is an ideal

of R, then√I = I

diffif and only if R is a simple DR−module.

2. Background

In order to study the differential power and differential closure, we must familiarize ourselves with def-initions of differential operators on a commutative k-algebra. An operator on the ring R is an element ofHomk(R,R). Recall the following definitions regarding differential operators. In these definitions, let R bea commutative k-algebra.

Definition 2.1 ([Cou95]). A derivation of R is a linear operator D of R which satisfies Leibniz’s Rule. (i.e.D(ab) = aD(b) + bD(a) for all a, b ∈ R)

DIFFERENTIAL CLOSURE OF IDEALS 3

Definition 2.2 ([Cou95]). The commutator of two differential operators P,Q is defined as

[P,Q] = P ·Q−Q · P

Consider the following examples of a derivation and commutator in a polynomial ring.

Example 2.3. Let R = C[x, y]. The operator ∂∂y is a derivation of R. We know from calculus partial

derivatives follow Leibniz’s Rule.

Example 2.4. Let R = C[x, y], P = ∂∂x , and Q = x ∂2

∂y2 . We can compute the commutator of P and Q:

[P,Q] =∂

∂x· x ∂

2

∂y2− x ∂

2

∂y2· ∂∂x

= x∂3

∂x∂y2+

∂2

∂y2− x ∂3

∂x∂y2

With these definitions we have the information we need in order to define a differential operator in acommutative k-algebra.

Definition 2.5 ([Cou95]). The order of a differential operator is defined inductively. An operator P hasorder zero if [a, P ] = 0 for every a ∈ R. If we have defined all differential operators of order less than n,define the set of operators of order n

Dn := {P ∈ Homk(R,R) | P does not have order < n and [a, P ] has order < n for all a ∈ R}

Applying this inductive definition, we can explicitly write out the operators of order less than or equal toone in terms of derivations and multiplication by a ring element.

Lemma 2.6 (Chapter 3, Lemma 1.1 [Cou95]). The operators of order ≤ 1 correspond to the elementsDerk(R) +R. The elements of order zero are the elements of R. Derk(R) denotes the derivations of R, andelements of R are considered differential operators through multiplication by that element.

In more general rings, this inductive notion of differential operators can appear abstract. For many ofour results in this paper, we focus on polynomial rings. In characteristic zero, derivations are of the form∑(

a · ∂∂xi

)for xi a variable in the ring, and a a ring element. A general differential operator is a linear

combination of partial derivatives and multiplication by a ring element. In characteristic p > 0, derivationsare of the same form as characteristic 0. When discussing more general differential operators in positive

characteristic, however, we must consider linear combinations which include generators of the form 1pe!

∂pe

∂xpe

i

for e a nonnegative integer. These generators are called divided powers.

Example 2.7. Let R = C[x, y]. The differential operator ∂∂x + y2 ∂

∂y acts on x+ y2 as follows:(∂

∂x+ y2 ∂

∂y

)(x+ y2) =

∂

∂x(x+ y2) +

(y2 ∂

∂y

)(x+ y2)

= 1 + y2(2y)

= 1 + 2y3

3. Properties of Differential Closure and Differential Powers

For this section, let R = k[x1, . . . , xz] be a polynomial ring where k is a field, and let I ⊆ R be an ideal.

3.1. Differential Closure is a Closure Operation. Before we discuss properties of the differential closure,we need to justify calling it a closure operation. To do this, we need some facts about differential powersand differential operators.

Proposition 3.1 (Proposition 3.8 [BJNB19]). I〈n〉 is an ideal.


Also, we note that I〈n〉 ⊆ I〈m〉 for all n > m. Instead of checking all differential operators of a givenorder, we write the following two lemmas to simplify which differential operators are sufficient to check whendetermining if an element is in a given differential power.

Lemma 3.2. Define Dnα to be a differential operator of order less than n which is a composition of partial

derivatives in characteristic k, and a composition of partial derivatives and divided powers in characteristick.

If y ∈ R satisfies Dnα · y ∈ I for all possible compositions of order less than n, then y ∈ I〈n〉.

Recall that divided powers are operators of the form

1

pl!

∂pl

∂xpl

i

for xi ∈ R and l ∈ N.

Proof. Let D =∑α rαD

nα, where rα denotes multiplication by an element of R. Since R is a polynomial

ring, D is an arbitrary differential operator of order less than n.

Let yα = Dnα(y) and we know that yα ∈ I. Consider

D(y) =

(∑α

rαDnα

)(y) =

∑α

rα(Dnα(y)) =

∑α

rαyα

where each rα are elements of R. Since yα ∈ I and rα ∈ R for every possible α, then D(y) ∈ I which impliesthat y ∈ I〈n〉. �

In the characteristic zero case, we need the following result on multiplying differential powers.

Lemma 3.3. If k characteristic zero, I〈n〉 · I〈m〉 ⊆ I〈m+n〉.

Proof. We know the elements of I〈n〉 · I〈m〉 are of the form∑i,j sitj where si ∈ I〈n〉 and ti ∈ I〈m〉. Without

loss of generality we can consider only one term. If st ∈ I〈n〉 · I〈m〉, then we want to show st ∈ I〈n+m〉.We first note that D<n(s) ∈ I, where D<n denotes an arbitrary differential operator of order less than n.Similarly D<m(t) ∈ I. We want to show D<m+n(st) ∈ I because this would imply our desired result. Letα ∈ Nz such that |α| ≤ m + n − 1, where |α| denotes the sum of the entries in α. Define an operator

Dα =∏αi∈α

∂αi

∂xαii

. Lemma 3.2 says in characteristic zero it is sufficient to consider differential operators as

compositions of partials, so it is sufficient to check that Dα(st) ∈ I for all α satisfying the above condition.We can use the General Leibniz rule to write

Dα(st) =∑

ω : ω≤α

(α

ω

)Dω(s)Dα−ω(t)

where

(α

β

)=

(α1

β1

)(α2

β2

)· · ·(αzβz

)and we say ω ≤ α for ω ∈ Nz if ωi ≤ αi for all i.

Note, we are only considering the order of the differential operators. In this case, we notice that if eitherDω(s) is in I or Dα−ω(t) is in I for all ω ≤ α then Dα(st) ∈ I. Equivalently, we need to show |α−ω| ≤ m−1or |ω| ≤ n − 1. We know |α − ω| = |α| − |ω| ≤ m + n − 1 − |ω|. If |α − ω| > m − 1, then |ω| ≤ n − 1.If |ω| > n − 1, then |α − ω| ≤ m − 1. So, either |α − ω| ≤ m − 1 or |ω| ≤ n − 1. Thus we have shownDω(t) ∈ I or Dα−ω(s) ∈ I for all ω : ω ≤ α. From this, we find that Dα(st) ∈ I, so st ∈ I〈n+m〉. Thus,I〈n〉 · I〈m〉 ⊆ I〈n+m〉. �


The following example illustrates Lemma 3.3.

Example 3.4. Let R = Q[x, y, z], I = (x2y3), n = 4, and m = 3. Lemma 3.3 states that I〈4〉 · I〈3〉 ⊆ I〈7〉.Indeed, we see

I〈4〉 · I〈3〉 = (x5y6) · (x4y5) = (x9y11)

I〈7〉 = (x8y9)

We can see I〈4〉 · I〈3〉 ⊆ I〈7〉 since xy2 · x8y9 = x9y11.

The following proposition is intuitive in the case of polynomial rings of characteristic zero. Generalizingthis fact to polynomial rings of positive characteristic (and, in fact, even more general rings) allows for ournext propositions to hold for all polynomial rings.

Proposition 3.5. Let r ∈ R. For all m > l, r|D≤l(rm) where 0 ≤ l < m and D≤l is any operator of orderless than or equal to l.

Proof. Proof by induction on l.

We begin by considering the base case where l = 0. Let D≤0 be an operator of order zero. We want toshow for all m > 0 we have r divides D≤0(rm). The only differential operators of order zero are multiplicationby a ring element so we may assume D≤0 is multiplication by a. Then, D≤0(rm) = arm and since m > 0,we know r divides D≤0(rm).

Next consider the inductive hypothesis. Assume for all m > l, r divides D≤l(rm) for every operator D≤l

of order less than or equal to l. We want to show for all m′ > l + 1, r divides D≤l+1(rm′) for any operator

D≤l+1 of order order less than or equal to l + 1. The inequality m > l implies m + 1 > l + 1. So we canrewrite m′ as m+ 1 to use the same m as the inductive hypothesis. After this substitution we want to showfor all m > l, r divides D≤l+1(rm+1).

Pick a differential operator D and without loss of generality let the order of D be l+ 1. Then, [D, r] is adifferential operator of order less than or equal to l. Use the inductive hypothesis to write [D, r](rm) = r(A)for some A in R. Then,

r(A) = [D, r](rm)

= (D · r)(rm)− (r ·D)(rm)

= D(rm+1)− r · (D(rm))

Since r divides the left side of the equation and r divides rD(rm), we have r divides D(rm+1). Since D wasan arbitrary differential operator, we have proven r divides D(rm+1) for all differential operators of orderless than or equal to l + 1. �

We can see this proposition more clearly through the use of the following example.

Example 3.6. Let R = Z/3Z[x, y, z], r = x2z. Consider the differential operators of order less than or

equal to five. We will look at m = 7 and the differential operator(

13!

∂3

∂x3

)∂∂x which has order four.(

1

3!

∂3

∂x3

)∂

∂x((x2z)7) =

(1

3!

∂3

∂x3

)∂

∂x(x14z7)

=

(1

3!

∂3

∂x3

)(2x13z7)

= 2x10z7

We can see that x2z divides 2x10z7.

