the ratliff-rush operation for certain monomial ideals in...

U.U.D.M. Project Report 2017:32

Examensarbete i matematik, 15 hpHandledare: Veronica Crispin Quinonez Examinator: Jörgen ÖstenssonAugusti 2017

The Ratliff-Rush Operation for Certain Monomial Ideals in K[x, y]

Petter Restadh

Department of MathematicsUppsala University

THE RATLIFF-RUSH OPERATION FOR CERTAIN MONOMIAL

IDEALS IN K[x, y]

PETTER RESTADH

Abstract. Given a regular ideal a in a Noetherian ring we can define the

Ratliff-Rush ideal associated to a as the union a =⋃

an+1 : an. Moreover,this is the largest ideal with the same high powers as a. We explain why that is

so and some of its consequenses. Then, with the use of results about numerical

semigroups we will see how we can compute a for some ideals in K[x, y] andgive examples of some properties that a possesses in this case.

Contents

1. Introduction 32. Basic Concepts 3

2.1. Basic Definitions 32.2. Arithmetics on Ideals 52.3. Monomial Ideals 72.4. Noetherian Rings 9

3. Numerical Semigroups 104. Ratliff-Rush Ideal 12

4.1. Ratliff-Rush Ideals in Noetherian Rings 134.2. Ideals in K[x, y] Generated by Monomials of the Same Degree 14

5. Examples and Various Properties 175.1. Examples 175.2. Some Interesting Properties 18

References 24

THE RATLIFF-RUSH OPERATION FOR CERTAIN MONOMIAL IDEALS IN K[x, y] 3

1. Introduction

Let R be a ring. On the set of ideals in R we can define the four operationsaddition, a+b, multiplication, a·b, intersection, a∩b, and ideal quotient, a : b. Usingthe multiplication we can define powers of ideals and using the other operations wecan study how these powers behave.

Ratliff and Rush [5] were the first ones who studied the union a =⋃k≥0(ak+1 :

ak) for regular ideals, that is containing a non-zero-divisor, in Noetherian rings.The ideal a is called the Ratliff-Rush ideal associated to a and we will see that, forregular ideals, a has the same high powers as a. Moreover it is the largest ideal,with respect to inclution, with this property.

In some literature the associated Ratliff-Rush ideals are called Ratliff-Rush clo-sures. This is not surprising. We have both a ⊆ a and ˜a = a. Although in general

a ⊆ b does not imply a ⊆ b, so we will not name it as such.The Ratliff Rush operation behaves, in general, quite unpredictble and errad-

ically. For example, ab is not always equal to ab, and as just mentioned, a ⊆ b

does not imply a ⊆ b and a + b is not always equal to a + b. This often makes thecomputations hard.

In this paper we will begin with some basic concepts about algebra in general,working our way up to arithmetric on ideals and Noetherian rings. We will after thatwork our way through some relevant results from [5]. Then we will present someresults about numerical semigroups from [2]. We will need this work to prove thatfor some special cases the associated Ratliff-Rush ideals can be described in a mucheasier way. This will allow us to use a computer to assist us with the computationsto produce examples so we can look at some properties of the associated Ratliff-Rush ideal.

2. Basic Concepts

Something many of the later results will revolve around is numerical semigroups.In fact, the study of the associated Ratliffe-Rush ideals on monomial ideals inK[x, y] boils down to just that, as we will see when we work our way through resultsfrom [2]. All results from this section can be found in [3]. We will sometimes usesligtly different definitions than the ones used in [3] but even so they are equivalent.

Definition 2.1 (Numerical semigroups). A numerical semigroup is a set S ⊆ Nsuch that:

i) 0 ∈ Sii) N\S is finiteiii) a, b ∈ S ⇒ a+ b ∈ S

Notice that the operation, ”+”, is the usual addition on the natural numbers,so we know it is commutative and associative. To be more precise, any numericalsemigroup is in fact a commutative monoid.

More about this later on. For now we will begin with the basics.

2.1. Basic Definitions. Here we will recall the most fundamental concepts. Wewill not discuss them in detail, but when we move on to the later chapters it isexpected of the reader to have a understanding of them.

Definition 2.2 (Binary Operation). A binary operation on a set A is a function∗ : A×A→ A. We will write a ∗ b instead of ∗(a, b).

4 PETTER RESTADH

Definition 2.3 (Monoid). A monoid is a set S and a binary operation ∗ such that:

i) The operation is associative, (a ∗ b) ∗ c = a ∗ (b ∗ c)ii) There exist a unit element e ∈ S such that a = a ∗ e = e ∗ a

A semigroup is a monoid that does not necessarily fulfill ii).A group is a monoid such that for each a there exits b such that a∗ b = b∗a = e.

This b is usually written as a−1 and called a’s inverse.If the operation is commutative, that is a ∗ b = b ∗ a, the group or monoid is

called commutative or abelian. Then we usally use ”+” instead of ”*” and writethe inverses as ”−a”.

Definition 2.4 (Ring). A ring is a set R and two binary operations, + and · suchthat:

i) The pair R and + is an abelian groupii) The pair R and · is a monoidiii) The multiplication is distributive over the multiplication, that is a·(b+c) =

a · b+ a · c and (b+ c) · a = b · a+ c · aAny ring can be seen as a quintuple (R,+, ·, 0, 1), where 0 is the additive unit and1 is the multiplicative unit.

If R and · are a commutative monoid the ring is called commutative.

Worth noting is that sometimes rings are defined without a mutiplicative unitand rings in the sense defined above is called rings with unit element, but we willonly work with rings of the second kind. Moreover, the following holds in all rings:

• For all r ∈ R we have r · 0 = 0.• The multiplicative and additive unit are unique.• For all r ∈ R we have (−1) · r = −a.

Example 2.5. Z with the usual ′′+′′ and ′′·′′ is a commutative ring.

Example 2.6. R = {0} with 0 + 0 = 0 and 0 · 0 = 0 is called the trivial ring orthe zero ring.

Definition 2.7 (Subring). A subring to a ring R is any subset R′ ⊆ R such that(R′,+, ·, 0, 1) is a ring.

In other words it is a ring within a ring.

Definition 2.8 (Field). A field is a non-trivial ring such that R\{0} and · is anabelian group. The multiplicative inverse for a is denoted a−1 and we often writeab instead of a · b.Definition 2.9 (Polynomial ring). Let R be a commutative ring. The polynomialring over R, written as R[x], is defined with the underlying set

R[x] = {p = p0 + p1x+ p2x2 + · · ·+ pnx

n | n ∈ N}with pi ∈ R. Then we have the addition p + q = r where ri = pi + qi andmultiplication p · q = r where ri =

∑k+l=i pk · ql.

Notice that with R[x, y] we mean R[x][y]. Sometimes R[x, y] can be defined inother ways, but they will behave in exactly the same way, or be isomorphic. Wewill not define being ”isomorphic”, but we can think of it as ”being the same, butwith different names”.

An important number associated with every element in a polynomial ring is thedegree. It is the same as in most other places, namely:


Remark 2.10 (Degree). The degree of a non-zero element p in R[x] is the greatestn such that pn 6= 0. The degree of 0 is −∞. We will denote the degree of p bydeg(p).

Definition 2.11 (Ideal). An ideal a in a ring R is a subset of R that is closedwith respect to the addition with any element in a and multiplication with any ringelement. That is

i) if a, b ∈ a then a+ b ∈ aii) if a ∈ a, r ∈ R then r · a ∈ aiii) if a ∈ a, r ∈ R then a · r ∈ a

Noteworthy is that ii) and iii) are equivalent if R is commutative.

