U.U.D.M. Report 2019:3

Department of MathematicsUppsala University

Properties of powers of monomial ideals

Oleksandra Gasanova

Filosofie licentiatavhandling

i matematik

som framläggs för offentlig granskning

den 9 december 2019, kl 10.15, Polhemsalen,

Ångströmlaboratoriet, Uppsala

Properties of powers of monomial ideals

Introduction

Monomial ideals form an important link between commutative algebra and combina-torics. A monomial ideal is an ideal generated by monomials in a multivariate polynomialring K[x1, . . . , xn]. One nice basic property of such ideals is existence of a unique minimalmonomial generating set, denoted by G(I), that is, a finite set of monomials generating Isuch that no monomial divides any other monomial. In both of the papers we only considerm-primary monomial ideals, where m = 〈x1, . . . , xn〉. In other words, a monomial ideal ism-primary if there exist positive integers d1, . . . , dn such that {xd1

1 , . . . , xdnn } ⊆ G(I).

In paper 1 we adress the following problem. Let I be a monomial ideal in thepolynomial ring K[x1, . . . , xn]. The Ratliff–Rush ideal associated to I is defined asI = ∪k≥0(I

k+1 : Ik), where I : J := {r ∈ K[x1, . . . , xn] | rJ ⊆ I}. In other words, Iis the unique largest ideal which contains I and such that Ik = Ik for all k large enough(see [1] for more details). Finding the Ratliff-Rush ideal of I is known to be a hard prob-lem, even when I is a monomial ideal. However, if I belongs to a special class of ideals,so-called good ideals, the problem becomes much easier. We also introduce the notion ofvery good ideals and discuss their connection to Freiman ideals. A monomial ideal (notnecessarily m-primary) is called a Freiman ideal if it is equigenerated (that is, minimallygenerated by monomials of the same degree) and

|G(I2)| = l(I)|G(I)| −(l(I)

2

),

where l(I) denotes the analytic spread of I. Since we only consider m-primary ideals, wehave l(I) = n.

In paper 2 we study the number of elements in G(I i). It is known that the functionf(i) = |G(I i)|, for large i, is a polynomial in i with a positive leading coefficient. Inparticular, for all i large enough we have |G(I i+1)| > |G(I i)| unless I is a principal ideal.But if i is small, different kinds of pathologies can occur. We explore these pathologiesand show that for any n ≥ 2 and d ≥ 2 there exists an m-primary monomial idealI ⊂ K[x1, . . . , xn] such that |G(I)| > |G(I i)| for all i ≤ d. In particular, we provide thefollowing examples:

1. I ⊂ K[x, y] with |G(I)| = 26, |G(I2)| = 9, |G(I3)| = 13, |G(I4)| = 17, |G(I5)| = 21,|G(I6)| = 25 (here n = 2 and d = 6);

2. I ⊂ K[x, y, z] with |G(I)| = 43, |G(I2)| = 18, |G(I3)| = 34 (here n = 3 and d = 3).

The question about pathological behaviour of G(I i) for small i arises from [2]. Here theauthors show existence of monomial ideals with tiny squares in K[x, y], that is, ideals forwhich |G(I)| > |G(I2)|. We generalise this result to any number of variables n and anypower d and provide an improved version of their main theorem.

1

Acknowledgements

I am grateful to my supervisor Veronica Crispin Quinonez for introducing me to thetopic, her help, suggestions and useful remarks during the preparation of my papers. Iam also thankful to the participants of the problem solving seminar in Stockholm and inparticular Samuel Lundqvist who drew my attention to the problem discussed in paper 2.

I also want to thank Love Forsberg who helped me prove the main theorem of paper 1,Johan Asplund who basically taught me to draw pictures in LATEX, and my other friendsfrom the math department who shared lots of great and enjoyable moments with me.

References

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

[2] S. Eliahou, J. Herzog and M. Mohammadi Saem, Monomial ideals with tiny squares,Journal of Algebra, 514 (2018), 99-112.

2

Powers of monomial ideals and the Ratliff–Rushoperation

Oleksandra Gasanova

Abstract

Powers of (monomial) ideals is a subject that still calls attraction in various ways.In this paper we present a nice presentation of high powers of ideals in a certainclass in K[x1, . . . , xn] and K[[x1, . . . , xn]]. As an interesting application it leads to analgorithm for computation of the Ratliff–Rush operation on ideals in that class. TheRatliff–Rush operation itself has several applications, for instance, if I is a regularm-primary ideal in a local ring (R,m), then the Ratliff–Rush associated ideal I isthe unique largest ideal containing I and having the same Hilbert polynomial as I.

1 Introduction

Let R be a commutative Noetherian ring and I a regular ideal in it, that is, an idealcontaining a non-zerodivisor. The Ratliff–Rush ideal associated to I is defined as I =∪k≥0(Ik+1 : Ik). For simplicity we will call it the Ratliff–Rush closure of I, even thoughit does not preserve inclusion, as shown in [14]. In [13] it is proved that I is the uniquelargest ideal that satisfies I l = I l for all large l. An ideal I is called Ratliff–Rush if I = I.Properties of the Ratliff–Rush closure and its interaction with other algebraic operationshave been studied by several authors, see [6], [7], [13], [14]. In particular, we would like tomention the following two results. If I is an m-primary ideal in a local ring (R,m), thenI is the unique largest ideal containing I with the same Hilbert polynomial (the lengthof (R/I l) for sufficiently large l) as I. It is also known that the associated graded ring⊕k≥0Ik/Ik+1 has positive depth if and only if all powers of I are Ratliff–Rush (see [7]for a proof). Several unexpected connections of the Ratliff–Rush closure are discussed in[10], [11] and most recently [15] and [16]. In general, the Ratliff–Rush closure is hard tocompute. In [5] the author presents an algorithm for Cohen-Macaulay Noetherian localrings, which, however, relies on finding generic elements. In this article we describe aconstructive algorithm for computing the Ratliff–Rush closure of m-primary monomialideals of a certain class (we will call it a class of good ideals) in K[x1, . . . , xn], which alsoworks in the local ring K[[x1, . . . , xn]]. This is a generalization of algorithms described in[1] and [12].

In Section 3 we introduce the notion of a good ideal. The idea is as follows: any m-primary monomial ideal has some xd11 , . . . , x

dnn as minimal generators and therefore defines

a (non-disjoint) covering of Nn with rectangular ”boxes” Ba1,...,an of sizes d1, . . . , dn, wherea1, . . . , an are nonnegative integers and

Ba1,...,an := ([a1d1, (a1 + 1)d1]× . . .× [andn, (an + 1)dn]) ∩ Nn.

Then I is called a good ideal if it satisfies the so-called box decomposition principle,namely, if for any positive integer l any minimal generator of I l belongs to some box

1

Ba1,...,an with a1 + . . . + an = l − 1. We will also discuss a necessary and a sufficientcondition for being a good ideal. From this point, unless specifically mentioned, we willwork with good ideals.

In Section 4 we will associate an ideal to each box in the following way: if I is a goodideal and Ba1,...,an is some box, then it contains some of the minimal generators of I l, wherel = a1 + . . . + an + 1. Since they are in Ba1,...,an , they are divisible by (xd11 )a1 · · · (xdnn )an .Therefore, we can define

Ia1,...,an :=

⟨m

(xd11 )a1 · · · (xdnn )an| m ∈ Ba1,...,an ∩G(I l)

⟩.

We will conclude this section by showing that

Ia1,...,an = I l : 〈(xd11 )a1 · · · (xdnn )an〉,

which immediately implies the following property: if (a1, . . . , an) ≤ (b1, . . . , bn), thenIa1,...,an ⊆ Ib1,...,bn .

In Section 5 we will study the asymptotic behaviour of Ia1,...,an . Now that we knowthat Ia1,...,an grows when (a1, . . . , an) grows, and given that ideals can not grow forever,we are expecting some sort of stabilization in Ia1,...,an when (a1, . . . , an) is large enough.In other words, we are expecting some pattern on I l for large l. In Section 6 we will provethe main theorem of this paper, namely, the following: if I is a good ideal, then

I = Iq1,0,...,0 ∩ I0,q2,...,0 ∩ . . . ∩ I0,...,0,qn ,

where Iq1,0,...,0 is the stabilizing ideal of the chain I0,0,...,0 ⊆ I1,0,...,0 ⊆ I2,0,...,0 ⊆ . . ., andI0,q2,...,0 is the stabilizing ideal of the chain I0,0,...,0 ⊆ I0,1,...,0 ⊆ I0,2,...,0 ⊆ . . ., and so on.The pattern eatablished in Section 5 will play an important role in the proof of the maintheorem. In Section 7 we will show that computation of I0,0,...,qi,0,...,0 is much easier than itseems. In particular, we will show that the corresponding chain stabilizes immediately assoon as we have two equal ideals. Section 8 contains examples and explicit computationsof I. We use Singular ([3]) for all our computations.