Finally, we have the information needed to prove the differential closure is a closure operation. In fact,

we will later show in Proposition 5.12 that Idiff

=√I for any ideal I in a polynomial ring of characteristic


zero and all of the properties below follow from this fact. However, we include a direct proof here for someproperties which hold more generally.

Proposition 3.7. The differential closure is a closure operation on ideals in a polynomial ring of charac-teristic zero.

Equivalently, Idiff

satisfies the following properties where I, J are ideals in a polynomial ring of charac-teristic zero.

(1) Idiff

is an ideal

(2) I ⊆ Idiff

(3) If I ⊆ J , then Idiff ⊆ Jdiff

(4) Idiff

diff

= Idiff

Proof.

Proof of (1):

First, we want to show that the differential closure is closed under multiplication by a ring element. If

a ∈ Idiff, want to show ra ∈ Idiff

. This means we want to find a nonzero c ∈ R such that c(ra)n ∈ I〈n〉 forall n sufficiently large. We have (ra)n = rnan. We know there exists a nonzero c such that can ∈ I〈n〉. Sincern is a ring element and I〈n〉 is an ideal, rn(can) ∈ I〈n〉.

Second, we want to show that the differential closure is closed under addition. For a, b ∈ Idiff, we want to

show that a+b ∈ Idiff. We know from our definitions that there exist nonzero ca, cb ∈ R such that caa

n ∈ I〈n〉and cbb

n ∈ I〈n〉 for n sufficiently large. We want to show c(a+b)(a+ b)m ∈ I〈m〉 for some nonzero c(a+b) ∈ Rfor all m sufficiently large. We know by the binomial theorem that

c(a+b)(a+ b)m = c(a+b)

m∑k=0

(m

k

)akbm−k =

m∑k=0

(m

k

)c(a+b)a

kbm−k

When the characteristic of k is zero, we can invoke Lemma 3.3, so it suffices to show that there existssome c(a+b) such that for each k, c(a+b)a

kbm−k ∈ I〈k〉 · I〈m−k〉 ⊆ I〈m〉. By the definition of differential

power, we know there exists an N such that caan ∈ I〈n〉 and cbb

n ∈ I〈n〉 for all n ≥ N , so we know that(caa

N ) · ai ∈ I〈i+N〉 ⊆ I〈i〉 and (cbbN ) · bi ∈ I〈i+N〉 ⊆ I〈i〉 for all i greater than zero. In this case, let

c(a+b) = cacbaNbN then we have c(a+b)a

kbm−k = caak+Ncbb

m−k+N . We know caak+N ∈ I〈k〉 from our

choice of N . Similarly, cbbm−k+N ∈ I〈m−k〉. So, c(a+b)a

kbm−k ∈ I〈k〉 · I〈m−k〉 ⊆ I〈m〉 and we are done.

Proof of (2):

Suppose r ∈ I. We want to show there exists a nonzero c such that crn ∈ I〈n〉 for all n sufficiently large.This means D(crn) ∈ I for all differential operators D of order less than n. Choose c = 1. Using Proposition3.5, D(rn) = c1r where c1 might be zero. If c1 is zero, then c1r ∈ I since 0 ∈ I. If c1 is nonzero, then c1r ∈ Isince r ∈ I.

Proof of (3):

Let a ∈ Idiff. By definition there exists a nonzero c ∈ R such that can ∈ I〈n〉 for all positive integers n

sufficiently large. This implies that D · can ∈ I for all differential operators D with order less than n. SinceI ⊆ J , then D · can ∈ I ⊆ J . This means that can ∈ J〈n〉 for all positive integers n sufficiently large. By

definition a ∈ Jdiff. Thus, I

diff ⊆ Jdiff.

Proof of (4):


Later we will prove that Idiff

=√I if and only if R is a simple D-module. Since polynomial rings are

simple D-modules, this equality holds. By properties of the radical, we know√√

I =√I (Exercise 1.13,

[AM69]), so we have the idempotence property for the differential closure.

�

3.2. Other Properties. Now we can discuss other properties of the differential closure and differentialpower. As it is often useful to consider decompositions of ideals, we want to prove that the differential powerand differential closure respect intersections.

Lemma 3.8 (Exercise 2.13 [DDSG+18]). Let {Iα}α∈A be an indexed family of ideals. Then,

⋂α∈A

Iα〈n〉 =

(⋂α∈A

Iα

)〈n〉for every positive integer n.

Proof. First, we want to show that⋂α∈A Iα

〈n〉 ⊆(⋂

α∈A Iα)〈n〉

. Let s be an element of⋂α∈A Iα

〈n〉. By

definition, we have s in Iα〈n〉 for all α ∈ A. By the definition of differential power, we have D · s ∈ Iα for all

α ∈ A and for all differential operators D with order less than n. This implies that D · s is an element of⋂α∈A Iα. Thus, s is an element of

(⋂α∈A Iα

)〈n〉.

Now let t be an element of(⋂

α∈A Iα)〈n〉

. By the definition of differential power, we have D · t ∈⋂α∈A Iα

for all differential operators D with order less than n. By the definition of intersection, we have D · t ∈ Iαfor all α ∈ A. This implies that t is an element of Iα

〈n〉 for all α ∈ A. Thus, t is an element of⋂α∈A Iα

〈n〉.We have shown both direction of containment, so this proves

⋂α∈A

Iα〈n〉 =

(⋂α∈A

Iα

)〈n〉for every positive integer n. �

Lemma 3.9. Let {Iα | α ∈ A} be a family of ideals with an index set A. Then⋂α∈A

Iαdiff ⊇

⋂α∈A

Iαdiff

,

and when A is finite, we have ⋂α∈A

Iαdiff

=⋂α∈A

Iαdiff

.

Proof. First, we want to show that⋂α∈A Iα

diff⊆⋂α∈A Iα

diff. We have

⋂α∈A Iα ⊆ Iα for every α ∈ A.

By the property of closure operations, this implies that⋂α∈A Iα

diff⊆ Iα

difffor every α ∈ A. Thus, by the

definition of intersection, we have ⋂α∈A

Iαdiff

⊆⋂α∈A

Iαdiff.

Now we want to show that⋂α∈A Iα

diff⊇⋂α∈A Iα

diff. Let b ∈

⋂α∈A Iα

diffbe an arbitrary element. This

means that b ∈ Iαdiff

for all α ∈ A. By definition, for all α ∈ A there exists an element cα ∈ R such that

cαbn ∈ Iα〈n〉 for all n sufficiently large. This implies that (

∏α∈A cα)bn ∈ Iα〈n〉 for all α ∈ A and n sufficiently

large. By definition of intersection and Lemma 3.8, we have(∏α∈A

cα

)bn ∈

⋂α∈A

Iα〈n〉 =

(⋂α∈A

Iα

)〈n〉.


By the definition of differential closure, b ∈⋂α∈A Iα

diff. This implies that

⋂α∈A Iα

diff⊇⋂α∈A Iα

diff.

Thus,⋂α∈A Iα

diff=⋂α∈A Iα

diff. �

Next, we look at an interesting application of Lemma 3.2.

Proposition 3.10. If k is of characteristic zero, I〈n〉〈m〉

= I〈n+m−1〉 for all n,m.

If k is of characteristic p > 0, then I〈n〉〈m〉 6= I〈n+m−1〉 if, for some e > 0, n ≤ pe and m ≤ pe and

n+m− 1 > pe.

Proof. Let r be an arbitrary element in I〈n+m−1〉. This means that D(r) ∈ I for all differential operatorsD with order less than or equal to n + m− 2. Let D1 be a differential operator of order less than or equalto n− 1 and let D2 be a differential operator of order less than or equal to m− 1. Then D = D1 ◦D2, has

order less than n+m−2, so (D1 ◦D2)(r) = D(r) ∈ I. Since D1 and D2 were arbitrary, we have r ∈ I〈n〉〈m〉.This implies I〈n+m−1〉 ⊆ I〈n〉〈m〉.

To prove I〈n〉〈m〉 ⊆ I〈n+m−1〉, we want to show for all D of order less than or equal to n + m − 2, there

exists a D1 and D2 of orders less than n and m respectively which satisfy the equality D(r) = (D1 ◦D2)(r)for all r ∈ R.

First consider the case where k is of characteristic zero. Recall for polynomial rings of characteristiczero, Lemma 3.2 states that when considering differential operators of order N it is sufficient to considercompositions of N first-order partial derivatives. Thus, write D =

∏i∈[z]

∂ci

∂xcii

where each ci is a nonnegative

integer, 0 ≤∑i∈[z] ci ≤ n + m − 2. If we restrict D1 and D2 to the forms D1 =

∏i∈[z]

∂ai

∂xaii

and D2 =∏i∈[z]

∂bi

∂xbii

where each ai, bi is a nonnegative integer, 0 ≤∑i∈[z] ai ≤ n− 1, and 0 ≤

∑i∈[z] bi ≤ m− 1, we

can find a D1 and D2 such that D = D1 ◦D2, since partial derivatives commute.

Second consider the case where k is of characteristic p > 0. Recall for polynomial rings of characteristicp, Lemma 3.2 states that when considering differential operators of order N , it is sufficient to consider

compositions of partial derivatives and operators of the form 1pe!

∂pe

∂xpe

i

where the sum of orders of operators

in the composition equals N . Partial derivatives all have order 1, and the other operators have order pe.

So, we consider all the values of e such that n+m− 1 > pe. Let D = 1pe!

∂pe

∂xpe

i

. We want to write 1pe!

∂pe

∂xpe

i

as a composition of an operator of order less than n and an operator of order less than m. But since 1pe!

∂pe

∂xpe

i

is a generator in the ring of differential operators, the only way we can write it as a composition is 1◦ 1pe!

∂pe

∂xpe

i

or 1pe!