Example 2.12. In any ring R, both R, the ring itself, and {0}, the zero ideal, areideals. These are called the trivial ideals. We will denote the zero-ideal with 0.

Remark 2.13. An ideal a contains the multiplicative unit if and only if a is the ring.If you are not used to ideals, showing this is a good exercise.

In fact, these two ideals are special in such a way that they are the only two thatalways exist in a ring. Studying ideals is interesting because they tell us how thering behaves, as the following result will show.

Theorem 2.14. For a commutative ring R the following two are equivalent:

(1) R is a field.(2) R has only two ideals, namely trivial ones.

Proof. Let R be a field. Take any non-zero ideal, a, then there exist a non-zeroelement in a, call it a. Since R is a field a−1 exists in R. Then by iii) in thedefinition 2.11 a · a−1 = 1 ∈ a. So by remark 2.13 a = R.

The other way around, assume R only has two ideals. We only need to verifythat aa = {r · a : r ∈ R} is indeed a non-zero ideal for all non-zero a ∈ R. Sincethen it follows that aa = R 3 1 so there must exist x ∈ R such that x · a = 1 andtherefore a has an inverse. Since a was chosen arbitrary, it follows that R is indeeda field.

Firstly, 1 · a = a ∈ aa so aa is non-zero. Left to do is verifying the conditions indefinition 2.11 and we are done.

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2.2. Arithmetics on Ideals. In this section we will continue with some moredefinitions and core concepts. We will talk about generators for ideals and definesome operators that will be useful later on.

Definition 2.15 (Generating set). Given any set A in a ring R. We call thesmallest, ordered by inclusion, ideal containing A for the ideal generated by A. Wedenote this as 〈A〉. We say that a set A generates an ideal I, or I is generated byA, if 〈A〉 = I.

The definition of a set generating a subgroup or semigroup is analogous.

Remark 2.16. If it were not for the fact that for any two ideals their intersection isalso an ideal, this would not be well-defined.

Remark 2.17. If at any point we speak about generators for anything other thanthe structures explicitly mentioned in definition 2.15 without defining generators on

6 PETTER RESTADH

that object, the definition is analogous to 2.15. That is, the smallest such structurewhich contains the generators.

Generating sets is a core concept. Many proofs, theorems and definitions usegenerators and it is often easier to work with a generating set than the structureas a whole.

Example 2.18. In a ring R, the ring generated by 〈1〉 = R.

Example 2.19. In the group Z the subgroup generated by 2 is 〈2〉 = {. . . ,−2, 0, 2, 4, . . . }.But Z seen as a monoid the monoid generated by 2 is 〈2〉 = {0, 2, 4, . . . } .

Let us now use what we have discussed by studying some operations on the setof ideals in some ring R.

Definition 2.20. Define the following operations on some pair of ideals a and bas:

i) a + b = {a+ b | a ∈ a, b ∈ b}ii) a · b = {

∑ki=0 aibi | ai ∈ a, bi ∈ b}

iii) a ∩ b = {r ∈ R | r ∈ a and r ∈ b}If the ring is commutative we can also define:

iv) a : b = {r ∈ R | r · b ∈ a ∀b ∈ b}The third operation is recognizible as the usual intersection between two sets.

Left to show is that these are in fact well defined operations on the set of ideals.Or, in other words, that a + b, a · b, a ∩ b and a : b are all ideals. We will showthis for a : b since that is the hardest one. The others will be left as an exercise forthe reader.

Proof. We will go through the conditions from definition 2.11.For any two elements r1 and r2 in a : b, any k ∈ R and any element b in b we

have:

i) (r1 + r2)b = r1b+ r2b lies in a since both r1b and r2b do.ii) (kr1)b = k(r1b) lies in a since r1b does.iii) Follows commutativity of the ring and by ii).

So a : b is indeed an ideal.�

First we will look at how these operations interact with eachother. Firstly somethings that come directly from the ring. For any ideals a, b and c

• a + b = b + a• (a + b) + c = a + (b + c)• a · (b + c) = a · b + a · c• (b + c) · a = b · a + b · a

And clearly if the ring is abelian:

• ab = ba

If any of the above seems unclear it is recommend proving it. It is really moreof an exercise in writing than dealing with rings and ideals.

This area is full of small theorems and different ways these operations interact.We will later use some but far from all. Therefore the ones we will use, apart from


the ones mentioned above, are summed up in this next theorem. Most of them canbe found in chapter 1.3.5 in [3].

Theorem 2.21. The following holds for any ideals in a ring R:

i) If a = 〈S〉 and b = 〈T 〉 then we have:• a + b = 〈S ∪ T 〉• a · b = 〈st | s ∈ S, t ∈ T 〉

Moreover if R is commutative

ii) If a ⊆ b then a : c ⊆ b : c and c : b ⊆ c : aiii) (a : b) : c = a : cbiv) (a : b)(c : d) ⊆ ac : bdv) a : b ⊆ ac : bc

Proof. We will show them one by one.

i) We will show that a + b ⊆ 〈S ∪ T 〉 and ab ⊆ 〈st | s ∈ S, t ∈ T 〉. The otherinclusion follows from the minimality condition in the definition 2.15 andthe fact that a ⊆ a + b since 0 ∈ b.• Take any element a + b ∈ a + b. By definition a ∈ 〈S ∪ T 〉 andb ∈ 〈S ∪ T 〉. Then, since 〈S ∪ T 〉 is an ideal a+ b ∈ 〈S ∪ T 〉.

• Take any element∑ki=0 aibi ∈ ab. Clearly aibi ∈ 〈st | s ∈ S, t ∈ T 〉.

Then, since 〈st | s ∈ S, t ∈ T 〉 is an ideal∑ki=0 aibi ∈ 〈st | s ∈ S, t ∈

T 〉.ii) We have r ∈ a : c⇔ rc ⊆ a⇒ rc ⊆ b⇔ r ∈ b : c and r ∈ c : b⇔ rb ⊆ c⇒

ra ⊆ c⇔ r ∈ c : a.iii) r ∈ (a : b) : c⇔ rc ⊆ a : b⇔ rcb ⊆ a⇔ r ∈ a : cbiv) Take any element

∑aibi in (a : b)(c : d). Then we see that for any aibi we

have aibidb ⊆ aicb = aibc ⊆ ac. The result follows.v) a : b = (a : b)R = (a : b)(c : c) ⊆ ac : bc

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2.3. Monomial Ideals. Now that we have talked about ideals in general we willfocus on monomial ideals.

Definition 2.22 (Monomial). A monomial in R[x0, . . . , xn] is any element on theform kxα0

0 . . . xαnn with k ∈ R. We usually write kxα where α = (α0, . . . , αn). If

there exists a set of monomials that generates an ideal we will call such an ideal fora monomial.

Remark 2.23 (Degree). On monomials we will use a slightly different defenition ofa degree, namely

∑ni=0 αi.

Why we use this degree will become more obvious later on.So why do we look at these ideals? Why do they have their own subsection?

The simple answer is ”because they behave nicely”, as we will see now. Let us firstsee how they behave under our operations defined in 2.20.

Let a and b be generated by S and T respectively. Here we let S and T be somesets of monomials. Then by 2.21 we have

a + b = 〈S ∪ T 〉a · b = 〈st|s ∈ S, t ∈ T 〉.

8 PETTER RESTADH

Since the right hand sides are sets of monomials, these ideals must be monomialideals too.