In Section 9 we discuss how to detect whether a given ideal is a good one if it satisfiesthe necessary condition and does not satisfy the sufficient condition from Section 3. InSection 10 we discuss the following question: are powers of good ideals also good? Un-fortunately, the answer is negative in most cases. We also give a definition of a very goodideal: if Ia1,...,an = I for all (a1, . . . , an), then such an ideal is called a very good ideal andall its powers are Ratliff–Rush.

In Section 11 we discuss the connection of the above results to Freiman ideals thathave been studied in [9]. In particular, we will show that for an m-primary equigeneratedmonomial ideal being Freiman is equivalent to being very good.

2 Preliminaries and notation

Throughout this paper we will work with R = K[x1, . . . , xn], although all the results willalso hold in the local ring K[[x1, . . . , xn]]. We will be dealing with monomial ideals I inR. We start by listing a few basic properties of monomial ideals that will be used later.

1. For each monomial ideal there is a unique minimal generating set consisting ofmonomials. For an ideal I we denote G(I) to be its minimal monomial generatingset.

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2. If m ∈ I = 〈m1, . . . ,mk〉, where m and all mi are monomials, then there is some isuch that mi divides m.

3. If I = 〈m1, . . . ,mk〉, J = 〈n1, . . . , nl〉, then IJ = 〈m1n1, . . . ,m1nl, . . . ,mkn1, . . . ,mknl〉, but this generating set is not minimal in general.

4. There is a natural bijection between monomials in K[x1, . . . , xn] and points inNn in the following way: xα1

1 xα22 · · ·xαnn ↔ (α1, α2, . . . , αn). We will say that

(β1, β2, . . . , βn) ≤ (α1, α2, . . . , αn) if βi ≤ αi for all i ∈ {1, 2, . . . , n}. Then itis clear that xβ11 x

β22 · · ·xβnn divides xα1

1 xα22 · · ·xαnn if and only if (β1, β2, . . . , βn) ≤

(α1, α2, . . . , αn) and that multiplication of monomials corresponds to addition ofpoints. We will often say that some monomial belongs to some subset of Nn, mean-ing that the corresponding point belongs to that subset. Sometimes we will also saythat some point belongs to some ideal I, meaning that the corresponding monomialbelongs to I.

5. 〈m1〉 : 〈m2〉 =⟨

m1

gcd(m1,m2)

⟩.

6. I : (J1 + J2) = (I : J1) ∩ (I : J2) and (I1 + I2) : 〈m〉 = I1 : 〈m〉+ I2 : 〈m〉.

Let I be an m-primary monomial ideal of R, where m = 〈x1, x2, . . . , xn〉, that is, for somepositive integers d1, . . . , dn we have {xd11 , . . . , xdnn } ⊆ G(I). Henceforth, by I we alwaysmean an m-primary monomial ideal and denote µi := xdii , 1 ≤ i ≤ n. Also, in this paperwe do not consider any polynomials other than monomials since it will always be sufficientto prove statements for monomials only.

3 Good and bad ideals

In this section we will introduce the notion of a good ideal, prove a necessary and asufficient condition for being a good ideal and give some examples.

Definition 3.1. Let I be an ideal. Recall that {µ1, . . . , µn} ⊆ G(I), where µi = xdii forsome di. Let a1, . . . , an be nonnegative integers and denote

Ba1,...,an := ([a1d1, (a1 + 1)d1]× . . .× [andn, (an + 1)dn]) ∩ Nn.

Ba1,...,an will be called the box with coordinates (a1, . . . , an), associated to I. Points ofthe type (k1d1, . . . , kndn) and the corresponding monomials, where all ki are nonnegativeintegers, will be called corners. We will mostly work with one ideal at a time, thus thereis no need to use any additional index to show that Ba1,...,an depends on I. Note that allminimal generators of I lie in B0,...,0.

Definition 3.2. We will say that an ideal I satisfies the box decomposition principleif the following holds: for every positive integer l, every minimal generator of I l belongs tosome box Ba1,...,an such that a1 + . . .+ an = l− 1. Ideals satisfying the box decompositionprinciple will be called good, otherwise they will be called bad.

Example 3.3. Consider the ideal I = 〈x3, y3, z3, xyz〉 in K[x, y, z]. Then x2y2z2 is aminimal generator of I2, but it only belongs to B0,0,0 and 0 + 0 + 0 6= 1. Therefore, I is abad ideal.

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Example 3.4. Let I = 〈x3, y3, z3, x2y2z2〉 in K[x, y, z]. Then

G(I2) = {x6, y6, z6, x3y3, x3z3, y3z3, x5y2z2, x2y5z2, x2y2z5}.

Note that the square of x2y2z2 is not a minimal generator, thus we are not examiningit. Below we list all the possible boxes with sums of coordinates equal to 1 and minimalgenerators of I2 that belong to these boxes:

B1,0,0 : x6, x3y3, x3z3, x5y2z2,

B0,1,0 : y6, x3y3, y3z3, x2y5z2,

B0,0,1 : z6, x3z3, y3z3, x2y2z5.

Note that each minimal generator of I2 belongs to at least one such box. For simplicity,we denote

S1,0,0 := {x6, x3y3, x3z3, x5y2z2}(minimal generators of I2 that belong to B1,0,0) and we similarly define S0,1,0 and S0,0,1.We see that elements in S1,0,0 are multiples by µ1 = x3 of the minimal generators of I(similarly for S0,1,0 and S0,0,1), that is,

I2 = 〈S1,0,0, S0,1,0, S0,0,1〉 = µ1I + µ2I + µ3I.

Geometrically it means that I2 is minimally generated by all appropriate translations ofI.

What happens in I3 and higher powers? It is easy to see that the situation is quitesimilar there as well. Say, for I3 we take products of minimal generators of I with minimalgenerators of I2 (which are translations of the minimal generators of I). Obviously, wewill get nothing but larger translations of I, that is, I3 = µ2

1I + µ22I + µ2

3I + µ1µ2I +µ1µ3I + µ2µ3I. The first summand corresponds to the minimal generators in B2,0,0, thesecond one – to those in B0,2,0, the third one – to those in B0,0,2, the fourth one – to thosein B1,1,0, the fifth one – to those in B1,0,1, the sixth one – to those in B0,1,1. Clearly, thepattern repeats in all powers of I: for every l ≥ 1 we have

I l =∑

l1+...+ln=l−1

µl11 . . . µlnn I.

Therefore, I is a good ideal.

Proposition 3.5. The following are equivalent:

(1) I is a good ideal;

(2) for any l ≥ 1 and for any m ∈ I l there exist a1, . . . , an such that m ∈ Ba1,...,an anda1 + . . .+ an ≥ l − 1;

(3) for any l ≥ 1 and for any m1, . . . ,ml ∈ G(I) there exist a1, . . . , an such thatm1 · · ·ml ∈ Ba1,...,an and a1 + . . .+ an ≥ l − 1.

Proof.

(1)⇒(2): Let I be a good ideal and let l ≥ 1 and m ∈ I l. Then m is divisible by somem1 ∈ G(I l) and m1 ∈ Bb1,...,bn for some b1, . . . , bn with b1 + . . . + bn = l − 1. Thenthere exist a1, . . . , an such that (a1, . . . , an) ≥ (b1, . . . , bn) (thus a1 + . . .+an ≥ l−1)and m ∈ Ba1,...,an .

4

(2)⇒(3): Obvious.