∂pe

∂xpe

i

◦ 1. So, either pe < n or pe < m, or both. So, for the two ideals not to be equal, we have pe ≥ nand pe ≥ m. �

4. Differential Powers and Closures of Monomial Ideals

In this section, let R = k[x1, . . . , xz] where k is a field of characteristic 0. The goal of this section is tounderstand the properties of differential power and closure when the ideal is a monomial ideal. We want

to know whether or not I being a monomial ideal implies Idiff

and I〈n〉 are monomial ideals. We are alsointerested in studying the conditions of containment problems between differential power I〈n〉 and ordinarypower In.

We recall some facts about monomial ideals. First, we know that every monomial ideal has a decompositionin terms of ideals which are pure powers of variables. In particular,


Theorem 4.1 (Theorem 1.3.1 [HH11]). Let I ⊂ S = K[x1, . . . , xn] be a monomial ideal. Then I =⋂mi=1Qi where each Qi is generated by pure powers of the variables. In other words, each Qi is of the form

(xa1i1 , . . . , xakik

). Moreover, an irredundant presentation of this form is unique.

In the particular case of squarefree monomial ideals, we see that this decomposition consists of idealsgenerated by variables.

Corollary 4.1.1. If I is a squarefree monomial ideal, then I =⋂mi=1Qi where each Qi is of the form

(xi1 , . . . , xik).

Furthermore, the any finite intersection of monomial ideals is still a monomial ideal, and we can explicitlydescribe its generators.

Definition 4.2. Let I be an ideal of R. G(I) is the set of minimal generators of I.

Proposition 4.3 (Proposition 1.2.1 [HH11]). Let I and J be monomial ideals. Then I ∩ J is a monomialideal, and {lcm(u, v) : u ∈ G(I), v ∈ G(J)} is a set of generators of I ∩ J .

Corollary 4.3.1.⋂ni=1 Ii is monomial ideal if Ii is a monomial ideal and n is a finite positive integer.

From these results, it is clear that understanding differential powers of ideals generated by pure powersof variables is the first step to understanding differential powers of monomial ideals.

4.1. Ideals Generated by Pure Powers of Variables. In this section, we are studying properties ofideals generated by pure powers of variables, which is equivalent to studying irreducible monomial ideals.Consider an ideal I = (xβ1

α1, xβ2α2, . . . , xβmαm) ⊂ R = k[x1, . . . , xz] where each βi is a positive integer. We can

explicitly write the generators of I〈n〉 for all positive integers n and Idiff

.

Proposition 4.4. I〈n〉 is generated by the elements of the forms xγ1α1xγ2α2

. . . xγmαm where γi is either 0 orgreater than or equal to βi that satisfies

∑i∈[m],γi≥βi γi =

∑i∈[m],γi≥βi(βi − 1) + n.

Proof. Consider an arbitrary element X of the form xγ1α1xγ2α2· · ·xγmαm where γi which satisfies the given con-

dition. This means that there is an induced set S where S = {i ∈ [m] | γi ≥ βi}. Let D be an arbitrarydifferential operator of order less than n. By Lemma 3.2, we only need to consider the case where D is acomposition of at most n− 1 differential operators of the form ∂

∂xi. Now we can let

D =∏i∈S

∂ωi

∂xωii

∏j∈[m],j /∈S

∂θj

∂xθjj

where ωi, θj are positive integers such that∑i∈S ωi +

∑j∈[m],j /∈S θj ≤ n− 1. If there exists an index j ∈ [m]

such that ωj > γj or θj > 0, then D(X) = 0. Now we can assume that ωi ≤ γi and θj = 0 for all i ∈ Sand j ∈ [m] \ S. Since θjs are 0, then we can assume that D =

∏i∈S

∂ωi

∂xωii

. Now we have the inequality∑i∈S ωi ≤ n− 1. Consider

D(X) = c∏i∈S

xγi−ωiαi

where c is some constant in k. Since∑i∈S

(γi − ωi) =∑i∈S

γi −∑i∈S

ωi =∑i∈S

(βi − 1) + n−∑i∈S

ωi ≥∑i∈S

(βi − 1) + 1,

then by the Pigeonhole Principle, there exists an index l ∈ S such that γl − ωl ≥ βl. This implies thatD(X) ∈ I. By Lemma 3.2, X ∈ I〈n〉.

Suppose there exists an element Y = xγ1α1xγ2α2· · ·xγmαm where γi is either 0 or greater than or equal to βi

such that∑i∈S γi ≤

∑i∈S(βi− 1) +n− 1 where S = {i ∈ [m] | γi ≥ βi}. We can find a differential operator

of the form

D =∏i∈S

∂ωi

∂xωiαi


where ωi = γi− βi + 1 for all i. We need to check that this operator is of order less than n. We can sum thepowers of each partial in the product. Doing this gives us∑

i∈S(γi − βi + 1) =

∑i∈S

γi −∑i∈S

βi + |S|

≤∑i∈S

(βi − 1) + n− 1−∑i∈S

βi + |S|

= n− 1

so D is a differential operator of order less than or equal to n− 1, which is less than n.

Calculating D(Y ) gives us

D(Y ) = c∏i∈S

xγi−(γi−βi+1)αi

= c∏i∈S

xβi−1αi

where c ∈ k.

We can see that the power of each xαj appearing in the sum is less than βj for all j ∈ [m]. This implies

that D(Y ) is not in I and Y 6∈ I〈n〉. Therefore, I〈n〉 is generated by the elements of the forms xγ1α1xγ2α2

. . . xγmαmwhere γi is either 0 or greater than or equal to βi that satisfies

∑i∈[m],γi≥βi γi =

∑i∈[m],γi≥βi(βi−1)+n. �

Proposition 4.5. Idiff

= (xα1, . . . , xαm).

Proof. Let i be an arbitrary index in [m], by Proposition 4.4, we know that xβi−1+nαi is an element of I〈n〉.

Notice that if we let c = xβi−1αi 6= 0 which is a constant, we have

c(xαi)n = xβi−1+n

αi ∈ I〈n〉

for all positive integer n sufficiently large. This implies that xαi ∈ Idiff

for all i ∈ [m]. Now we must

show that every element not in (xα1, . . . , xαm) is also not in I

diff. First, we show a field element is not in

Idiff

. We just check that 1 is not in Idiff

. Suppose 1 ∈ Idiff. Then, there exists a nonzero c ∈ R such that

c(1)n ∈ I〈n〉 for n sufficiently large. This implies that c ∈ I〈n〉 for all n sufficiently large, we need to showthat there exists at least one monomial term b of c such that b 6∈ I〈n〉 for all n sufficiently large. we canwrite b = r

∏i∈[m] x

γiαi where each γi is a constant positive integer, and r is a monomial with variables not

in {xα1, . . . , xαm}. Let S = {i ∈ [m] | γi ≥ βi}, and suppose that b ∈ I〈n〉. By Proposition 4.4, this implies

that∑i∈S γi ≥

∑i∈S(βi − 1) + n. Here we can see that b 6∈ I〈n〉 for all n >

∑i∈S(γi − βi + 1). This shows

that b 6∈ I〈n〉 for all n sufficiently large, which implies that c 6∈ I〈n〉 for all n sufficiently large. This is a

contradiction, so 1 6∈ Idiff.

Next, we show that a linear combination of variables which do not appear in I are not in Idiff

. It sufficesto check monomial terms. Let X =

∏j∈[z],j /∈α x

ωjj where ωj are positive integers. Suppose that there exists

c ∈ R such that cXn ∈ I〈n〉 for all n sufficiently large. Consider a monomial term b of c and we can showthat Xnb 6∈ I〈n〉 for all n sufficiently large for the same reason. Thus, a linear combination of variables not

in {xα1, . . . , xαm} is not in I

diff. We showed that I

diff= (xα1

, . . . , xαm). �

Remark 4.6. Notice that if we have any irreducible monomial ideal I, then Idiff

and I〈n〉 will be monomialideals for all n.

Example 4.7. Let R = Q[x, y], I = (x3, y3). We find I〈3〉 by considering differential operators. Accordingto Lemma 3.2, we only need to consider operators which are compositions of partial derivatives. In this case,

that means we need to check ∂∂x ,

∂∂y ,

∂2

∂x∂y ,∂2

∂x2 ,∂2

∂y2 . Since constants are not affected by partial derivatives,


we check the behavior of these partials on an element xiyj .

∂

∂x(xiyj) = ixi−1yj

∂

∂y(xiyj) = jxiyj−1

∂2

∂x2= i(i− 1)xi−2yj

∂2

∂y2= j(j − 1)xiyj−2

∂2

∂x∂y= ijxi−1yj−1

From these equations, we can see that xiyj is in I〈3〉 if i ≥ 5 or j ≥ 5 or (i = 4 and j = 3) or (i = 3 andj = 4). Thus, I〈3〉 = (x5, y5, x3y4, x4y3), which agrees with Proposition 4.4.

Example 4.8. Let R = C[x, y, z], I = (x2, y3). Applying Proposition 4.4 and Proposition 4.5, we have

I〈2〉 = (x3, y4, x2y3), I〈3〉 = (x4, y5, x3y3, x2y4), and Idiff

= (x, y).

Example 4.9. Let R = C[x, y, z], I = (x3, y3, z5). Applying Proposition 4.4 and Proposition 4.5, we haveI〈2〉 = (x4, y4, z6, x3y3, x3z5, y3z5), I〈3〉 = (x5, y5, z7, x4y3, x3y4, x4z5, x3z6, y4z5, y3z6, x3y3z5), and

Idiff

= (x, y, z)

4.2. Monomial Ideals. Since monomial ideals can be written as an intersection of finitely many idealsgenerated by pure powers of variables by Theorem 4.1, we can use our results in the previous section tostudy more general monomial ideals.

Theorem 4.10. Given a monomial ideal I in R, I〈n〉 is also a monomial ideal for every positive integer n.