Example 2.24. Let a = 〈x2, xy2, y3〉 and b = 〈xy3, y5〉 in K[x, y] for some fieldK. Then

a + b = 〈x2, xy2, y3, xy3, y5〉.

But since y3 divides both xy3 and y5, the right hand side is in fact 〈x2, xy2, y3〉 =a. We also have

a · b = 〈x3y3, x2y5, x2y5, xy7, xy6, y8〉.

But as before we see that xy7 is not needed. So a · b = 〈xy6, x2y5, x3y3, y8〉.

In fact, both a : b and a∩b are also monomial ideals given that a and b are. Thefollowing theorem is a good starting point to understanding how monomial idealswork. It is theorem 1 in the second chapter in [3].

Theorem 2.25. Let a be a monomial ideal. A element∑aix

αi is in a if and onlyif aix

αi is in a for all i.

We refer to [3] for the proof, we will instead discuss a little about monomialideals in K[x, y]. As we will see later, all ideals in K[x, y] are finitely generated andtherefore we will only discuss these ideals.

Firstly, an element kxayb is in a monomial ideal a if and only if xayb is in a.This follows since K is a field and therefore all elements k ∈ K are invertible. Fromthis and theorem 2.25 the following can be proved:

Theorem 2.26. Let a and b be monomial ideals in K[x, y]. If a monomial m liesin their sum, it lies in one of them. That is if m ∈ a + b then m ∈ a or m ∈ b.

Secondly, the generating elements can be ordered as mi = xaiybi where ai > ai+1

and bi < bi+1. This can be done because otherwise some of the generating elementswould be superfluous as in example 2.24. This is something unique to polynomialrings in two variables. How to order them, for example in K[x, y, z], to get quickercalculations is the beginning of the study of Grobner bases. This is something wewill not mention any more in this paper, but more informaion about this can befound in [3].

Moreover, for monomials m the following holds:

(ma : mb) = a : b

and

(ma : b) = m(a : b)

This follows from the injectivity of the map given by x 7→ mx. This gives us:

(2.1) (ma)n+1 : (ma)n = mn+1an+1 : mnan = m(an+1 : an)

This equality can be found in [2] as (1.1). With this we will only need to considersome specific cases later on. But for now we will leave it as it is.


2.4. Noetherian Rings. Noetherian rings have been studied extensively and itis still a big research area within algebra. Indeed, this paper could very well onlyhave been about Noetherian rings. The contents of this chapter and ideas for theproofs can be found in [1]. Since we will not be using much of the properties thatfollows from a ring being Noetherian we will only show a few of them.

Definition 2.27 (Noetherian ring). A Noetherian ring is a commutative ring Rwhich satisfies one of the following conditions:

i) Any infinite chain of increasing ideals has a maximal element. In otherwords, if a1 ⊆ a2 ⊆ . . . then there exists a natural number N such thatan = aN for all n ≥ N . This is known as the ascending chain condition.

ii) All ideals in R are finitely generated.iii) Any family of ideals in R ordered by inclusion has a maximal element.

It is also possible to define what it means for a non-commutative ring to beNoetherian, but we will only work with commutative Noetherian rings. In this casethe definitions coincide.

Theorem 2.28. The above three properties are equivalent.

Proof. We will, in order, show that iii) ⇒ i) ⇒ ii) ⇒ iii).‘iii)⇒ i)’Take any increasing chain a1 ⊆ a2 ⊆ . . . and create {an}∞n=1. This set has a

maximal element by iii), say aN . Then for n ≥ N aN ⊆ an, but since aN maximal,an = aN .

‘i)⇒ ii)’Assume R has an ideal a which is not generated by some finite set. Take any

element a1 ∈ a, then let a1 = 〈a1〉 ( a. Define an+1 recursively by taking an+1 ∈a\an and defining an+1 = an + 〈an+1〉 = 〈a1, . . . , an+1〉. Note that an+1 must existsince no finite set generates a but an = 〈a1, . . . , an〉.

We have an ⊆ an+1 by construction. Furthermore an+1 ∈ an+1 but an+1 6∈ an,again from construction, so an ( an+1. Then we have a counterexample to i) andthus the assumption is wrong.

‘ii) ⇒ iii)’We will show that for any set of ideals all chains have a maximal element, the

result then follows by Zorn’s lemma. Let a1 ⊆ a2 ⊆ . . . be a chain of ideals. Definea =

⋃ai. Then a will be an ideal, this can be verified by checking the conditions

in definition 2.11. By assumption a is finitely generated, say a = 〈x1, . . . , xn〉.For each xi there must exist ni such that xi ∈ ani . Then define n = max{ni}. Itfollows that a = an and since ai ⊆ a for all i it follows that the chain has a maximalelement, namely an. We are done.

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Example 2.29. All fields are Noetherian, simply because the two ideals are 〈1〉and 〈0〉 and so they indeed fulfill the second condition in definition 2.27.

In this paper we will, for the most parts, only need the ascending chain condition.Although we do not yet know which rings are Noetherian. The next theorem willtell us something about that.

Theorem 2.30. If R is Noetherian then R[x] is Noetherian.

10 PETTER RESTADH

This theorem is very useful. For example it tells us that Z[x1, . . . , xn] is Noe-therian, simply because Z is. Though we will only use it to conclude that K[x, y]is Noetherian for any field K.

Proof. Take any ideal a ⊆ R[x] and consider the set b = {an |∑ni=0 aix

i ∈ a},that is the coefficients of the highest degree terms of all polynomials in a. This setwill be an ideal in R. Since R is Noetherian there exists a finite set generating thisideal, call it {pi} ⊆ R.

Clearly it must exist a set of polynomials in a on the form fi = prixri +∑ri−1

k=0 akxk. Set r = max{ri}.

We can now see that in the same way bk = {an | n ≤ k and∑ni=0 aix

i ∈ a}, orthe leading coefficients of all polynomials with degree k or less, will be ideals. Thencreate {fk,j} from each bk in the same way as we made {fi} from b. Define

a′ =⟨{fi} ∪ {fk,j}k≤r

⟩.

Clearly, since {fi} and each of {fk,j} are finite, a′ is finitely generated. Moreoversince {fi} ⊆ a and {fk,j} ⊆ a we have a′ ⊆ a.

Left to show is a ⊆ a′. Take any element a =∑ni=0 aix

i ∈ a. If n ≥ r then thereexits ui ∈ R[x] such that

∑uifi has leading coefficient an and degree n. So then

a ∈ a if and only if a −∑uifi ∈ a. But a −

∑uifi must by construction have

degree smaller or equal to n−1. So it is enough to show that all elements of degree≤ r in a lie within a′.

In the same way using elements from bk instead of b we conclude that a lie in a′

if and only if all ring elements in a lies in a′. But using b0 and the same methodthis is the same as saying that the additive inverse lies in both ideals which holdsfor all ideals and we are done.

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This is all we need to know about Noetherian rings for now.

3. Numerical Semigroups

Most of the results in this section are taken from [2]. Recall definition 2.1 ofa numerical semigroup from before. We will later apply these results to concludesome very nice results about the Ratliff-Rush ideals.

Throughout this chapter, unless specified otherwise, all numbers are naturalnumbers and all semigroups S ⊆ N.

We will often assume a0 < a1 < · · · < an when we speak of a generating set{a0, . . . , an}. We can do this without loss of generality since we only change theindex.

Remark 3.1. The following is noteworthy before we begin:

(1) Any set {a0, . . . , an} ⊆ N generates a numerical subgroup if and only ifgcd(ai) = 1.