(3)⇒(1): Let l ≥ 1 and m ∈ G(I l). We want to show that there is a box Bb1,...,bn

such that m ∈ Bb1,...,bn and b1 + . . . + bn = l − 1. Since m ∈ G(I l), we havem = m1 · · ·ml for some m1, . . . ,ml ∈ G(I). Then there exist a1, . . . , an such thatm = m1 · · ·ml ∈ Ba1,...,an and a1+. . .+an ≥ l−1. If we assume that a1+. . .+an ≥ l,then m is divisible by µa11 · · ·µann ∈ I l. We have two cases:

1. If m 6= µa11 · · ·µann , it contradicts m ∈ G(I l) and thus a1 + . . .+ an = l− 1 andwe can set bi := ai for all i.

2. If m = µa11 · · ·µann , then a1 + . . . + an = l since m can not possibly belong toG(I l) if a1 + . . .+an > l. Note that we are not proving that µa11 · · ·µann ∈ G(I l)if a1 + . . . + an = l (this will be done later in this paper). In any case, wewill find an appropriate box for m. At least one of ai is different from 0.Without loss of generality, we can assume a1 ≥ 1. Then m ∈ Ba1−1,a2,...,an and(a1 − 1) + a2 + . . .+ an = l − 1.

Now we are interested in some necessary and sufficient conditions on G(I) for an idealI to be good. Note that every monomial ideal in K[x] is a good one.

Theorem 3.6. (A necessary condition for being a good ideal) Let I be an ideal inK[x1, . . . , xn]. If I is a good ideal, then for any minimal generator xα1

1 xα22 · · · xαnn of I

the following holds:α1

d1+ · · ·+ αn

dn≥ 1.

Proof. Assume that there is a minimal generator for which the above condition fails, thatis, m = xα1

1 xα22 · · · xαnn with α1

d1+ · · ·+ αn

dn= 1− ε for some 0 < ε < 1. Let l be a positive

integer such that l > 1ε. We will show that the box decomposition principle fails for

I l. Consider ml = xlα11 xlα2

2 · · ·xlαnn . The first coordinate of ml is lα1, therefore, the firstcoordinate of the box where ml belongs is at most b lα1

d1c and similar inequalities hold for

the other coordinates. Therefore, the sum of coordinates of any box containing ml is lessor equal to b lα1

d1c+. . .+b lαn

dnc ≤ lα1

d1+. . .+ lαn

dn= l(α1

d1+· · ·+ αn

dn) = l(1−ε) < l(1− 1

l) = l−1.

Thus all boxes containing ml have the sum of coordinates strictly less than l − 1 and weare done by Proposition 3.5.

Theorem 3.7. (A sufficient condition for being a good ideal)Let I be an ideal in K[x1, . . . , xn]. Assume that for any minimal generator

xα11 x

α22 · · ·xαnn of I which is not a corner the following holds:

α1

d1+ · · ·+ αn

dn≥ n

2.

Then I is a good ideal.

Proof. The claim is trivial for n = 1, thus assume n ≥ 2. Let m1,m2 ∈ G(I), wherem1 = xα1

1 · · ·xαnn , m2 = xβ11 · · ·xβnn with α1

d1+ · · · + αn

dn≥ n

2and β1

d1+ · · · + βn

dn≥ n

2.

By Proposition 3.5, it suffices to show that m1m2 = µixγ11 · · · xγnn for some i and with

γ1d1

+ · · · + γndn≥ n

2. Note that α1+β1

d1+ · · · + αn+βn

dn≥ n, thus we must have αi+βi

di≥ 1

for some i. We can assume i = 1, then α1+β1−d1d1

+ · · · + αn+βndn

≥ n − 1 ≥ n2. Setting

γ1 = α1 + β1 − d1 and γi = αi + βi for 2 ≤ i ≤ n finishes the proof.

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Remark 3.8. For n = 2 the necessary condition is equivalent to the sufficient condition.

Example 3.9. (A good ideal that does not satisfy the sufficient condition)Let I = 〈µ1, µ2, µ3,m〉 = 〈x5, y5, z5, xyz4〉 ⊂ K[x, y, z]. The ideal satisfies the nec-

essary condition, but not the sufficient condition, so we will explore it by hand. Whatkinds of generators do we have in I l? First of all, we notice that m5 = x5y5z20 is divisibleby, say, µ1µ2µ

33 ∈ I5, therefore, it is not a minimal generator of I5. Therefore, for any l,

the minimal generators of I l will be of the form µk11 µk22 µ

k33 m

k, where k1 + k2 + k3 + k = land k ≤ 4. If k = 0, the monomial is just a corner and this case is trivial, so let k ≥ 1.Clearly, such a monomial belongs to a box whose sum of coordinates is l − 1 if and onlyif mk belongs to a box whose sum of coordinates is k − 1. So the only thing we needto check is whether mk belongs to a box whose sum of coordinates is k − 1, 2 ≤ k ≤ 4(this is always true for k = 1). We see that m2 = x2y2z8 ∈ B0,0,1, m

3 = x3y3z12 ∈ B0,0,2,m4 = x4y4z16 ∈ B0,0,3. Therefore, I is a good ideal.

Example 3.10. (A bad ideal that satisfies the necessary condition)Let I = 〈x5, y5, z5, x2y2z2〉 ⊂ K[x, y, z]. The ideal satisfies the necessary condition,

but not the sufficient condition. We see that x4y4z4 is a minimal generator of I2 and itonly belongs to B0,0,0. Since 0 + 0 + 0 6= 1, I is a bad ideal.

Ideals that satisfy the necessary condition, but do not necessarily satisfy the sufficientcondition, will be discussed further in Section 9 of this paper. There is a way to detectwhether an ideal is good or bad and it basically uses the ideas from the two examplesabove.

4 Ideals inside boxes and their connection to each

other

We will start this section with an example aimed to give a motivation for the futureconstructions.

Example 4.1. Let I = 〈x5, y5, xy4, x4y〉 ⊂ K[x, y]. I is a good ideal since it satisfiesthe sufficient condition. In this case the associated boxes have sizes 5 × 5. Figure 1represents powers of I up to I4. Consider the box B1,0. Inside this box we see some ofthe minimal generators of I2, namely, {x5y5, x6y4, x8y2, x9y, x10}. Since they are in B1,0,they are divisible by µ1

1µ02 = x5. If we divide all these monomials by µ1

1µ02, we will get

{y5, xy4, x3y2, x4y, x5}.Define I1,0 := 〈y5, xy4, x3y2, x4y, x5〉. Geometrically, this means viewing monomi-

als in B1,0 as if the lower left corner of B1,0 was the origin. In this particular ex-ample we have I0,0 = I, I1,0 = 〈y5, xy4, x3y2, x4y, x5〉, I0,1 = 〈y5, xy4, x2y3, x4y, x5〉,Ia,b = 〈y5, xy4, x2y3, x3y2, x4y, x5〉 for all other (a, b).

6

00

1

1

2

2

3

3

4

4

Figure 1: powers of I: I, I2, I3 and I4

This gives rise to a more general definition.

Definition 4.2. Let I be a good ideal and a1, . . . , an nonnegative integers. We define

Ia1,...,an :=

⟨m

µa11 · · ·µann| m ∈ Ba1,...,an ∩G(I l)

⟩,

where l = a1 + . . .+ an + 1. Note that this a minimal generating set of Ia1,...,an .

A priori it is not clear why, given a good ideal I, any box Ba1,...,an has a nonemptyintersection with G(I l), where l = a1 + . . . + an + 1. The next remark will in particularshow that intersections of this type are never empty.

Remark 4.3. (Corners are needed) Let I be a good ideal, let m = µk11 · · ·µknn be somecorner and put l := k1 + k2 + . . . + kn. Then m ∈ G(I l). Indeed, assume that m isnot a minimal generator of I l, which means that there exists a strictly smaller generators = xs11 · · ·xsnn . Note that s is a product of l minimal generators of I, and since I is agood ideal, the necessary condition holds and therefore s1

d1+ . . . + sn

dn≥ l. As for m, we

have k1d1d1

+ . . .+ kndndn

= k1 + . . .+ kn = l, which is a contradiction, since s strictly dividesm.

Now we see that, given a good ideal I and a box Ba1,...,an , the box necessarily contains,for instance, all monomials of the type {µj

∏ni=1 µ

aii | 1 ≤ j ≤ n}. All these monomials

are corners, therefore, they are minimal generators of I l, where l = a1 + . . . + an + 1.Thus we conclude that any box Ba1,...,an has a nonempty intersection with G(I l), sincethis intersection contains n corners, mentioned above. As a consequence, µ1, . . . , µn areminimal generators of any Ia1,...,an .