Proof. By Theorem 4.1, we have

I =

m⋂i=1

Qi.

Consider the n-th differential power of I and apply Lemma 3.8,

I〈n〉 =

(m⋂i=1

Qi

)〈n〉=

m⋂i=1

Qi〈n〉.

Since Qi is an irreducible monomial ideal for all i, then by Proposition 4.4, Qi〈n〉 is a monomial ideal for all

i. By Proposition 4.3, we know that⋂mi=1Qi

〈n〉 is a monomial ideal. Thus, I〈n〉 is a monomial ideal. �

Since Lemma 3.8 tells us the differential power respects intersections, we are able to compute the n-thdifferential power of any monomial ideal by breaking it up into irreducible monomial ideals. When I isgenerated by a single monomial, we can explicitly write the generators of I〈n〉 for all n.

Proposition 4.11. Consider an ideal I = (∏i∈[m] x

βiαi) ⊂ k[x1, . . . , xz] where each βis are positive integers.

Then I〈n〉 = (∏i∈[m] x

βi+n−1αi ).

Proof. First notice that I =⋂i∈[m](x

βiαi). From Lemma 3.8, we know that differential powers respect finite

intersections. Thus,

I〈n〉 =⋂i∈[m]

(xβiαi)〈n〉

We know how to compute each of these differential powers by Proposition 4.4. This gives us

I〈n〉 =⋂i∈[m]

(xβi−1+nαi )

Finally, we rewrite this intersection as a single ideal

I〈n〉 =

∏i∈[m]

xβi+n−1αi


�

Now we know that the n-th differential power of a monomial ideal is a monomial ideal, the next questionwill be whether or not the differential closure of a monomial ideal is a monomial ideal. Since we showed thatdifferential closure commutes with finite intersections in Lemma 3.9, then we can show this property usinga very similar proof.

Theorem 4.12. I is a monomial ideal in R implies that Idiff

is a monomial ideal.

Proof. By Theorem 4.1, we have

I =

m⋂i=1

Qi

where each Qi is an irreducible monomial ideal. Consider the differential closure of I and apply Lemma 3.9,

Idiff

=

(m⋂i=1

Qi

)diff

=

m⋂i=1

Qidiff.

Since Qi is an irreducible monomial ideal for all i, then by Proposition 4.5, Qidiff

is a monomial ideal for all

i. By Theorem 4.3, we know that⋂mi=1Qi

diffis a monomial ideal. Thus, I

diffis a monomial ideal. �

Example 4.13. Let R = C[x, y, z] and I = (x2y, xz) = (x) ∩ (x2, z) ∩ (y, z).

I〈2〉 = (x)〈2〉 ∩ (x2, z)

〈2〉 ∩ (y, z)〈2〉

= (x2) ∩ (x3, z2, x2z) ∩ (y2, z2, yz)

= (x3y2, x2z2, x2yz)

Idiff

= (x)diff∩ (x2, z)

diff∩ (y, z)

diff

= (x) ∩ (x, z) ∩ (y, z)

= (xy, xz).

4.3. Containment Problem on Irreducible Monomial Ideals. This section is inspired by previouswork on containment problems of symbolic powers, namely Theorem 2.2 from [ELS01] by Ein, Lazarsfeld,and Smith which states that in the ring R = k[x1, . . . , xz] where k is a field with characteristic 0, wehave I(zn) ⊆ In for all positive integer n and radical ideals I. This leads us to ask the same questionbetween differential powers and ordinary power. When is the differential power contained in the ordinarypower (I〈a〉 ⊆ Ib)? When is the differential power containing the ordinary power(I〈a〉 ⊇ Ib)? Since wehave an explicit way (Proposition 4.4) to write out the generators of the n-th differential power of anirreducible monomial ideal, then we want to approach these two questions by restricting to I being anirreducible monomial ideal first before generalizing to other ideals. Recall that in the irreducible monomialideal section, we defined I = (xβ1

α1, xβ2α2, . . . , xβmαm) where each βis are positive integers. First, we want to

study the conditions in which Ia ⊆ I〈b〉.

Proposition 4.14. In ⊆ I〈n〉.

Proof. We can write the set of generators of In as {xβ1ω1α1· · ·xβmωmαm |

∑mi=1 ωi = n, ωi ∈ Z≥0∀i ∈ [m]}. Let

xβ1ω1α1· · ·xβmωmαm be an arbitrary element from In, S ⊆ [m] be the set of indexes such that ωj 6= 0 for all

j ∈ S. We can see that this S construction is same as Proposition 4.4 because when ωj = 0, ωjβj < βj , and


when ωj > 0, ωjβj ≥ βj . Consider∑i∈S

βiωi =∑i∈S

(βi − 1) + |S|+∑i∈S

βi(ωi − 1)

≥∑i∈S

(βi − 1) + |S|+∑i∈S

(ωi − 1)

=∑i∈S

(βi − 1) +∑i∈S

1 +∑i∈S

(ωi − 1)

=∑i∈S

(βi − 1) +∑i∈S

ωi

=∑i∈S

(βi − 1) + n.

By Proposition 4.4, we know that every element in In is in I〈n〉. Thus, In ⊆ I〈n〉. �

Next, we want to find a tighter bound on In using the differential power. We have found that In ⊆ I〈n+c〉

where c depends on n. This c is defined explicitly in the following proposition.

Proposition 4.15. Fix an arbitrary positive integer n. Let T be the subset of Zm≥0 such that ω is an element

of T if and only if∑i∈[m] ωi = n. Then In ⊆ I〈n+c〉 for all c ≤ min{

∑i∈[m],ωi>0(βi − 1)(ωi − 1) | ω ∈ T}.

Proof. With the same construction as the proof of Lemma 4.14, consider∑i∈S

βiωi =∑i∈S

(βi − 1) + |S|+∑i∈S

βi(ωi − 1)

=∑i∈S

(βi − 1) +∑i∈S

1 +∑i∈S

(ωi − 1) +∑i∈S

(βi − 1)(ωi − 1)

=∑i∈S

(βi − 1) + n+∑i∈S

(βi − 1)(ωi − 1)

≥∑i∈S

(βi − 1) + n+ c.

This implies that the generators of In are elements of I〈n+c〉. Thus, In ⊆ I〈n+c〉. �

From [ELS01], we have the containment is I(zn) ⊆ In for all positive integers n and radical ideals I. Ourgoal is to find a constant c that depends only on the ring R, such that I〈cn〉 ⊆ In for some certain type ofideal I. First, we want to simplify the condition by letting I be any irreducible monomial ideal, and we wantto find a constant c that depends on n and see if the sequence of c on n converges.

Proposition 4.16. Let I = (xβ1α1, xβ2α2, . . . , xβmαm). Then I〈cn

b〉 ⊆ In where c = d (n−1)βmax+m+1nb

e for allpositive integers b and βmax = max{βi | i ∈ [m]} for all n.

Proof. By Proposition 4.4, I〈cnb〉 is generated by

∏i∈[m] x

γii where γi is either 0 or greater than or equal to

βi which induces a subset S ∈ P([m]) for each generator where S = {s ∈ [m] | γs ≥ βs}, and the generators

satisfy∑i∈S γi =

∑i∈S(βi − 1) + cnb. We also know that In is generated by

∏i∈[m] x

βiωii where each ωi is

a nonnegative integer and∑i∈[m] ωi = n.

We want to show that given any generator∏i∈[m] x

γii from I〈cn

b〉, this element is also in In. In other

words, we want to show that there exists ω ∈ Zm≥0 such that∑i∈[m] ωi = n, and it satisfies that γi ≥

βiωi for all i ∈ S where S is the induced set from earlier. Let γi = βi + δi for every i ∈ S, and letβmax = max{βi | i ∈ [m]}. Since γi ≥ βi for all i ∈ S, then δi ≥ 0 for all i ∈ S. By Proposition 4.4, weknow that

∑i∈S γi =

∑i∈S(βi − 1) + cnb. This implies that

∑i∈S δi = cnb − |S|. Since γi ≥ βiωi and


γiβi

= 1 + δiβi≥ 1 + δi

βmax, then it is sufficient to show that there exist ωi’s that satisfy 1 + δi

βmax≥ ωi. Let

ωi = d δiβmaxe ≤ 1 + δi

βmax, we want to show that

∑i∈S ωi ≥ n. Consider∑

i∈Sωi =

∑i∈S

⌈δi

βmax

⌉

≥

⌈∑i∈S

δiβmax

⌉

=

⌈cn− |S|βmax

⌉≥

⌈(n−1)βmax+m+1

nbnb − |S|

βmax

⌉

=

⌈(n− 1)βmax +m+ 1− |S|

βmax

⌉=

⌈n− 1 +

m+ 1− |S|βmax

⌉≥ n.

Therefore, we showed I〈cnb〉 ⊆ In where c = d (n−1)βmax+m+1

nbe for all n. �

Remark 4.17. When b = 1, c = d (n−1)βmax+m+1n e. Notice that this sequence of c depending on n is converging

to βmax, so we can find a constant c that is only depending on the ideal I.

Theorem 4.18. I〈cn〉 ⊆ In where c = max{β1, . . . , βm,m+ 1}.

Proof. By Proposition 4.4, I〈cn〉 is generated by∏i∈[m] x

γii where γi is either 0 or greater than or equal to

βi which induces a subset S ⊆ [m] for each generator where S = {s ∈ [m] | γs ≥ βs}, and each generator

satisfies∑i∈S γi =

∑i∈S(βi − 1) + cn. We also know that In is generated by

∏i∈[m] x

βiωii where ωis are

nonnegative integers that satisfies∑i∈[m] ωi = n.