(2) Any semigroup generated by {a0, . . . , an} ⊆ N is isomorphic to the numer-ical semigroup generated by { a0

gcd(ai), . . . , an

gcd(ai)} ⊆ N.

(3) The semigroup generated by {a0, . . . , an} is the set of linear combinations∑ni=0 λiai, with λi ∈ N. Notice though that the coefficients, the λi’s, are

not necessarily unique.(4) All semigroups are finitely generated.


Definition 3.2 (Frobenius number). Let S = 〈a0, . . . , an〉 be a semigroup and leth = gcd(ai). Define the Frobenious number g(S) as the greatest multiple of h notin S.

If S is a numerical semigroup then g(S) is simply the largest number in N notin S.

There does not exist a closed formula for the Frobenius number yet. Althoughthere does exist a formula for the special case when S = 〈a, b〉 and gcd(a, b) = 1,which is g(s) = (a− 1)(b− 1)− 1. This was shown in [7]. This can be generalizedby using (2) in remark 3.1. Then we get:

g(〈a, b〉) = gcd(a, b)((a

gcd(a, b)− 1)(

b

gcd(a, b)− 1)− 1).

It is worth noting that we do not have g(〈A〉) ≥ g(〈A ∪ {a}) unless gcd(A) =gcd(A ∪ {a}). This can be seen in the next example.

Example 3.3. Let A = {4, 6}. Then gcd(4, 6) = 2 and 〈A〉 = {0}∪{4, 6, 8, . . . }so g(〈A〉) = 2. But set a = 9. Then gcd(A ∪ {a}) = gcd(4, 6, 9) = 1 and〈A ∪ {a}〉 = {0, 4, 6} ∪ {8, 9, 10, . . . } so g(〈A ∪ {a}〉) = 7.

Now to something else.

Definition 3.4. Let S = 〈a0, . . . , an〉. Then define λ : S → N as

λ(s) = min{∑

λi | s =∑

λiai}.

This λ can be seen as a measurement of how far from the generators s is.

Definition 3.5. For each semigroup S = 〈a0, . . . , an〉 define

Λ(S) = max{λ(s)| s ∈ S, s ≤ an + g(S)}.

We will use this Λ as a ”worst case scenario”, but more to that later. Noticethat both λ and therefore Λ depends on which generators we choose. This nextexample will show this.

Example 3.6. Let

S = 〈3, 5, 6, 8, 9〉 = {3} ∪ {5, 6} ∪ {8, 9, . . . }

and

T = 〈3, 5〉 = {3} ∪ {5, 6} ∪ {8, 9, . . . }.Then g(S) = g(T ) = 7 but Λ(S) = 2 and Λ(T ) = 4. This despite that as

semigroups S and T are the same.

Although this behavior might seem somewhat unorthodox it is a quite usefulproperty. This way we can consider any set of generators instead of only somespecific set. In some cases we are unable to choose generators and instead they aregiven to us as a necessity. Then a flexible definition, like this one, is easier to use.

Let us now see to some of the properties of λ.

Theorem 3.7. Let S = 〈a0, . . . , an〉. Moreover let α < 1 and β be real nonnegativenumbers. Then if s ∈ S and s ≤ αanl + β then λ(s) ≤ l if l large enough.

12 PETTER RESTADH

Proof. For all s > g(S) there exits a unique positive integer k such that g(S) +ank + 1 ≤ s ≤ g(S) + an(k + 1). From this we clearly have λ(s) ≤ Λ(S) + k ≤Λ(S) + s−g(S)−1

an. Using s ≤ αanl + β we get:

λ(s) ≤ Λ(S) +s− g(S)− 1

an≤ Λ(S) + αl +

β − g(S)− 1

an.

If we now can find large enough l such that Λ(S) + αl + β−g(S)−1an

≤ l we are

done. This is true for all l ≥ anΛ+β−g(S)−1an(1−α) .

�

What this says is that λ(s) cannot become too large in proportion to s. This ishowever in some sense the best one because the condition α < 1 is necessary. Thiscan be seen in remark 2.5 in [2], it is as follows:

Remark 3.8. Let S = 〈2, 5〉. Choose α = 1 and β = 4. Then for s = 5l + 4 =5l + 2 · 2 ∈ S clearly λ(s) = l + 2 for all l. So α < 1 is necessary.

Though the following result tells us that all α < 1 are in some sense ’relevant’.

Theorem 3.9. Let S = 〈a0, . . . , an〉. Then lims→∞s

λ(s) = an.

Proof. For all s > g(S) there exits a unique non-negative integer k such that g(S)+ank + 1 ≤ s ≤ g(S) + an(k + 1). Then we have k ≤ λ(s) ≤ Λ(S) + k. Putting thistogether we get:

g(S) + ank + 1

Λ(S) + k≤ s

λ(s)≤ g(S) + ank + an

k.

From this the limit follows directly since k →∞ as s→∞.�

Our work thus far leads up to the following theorem, which is why we have achapter about numerical semigroups.

Theorem 3.10. If S = 〈a0, . . . , an〉 and T = 〈b0, . . . , bn〉 such that a0 = bn = 0and ai + bi = d for all i. Then for some fixed β and all large enough l the followingholds:

if s ∈ S and s ≤ αanl + β ≤ dl, then there exist λ0, . . . , λn such that s =∑λiai ∈ S and

∑λi = 0. Furthermore dl − s =

∑λibi ∈ T .

Proof. By theorem 3.7, if l is large enough then, since s ≤ αanl + β, we haveλ(s) ≤ l. By definition of λ and the fact that a0 = 0 there exist λ1, . . . , λn suchthat s =

∑λiai with

∑λi ≤ l. Then let λ0 = l −

∑λi. Then s =

∑ni=0 λiai with∑

λi = l. By assumptions we get dl − s =∑dλi −

∑λiai =

∑λibi ∈ T .

�

4. Ratliff-Rush Ideal

The study of powers of ideals can be used in a couple of different areas withinmathematics. How these powers behave in general is something we know very littleabout. Let a be a regular ideal, that is contaning a non-zero divisor. The uniona =

⋃l∈Z+(al+1 : al) was studied by Ratliffe and Rush in [5] and was found to have

some nice properties. All ideals in this chapter are assumed to be regular.


4.1. Ratliff-Rush Ideals in Noetherian Rings. We will now repeat the impor-tant results from [5].

Definition 4.1 (The Ratliff-Rush ideal associated to a). Let a be a regular ideal.Define the Ratliff-Rush ideal associated to a as a =

⋃al+1 : al.

Remark 4.2. An ideal is said to be Ratliff-Rush if a = a.

Theorem 4.3. In Noetherian rings, a =⋃l∈Z+(al+1 : al) is the largest ideal such

that an = an for all large enough n.

Theorem 4.3 can be found as Theorem (2.1) in [5]. Before we sketch a proof wewill study some properties of a =

⋃l∈Z+(al+1 : al). First of all

(a2 : a1) ⊆ (a3 : a2) ⊆ · · · ⊆ (al+1 : al) ⊆ (al+2 : al+1) ⊆ . . .This follows from xal ⊆ al+1 ⇒ xal+1 ⊆ al+2. Since the ring is Noetherian

there exists n such that an+1 : an = am+1 : am for all m ≥ n. It follows thata = an+1 : an.

Definition 4.4 (Ratliff-Rush reduction number). The Ratliff-Rush reduction num-ber is defined as

r(a) = min{n | a = an+1 : an}.