Proposition 4.4. Let I be a good ideal and a1, . . . , an nonnegative integers. Then

Ia1,...,an = I l : 〈µa11 · · ·µann 〉,

where l = a1 + . . .+ an + 1.

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Proof. It is clear from the definition that Ia1,...,an ⊆ I l : 〈µa11 · · ·µann 〉. For the otherinclusion, let m ∈ I l : 〈µa11 · · ·µann 〉. Then mµa11 · · ·µann ∈ I l, that is, mµa11 · · ·µann is amultiple of some g ∈ G(I l), say, mµa11 · · ·µann = gg1. Being a minimal generator of I l,g belongs to some box, say, Bb1,...,bn , with b1 + . . . + bn = l − 1 = a1 + . . . + an. If(a1, . . . , an) = (b1, . . . , bn), then m is a multiple of g

µa11 ···µ

ann

, which is a generator of Ia1,...,anand thus we are done. If (a1, . . . , an) 6= (b1, . . . , bn), then there is some ai < bi. Withoutloss of generality, we assume that a1 < b1. Then the right hand side of mµa11 · · ·µann = gg1is divisible by µb11 , thus m is divisible by µ1, and µ1 is a minimal generator of Ia1,...,an byRemark 4.3. Therefore, m ∈ Ia1,...,an .

Let a1, a2, . . . , an and b1, b2, . . . , bn be nonnegative integers such that(a1, . . . , an) ≤ (b1, . . . , bn). Since Ia1,...,an = Ia1+...+an+1 : 〈µa11 · · ·µann 〉 and Ib1,...,bn =Ib1+...+bn+1 : 〈µb11 · · ·µbnn 〉, we immediately conclude the following:

Corollary 4.5. Let I be a good ideal and let a1, a2, . . . , an and b1, b2, . . . , bn be nonnegativeintegers such that (a1, . . . , an) ≤ (b1, . . . , bn). Then Ia1,...,an ⊆ Ib1,...,bn.

5 Asymptotic behaviour of Ia1,...,an

Now we know that Ia1,...,an grows as (a1, . . . , an) grows. Since Ia1,...,an can not increaseforever, one expects some pattern on high powers of I, which is indeed the case. Let ustake a closer look at the situation.

Definition 5.1. Let a1, . . . , an be nonnegative integers. We will use the following nota-tion:

Ca1,a2,...,ak,ak+1,ak+2,...,an := {(b1, . . . , bn) ∈ Nn |b1 = a1, . . . , bk = ak, bk+1 ≥ ak+1, . . . , bn ≥ an}.

We will use a similar notation for any configuration of fixed and non-fixed coordinates.Sets of this type will be called cones, for any cone the number of non-fixed coordinates willbe called its dimension and (a1, . . . , an) will be called its vertex. Note that Nn = C0,0,...,0.

Example 5.2. Let n = 3 and a1 = 1, a2 = 2, a3 = 3. Then C1,2,3 = {(b1, 2, b3) | b1 ≥1, b3 ≥ 3} and the dimension of this cone is 2.

Definition 5.3. Let a1, . . . , an be nonnegative integers. By Aa1,...,an we denote the set ofall cones that satisfy the following conditions:

1. if (b1, . . . , bn) is the vertex of a cone in Aa1,...,an , then bi ≤ ai for all 1 ≤ i ≤ n;

2. for all 1 ≤ i ≤ n the following holds: if bi = ai, then bi is not underlined and ifbi < ai, then bi is underlined.

Note that the unique cone of dimension n in Aa1,...,an is Ca1,...,an .

Example 5.4. Let n = 2, a1 = 2, a2 = 1. We would like to find all the cones in A2,1. Forany cone in A2,1 the first coordinate of its vertex can only be chosen from the set {0, 1, 2};we underline it if we choose 0 or 1 and do not underline it if we choose 2. Independently,the second coordinate can only be chosen from the set {0, 1} and we underline it if wechoose 0 and do not underline it if we choose 1. Therefore, we will get six cones in total:

8

A2,1 = {C0,0, C0,1, C1,0, C1,1, C2,0, C2,1}.

00

1

1

2

2

3

3

4

4

5

5

6

6

7

7

Figure 2: cones of A2,1

Figure 2 represents the six cones from A2,1. The boundary lines are only drawn forbetter visibility. Clearly, the number of boundary lines equals the dimension of the cone.

Lemma 5.5. Let a1, . . . , an be nonnegative integers. Then cones in Aa1,...,an form a dis-joint covering of Nn.

Proof. Let b = (b1, . . . , bn) be a point in Nn. We will find a unique cone in Aa1,...,an thatcontains this point. First of all, we compare a1 and b1.

1. If b1 ≥ a1, then the first coordinate of our future cone containing b is non-fixed sinceotherwise Aa1,...,an contains Cb1,..., which is a contradiction with b1 ≥ a1. Therefore,our first coordinate has to be non-fixed, hence equal to a1.

2. If b1 < a1, then the first coordinate can only be a fixed one since otherwise it is equalto a1, but b can not belong to Ca1,... since b1 < a1. Therefore, the first coordinatehas to be a fixed one, hence equal to b1.

Proceeding in the same way we construct a cone in Aa1,...,an that contains (b1, . . . , bn).From the construction it is clear that this cone is unique, which finishes our proof.

We have seen that given Nn = C0,0,...,0 and a point (a1, . . . , an) ∈ C0,0,...,0, we candecompose C0,0,...,0 into a disjoint union of cones, associated to this point, where the uniquecone of dimension n is Ca1,...,an and all other cones have strictly lower dimensions. It is nothard to see that we can replace Nn = C0,0,...,0 with any other cone and replace (a1, . . . , an)with any point in this cone and have a similar decomposition. First of all, assume that allcoordinates of this cone are non-fixed, say, we have Cs1,...,sn and a point (s1 + k1, . . . , sn +kn) ∈ Cs1,...,sn for some nonnegative integers k1, . . . , kn. Clearly, points in Cs1,...,sn are inbijection with points in C0,0,...,0 under the obvious shift. We can find the decompositionof C0,0,...,0 with respect to (k1, . . . , kn) as in the proposition above and then shift all thecones in the decomposition by (s1, . . . , sn) to get new cones. This will give us the desireddecomposition of Cs1,...,sn . Again, the unique cone of dimension n in this decompositionis Cs1+k1,...,sn+k1 . Now assume that some coordinates of our cone are fixed, say, we haveCs1,...,sm,sm+1...,sn (without loss of generality, we can assume that fixed coordinates are the

last (n−m) coordinates) and (s1 +k1, . . . , sm+km, sm+1, . . . , sn) ∈ Cs1,...,sm,sm+1...,sn . Notethat this is an m-dimensional cone and points in this cone are in bijection with points in

9

Nm, in particular, (s1 + k1, . . . , sm + km, sm+1, . . . , sn) ↔ (k1, . . . , km). Thus we can findthe decomposition of Nm with respect to (k1, . . . , km), then shift all cones by (s1 . . . , sm)(this will give us the first m coordinates of each cone) and the last (n −m) coordinatesof each cone in this decomposition are sm+1, . . . , sn. The unique m-dimensional cone inthis decomposition is Cs1+k1,...,sm+km,sm+1,...,sn , others have lower dimensions. Therefore,the previous proposition can be restated in a more general context:

Theorem 5.6. Given any cone C in Nn of dimension k and a point a ∈ C, we candecompose C into a disjoint union of finitely many cones, where exactly one cone hasdimension k and vertex a, and all other cones have strictly lower dimensions.

Example 5.7. Let n = 5 and consider C5,7,4,2,3. Consider (a1, . . . , a5) = (5, 9, 4, 3, 3) ∈C5,7,4,2,3. The first, the third and the fifth coordinates are fixed once and forever, thatis, all cones that we will find have the form C5,?,4,?,3. We are left with the second andthe fourth coordinate, that is, (7, 2) for the cone and (9, 3) for the point. Shifting in thenegative direction by (7, 2), we will get (0, 0) and (2, 1) respectively. Thus, it is enoughto find the decomposition of N2 with respect to (2, 1). This is exactly what we did inExample 5.4. We obtained A2,1 = {C0,0, C0,1, C1,0, C1,1, C2,0, C2,1}. Shifting in the positivedirection by (7, 2) gives us {C7,2, C7,3, C8,2, C8,3, C9,2, C9,3} and inserting back the first, thethird and the fifth coordinates gives us

{C5,7,4,2,3, C5,7,4,3,3, C5,8,4,2,3, C5,8,4,3,3, C5,9,4,2,3, C5,9,4,3,3}.