We want to show that if∑i∈S γi =

∑i∈S(βi − 1) + cn with c = max{β1, . . . , βm, z + 1}, then there exist

ωi’s such that γi ≥ βiωi for all i ∈ S where∑i∈S ωi = n. It is sufficient to show there exists a set of ωi’s

such that∑i∈S ωi ≥ n because we can always decrease the power to equal n. This is equivalent in showing

γiβi≥ ωi. Since γi ≥ βi for all i ∈ S by the construction of S, then γi = βi + δi for some nonnegative integer

xi for all i ∈ S. This means that∑i∈S δi = cn− |S|. Notice

γiβi

= 1 +δiβi≥ 1 +

δic≥⌈δic

⌉.

Choose each ωi = d δic e, we need to check that∑i∈S ωi ≥ n. Now we have

∑i∈S

ωi =∑i∈S

⌈δic

⌉≥

⌈∑i∈S

δic

⌉=

⌈n− |S|

c

⌉= n

because |S| < m+ 1 ≤ c. We have shown that each generator of I〈cn〉 is in In. Thus, I〈cn〉 ⊆ In. �

4.4. Containment Problem on Other Monomial Ideals. Here are some more results on other monomialideals. First, the goal is to study whether there can exist a constant c which does not depend on the idealI such that I〈cn〉 ⊆ In. We can construct a counterexample with a monomial ideal to show that there can’texist such constant c.

Proposition 4.19. There does not exist a constant c where c is only depending on R such that I〈cn〉 ⊆ In

for all ideals I in R and for all n ∈ Z>0.


Proof. Suppose that there exists such c, then we can find this counterexample with I = (xc+11 ). By Proposi-

tion 4.4, I〈cn〉 is generated by the element xc+cn1 . We know that In is generated by xcn+n1 . Since I〈cn〉 ⊆ In

implies that c+ cn ≥ cn+ n, then we have c ≥ n which implies that c depends on n. This contradicts withthe assumption that c only depend on the ring R. Therefore, such c does not exist. �

On the other hand, we can also use the same counterexample to show the following result.

Proposition 4.20. The only constant c such that In ⊆ I〈cn〉 for all ideal I in R and for all n ∈ Z>0 isc = 1.

Proof. Suppose that there exists such c where c > 1, then we have the counterexample I = (xc+11 ). By

Proposition 4.4, I〈cn〉 is generated by the element xc+cn1 . We know that In is generated by xcn+n1 . Since

In ⊆ I〈cn〉 implies that cn + n ≥ c + cn, then we have n ≥ c for all n ∈ Z>0 which implies that n > 1.Therefore, only c = 1 satisfies that condition. �

Since we can’t find a constant c which does not depend on I or on n such that I〈cn〉 ⊆ In, then the next

idea would be considering if the statement is true for I〈cn2〉 ⊆ In, more generally whether does the statement

still hold for I〈p(n)〉 ⊆ In where p(n) is a polynomial with constant coefficient which does not depend on Iand with variable n. Now we have this new statement.

Proposition 4.21. There does not exist a polynomial p(n) = cmnm + cm−1n

m−1 + . . .+ c1n+ c0 with a setof constant positive integers ci and a nonnegative integer m such that I〈p(n)〉 ⊆ In for all n ∈ Z≥0.

Proof. Suppose that there exists a polynomial p(n) such that I〈p(n)〉 ⊆ In, then we have the counterexample

I = (xβ1 ) where β = p(2). By Proposition 4.4, we have this inequality,

β − 1 + p(n) ≥ βnp(n) ≥ β(n− 1) + 1.

Notice that when n = 2 the inequality is p(2) ≥ p(2) + 1 which leads to a contradiction. Therefore, theredoes not exist a polynomial p(n) such that I〈p(n)〉 ⊆ In for all n ∈ Z≥0. �

The natural next step is finding a value of c such that I〈cn〉 ⊆ In, where c necessarily depends on I.

Proposition 4.22. Let α be an ordered subset of {1, 2, . . . , z} = [z] with |α| = m ≤ z where z,m ∈ Z>0.Denote the i-th element in the set α as αi. Consider an ideal I of R where I = (

∏i∈[m] x

βiαi) where each βis

are positive integers. I〈cn〉 ⊆ In where c = (n−1)βmax+1n and βmax = max{βi | ∀i ∈ [m]} for all n.

Proof. By Proposition 4.11, I〈cn〉 = (∏i∈[m] x

βi+cn−1αi ), and since In = (

∏i∈[m] x

βinαi ), we want to show that∏

i∈[m] xβi+cn−1αi ∈ In. This is equivalent in showing that βi + cn− 1 ≥ βin for all i ∈ [m]. Consider

βi + cn− 1 = βi +(n− 1)βmax + 1

nn− 1

= βi + (n− 1)βmax

≥ βin.

Thus, I〈cn〉 ⊆ In. �

Remark 4.23. Take the same ideal I from Proposition 4.22, I〈βmaxn〉 ⊆ In for all n ∈ Z>0.


4.5. Differential Powers which are Eventually Principal. In this section, we let the ring R = k[x, y].Notice that we can create a graphical representation for every monomial ideal in R using an integer latticewhere the x-axis is the power of x and y-axis is the power of y. For example, the point (2, 3) would representthe set of monomials cx2y3 where c ∈ k and the ideal I = (xy2, x3) would be all the points inside the shadedarea.

(3, 0)

(1, 2)

power of y

power of xI

Taking the partial derivative ∂/∂y would be a translation down in the y direction, and taking the partialderivative ∂/∂x would be a translation left in the x direction. Now we can also represent the differentialpowers of monomial ideals graphically. For example, the 2nd and 3rd differential power of the ideal I =(xy2, x3) can be represented as

(3, 0)

(1, 2)

power of y

power of x

∂/∂y

∂/∂x

I〈2〉I I〈3〉

For the rest of this section, let I = (xβ11yβ12 , . . . , xβm1yβm2).

Remark 4.24. Notice that we can rearrange the generators of I to satisfy the ineqality β11 < · · · < βm1

because βi1 = β(i+1)1 implies that one of them is redundant. This inequality implies another ineqality


β12 > · · · > βm2 because βj2 ≤ β(j+1)2 and βj1 < β(j+1)1 implies that xβ(j+1)1yβ(j+1)2 is redundant. For therest of this section, we assume that the ideal I has this structure that satisfies both inequalities.

We claim that we can find a positive integer N for all such ideal I such that I〈N〉 is principal. We needto first show a lemma.

Lemma 4.25. Let β(i+1)1, βi2 be the pair of βs such that β(i+1)1 + βi2 = max{β(j+1)1 + βj2 | 1 ≤ j < m}.If a+ b ≥ β(i+1)1 + βi2 − 1 and a ≥ β11, b ≥ βm2, then xayb ∈ I.

Proof. Suppose there exists such a, b satisfying the condition, and xayb 6∈ I. Since a ≥ β11 and b ≥ βm2,then there exists an index k such that βk1 ≤ a < β(k+1)1 and βk2 > b. Since βk1, β(k+1)1, βk2, a, b are positiveintegers, then we have β(k+1)1 ≥ a + 1 and βk2 ≥ b + 1. This implies that β(k+1) + βk2 ≥ a + b + 2 ≥β(i+1)1 + βi2 + 1 which contradicts the assumption that β(i+1)1 + βi2 = max{β(j+1)1 + βj2 | 1 ≤ j < m}.Thus, xayb ∈ I. �

Theorem 4.26. For all such ideals I, there exists a positive integer N such that I〈N〉 is principal.

Proof. If m = 1, then by Proposition 4.11, I〈n〉 is always a principal monomial ideal for all positive integer n.Now assume m > 1. Let β(i+1)1, βi2 be a pair of βs such that β(i+1)1 +βi2 = max{β(j+1)1 +βj2 | 1 ≤ j < m}.We want to show that I〈N〉 is principal where N = β(i+1)1 +βi2−βm2−β11. Let a = β(i+1)1 +βi2−βm2−1,

and b = β(i+1)1 + βi2 − β11 − 1. We claim that I〈N〉 = (xayb). We want to show that xayb is an element of

I〈N〉 and xa−1ye and xfyb−1 are not elements of I〈N〉 for all positive integer e, f . To show that xayb ∈ I〈N〉,it is sufficient to check the differential operators D = ∂N−1

∂xd∂yN−1−d for all nonnegative integers d less than or

equal to N − 1 because ∂N−1

∂xN−1 (xayb), ∂N−1

∂yN−1 (xayb) 6∈ k. Since β12 > βm2 and β11 < βm1, then N > 0. Notice

D(xayb) =∂N−1

∂xd∂yN−1−d (xayb)

= c1xa−dc2y

b−N+1+d

= c1c2xa−dyβm2+d

where c1, c2 are some elements in k. Now we need to check when 0 ≤ d ≤ N − 1. By Lemma 4.25, sincea− d + βm2 + d = a + βm2 = β(i+1)1 + βi2 − 1 and a− d ≥ a−N + 1 = β11, and βm2 + d ≥ βm2, then we

know that D(xayb) ∈ I.

Since ∂N−1

∂xN−1 (xa−1ye) 6∈ I and ∂N−1

∂yN−1 (xfyb−1) 6∈ I for all positive integers e, f which implies that

xa−1ye, xfyb−1 6∈ I〈N〉 for all positive integers e, f , then I〈N〉 = (xayb). This shows that there exists apositive integer N such that I〈N〉 is principal. �

Remark 4.27. Not only we know that there exists an integer N , we know that N = β(i+1)1 +βi2−βm2−β11

where β(i+1)1 and βi2 is a pair of β such that β(i+1)1 + βi2 = max{β(j+1)1 + βj2 | 1 ≤ j < m}.