Furthermore a ⊆ a. This follows trivially since a ⊆ a2 : a1. From this andtheorem 2.21 ii) we get

ak ⊆ ak−1(a)1 ⊆ · · · ⊆ a1(a)k−1 ⊆ (a)k

for all k. We will be using this when showing 4.3. In fact, what we will be showingis that for large enough k

(4.1) ak ⊆ ak−1(a)1 ⊆ · · · ⊆ a1(a)k−1 ⊆ (a)k ⊆ ak.

From this ak = (a)k follows. Now to proving equation 4.1. We will as earliermentioned only sketch some parts of the proof. This because it contains someconcepts we will not need for anything other than this proof and those conceptswould deserve their own chapter. For more details about these parts of the theoremsee proof of theorem (2.1) in [5] and, (22.1) and (22.2) in [4].

Proof. What really is left to show of equation 4.1 is (a)k ⊆ ak for large enough k.There will exist two positive natural numbers h and c and a non-zero divisor

a ∈ ah such that ak : aR ∩ ac = ak−h for all large enough k. Or put in anotherway, a superficial regular element a of a of degree h. This is ensured by (22.1) and(22.2) in [4].

Moreover, by (3.12) in [4], for even larger n, say n ≥ n∗, we will have an : aR =an−h. Using the fact that a ∈ ah we have aR ⊆ ah and therefore

an∗−h+i ⊆ an

∗−h+i+1 : a ⊆ · · · ⊆ an∗−h+i+h : ah ⊆ an

∗+i : aR = an∗−h+i

for all i ∈ N.From this it follows that ak+1 : a = ak for all k ≥ n∗ − h. Using that (a : b) :

c = a : bc we get

ak+m : am = (ak+m : a) : am−1 = ak+m−1 : am−1 = · · · = ak+1 : a = ak

14 PETTER RESTADH

for all m ≥ 1. Let l be a number ≥ k such that a = al+1 : al. Using the fact that

(a : b)(c : d) ⊆ ac : bd we now get al ⊆ al ⊆ (al+1 : al)l ⊆ al+l2

: al2

= al. So forlarge enough l we have al = al.

For the maximality, assume ak = bk for all k ≥ k′. Then we must have (a +

b)2k′ =∑

aib2k′−i ⊆ a2k′ = b2k′ since each term must be either a2k′ (the ones

where 2k′ − i ≥ k′) or b2k′ (the ones where i ≥ k′) and ak = bk for all k ≥ k′.So then for all b ∈ b and n ≥ 2k′ we must have an ⊆ 〈a, b〉n ⊆ (a + b)n = an.This gives us ban−1 ⊆ an and hence b ∈ an : an−1 ⊆ a and by symmetry the resultfollows.

�

Corollary 4.5. If we have ak = ak for some k, then ak′

= ak′

for all k′ ≥ k.

Proof. By equation 4.1 we must then have ak = ak−lal for all l ≤ k. So ak+1 =aak = aak−1a = aka = ak+1. The result follows by induction over k.

�

From this we get an upper limit to the Ratliff-Rush reduction number, namelyas:

Corollary 4.6. If we have ak = ak for some k, then r(a) ≤ k.

Proof. For all l ≥ k we have a ⊆ al+1 : al = al+1 : al ⊆ a.�

Though this inequality does not tell us much, for how will we know when ak = ak

without calculating a and thus in turn calculating r(a)? A couple of other thingscan also be proven from theorem 4.3, here are a couple. Most of them are fromRemark (2.3) in [5].

Corollary 4.7. The following holds:

i) The associated Ratliff-Rush ideals are Ratliff-Rush, that is ˜a = a.

ii) Any high enough power of an ideal is Ratliff-Rush, that is an = an for largeenough n.

iii) If a ⊆ b ⊆ a then b = a.

iv) In general ab 6= ab.

We will later see examples of iv) when we actually can calculate a. Becauseeven if we have theorem 4.3, a : b is not always easily calculated. Even if it were,an+1 : an = an+2 : an+1 does not imply an+2 : an+1 = an+3 : an+2, so calculatinga is not something easily done in general. But with the right equipment it cansometimes be done.

4.2. Ideals in K[x, y] Generated by Monomials of the Same Degree. Herewe will continue to study results from [2] and also try to tie the knot together.We will see how we can calculate the Ratliff-Rush ideal associated to a if a is on acertain form.

Let a be generated by monomials of the same degree. That is a = 〈xaiybi〉ni=0

where ai + bi = d for all i. As discussed previously we can assume ai > ai+1 andbi < bi+1. Moreover equation 2.1 shows that it suffices to study 〈x, y〉-primaryideals, or, equivalently, ideals such that a0 = bn = 0. With the degree of themonomial we will refer to d. We will only work with ideals on this form in thissection.


For any ideal a = 〈xaiybi〉 define the two semigroups S = 〈ai〉 and T = 〈bi〉. Wecan see that

al = 〈∏

∑λi=l

(xaiybi)λi〉 = 〈∏

∑λi=l

(x∑λiaiy

∑λibi)〉

Or put in another way:

al = 〈xsyt | s ∈ S, t ∈ T such that λ(s) ≤ l, λ(t) ≤ l and s+ t = dl〉

The next corollary will more or less say that for large enough l the conditionsλ(s) ≤ l and λ(t) ≤ l are superfluous. This way, high enough powers of a will reacha ’maximal state’.

Theorem 4.8. For sufficienly large l the following holds:

(1) If s ∈ S, s ≤ u and s+ u ≥ dl for some u ∈ N, then xsyu ∈ al

(2) If t ∈ T , t ≤ v and t+ v ≥ dl for some v ∈ N, then xvyt ∈ al

Proof. We will show them one at the time.

(1) Since s+u ≥ dl there exists j, b ∈ N such that d(l+j) ≤ s+u = d(l+j)+b ≤d(l + j + 1)− 1. We want to show that xsyu ∈ al+j ⊆ al.

Since s + u ≤ d(l + j + 1) − 1 and s ≤ u we have s ≤ d(l+j)2 + d−1

2 . Sothe prerequisites for theorem 3.10 are fulfilled. So if l+ j are large enough,or just l since l ≤ l + j, we can write s =

∑λiai and u = b+

∑λibi with∑

λi = l + j. Therefore xsyu = yb∏

(xaiybi)λi ∈ al+j ⊆ al.(2) This part is proved in the same way.

�

Corollary 4.9. For large enough l we have al = 〈xsyt | s ∈ S, t ∈ T such that s+t = dl〉.

Proof. Clearly al = 〈xsyt | s ∈ S, t ∈ T such that λ(s) ≤ l, λ(t) ≤ l and s + t =dl〉 ⊆ 〈xsyt | s ∈ S, t ∈ T such that s+ t = dl〉.

The other inclusion follows from the fact that if s + t = dl then either s ≤ t ort ≤ s. The rest follows from the previous theorem.

�

This would be the ’maximal state’ we mentioned earlier. Because clearly no otherelements than the ones on the form xsyt such that s ∈ S, t ∈ T and s+ t = dl willever be a minimal generator to al and with this all of them will.

Moreover, when the power of an ideal starts behaving this way it will, possiblya bit later, become Ratliff-Rush. Although we still need a few more theorems toprove this.

So as we can see, how powers of ideals behave depends on how their respectivesemigroups behave. With this in mind, let us define some ideals.

Definition 4.10. For an ideal a = 〈xaiybi〉 let S = 〈ai〉 and T = 〈bi〉. DefineaS = 〈xsyd−s | s ∈ S, s ≤ d〉 and aT = 〈xd−tyt | t ∈ T, t ≤ d〉 where d is thedegree of a.