Therefore, C5,7,4,2,3 is a disjoint union of these six cones.

Now we will use these results on monomial ideals. Let I be a good ideal. Then for anyvector of nonnegative integers (a1, . . . , an) we have defined a box Ba1,...,an and the corre-sponding ideal Ia1,...,an . Clearly, there is a bijection between points in Nn and boxes/ideals;recall that if (a1, . . . , an) ≤ (b1, . . . , bn), then Ia1,...,an ⊆ Ib1,...,bn by Corollary 4.5.

Theorem 5.8. For any good ideal I there exists a finite coloring of Nn such that if(a1, . . . , an) has the same color as (b1, . . . , bn), then Ia1,...,an = Ib1,...,bn and for each colorthe set of points of this color forms a cone.

Proof. We use induction on the highest dimension of uncolored cones. We are startingwith an n-dimensional cone Nn. We will show how to obtain finitely many cones of strictlylower dimensions, each of which will then be treated similarly in a recursive way. Firstof all, note that it is possible to find a point (a1, . . . , an) such that the following holds:if (b1, . . . , bn) ≥ (a1, . . . , an), then Ia1,...,an = Ib1,...,bn . Indeed, if we assume the converse,then for every point of Nn there exists a strictly larger point that corresponds to a strictlylarger ideal, therefore, we can build an infinite chain of strictly increasing ideals, whichis impossible, for example, by Noetherianity of the polynomial ring. So existence of sucha point (a1, . . . , an) is justified. Then from the Theorem 5.6, Nn can be covered witha disjoint union of (finitely many) cones in Aa1,...,an . The unique n-dimensional cone inAa1,...,an is Ca1,...,an and, as we have just figured out, we may paint all points in this conewith the same color. Now we are left with a finite disjoint union of cones of dimensionsat most n−1 which need to be painted and we apply induction on each of them, loweringthe maximal dimension by 1 again. Since it is a finite process, in the end we will obtaina finite coloring of Nn.

We remark that the coloring described above is not unique since it depends on thechoice of (a1, . . . , an) and its lower dimensional analogues.

10

Example 5.9. Let I be the ideal in Example 4.1. We can choose (a1, a2) = (1, 1) sinceIb1,b2 = I1,1 for all (b1, b2) ≥ (1, 1). Then N2 is a disjoint union of C1,1, C0,1, C1,0 andC0,0. Now consider C0,1. We see that I0,b = I0,2 for all b ≥ 2. Therefore, we considerthe decomposition of C0,1 with respect to (0, 2): C0,1 is a disjoint union of C0,2 and C0,1.Similarly, C1,0 is a disjoint union of C2,0 and C1,0. The left picture in Figure 3 describesthe coloring we have just discussed. The picture on the right describes another possiblecoloring if, for instance, we choose (a1, a2) = (0, 2).

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4

0 1 2 3 40

1

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4

Figure 3: two examples of possible colorings of N2, associated to I

Given a good ideal I, any coloring as in Theorem 5.8 represents a finite disjoint unionof cones. Each cone has a vertex. Let L denote the maximum of sums of coordinates ofthese vertices. This number depends on I and on the coloring we choose, but we will notput any additional indices: as soon as we found some coloring (which exists according toTheorem 5.8), we simply work with it once and forever. For example, for both coloringsin Figure 3 we have L = 2.

Remark 5.10. Note that from the construction in Theorem 5.8 it is clear that L can notbe attained at a zero dimensional cone, in other words, for any zero dimensional cone, thesum of coordinates of its vertex is strictly less than L.

The geometric meaning of this number is the following: starting from IL+1, powers ofI look similar to each other in some sense. For instance, for the left coloring in Figure 3we know that every power of I starting from I3 consists of a green box, an orange boxand several red boxes and we exactly know where each of them is. This means, there is apattern on high powers of I, and this is a key point for finding the Ratliff–Rush closureof I.

6 The main result

Now we are ready to prove our main theorem, but first we need a preliminary lemma.

Lemma 6.1. Let I be a good ideal and let Q be any nonnegative integer. Then thereexists a number L(Q) such that for any l ≥ L(Q) the following holds: for every minimalgenerator m of I l there is an i such that m = m′µQi and m′ is a minimal generator ofI l−Q.

Proof. If Q = 0, the claim is trivial. Let Q > 0 and let L be the number defined inthe end of Section 5. Take L(Q) = L + nQ − n + 2 and let l ≥ L(Q). Let m be

11

a minimal generator of I l, then it belongs to some box Bb1,...,bn with b1 + . . . + bn =l − 1 ≥ L + nQ − n + 1. We also know that (b1, . . . , bn) belongs to one of the conesfrom our coloring; assume that the vertex of this cone is (a1, . . . , an) (some coordinatesare underlined, some are not underlined). Now we want to find a coordinate bi such that(b1, . . . , bi−1, bi−Q, bi+1, . . . , bn) belongs to the same cone. Assume that it is not possible.Then it follows that b1 − Q ≤ a1 − 1, . . . , bn − Q ≤ an − 1. These inequalities yield acontradiction L < b1 + . . . + bn − nQ + n ≤ a1 + . . . + an ≤ L, where the last inequalityfollows from the definition of L. So we can find an index i such that bi − Q ≥ ai (inparticular, this implies that ai is not underlined). Without loss of generality we assumethat i = 1. That means, (b1, . . . , bn) and (b1 − Q, b2, . . . , bn) are both in the same cone.This implies that their colors are equal, which means Ib1,...,bn = Ib1−Q,b2,...,bn . In otherwords, the set of monomials in Bb1,...,bn ∩ G(I l) coincides with the set of monomials in

Bb1−Q,b2,...,bn ∩ G(I l−Q) up to a shift by µQ1 . Therefore, if m ∈ Bb1,...,bn is a minimalgenerator of I l, then m

µQ1∈ Bb1−Q,b2,...,bn is a minimal generator of I l−Q, as desired.

Now let us consider the following line of boxes which is in bijection with nonnegativeinteger points on the x1-axis: B0,0,...,0, B1,0,...,0, B2,0,...,0 etc. Let Bq1,0...,0 be the stabilizingbox of this sequence in a sense that q1 is the smallest nonnegative integer such thatIt,0,...,0 = Iq1,0,...,0 for all t ≥ q1. Similarly, considering lines of boxes going along the othercoordinate axes, we will get q2, q3, . . . , qn. Denote q := max{q1, . . . , qn}.

Theorem 6.2. Let I be a good ideal, let L, qi and q be as above. Then I = Iq1,0,...,0 ∩I0,q2,...,0 ∩ . . . ∩ I0,...,0,qn.

Proof. ⊆ Let l ≥ q. We will show that I l+1 : I l ⊆ Iq1,0,...,0 ∩ I0,q2,...,0 ∩ . . . ∩ I0,...,0,qn .In fact, we will show that I l+1 : I l ⊆ Iq1,0,...,0, other inclusions are analogous. SinceI l+1 : I l ⊆ I l+1 : 〈µl1〉, it is sufficient to show that I l+1 : 〈µl1〉 ⊆ Iq1,0,...,0. By Proposition4.4, I l+1 : 〈µl1〉 = Il,0,...,0 which equals Iq1,0,...,0, given the way Iq1,0,...,0 was defined andgiven that l ≥ q ≥ q1 . Therefore, everything follows.⊇ Let m ∈ Iq1,...,0 ∩ I0,q2,...,0 ∩ . . . ∩ I0,...,0,qn , let l ≥ L(q) = L + nq − n + 2 (as in

Lemma 6.1). We will show that for every ml ∈ I l we have mml ∈ I l+1. It is enoughto consider ml to be minimal generators of I l. First of all, from Lemma 6.1 we knowthat we can factor out some µqi from ml and get a minimal generator of I l−q, that is,ml = µqiml−q for some index i and ml−q a minimal generator in I l−q. Also, since mbelongs (in particular) to I0,...,0,qi,0...,0 = Iqi+1 : 〈µqii 〉, it means, mµqii ∈ Iqi+1.

Therefore, mml = mµqii µq−qii ml−q ∈ I l+1 since mµqii ∈ Iqi+1, µq−qii ∈ Iq−qi , ml−q ∈

I l−q.