Example 4.28. Let I = (x2y5, x4y3, x5y), then we know that I〈N〉 is principal where N = 5 + 4−1−2 = 6.This means that I〈6〉 = (x7y6). We can check the result with the integer lattice


power of y

power of x

I

(7,6)

I〈2〉

I〈3〉

I〈4〉

I〈5〉

I〈6〉

Notice that from our example, N is the smallest n such that I〈n〉 is principal. Here we can prove that Nis always the smallest n such that I〈n〉 is principal for all I.

Proposition 4.29. Let I = (xβ11yβ12 , . . . , xβm1yβm2) which follows the ordering of β’s specified at thebeginning of the section. Let β(i+1)1 and βi2 be a pair of β such that β(i+1)1 + βi2 = max{β(j+1)1 + βj2 | 1 ≤j < m}, then N = β(i+1)1 + βi2 − βm2 − β11 is the smallest n such that I〈n〉 is principal.

Proof. This is equivalent in showing that I〈N−1〉 is not principal because by Proposition 4.11 I〈m〉 is principalimplies that I〈m+1〉 is also principal. Let a = β(i+1)1 + βi2 − βm2 − 1, and b = β(i+1)1 + βi2 − β11 − 1, we

can show that xa−1yb, xayb−1 ∈ I〈N−1〉 and xa−1yb−1 6∈ I〈N−1〉 which implies that I〈N−1〉 is not principal.

By Lemma 3.2, we only need to check the differential operators of the form ∂d

∂xd1∂yd2where d ≤ N − 2 and

d1 + d2 = d. Now consider

a− 1− d1 + b− d2 ≥ a+ b− 1−N + 2

= β(i+1)1 + βi2 − 1.

By Lemma 4.25, ∂d

∂xd1∂yd2(xa−1yb) ∈ I which implies that xa−1yb ∈ I〈N−1〉. Now consider

a− d1 + b− 1− d2 ≥ a+ b− 1−N + 2

= β(i+1)1 + βi2 − 1.

By Lemma 4.25, ∂d

∂xd1∂yd2(xayb−1) ∈ I which implies that xayb−1 ∈ I〈N−1〉. Since

∂N−2

∂xβi2−βm2−1∂yβ(i+1)1−β11−1(xa−1yb−1) = cxβ(i+1)1−1yβi2−1 6∈ I,


then xa−1yb−1 6∈ I〈N−1〉. Since I〈N−1〉 is not principal and I〈N〉 is principal, then N = β(i+1)1+βi2−βm2−β11

is the smallest n such that I〈n〉 is principal. �

From Remark 4.23, we know the rule of containment for the ideals generated by single monomials. Wecan apply Theorem 4.26 to find the containment for the special type of monomial ideal I.

Proposition 4.30. Let β(i+1)1, βi2 be a pair of βs such that β(i+1)1 +βi2 = max{β(j+1)1 +βj2 | 1 ≥ j < m}.Define a = β(i+1)1 + βi2 − βm2 − 1, b = β(i+1)1 + βi2 − β11 − 1, N = β(i+1)1 + βi2 − βm2 − β11, and

βmax = max{a, b}. Then I〈βmaxn+N−1〉 ⊆ In

Proof. By Theorem 4.26, I〈N〉 is a principal monomial ideal generated by xayb. By Remark 4.23, we have

I〈N〉〈βmaxn〉 ⊆ (I〈N〉)n. Since I〈N〉 ⊆ I, then we have (I〈N〉)n ⊆ In which implies that I〈N〉

〈βmaxn〉 ⊆ In. By

Proposition 3.10, we have I〈N〉〈βmaxn〉

= I〈βmaxn+N−1〉. Thus, by transitivity of containment I〈βmaxn+N−1〉 ⊆In. �

4.6. Differential Powers which are Eventually Principal in 3 variables. In this section, we let thering R = k[x, y, z], I = (xβ11yβ12zβ13 , . . . , xβm1yβm2zβm3) with some positive integer m and each βij ispositive.

Theorem 4.31. For all such ideal I, there exists a positive integer N such that I〈N〉 is principal.

Proof. Let

βxmax = max{βi1 | i ∈ [m]} βymax = max{βi2 | i ∈ [m]} βzmax = max{βi3 | i ∈ [m]}βxmin = min{βi1 | i ∈ [m]} βymin = min{βi2 | i ∈ [m]} βzmin = min{βi3 | i ∈ [m]}

Define

Nxy = βxmax + βymax − βxmin − βyminNyz = βymax + βzmax − βymin − βzminNxz = βxmax + βzmax − βxmin − βzmin

Let N = max{Nxy, Nyz, Nxz} and

a = N − 1 + βxmin b = N − 1 + βymin c = N − 1 + βzmin

We need to show that xaybzc is an element of I〈N〉 and xa−1ydze, xdyb−1ze, xdyezc−1 6∈ I〈N〉 for all

positive integer d, e. Since ∂N−1

∂xN−1xaybzc = C1x

βxminybzc 6∈ k, ∂N−1

∂yN−1xaybzc = C2x

ayβyminzc 6∈ k, and∂N−1

∂zN−1xaybzc = C3x

aybzβzmin 6∈ k, then it is sufficient to check all of the differential operators in the form∂N−1

∂xk1∂yk2∂zk3where k1 + k2 + k3 = N − 1. Now suppose that xa−k1yb−k2zc−k3 6∈ I which is equivalent

in xaybzc 6∈ I〈N〉. Since any two of a − k1 ≥ βxmax, b − k2 ≥ βymax, c − k3 ≥ βzmax being true impliesxa−k1yb−k2zc−k3 ∈ I, then by the contrapositive xa−k1yb−k2zc−k3 6∈ I implies that two of a−k1 < βxmax, b−k2 < βymax, c− k3 < βzmax are true. Without the loss of generality, let a− k1 < βxmax and b− k2 < βymax.This implies that a−k1 ≤ βxmax−1 and b−k2 ≤ βymax−1. This means that a−k1+b−k2 ≤ βxmax+βymax−2.However, we know that

a− k1 + b− k2 ≥ a+ b− k1 − k2 − k3

= a+ b−N + 1

= 2N − 2 + βxmin + βymin −N + 1

= N − 1 + βxmin + βymin

≥ βxmax + βymax − 1.

This is a contradiction, so xa−k1yb−k2zc−k3 ∈ I which implies that xaybzc ∈ I〈N〉.


Since ∂N−1

∂xN−1xa−1ydze = C1x

βxmin−1ydze 6∈ I, ∂N−1

∂yN−1xdyb−1ze = C2x

dyβymin−1ze 6∈ I, and ∂N−1

∂zN−1xdyezc−1 =

C3xdyezβzmin−1 6∈ I, then we know that I〈N〉 = (xaybzc). �

We conclude this subsection with a list of examples with omitted computations.

Example 4.32.

(1) I = (x5y4z, x2yz10, xy7z3). N = 15, smallest actual N = 12. I〈12〉 = (x12y12z12).(2) I = (x4yz, xy4z, xyz4). N = 6, smallest actual N = 5. I〈5〉 = (x5y5z5).(3) I = (x2y7z3, x7y2z5, xy5z7). N = 11, smallest actual N = 9. I〈9〉 = (x9y10z11).(4) I = (x7y9z3, x5y5z7, x10y4z8). N = 10, smallest actual N = 9. I〈9〉 = (x13y12z11).

Remark 4.33. Notice that this N that we get from Theorem 4.31 is not the smallest n such that I〈n〉 isprincipal.

Remark 4.34. Define Ixy = (xβ11yβ12 , . . . , xβm1yβm2), Ixz = (xβ11zβ13 , . . . , xβm1zβm3), andIyz = (yβ12zβ13 , . . . , yβm2zβm3). To study the ideal I, it is not sufficient to study only the ideals Ixy, Ixz, Iyz.For example, take I = (x3y2z7, x5y2z2, x2y6z3, x4y5z4) and the point x4y4z5. Notice that x4y4 ∈ Ixy,x4z5 ∈ Ixz, and y4z5 ∈ Iyz, but x4y4z5 6∈ I.

5. General Results

5.1. Results on More General Ideals. To begin studying results on more general ideals, we can considerideals which are isomorphic to monomial ideals through some k-algebra homomorphism. First, we look at aspecial type of isomorphism.

Lemma 5.1. Let k be a field of characteristic zero, and ϕ be a ring isomorphism from R = k[x1, . . . , xz] toS = k[a1, . . . , az] where ϕ is defined by ϕ(xi) = ci1ai + ci2 for all i where ci1, ci2 are elements of k. Then,

ϕ

((d∑i=1

Dαi

)r

)=

d∑i=1

(D′

C

)αi(ϕ(r))

where D =∏zj=1

∂∂xj

, D′ =∏zj=1

∂∂aj

, αi ∈ Zz≥0 and∑zj=1 αij < n for all i ∈ [d], C =

∏zj=1 cj1. Let Dαi

denote∏zj=1

∂αij

∂xαijj

.

Proof. Since ϕ is a ring homomorphism which preserves addition, this means that we only need to check

ϕ (Dα · r) =

(D′

C

)α(ϕ(r))

where D =∏zj=1

∂∂xj

, D′ =∏zj=1

∂∂aj

, α ∈ Zz≥0 and∑zj=1 αj < n, C =

∏zj=1 cj1.

Since r ∈ R, then we can write r =∑mi=1 uix

βi for some positive integer m, where x = x1 · · ·xz, ui ∈ k,and βi ∈ Zz≥0 for all i. Since ϕ is a ring homomorphism which preserves addition, then we only need to check

ϕ(Dα · uxβ

)=

(D′

C

)α(ϕ(uxβ))

where x = x1 · · ·xz, u ∈ k, and β ∈ Zz≥0. If there exists an index j ∈ [k] such that αj > βj , then Dα ·uxβ = 0

which means the equality holds, so we can assume that αj ≤ βj for all j ∈ [z]. Now consider the left handside

ϕ(Dα · uxβ

)= ϕ

(u

z∏i=1

βi!