The next theorem will prove their relevance.

16 PETTER RESTADH

Theorem 4.11. For sufficiently large l and some ideal aM,l we have

al = (xd(l−1))aT + (yd(l−1))aS + (xdyd)aM,l

Proof. First of, let l ≥ max {g(S), g(T )}. Then any element yd(l−1)(xsyd−s) =xsydl−s with s ≤ S generates al if and only if s ∈ S and λ(s) ≤ l. By our choiceof l we have λ(s) ≤ l for all s ≤ d. So by definition of aS we have that xsyt withs ≤ d lies in yd(l−1)aS if and only if it lies in al.

Using symmetry, xsyt with t ≤ d lies in xd(l−1)aT if and only if it lies in al.The only elements left are xd+syd+t. Choose aM,l = al : xdyd and the result

follows.�

Even though we now can describe powers of a we still do not know how anythingabout how a looks like. The next theorem takes care of that for us. It allows us toactually compute the Ratliff-Rush associated ideal, and this relatively easy.

Theorem 4.12. Let a, aS and aT be as in definition 4.10. Then the Ratliff-Rushideal associated to a is

a = aS ∩ aT .

Proof. Since a = al+1 : al for large enough n it is enough to show that aS ∩ aT =al+1 : al for large enough l. Since a monomial al is monomial and by theorem 2.25any polynomial

∑kix

aiybi lies in al if and only if each term kxayb lies in al. So itenough to consider monomial elements m.

If m ∈ aS∩aT then by definition there exist s′ ∈ S, t′′ ∈ T and m′, m′′ ∈ K[x, y]

such that m = m′xs′yd−s

′= m′′xd−t

′′yt′′. By corollary 4.9 al is generated by xsyt

such that s + t = dl, if l is large enough. We want to show that for all of thesegenerators mxsyt ∈ al+1 and thus m ∈ al+1 : al. Since s + t = dl either s ≤ dl

2 or

t ≤ dl2 .

Assume s ≤ dl2 . Then by the first equality mxsyt = m′xs+s

′yd(l+1)−(s+s′). Since

s ≤ dl2 and s′ ≤ d we get s+s′ ≤ dl

2 +d = d(l+1)2 + d

2 . Using theorem 3.10 we get thatif l is indeed large enough we have s+ s′ =

∑λiai and d(l+ 1)− (s+ s′) =

∑λibi

where∑λi = l+ 1. Using these equalities we obtain mxsyt =

∏(xaiybi)λi ∈ al+1.

If t ≤ dl2 we need only to use the second equality (m = m′′xd−t

′′yt′′) and we obtain

the same result. Hence aS ∩ aT ⊆ al+1 : al for all large enough l.The other way around assume m 6∈ aS ∩ aT , that is m 6∈ aS or m 6∈ aT . Begin

with m 6∈ aS . Then mydl 6∈ ydlaS and by theorem 4.11 mydl 6∈ al+1 and hencem 6∈ al+1 : al since ydl ∈ al. If m 6∈ aT then in the same way mxdl 6∈ al+1 andm 6∈ al+1 : al, which completes the proof.

�

Remark 4.13. No step in calculating aS ∩aT is complicated. The author has in factmade a program available at [6]. This way, producing examples is very easy andit is even possible to check some specific property for all relevant monomial idealswith d ≤ 20. Although it does take some time.


5. Examples and Various Properties

Now that we know how to calculate the associated Ratliff-Rush ideals for someideals we can produce examples and see how they behave. We will begin withsome examples of earlier mentioned properties and then look closer at some specificclasses and state a few things about them.

All Ideals are presumed to have the same form as in the previus chapter. Thatis a is generated by monomials {xsiyti}ni=0 such that sn = t0 = 0, si + ti = d andsi > si+1. Moreover we have the related semigroups S = 〈si〉 and T = 〈ti〉.

5.1. Examples. Here we will take examples of earlier mentioned properties.

Example 5.1. Let a = 〈x10, x8y2, x5y5, x3y7, y10〉. Then a2 : a1 = a3 : a2 =〈x10, x8y2, x7y4, x5y5, x3y7, y10〉 but a4 : a3 = 〈x10, x8y2, x6y4, x5y5, x3y7, y10〉.Thus an+1 : an = an+2 : an+1 does not imply an+2 : an+1 = an+3 : an+2.

Example 5.2. Let a = 〈x5, x4y, y5〉 and b = 〈x5, x3y2, y5〉.We see that a = a = 〈x5, x4y, y5〉, b = b = 〈x5, x3y2, y5〉 and so a · b =

ab = 〈x10, x9y1, x8y2, x7y3, x5y5, x4y6, x3y7, y10〉. But on the other hand

ab = 〈x10, x9y, x8y2, x7y3, x6y4, x5y5, x4y6, x3y7, y10〉. So in general a · b 6= ab.

Example 5.3. Let a = 〈x5, y5〉, b = 〈x4y, xy4〉 = xy〈x3, y3〉. Then we have

a = a and b = ˜(xy〈x3, y3〉) = xy ˜(〈x3, y3〉) = xy〈x3, y3〉 = b. So a + b = a + b =

〈x5, x4y, xy4, y5〉, but a + b = 〈x, y〉5. So in general a + b 6= a + b.

Example 5.4. If l is large enough al is Ratliff-Rush.

This we already know from corollary 4.7 but we will show it using some of thenew methods presented to us in the last chapter.

Proof. Let l be large enough such that dl > g(S) + g(T ) and al is on the form incorollary 4.9. Then alS = al + xg(S)+1〈x, y〉dl−g(S)−1. Indeed it must be so sincefor all xsyt ∈ 〈x, y〉dl−g(S)−1 we have xg(S)+1xsyt = xg(S)+s+1yt ∈ alS becauseg(S) + s+ 1 ∈ S.

The other way around we see that alS = 〈xsydl−s | s ∈ S〉. Therefore for any

xsyt generating alS , if s ≥ g(S) + 1 xsyt ∈ xg(S)+1〈x, y〉dl−g(S)−1 an if s ≤ g(s)then t = dl− s ≥ g(T ) + 1 and so by corollary 4.9 xsyt ∈ al since we assumed l waslarge enough for corollary 4.9 to hold.

Using this consider xsyt ∈ al = alS ∩ alT we get three cases,

i) s ≤ g(S),ii) t ≤ g(T ),iii) g(S) + 1 ≤ s and g(T ) + 1 ≤ t.

Let us do them one by one.

i) We get xsyt ∈ alS = al + xg(S)+1〈x, y〉dl−g(S)−1. Since s ≤ g(S) we clearly

have xsyt 6∈ xg(S)+1〈x, y〉dl−g(S)−1 and theorem 2.26 we get xsyt ∈ al.ii) Follows by symmetry.iii) If s+ t ≥ dl we have s+ t ≥ dl > g(S) + g(T ) and so there exists s′ and t′

in N such that s′ + t′ = dl with s ≥ s′ > g(S) and t ≥ t′ > g(T ). Clearly

xs′yt′ ∈ al since al is on the form of corollary 4.9. If s + t ≤ dl clearly

xsyt 6∈ alS since alS is generated by monomials of degree dl.

18 PETTER RESTADH

It follows that al ⊆ al and we are done.�

5.2. Some Interesting Properties. We will now look closer at some specificclasses of ideals and prove some things about them. We still assume that all idealsare in K[x, y], monomial and the generators are on the form xsiyti where sn =t0 = d, si + ti = d and si > si+1.