7 Explicit computation of I0,...,0,qi,0,...,0

We have seen that, given a good ideal I, its Ratliff–Rush closure is computed as I =Iq1,0,...,0 ∩ I0,q2,...,0 ∩ . . . ∩ I0,...,0,qn . Surprisingly, no other boxes affect the Ratliff–Rushclosure, but only those going along coordinate axes. Therefore, we would like to knowmore about I0,...,0,qi,0...,0. Let i = 1, other cases are analogous. So far we only know thatIt,0,...,0...,0 = I t+1 : 〈µt1〉. Computation of I t might take much time if t is large enough.In addition, we do not know yet at which moment the line has stabilized. So far theprocess seems more complicated than it is. We will state a few results that will make thiscomputation easier.

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Lemma 7.1. If I is a good ideal, then It+1,0,...,0 = (It,0...,0 · I) : 〈µ1〉 for all t ≥ 0.

Proof. ⊆ Let m ∈ It+1,0,...,0 = I t+2 : 〈µt+11 〉. Then mµt+1

1 = cg1 · · · gt+2 ∈ I t+2, whereg1, . . . , gt+2 are minimal generators of I and c is some monomial. We claim that cg1 · · · gt+1

is divisible by µt1. Indeed, if it is not the case, then the x1-exponent of gt+2 is strictlygreater than d1, which is impossible since gt+2 is a minimal generator of I. Therefore,mµ1 = cg1···gt+1

µt1gt+2. Since cg1···gt+1

µt1∈ It,0,...,0 and gt+2 ∈ I, we are done.

⊇ Let m ∈ (It,0...,0 · I) : 〈µ1〉. Then mµ1 = gm1, where g ∈ I, m1 ∈ It,0...,0. Thenmµt+1

1 = mµ1µt1 = gm1µ

t1 ∈ I t+2 since g ∈ I and m1µ

t1 ∈ I t+1.

Remark 7.2. We would like to point out that if It,0,...,0 = It+1,0,...,0, then the line hasstabilized, that is, Ik,0,...,0 = It,0,...,0 for all k ≥ t. This is a direct corollary of Lemma 7.1.

Remark 7.3. For all t ≥ 0 let Et := G(It,0,...,0) and for all t ≥ 1 let

Ft := {m ∈ Et = G(It,0...,0) | m 6∈ It−1,0,...,0}.

We also set E−1 = ∅, F0 = G(I). We will show how, given {Et−1, Ft}, one can obtain{Et, Ft+1} for any t ≥ 0. Clearly, Et is the reduced union of Et−1 and Ft. From Lemma 7.1we remember that It+1,0,...,0 = (It,0,...,0 · I) : 〈µ1〉 = (〈Et−1 ∪ Ft〉 · I) : 〈µ1〉 = ((It−1,...,0 +〈Ft〉) · I) : 〈µ1〉 = (It−1,...,0 · I + 〈Ft〉 · I) : 〈µ1〉 = (It−1,...,0 · I) : 〈µ1〉 + (〈Ft〉 · I) : 〈µ1〉 =It,0,...,0 + (〈Ft〉 · I) : 〈µ1〉. Therefore, we conclude that minimal generators of It+1,0,...,0

which are not in It,0,...,0 (our future Ft+1) could only be among the minimal generatorsof (〈Ft〉 · I) : 〈µ1〉, that is, only new monomials from the previous iteration can giverise to new monomials in the next iteration. Therefore, in order to compute Ft+1 we

need to compute At :={

fmgcd(fm,µ1)

| f ∈ Ft,m ∈ G(I)}

, reduce this set and throw away

monomials that are already in 〈Et〉 = It,0,...,0. If Ft+1 = ∅, it means that It+1,0...,0 = It,0...,0and Et is the desired generating set.

Remark 7.4. Now we know that in order to compute Ft+1 we need to compute At ={fm

gcd(fm,µ1)| f ∈ Ft,m ∈ G(I)

}, reduce this set and throw away monomials that are al-

ready in 〈Et〉 = It,0,...,0, where Et has already been computed. This is already quitestraightforward, but we can simplify the calculations a bit more. First of all note that ifm = µ1, then for any f ∈ Ft we have fm

gcd(fm,µ1)= f ∈ 〈Ft〉 ⊆ 〈Et〉 and thus this monomial

will be thrown away anyway. If m = µi for some 2 ≤ i ≤ n, say, m = µ2, then for anyf ∈ Ft we get that fm

gcd(fm,µ1)is divisible by µ2 ∈ 〈Et〉 (recall that all µi are minimal gen-

erators of all Ia1,...,an) and will thus be thrown away. In particular, this implies that in thedefinition of At one can replace G(I) with P (I) := G(I)\{µ1, . . . , µn}. Another observa-tion is the following. Assume that f ∈ Ft, m ∈ P (I). Write fm = xα1

1 · · ·xαnn . If αi ≥ difor some 2 ≤ i ≤ n, say, α2 ≥ d2, then fm

gcd(fm,µ1)is a again a multiple of µ2 ∈ 〈Et〉). There-

fore, in the definition of At we may force that degxi(fm) < di for all 2 ≤ i ≤ n, that is, we

may take At := { fmgcd(fm,µ1)

| f ∈ Ft,m ∈ P (I), degxi(fm) < di, 2 ≤ i ≤ n}. Finally, con-

sider xα11 · · ·xαnn µt1 = fµt1m ∈ I t+2 since fµt1 ∈ I t+1 according to Proposition 4.4. Since we

have imposed αi < di for all 2 ≤ i ≤ n, we must have α1 ≥ d1, since otherwise fµt1m wouldbelong to a box with the sum of coordinates at most t. Thus gcd(fm, µ1) = µ1. Therefore,

we conclude that we can write At ={fmµ1| f ∈ Ft,m ∈ P (I), degxi(fm) < di, 2 ≤ i ≤ n

}.

In order to compute Ft+1 one still needs to reduce this set and throw away monomialswhich are already in 〈Et〉, if needed.

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The algorithm below produces E = G(Iq1,0,...,0) given the input G(I).

beginE := ∅, F := G(I);while F 6= ∅ do

E := reduce(E ∪ F );F1 := ∅;for m ∈ P (I) = G(I)\{µ1, . . . , µn}, f ∈ F do

if degxi(fm) < di for all 2 ≤ i ≤ n then

a := fmµ1

;

if a 6∈ 〈E〉 thenF1 := F1 ∪ {a};

end

end

endF := reduce(F1);

endprint E;

end

8 Examples

Example 8.1. Let R = K[x, y, z] and let

I = 〈µ1, µ2, µ3,m1,m2,m3〉 = 〈x29, y29, z29, x28y8z8, x8y28z8, x8y8z28〉 ⊂ R.

Since I satisfies the sufficient condition, it is a good ideal. Computations in Singular showthat

I2 : I = I + 〈x27y27z27〉,

I3 : I2 = I4 : I3 = I + 〈x26y27z27, x27y26z27, x27y27z26〉,

I5 : I4 = I6 : I5 = · · · = I10 : I9 = I + 〈x26y26z26〉.

It is natural to conjecture that I = I + 〈x26y26z26〉. Now let us see what we get if weapply the algorithm above. We start with E−1 = ∅, F0 = G(I). Then we obtain E0 byreducing E−1 ∪ F0, that is, E0 = G(I) (as it should be). In order to compute F1, wetake all products of F0 = G(I) with P (I), keeping in mind that y− and z− coordinatesof each product need to be less than 29, and divide each such product by µ1. If we takef = µ1 ∈ F0 = G(I) and any m ∈ P (I), then fm

µ1= m will be thrown away (and this

always happens in the first iteration, but never afterwards since in the higher iterationsµ1 never belongs to any of the considered sets). The only monomial that is not thrown

away ism2

1

µ1= x56y16z16

x29= x27y16z16. This monomial is not in 〈E0〉, therefore, we add it

to our set F1 (and this set is already reduced). Therefore, E0 = G(I), F1 = {x27y16z16}.Now E1 = E0 ∪ F1 = G(I) ∪ {x27y16z16} (this union is already reduced), and in orderto compute F2 we need to multiply x27y16z16 with monomials from P (I) (keeping inmind the condition on y− and z− coordinates) and divide the products by µ1. The only

possible monomial is x27y16z16·m1

µ1= x26y24z24. This monomial is not in 〈E1〉, therefore,

F2 = {x26y24z24}. E2 = E1 ∪ F2 = G(I) ∪ {x27y16z16, x26y24z24} (this set is already

14

reduced) and if we try to compute F3, we see that we can not get any new monomials.Therefore, F3 = ∅ and the stabilizing point is I2,0,0 = 〈E2〉 = I + 〈x27y16z16, x26y24z24〉

By symmetry,I0,2,0 = I + 〈x16y27z16, x24y26z24〉

andI0,0,2 = I + 〈x16y16z27, x24y24z26〉.