(βi − αi)!xβi−αii

)

= u

z∏i=1

βi!

(βi − αi)!(ci1ai + ci2)βi−αi ,


and the right hand side is(D′

C

)α(ϕ(uxβ)) =

1

CαD′αu

z∏i=1

(ci1ai + ci2)βi

=1

Cαu

z∏i=1

βi!

(βi − αi)!cαii1 (ci1ai + ci2)βi−αi

=1

CαuCα

z∏i=1

βi!

(βi − αi)!(ci1ai + ci2)βi−αi

= u

z∏i=1

βi!

(βi − αi)!(ci1ai + ci2)βi−αi

= ϕ(Dα · uxβ

).

Thus, ϕ (Dα · r) = (D′

C )α(ϕ(r)). �

Using this lemma, we can make a statement about differential powers of ideals of the form ϕ(I) for ϕ ofthe form above.

Proposition 5.2. Let ϕ be a ring isomorphism and k-algebra isomorphism from R = k[x1, x2, . . . , xz] to

S = k[a1, a2, . . . , az] defined by ϕ(xi) = ci1ai+ci2, where k is a field of characteristic zero. ϕ(I)〈n〉

= ϕ(I〈n〉).

Proof. We want to show that ϕ(I)〈n〉 ⊇ ϕ(I〈n〉). Take an element r from I〈n〉, by definition, we have

D ·r ∈ I for all differential operators D with order less than n. Let D be a composition of partial derivatives,D =

∏zj=1

∂αj

∂xαjj

for α = (α1, . . . , αz) ∈ Nz such that∑zj=1 αj < n. Since ϕ is an isomorphism, then we have

ϕ(D · r) ∈ ϕ(I). By Lemma 5.1, we know that

ϕ(D · r) =1

CD′ · ϕ(r) ∈ ϕ(I)

where D′ =∏zj=1

∂αj

∂αjaj, and C =

∏zj=1 cj1.

This implies that C · 1CD

′ · ϕ(r) = D′ · ϕ(r) ∈ ϕ(I) because ϕ(I) is an ideal. By Lemma 3.2, we know

that D′ is any differential operator in the ring S of order less than n. By definition, ϕ(r) ∈ ϕ(I)〈n〉

. Thus,

ϕ(I)〈n〉 ⊇ ϕ(I〈n〉).

Now we want to show that ϕ(I)〈n〉 ⊆ ϕ(I〈n〉). Take an element s ∈ ϕ(I)

〈n〉, by definition, we have D ·s ∈ ϕ(I)

for all differential operators D in S with order less than n. Since ϕ is an isomorphism, then there exists aring isomorphism ψ such that ψ is the inverse of ϕ which is defined by

ψ(ai) =1

ci1xi +

(−ci2ci1

)for all i ∈ [z]. By Lemma 5.1, we know that

ψ(D · s) = CD′ · ψ(s) ∈ ψ(ϕ(I)) = I.

This implies that D′ · ψ(s) ∈ I because I is an ideal. Then by Lemma 3.2, ψ(s) ∈ I〈n〉.Since we showed both ways of containment, we have ϕ(I)

〈n〉= ϕ(I〈n〉). �

This is especially important since it allows us to explicitly write out the differential power for non-monomial ideals which are isomorphic to a monomial ideal under a mapping satisfying the properties in theproposition.

Example 5.3. Let R = Q[a, b, c] and I = ((a + 1)3, (b − 3)5). Consider the map ϕ from Q[x, y, z] to Rwhich takes x to a+ 1, y to b− 3, and z to c. This map is a ring isomorphism and k-algebra homomorphism.

By Proposition 5.2, we can say I〈n〉 = ϕ((x, y))〈n〉

= ϕ((x, y)〈n〉

). Performing this calculation gives usI〈n〉 = ((a+ 1)n+2, (b− 3)n+4, {(a+ 1)i(b− 3)j : i+ j = n+ 6}).


We can also show that, in general, ring isomorphisms which are also k-algebra homomorphisms respectdifferential operators and their order.

Proposition 5.4. For any differential operator D of order η and any ring isomorphism and k-algebrahomomorphism ϕ, Ψ = ϕ ◦D ◦ ϕ−1 is a differential operator of order η.

Proof. Proof by Induction.

Base case, η = 1. We want to check that [a,Ψ] has order zero for all a ∈ R.

[a,Ψ](f) = (a ◦Ψ)(f)− (Ψ ◦ a)(f)

= (a ◦ ϕ ◦D ◦ ϕ−1)(f)− (ϕ ◦D ◦ ϕ−1 ◦ a)(f)

= (a ◦ ϕ ◦D ◦ ϕ−1)(f)− (ϕ ◦D ◦ ϕ−1)(af)

= (a ◦ ϕ ◦D ◦ ϕ−1)(f)− (ϕ ◦D)(ϕ−1(a) · ϕ−1(f))

= (a ◦ ϕ ◦D ◦ ϕ−1)(f)− ϕ(ϕ−1(a) · (D ◦ ϕ−1)(f) + (D ◦ ϕ−1)(a) · ϕ−1(f))

= (a ◦ ϕ ◦D ◦ ϕ−1)(f)− (a · (ϕ ◦D ◦ ϕ−1)(f) + (ϕ ◦D ◦ ϕ−1)(a) · f)

= −(ϕ ◦D ◦ ϕ−1)(a) · f

This last line is a ring element −(ϕ ◦D ◦ϕ−1)(a) multiplied with f . All ring elements are order 0, so we aredone.

Inductive hypothesis: Assume that for all differential operators D′ of order η − 1, Ψ′ = ϕ ◦D′ ◦ ϕ−1 hasorder η − 1.

We want to show that this implies all differential operators D of order η, Ψ = ϕ ◦ D ◦ ϕ−1 has orderη. So we can show that [a,Ψ] has order less than η for all a ∈ R. We can write D = (D1 ◦ d) + D2 for adifferential operator D1 of order η − 1, a differential operator D2 of order less than or equal to η − 1, and adifferential operator d of order 1. Let Ψ1 = ϕ ◦D1 ◦ ϕ−1, Ψ2 = ϕ ◦D2 ◦ ϕ−1, and ψ = ϕ ◦ d ◦ ϕ−1. Then,Ψ = Ψ1 ◦ ψ + Ψ2. Calculating [a,Ψ],

[a,Ψ] = (a ◦ (Ψ1 ◦ ψ + Ψ2))− ((Ψ1 ◦ ψ + Ψ2) ◦ a)

= (a ◦Ψ1 ◦ ψ)− (Ψ1 ◦ ψ ◦ a) + (a ◦Ψ2)− (Ψ2 ◦ a)

= ((Ψ1 ◦ a) + [a,Ψ1]) ◦ ψ − (Ψ1 ◦ ψ ◦ a) + [a,Ψ2]

= (Ψ1 ◦ a ◦ ψ) + ([a,Ψ1] ◦ ψ)− (Ψ1 ◦ ψ ◦ a) + [a,Ψ2]

= Ψ1 ◦ ((ψ ◦ a) + [a, ψ]) + [a,Ψ1] ◦ ψ − (Ψ1 ◦ ψ ◦ a) + [a,Ψ2]

= (Ψ1 ◦ ψ ◦ a) + Ψ1 ◦ [a, ψ] + [a,Ψ1] ◦ ψ − (Ψ1 ◦ ψ ◦ a) + [a,Ψ2]

= Ψ1 ◦ [a, ψ] + [a,Ψ1] ◦ ψ + [a,Ψ2]

From our inductive hypothesis and base case, we have Ψ1 has order η − 1, Ψ2 has order less than or equalto η− 1, and ψ has order 1. By the definition of commutators, this implies [a,Ψ1] has order less than η− 1,[a,Ψ2] has order less than η− 1, and [a, ψ] has order 0. Thus, [a,Ψ] has order less than or equal to η− 1 forall a. This proves Ψ is a differential operator of order η. �

This helps us understand the relationship between differential powers and ring isomorphisms/k-algebrahomomorphisms.

5.2. Simple D-Modules. In this section, we generalize our results on polynomial rings to finitely generatedk-algebras that are either regular characteristic zero, or any characteristic p.

We begin with the necessary definitions.

Definition 5.5. A DR-module is a module over the ring of differential operators, DR over the ring R.


Definition 5.6. A simple DR-module is a DR-module that has no proper DR-submodule.

Note, we consider only regular rings of characteristic 0 because of the following lemma.

Lemma 5.7 ( [QG]). If R is a regular k-algebra, where k is a field of characteristic zero, then Dk(R) isgenerated by R and its derivations.

By this lemma, we generalize Proposition 3.5.1, so we know that Idiff

is an ideal when R is regularcharacteristic 0.

Our following proposition further generalizes our prior result, Proposition 3.5.

Proposition 5.8. Let R be a ring and r ∈ R. For all m > l, rm−l|D≤l(rm) where 0 ≤ l ≤ m and D≤ldenotes an arbitrary differential operator of order less than or equal to l.

Proof. Proof by induction on l.

We begin by considering the base case l = 0. The proof for this case is the same as Proposition 3.5.

Next we form our inductive hypothesis. Assume that for all m > l, rm−l divides D≤l(rm) (IH1). We want

to show for all m′ > l+ 1, rm′−l−1 divides D≤l+1(rm

′). The inequality m > l implies m+ 1 > l+ 1. So, we

can rewrite m′ as m+ 1 to use the same m as the inductive hypothesis. After this substitution, we want toshow for all m > l, rm−l divides D≤l+1(rm+1). To prove this statement, we use another proof by induction.

∗ ∗ ∗ ∗ ∗

Proof by induction on m when l is fixed.