Theorem 5.5. Let a = 〈xsiyti〉 be an ideal of degree d. Let tmin = min{ti > 0}. Iffor all i and all j either si+sj > d−tmin or there exists k such that si+sj = sk ∈ S,then a is Ratliff-Rush.

This is a slightly stronger version of corollary 3.8 in [2].

Proof. Clearly any element on the form xsyt+tmin is in a if and only if it is in aS ,that is a : ytmin = aS : ytmin . But since a ⊆ aT we also have a : ytmin ⊆ aT : ytmin .Thus we get

a : ytmin = (aS ∩ aT ) : ytmin

= aS : ytmin ∩ aT : ytmin

= a : ytmin ∩ aT : ytmin

= a : ytmin .

Therefore any element xsyt+tmin is in a if and only if it is in a. All other elementsxsyt must have t < tmin. Then they are in aT if and only if s ≥ d, but all suchelements must be in a since xd ∈ a and therefore also in a. So in conclusion anyelement xsyt is in a if and only if they are in a.

�

Example 5.6. Let a = 〈x7, x4y3, x3y4, x2y5, y7〉. Then we see that a fufills theprerequsites of theorem 5.2.

We have tmin = 3, aS = 〈x7, x6y, x5y2, x4y3, x3y4, x2y5, y7〉 and aT =〈x7, x4y3, x3y4, x2y5, xy6, y7〉.

In a = aS ∩ aT we clearly do not have any elements on xsyt where t ≤ tmin ands ≤ 7 since none of these elements are present in aT . The rest of the elements areshared between a and aS thus we get a = aS ∩ aT ⊆ a. So a is Ratliff-Rush.

Remark 5.7. Since we assumed that si and therefore ti where ordered tmin = t1and smin = sn−1. Hence we will start naming them as such.

Corollary 5.8. Let a be as in the previus theorem. If 2sn−1 + t1 ≥ d then a isRatliff-Rush.

Proof. We have si + sj + t1 ≥ 2sn−1 + t1 ≥ d for all si, sj ∈ S\{0}. If sj = 0 thensi + sj = si ∈ S. The rest follows from theorem 5.5.

�

Corollary 5.9. If a = 〈xd, xayd−a, yd〉, then a is Ratliff-Rush.

Proof. We have 2a+ (d− a) = d+ a ≥ d and thus corollary 5.8 gives us the result.�

We will now look closer on the generators for a and on how the generators for aaffect them.


Example 5.10. Let a = 〈x7, x5y2, x2y5, y7〉. Then we get S = T = 〈2, 5, 7〉 ={0} ∪ {2} ∪ {4, 5, . . . } and therefore aS = 〈x7, x6y, x5y2, x4y3, x2y5, y7〉 andaT = 〈x7, x5y2, x3y4, x2y5, xy6, y7〉. From theorem 4.12 we then obtain a =aS ∩aT = 〈x7, x5y2, x4y4, x2y5, y7〉 = a+ 〈x4y4〉. We see that a is not necessarilygenerated by monomials of the same degree.

With this in mind, define m(a) as the difference between the maximal degree ofa minimal generator for a and the degree of a. In the previous example we thenhave m(a) = 1.

In fact the previous example is the ideal of the smallest degree having m(a) ≥ 1,that is a has minimal generators of greater than the degree of a. This is easilychecked with a computer. Though something more interesting is that it is the firstone in a class of ideals where the associated Ratliff-Rush ideals are not generatedby monomials of the same degree, as we will see in the next example.

Example 5.11. Let a = 〈xd, xd−aya, xayd−a, yd〉 be an ideal such that gcd(a, d) =1. Let n be the unique integer such that an < d < a(n+ 1). Then if n ≥ 3 then ahas a minimal generator of degree a(n+ 1) and thus m(a) = a(n+ 1)− d.

Proof. We see that S = 〈a, d− a, d〉 so {s ∈ S : s ≤ d} = {0, a, 2a, . . . , a(n−1), d − a, an, d} and therefore aS = 〈xd, xanyd−an, xd−aya, . . . , xayd−a, yd〉.Symmetry then gives us aT = 〈xd, xd−aya, . . . , xayd−a, xd−nayna, yd〉. So

a = aS ∩ aT =

= 〈 lcm(yd, yd), lcm(yd, xd−nayna), lcm(xd−nayna, xayd−a), lcm(xayd−a, xayd−a),

lcm(xayd−a, xd−(n−1)ay(n−1)a), . . . , lcm(xd, xd)〉 =

= 〈yd, xd−nayd, xayna, xayd−a, xd−(n−1)ayd−a, . . . , xd〉

= 〈yd, xayd−a, x2ay(n−1)a, xd−(n−2)ayd−2a, . . . , xd−2ayd−(n−2)a, x(n−1)ay2a, xd−aya, xd〉 =

a + 〈xkay(n−k+1)a) | 1 < k < n〉+ 〈xd−kayd−(n−k)a | 1 < k < n− 1〉.

xkay(n−k+1)a has degree (n+ 1)a. So if there exists k such that 2 ≤ k ≤ n− 1 weare done. This is true for all n ≥ 3.

�

Remark 5.12. Let a be as in the previous example, and choose d = 3a+ 1. That islet a = 〈x3a+1, x2a+1ya, xay2a+1, y3a+1〉. Clearly gcd(d, a) = 1. Then we haveS = T = {0, a, 2a, 2a+ 1, 3a, 3a+ 1, . . . } so

aS = a + 〈x3ay, x2aya+1〉 =

〈x3a+1, x3ay, x2a+1ya, x2aya+1, xay2a+1, y3a+1〉and

aT = 〈x3a+1, x2a+1ya, xa+1y2a, xay2a+1, xy3a, y3a+1〉.Which gives us

a = aS ∩ aT =

= 〈x3a+1, x2a+1ya, x2ay2a, xay2a+1, y3a+1〉 = a + 〈x2ay2a〉

20 PETTER RESTADH

m(a)

deg(a)

1 2 3 4 5 6 7 8...

......

......

......

......

7 1 0 0 0 0 0 0 08 0 0 0 0 0 0 0 09 7 0 0 0 0 0 0 010 1 1 0 0 0 0 0 011 32 0 0 0 0 0 0 012 4 0 0 0 0 0 0 013 113 4 1 0 0 0 0 014 57 12 0 0 0 0 0 015 386 16 4 0 0 0 0 016 208 51 1 1 0 0 0 017 1318 50 15 0 0 0 0 018 720 132 4 4 0 0 0 019 3 962 204 46 1 1 0 0 020 2 717 395 19 11 0 0 0 021 11 446 5 024 130 740 0 0 0 022 8 687 1 195 55 33 1 1 0 023 35 946 1 500 365 24 12 0 0 024 24 151 3 109 173 89 4 4 0 025 102 925 4 595 944 68 36 1 1 026 80 609 8 079 514 234 11 11 0 027 281 698 11 513 2 366 208 96 4 4 0

Figure 5.1. A table of how big the generators for a can become

This class of ideals are interesting in a couple of ways. Firstly, they maximize them(a) in comparison to the degree of a, within the class of monomials in example5.11. Moreover they do this while still not being isomorphic to some monomialideal of smaller degree. We can, of course, get quite large differences if we onlydouble or triple the degree of all monomials generating a but, in some sense, theywill be uninteresting.

Secondly, they show that a linear relation between m(a) and the degree of theideal is possible for certain ideals. Quite trivially the relation cannot be ”greater”than linear. Otherwise, for large enough degrees, m(a) would grow faster than thedegree of the ideal, and in time be greater. This clearly cannot happen. So thequestion instead becomes how ”big” can this linear difference become?