According to the theorem, I = I2,0,0 ∩ I0,2,0 ∩ I0,0,2 = I + 〈x26y26z26〉, just as expected.

Example 8.2. Consider I = 〈x53, y56, z59, w61, x50y18z20w25, x15y54z22w24,x18y20z56w22, x16y19z23w60〉 ⊂ K[x, y, z, w].

Since I satisfies the sufficient condition, it is a good ideal. Then, applying the algo-rithm from Section 7, we will get

I1,0,0,0 = I + 〈x47y36z40w50〉 = I2,0,0,0 = I3,0,0,0 = I4,0,0,0 = · · ·

I0,1,0,0 = I + 〈x30y52z44w48〉 = I0,2,0,0 = I0,3,0,0 = I0,4,0,0 = · · ·

I0,0,1,0 = I + 〈x36y40z53w44〉 = I0,0,2,0 = I0,0,3,0 = I0,0,4,0 = · · ·

I0,0,0,1 = I + 〈x32y38z46w59〉 = I0,0,0,2 = I0,0,0,3 = I0,0,0,4 = · · ·

Then I = I1,0,0,0 ∩ I0,1,0,0 ∩ I0,0,1,0 ∩ I0,0,0,1 = I + 〈x47y52z53w59〉 which coincides with ourexpectations based on computations in Singular.

Example 8.3. Let I = 〈x41, y41, z41, x40y5z5, x5y40z5, x5y5z40〉 ⊂ K[x, y, z]. It will beproven in Example 9.3 that I is a good ideal. All the new monomials can only be obtainedfrom powers of non-corners:

I1,0,0 = I + 〈x39y10z10〉,

I2,0,0 = I1,0,0 + 〈x38y15z15〉,

. . .

I6,0,0 = I5,0,0 + 〈x34y35z35〉.

Here the line stabilizes. We similarly get I0,6,0 and I0,0,6. Each of these three ideals hastwelve minimal generators in total. Intersecting them we will get

I = I6,0,0 ∩ I0,6,0 ∩ I0,0,6 = I + 〈x34y35z35, x35y34z35, x35y35z34〉.

If we compute successive quotients via computer algebra, the result is the following: I2 : I1

has 7 minimal generators, that is, |G(I2 : I1)| = 7; |G(I3 : I2)| = 9; |G(I4 : I3)| = 12;|G(I5 : I4)| = 16; |G(I6 : I5)| = 21; |G(I7 : I6)| = 27; |G(I8 : I7)| = 31; |G(I9 : I8)| = 33;|G(I10 : I9)| = 33; |G(I11 : I10)| = 31; |G(I12 : I11)| = 24; |G(I13 : I12)| = 18; |G(I14 :I13)| = 13; |G(I15 : I14)| = 9. I15 : I14 finally coincides with the ideal obtained above(but we still can not be sure this is the Ratliff–Rush closure of I assuming that we are notusing our formula). It takes much time to perform these computations using computeralgebra, whereas the computation of I6,0,0, I0,6,0 and I0,0,6 and their intersection is mucheasier and can even be done by hand in this example.

15

9 A bit more about good and bad ideals

Let I be an ideal that satisfies the necessary condition, but does not satisfy the sufficientcondition. How could we possibly figure out whether it is a good or a bad ideal? We willstart with ideals that have only one extra generator except µi. Let I = 〈µ1, . . . , µn,m〉,where µi = xdii and m = xα1

1 · · ·xαnn with α1

d1+ . . .+ αn

dn≥ 1. Then the minimal generators

of I l will be of the form µk11 · · ·µknn mk, where k1 + . . .+kn+k = l. We will first of all showthat if d = lcm(d1, . . . , dn), then md is not needed as a minimal generator of Id. Indeed,

md = xdα11 · · ·xdαnn = µ

dα1d11 · · ·µ

dαndnn and dα1

d1+ . . .+ dαn

dn= d(α1

d1+ . . .+ αn

dn) ≥ d. Therefore,

md ∈ 〈µ1, . . . , µn〉d. Let K be the smallest positive integer such that mK ∈ 〈µ1, . . . , µn〉K .As we have seen, d = lcm(d1, . . . , dn) is an upper bound for K (but this upper bound canbe easily improved). Therefore, for any l, the minimal generators of I l will be of the formµk11 · · ·µknn mk, where k1 + . . .+ kn + k = l and k ≤ K − 1. Conversely, each monomial ofthis form is a minimal generator of I l. Clearly, such a monomial belongs to a box whosesum of coordinates is l − 1 if and only if mk belongs to a box whose sum of coordinatesis k − 1 (if k = 0, the monomial is just a corner and this case is trivial). Therefore, theconclusion is the following: for all k ≤ K − 1 we need to check if mk belongs to a boxwhose sum of coordinates is k − 1. If the answer is positive, I is good and otherwise it isbad.

Example 9.1. Let I = 〈x10, y10, z10, x2y2z8〉 ⊂ K[x, y, z]. Let m = x2y2z8. Then m5 ∈〈µ1, µ2, µ3〉5, but m4 6∈ 〈µ1, µ2, µ3〉4, thus K = 5. Further, m2 = x4y4z16 ∈ B0,0,1,m3 = x6y6z24 ∈ B0,0,2 and m4 = x8y8z32 ∈ B0,0,3. Therefore, I is a good ideal.

The situation does not change much if I has more generators.Let I = 〈µ1, . . . , µn,m1, . . . ,mt〉. Let Ki be the smallest positive integer such that

mKii ∈ 〈µ1, . . . , µn〉Ki , 1 ≤ i ≤ t. Then minimal generators of I l will be of the form

µk11 · · ·µknn mj11 · · ·m

jtt , where k1 + . . . + kn + j1 + . . . + jt = l and ji ≤ Ki − 1, 1 ≤ i ≤ t.

Note that the converse is not true: unlike in the case with only one additional generator,not every monomial of this form is a minimal generator. Therefore, we want to checkwhether each of them belongs to a box with the sum of coordinates at least l − 1. Asbefore, such a monomial belongs to a box whose sum of coordinates is at least l−1 if andonly if mj1

1 · · ·mjtt belongs to a box whose sum of coordinates is at least j1 + . . .+ jt−1 (if

j1 = . . . = jt = 0, the monomial is just a corner and this case is trivial). Therefore, it isenough to check whether all these monomials belong to boxes with the sums of coordinatesgreater or equal to the expected ones. If the answer is positive, the ideal is good, otherwiseit is bad. Finding Ki requires knowing the interaction of mi with µ1, . . . , µn. We couldhave defined Ki in a different way (and thus make them smaller), for instance,

Ki = min{K ∈ N | mKi ∈ 〈µ1, . . . , µn,m1, . . . ,mi−1, mi,mi+1, . . . ,mt〉K}

(mi means that mi is omitted), but this would require a bit more justification in general.

Example 9.2. Let I = 〈x5, y5, z5, x2y4z, x4y2z〉 ⊂ K[x, y, z]. Then K1 = 3 since m31 =

x6y12z3 is divisible by µ1µ22, but m2

1 = x4y8z2 6∈ 〈µ1, µ2, µ3〉2. The same holds for m2, soK2 = 3. Therefore, we would like to check all the monomials mj1

1 mj22 with 0 ≤ j1 ≤ j2 ≤ 2

(we might take j1 ≤ j1 due to the symmetry of the ideal). The case j1 = j2 = 0 is not tobe considered, as discussed earlier. If j1 = 0, j2 = 1, then m2 ∈ B0,0,0. If j1 = 0, j2 = 2,then m2

2 = x8y4z2 ∈ B1,0,0. Note, however, that this monomial is not a minimal generator

16

since it is strictly divisible by m1µ1 = x7y4z. If j1 = j2 = 1, then m1m2 = x6y6z2 ∈ B1,1,0.This is the only box where this monomial can be placed. In particular, this implies it isnot a minimal generator of I2. Indeed, x6y6z2 is strictly divisible by by µ1µ2. The lastcase to consider is j1 = 1, j2 = 2, but then m1m

22 can not be a minimal generator of I3

since m1m2 was not a minimal generator of I2, as we have already seen. Now, for anyl ≥ 1, minimal generators of I l (up to shifts by corners) can not be of types other thanthese. Therefore, I is a good ideal. Note that we could have chosen K1 and K2 in a lessnaive way, as discussed before this example: K1 = K2 = 2 since m2

1 is divisible by µ2m2

and m22 is divisible by µ1m1. Then the cases to consider would be (j1, j2) = (0, 1) and

(j1, j2) = (1, 1) (given the symmetry of the ideal).