We begin by considering the base case m = l + 1. In this case, we want to show r|D≤l+1(rm+1) whichhas been proved in Proposition 3.5.

Next we form our inductive hypothesis. Assume rm−l divides D≤l+1(rm+1) (IH2). We want to showrm+1−l divides D≤l+1(rm+2).

Let D be an operator of order less than or equal to l + 1. Then, [D, r] is a differential operator of orderless than or equal to l. Thus, we can use IH1 to say r(m+1)−l divides [D, r](rm+1). (We are replacing the min IH1 with m+1 because the hypothesis holds for all m > l). We also use IH2 to say rm−l divides D(rm+1).Let A,B ∈ R such that [D, r](rm+1) = rm+1−l ·A and D(rm+1) = rm−l ·B. Then,

rm+1−l ·A = [D, r](rm+1)

= D(rm+2)− rD(rm+1)

= D(rm+2)− r(rm−l ·B)

rm+1−l(A+B) = D(rm+2)

for some A,B ∈ R.

So, rm+1−l divides D(rm+2) for all differential operators D of order less than or equal to l + 1.

∗ ∗ ∗ ∗ ∗

Thus, we have shown for all m > l, rm−l divides D≤l+1(rm+1) and we are done. �

We are now able to prove one direction of containment for the radical and the differential closure.

Proposition 5.9. Let R be a ring, and let I ⊆ R be a ideal. Then√I ⊆ Idiff

.


Proof. Let r ∈√I. Then there exists a k > 0 such that rk ∈ I. We want to show r ∈ Idiff

, or equivalently,there exists a nonzero c such that for all n sufficiently large and for all D differential operators of order lessthan n D(crn) ∈ I.

If we are able to show rk|D(crn) then we are done. To do this, we use Proposition 5.8 to showrm−l|D≤l(rm) and then set m to satisfy m ≥ k + l. Let c = rk. So, we want to show rk|D(rk+n).

Since D order strictly less than n, apply inductive proof to say r(k+n)−n|D(crn).

�

We will now show the other containment, first by the fact that for maximal ideals equal their owndifferential closures. Using that fact, we will then show that for any radical ideal the differential closure isalso equivalent, leading us to the other containment.

Proposition 5.10. Let m ⊆ R a maximal ideal of the ring R, and let R is a regular k-algebra and a simpleD−module then m = mdiff.

Proof. Following same idea as for polynomial rings. We know m ⊆ mdiff, since it holds for all ideals. Sincem is maximal, mdiff must either be the whole ring or equal to m.

If mdiff is the whole ring, then it contains some unit u. Thus, there exists a nonzero c such that forall n sufficiently large, cun ∈ m〈n〉. This implies there exists a nonzero c suc that for all n sufficientlylarge and for all D differential operators of order less than n, D(cun) ∈ m note un is still a unit. Then,D(cun) = unD(c) ∈ m. R is a simple D−module if and only if for all r ∈ R\{0} there exists some D ∈ DR/k

such that D(r) = 1. Then unD(c) is a unit in m. This is not possible since m is maximal, so mdiff must beequal to m. �

Proposition 5.11. If R is a k−algebra and a regular Jacobson ring, and I ⊆ R is a radical ideal, then

Idiff

= I

Proof. Using Proposition 3.9, the fact that R is Jacobson, and Proposition 5.10 write

Idiff

=⋂I⊆mα

mαdiff

⊆⋂I⊆mα

mαdiff =

⋂I⊆mα

mα = I

Hence we see, Idiff ⊆ I. We already know I ⊆ Idiff

from Proposition 3.7. So we have equality of I = Idiff

for some radical ideal I. �

Idempotence of the differential operator follows directly from this proposition.

Corollary 5.11.1. If R is a k-algebra and a regular Jacobson ring, and I ⊆ R is a radical ideal, then

Idiff

diff

= Idiff

Proposition 5.12. For any regular characteristic 0 ring, that is is a simple DR−module, and any ideal

I ⊆ R, we have Idiff

=√I.

Proof. By definition of the radical, I ⊆√I. By properties of closure operations, I

diff ⊆√I

diff

. By Proposi-

tion 5.11, we have√I

diff

=√I. So, I

diff ⊆√I. From Proposition 5.9

√I ⊆ I

diff. Putting this all together,

we have√I = I

diff. �

Finally, we will show that when Idiff

is an ideal, then we have a direct correspondence between√I = I

diff

and our ring being a simple DR-module.


Theorem 5.13. Let R be a a characteristic 0 regular ring, and I is an ideal of R, then√I = I

diffif and

only if R is a simple DR−module.

Proof. One direction was proved in Proposition 5.12 For the other direction, we use the contrapositive.Assume R is not a simple DR-module, then there exists some proper DR−submodule, M , which implies that

for all D ∈ DR, D(M) ⊆ M. Note, M is an ideal of R. Now consider that r ∈ Mdiffif D(crn) ∈ M for D

some n-th differential operator where n is sufficiently large and some c ∈ R. Now, since M has the property

D(M) ⊆ M for all D, then all we need is that crn ∈ M in order to know r ∈ Mdiff. However, for all r, we

may allow c ∈ M , then it follows that for all r ∈ R, then r ∈ Mdiff, or M

diff= R. We know though, that√

M 6= R since M is a proper submodule. In this case, we can conclude that√I 6= I

diffwhen R is not a

simple DR−module. By the contrapositive, our proposition follows. �

We will close with one example, and one non-example.

Example 5.14. Consider that R = k[x, y, z]/(z − xy) is a regular ring, where k is of positive characteristic.By [Smi95], will be a simple DR-module since it is a domain and strongly F−regular.

Example 5.15. Non-example, consider the ring R = k[x, y]/(xy), we have that (0), (x), (y), (xy) are DR-ideals. In general, Stanley-Reisner Rings are not simple DR-modules [Jef20]. In this case, for example,√

(x) 6= (x)diff

.

6. Open Questions

Our final section consists of topics that could be explored in the future. We begin with the differentialRees Algebra.

Definition 6.1. The differential Rees Algebra is

〈R〉(It) =⊕n≥0

I〈n〉tn

We know the algebra is not necessarily finitely generated since there exists a prime ideal such that thesymbolic Rees Algebra is not finitely generated. By Zariski-Nagata, we then know that the differential Reesalgebra is not necessarily finitely generated. Hence, we have the following question.

Question 6.1. Under what conditions is the differential Rees algebra finitely generated?

We also believe that Theorem 5.13 can be generalized. In other words,

Question 6.2. For all rings R with an ideal I, is it true that,√I = I

diffif and only if R is a simple

DR-module?

In Theorem 4.26 we prove that there exists an N such that I〈N〉 is principal for ideals of a particular form.We also want to know when the differential power is principal in more than two variables, which promptsthe following question.

Question 6.3. Given an ideal I in a ring k[x1, . . . , xz] where each generator of the ideal contains each ofthe variables, what is the smallest n such that I〈n〉 is principal?

We also want to generalize our results of the generators of the differential power, to ideals besides principaland monomial ideals.

Question 6.4. When can we write an explicit formula for the differential power of a non-monomial ideal?

Moreover, in [BJNB19], they give an explicit formulation of the differential power in polynomial rings ofcharacteristic 0, so generalizing to broader rings is a future goal.


Question 6.5. Is there a formulation of the differential power in polynomial rings of characteristic p?Non-polynomial rings?

7. Acknowledgments

The authors would like to thank our mentors, Jennifer Kenkel, Janet Page, and Daniel Smolkin for theirinvaluable advice and guidance. They would also like to acknowledge the University of Michigan Departmentof Mathematics, and in particular David Speyer, for organizing the REU. The research was supported bythe executive committee funds, NSF grants DMS-1855135, and DMS-1801697.

References

[AM69] Michael Francis Atiyah and I. G. MacDonald. Introduction to commutative algebra. Addison-Wesley-Longman,

1969.[BJNB19] Holger Brenner, Jack Jeffries, and Luis Nunez-Betancourt. Quantifying singularities with differential operators.

Advances in Mathematics, 358:106843, 2019.

[Cou95] S. C. Coutinho. A primer of algebraic D-modules, volume 33 of London Mathematical Society Student Texts.Cambridge University Press, Cambridge, 1995.

[DDSG+18] Hailong Dao, Alessandro De Stefani, Eloısa Grifo, Craig Huneke, and Luis Nunez Betancourt. Symbolic powers of

ideals. In Singularities and foliations. geometry, topology and applications, volume 222 of Springer Proc. Math.Stat., pages 387–432. Springer, Cham, 2018.

[ELS01] Lawrence Ein, Robert Lazarsfeld, and Karen E. Smith. Uniform bounds and symbolic powers on smooth varieties.

Invent. Math., 144(2):241–252, 2001.[HH89] Melvin Hochster and Craig Huneke. Tight closure. In Commutative algebra (Berkeley, CA, 1987), volume 15 of

Math. Sci. Res. Inst. Publ., pages 305–324. Springer, New York, 1989.

[HH11] Jurgen Herzog and Takayuki Hibi. Monomial ideals, volume 260 of Graduate Texts in Mathematics. Springer-VerlagLondon, Ltd., London, 2011.

[Hoc07] Melvin Hochster. Foundations of tight closure theory, 2007.[HS06] Craig Huneke and Irena Swanson. Integral closure of ideals, rings, and modules, volume 336 of London Mathe-

matical Society Lecture Note Series. Cambridge University Press, Cambridge, 2006.

[Jef20] Jack Jefferies. Differential operators spring 2020, May 2020.[QG] Eamon Quinlan-Gallego. Rings of differential operators.

[Smi95] Karen E. Smith. The D-module structure of F -split rings. Math. Res. Lett., 2(4):377–386, 1995.

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DIFFERENTIAL CLOSURE OF IDEALS