In figure 5.1 the number in the i:th row column j represents how many ideals ofdegree i that have m(a) ≥ j.

Notice that only ideals that are not isomorphic to some ideal of lower degreewere tested. So for example, 〈x8, x6y2, y8〉 was not tested.

The encircled numbers represent the ideals in remark 5.12, and we see that withinour table these ideals maximises this difference. We can also see that the numbersseem to increase with higher degree and fewer divisors.

Proposition 5.13. Let a = 〈xsiyti〉ni=0 such that sn = t0 = 0, si + ti = d andsi > si+1. If m(a) ≥ a− 1 then a must have degree greater or equal to 3a+ 1.


1 2 3 4 5 6 7 8 9 10...

......

......

......

......

......

3 4 0 0 0 0 0 0 0 0 04 7 1 0 0 0 0 0 0 0 05 13 2 1 0 0 0 0 0 0 06 21 6 4 1 0 0 0 0 0 07 38 15 6 4 1 0 0 0 0 08 59 32 24 8 4 1 0 0 0 09 103 78 48 14 8 4 1 0 0 010 159 157 118 47 18 8 4 1 0 011 267 346 247 101 34 16 8 4 1 012 406 698 581 222 78 34 16 8 4 1

Figure 5.2. A table showing when ak and ak coincide.

Proof. For a to have a minimal generator on the form xs′yt′we must have xs

′yd−s

′, xsyd−s ∈

aS , by symmetry s < s′, and xd−t′yt′, xd−tyt ∈ aS , by symmetry t < t′, such that

lcm(xs′yd−s

′, xd−t

′yt′) = xs

′yt′. This simply because 〈ai〉 ∩ 〈bj〉 = 〈lcm(ai, bj)〉.

Moreover there cannot exist xs′′yd−s

′′ ∈ as or xd−t′′yt′′ ∈ at, such that s < s′′ < s′

or t < t′′ < t′ since then xs′yt′

would not be minimal. In terms of S and T thistranslates as there must exist s, s′ ∈ S and t, t′ ∈ T that fufill the followingproperties:

(1) s < d− t′ < s′ < d− t.(2) There does not exist s′′ ∈ S or t′′ ∈ T such that s < s′′ < s′ or t < t′′ < t′.

Since we assumed m(a) = a− 1 we can let s′ + t′ = d+ a− 1.Clearly (2) gives us s′− s ≤ sn−1 and t′− t ≤ t1. Otherwise s+ sn−1 would be a

counterexample. So then we have a− 1 = s′+ t′− d = s′− (d− t′) < s′− s ≤ sn−1,that is a ≤ sn−1. Symmetry gives us a ≤ t1.

Assume by symmetry sn−1 ≤ t1. Then corollary 5.8 states that if 2sn−1 + t1 ≥ dholds, a is Ratliff-Rush. This clearly cannot hold. Therefore we have 3a ≤ 3sn−1 ≤2sn−1 + t1 < d. So 3a+ 1 ≤ d.

�

Now we have looked at the degree of the generators, but a more striking questionis when ak = ak. Let n(a) = min{k ∈ Z+ | ak = ak}. Corollary 4.6 then simplystates r(a) ≤ n(a).

Example 5.14. Let a = 〈xd, xd−1y, xyd−1, yd〉. Then a = 〈x, y〉d and n(a) =d− 2.

In figure 5.2 we have encircled the ideals just mentioned in 5.14. The number inrow d column n represents the number of ideals of degree d with n(a) = n.

As we can see, the numbers at the end of each row are powers of 2. For example,row 10 ends with 8 4 1.Let us examine these ideals closer. The ideal of degree 10with n(a) = 8 is:

a0 =〈x10, x9y, xy9, y10〉.

22 PETTER RESTADH

The 4 ideals of degree 10 with n(a) = 7 are:

a1 =〈x10, x9y, x8y2, xy9, y10〉a2 =〈x10, x9y, x2y8, xy9, y10〉a3 =〈x10, x9y, x2y8, y10〉a4 =〈x10, x8y2, xy9, y10〉.

The 8 ideals of degree 10 with n(a) = 6 are:

a5 =〈x10, x9y, x8y2, x7y3, xy9, y10〉a6 =〈x10, x9y, x3y7, x2y8, xy9, y10〉a7 =〈x10, x9y, x7y3, xy9, y10〉a8 =〈x10, x9y, x3y7, x2y8, y10〉a9 =〈x10, x9y, x3y7, xy9, y10〉a10 =〈x10, x8y2, x7y3, xy9, y10〉a11 =〈x10, x9y, x3y7, y10〉a12 =〈x10, x7y3, xy9, y10〉

And as far as we have checked, for large enough d, the ideal of degree d withn(a) = d− 2 is:

a0 =〈xd, xd−1y, xyd−1, yd〉.

The 4 ideals of degree d with n(a) = d− 3 are:

a1 =〈xd, xd−1y, xd−2y2, xyd−1, yd〉

a2 =〈xd, xd−1y, x2yd−2, xyd−1, yd〉

a3 =〈xd, xd−1y, x2yd−2, yd〉

a4 =〈xd, xd−2y2, xyd−1, yd〉

and so on. What is interesting is how the generators are located. They seemto avoid numbers around d

2 . This is in line with the upper limits presented in [2],since this leads to larger values of Λ(S).

With 4.7 iii) in mind we can partially order the ideals as a < b ⇔ a ⊆ b ⊆ a.Then we get that if a < b then we have n(a) ≥ n(b).

By theorem 4.11 we see that if a and b are of the same degree with the sameRatliff-Rush associated ideal, then aS = bS and aT = bT , since they coinside afterlarge enough powers. With this in mind it is possible to construct an ideal a′ froma such that a′ < a as following:


(1) Let d be the degree of a. Take the minimal generators to S, call them ai.Create {xaiyd−ai} ∪ {xd}.

(2) Repeat with T and create {xd−biybi} ∪ {yd}.(3) Let a′ =

⟨{xaiyd−ai} ∪ {xd} ∪ {xd−biybi} ∪ {yd}

⟩.

Clearly a′ ⊆ a, Sa = Sa′ and Ta = Ta′ , and thus a′ < a. Moreover, by construc-tion, a′ must be the smallest ideal having the same associated semigroups, and isthus the smallest ideal, within the class of ideals we are discussing here with thesame high powers as a, keeping theorem 4.11 in mind, and is thus the smallest ideal,here with respect to inclution, with the same assosiated Ratliff-Rush ideal.

24 PETTER RESTADH

References

[1] M. F. Atiyah and I. G. Macdonald, Introduction to Commutative Algebra, Addison-Wesley,1969.

[2] V. Crispin Quinonez, Ratliff-Rush monomial ideals, Algebraic and geometric combinatorics,

43–50, Contemp. Math., 423, Amer. Math. Soc., Providence, RI, (2006).[3] R. Froberg, An Introduction to Grobner Bases, John Wiley Sons Ltd, Chichester, 1997.

[4] M. Nagata, Local Rings, Interscience Tracts 13, Interscience Publishers, New York, NY, 1962.

[5] L. J. Ratliff, Jr. and D. E. Rush, Two Notes on Reductions of Ideals, Indiana Univ. Math. J.27 (1978), no. 6, 929–934.

[6] P. Restadh http://user.it.uu.se/∼pere4728/[7] J. J. Sylvester, Question 7382, Mathematical Questions from the Educational Times 41, 21,

(1884).

E-mail address: [email protected]

http://user.it.uu.se/~pere4728/

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