Example 9.3. Let I = 〈x41, y41, z41, x40y5z5, x5y40z5, x5y5z40〉 ⊂ K[x, y, z]. It is not hardto notice that K1 = K2 = K3 = 9. Also we notice that m1m2, m2m3 and m1m3 are notminimal generators of I2. Thus the only monomials we need to check are powers of mi.We will check powers of m1, others are analogous. We see that m2

1 = x80y10z10 ∈ B1,0,0,m3

1 = x120y15z15 ∈ B2,0,0, m41 = x160y20z20 ∈ B3,0,0, m

51 = x200y25z25 ∈ B4,0,0, m

61 =

x240y30z30 ∈ B5,0,0, m71 = x280y35z35 ∈ B6,0,0 and m8

1 = x320y40z40 ∈ B7,0,0. Therefore, I isa good ideal. Note that m8

1 6∈ G(I8) since it is divisible by µ71 ·m2 and µ7

1 ·m3. Again, aless naive choice of Ki would result into K1 = K2 = K3 = 8 and we would not need toconsider the case (j1, j2, j3) = (8, 0, 0).

10 Powers of good ideals

A natural question to ask is whether powers of good ideals also good. The answer isclearly positive if n = 1, but in most of the other cases the answer is negative.

Let I be a good ideal. As before, for nonnegative integers a1, . . . , an, by Ba1,...,an

we denote the corresponding box, associated to I. But Ik will determine its own boxes:xkd11 , . . . , xkdnn are minimal generators of Ik, thus the new boxes will have sizes kd1, . . . , kdn.For nonnegative integers b1, . . . , bn we denote by Bk

b1,...,bnthe corresponding box, associated

to Ik, that is, Bkb1,...,bn

= ([b1kd1, (b1 + 1)kd1] × [b2kd2, (b2 + 1)kd2] × . . . × [bnkdn, (bn +1)kdn]) ∩ Nn.

Proposition 10.1. Let I be a good ideal and a1, . . . , an be nonnegative integers. Fork ≥ 1, let b1 = ba1

kc, . . . , bn = ban

kc. Then Ba1,...,an ⊆ Bk

b1,...,bn.

Proof. Clearly, bik ≤ ai and ai + 1 ≤ (bi + 1)k for all i ∈ {1, . . . , n}. Therefore, bikdi ≤aidi < (ai + 1)di ≤ (bi + 1)kdi, as desired.

Proposition 10.2. Let I ∈ K[x, y] be a good ideal. Then all powers of I are also good.

Proof. Let xd1 and yd2 be minimal generators of I, that is, d1 and d2 determine the sizeof the boxes associated to I. For k ≥ 1 consider (Ik)l = Ikl. Assume that m is a minimalgenerator of Ikl. Since I is a good ideal, we know that m ∈ Ba1,a2 for some a1 and a2such that a1 + a2 = kl − 1. From Proposition 10.1 we also know that m ∈ Bk

b1,b2, where

bi = baikc. We want to show that b1 + b2 = l − 1. On the one hand,

b1 + b2 =⌊a1k

⌋+⌊a2k

⌋≤⌊a1 + a2k

⌋=

⌊kl − 1

k

⌋= l − 1,

17

on the other hand,

b1 + b2 ≥a1 − k + 1

k+a2 − k + 1

k=a1 + a2 − 2k + 2

k=kl − 1− 2k + 2

k=

=k(l − 2) + 1

k> l − 2.

Since b1 + b2 is an integer, we get b1 + b2 = l − 1 which finishes the proof.

Remark 10.3. Another way to prove this proposition is to note that for n = 2 thenecessary and sufficient conditions for being a good ideal coincide: I is a good ideal ifand only if for any minimal generator xα1yα2 the following holds:

α1

d1+α2

d2≥ 1.

Since I is a good ideal, for each minimal generator of I the condition above holds. Minimalgenerators of Ik are products of k minimal generators of I, that is, if xβ1yβ2 is a minimalgenerator of Ik, we have β1

d1+ β2

d2≥ k, that is, β1

kd1+ β2

kd2≥ 1, which means that Ik is a

good ideal as well.

Let I be a good ideal, let m be some monomial. Note that among boxes containing mthere is always the largest box in the sense of partial order on coordinates of boxes: thereexists a box Bs1,...,sn containing m such that, if Bt1,...,tn also contains m, then si ≥ ti forall i.

Proposition 10.4. Let I, k, ai, bi be as in Proposition 10.1. Let m be a monomial. Thenthe following holds:

• if Ba1,...,an is the unique box containing m among boxes associated to I, then Bkb1,...,bn

is the unique box containing m among boxes associated to Ik;

• if Ba1,...,an is the largest box containing m among boxes associated to I, then Bkb1,...,bn

is the largest box containing m among boxes associated to Ik.

Proof.

• m is contained in a unique box, associated to I if and only if none of the coordinatesof m is divisible by the corresponding di. Then none of the coordinates of m isdivisible by the corresponding kdi or, equivalently, m is contained in a unique box,associated to Ik. The rest follows from Proposition 10.1.

• Let m = xα11 · · ·xαnn . We know that Ba1,...,an ⊆ Bk

b1,...,bn, but we assume that Bk

b1,...,bn

is not the largest one, that is, there is some Bkc1,...,cn

containing m with at leastone i such that ci > bi. Without loss of generality we assume that i = 1. Thenα1 ≥ c1kd1 ≥ (b1 + 1)kd1 = (b1k + k)d1 ≥ (a1 + 1)d1, that is, Ba1,...,an is not thelargest box containing m among boxes associated to I, which is a contradiction.

Henceforth, by old boxes we will mean boxes associated to I and by new boxes we willmean boxes associated to Ik. Clearly, a new box consists of kn old boxes.

18

Proposition 10.5. Let I ∈ K[x1, . . . , xn] be a good ideal and n ≥ 4. Then for all k ≥ 2,Ik is a bad ideal.

Proof. Consider Ik, k ≥ 2. We will show that the box decomposition principle failsalready in (Ik)2 = I2k. Let xd11 = µ1, . . . , x

dnn = µn be minimal generators of I. Then m =

µ1µ2µk−13 µk−14 ∈ I2k. The largest old box containing m is B1,1,k−1,k−1,0,...,0, therefore, by

Proposition 10.4, the largest new box containing m is Bkb 1kc,b 1

kc,b k−1

kc,b k−1

kc,0,0,...,0 = Bk

0,0,...,0

and 0 + 0 + . . .+ 0 < 1. Therefore, by Proposition 3.5, I is a bad ideal.

The case n = 3 is special in the following sense:

Proposition 10.6. Let I ∈ K[x, y, z] be a good ideal. Then the following holds:

• if I has exactly three generators, that is, I = 〈µ1, µ2, µ3〉, then I2 is a good ideal andall Ik, k ≥ 3, are bad;

• if I has more than three generators, then all Ik, k ≥ 2, are bad.

Proof.

• Assume that I = 〈µ1, µ2, µ3〉. Consider (I2)l = I2l. According to Proposition 3.5, itis enough to show that every m = µl11 µ

l22 µ

l33 , where l1+l2+l3 = 2l, can be placed into

a new box with the sum of coordinates greater or equal to l− 1. Since m ∈ Bl1,l2,l3 ,we conclude that m ∈ B2

b l12c,b l2

2c,b l3

2c. Note that at least one of li is even, say, l1 is

even. Then b l12c+ b l2

2c+ b l3

2c ≥ l1

2+ l2−1

2+ l3−1

2= l − 1. Therefore, I2 is a good

ideal.

Now we want to show that Ik, k ≥ 3, is a bad ideal. We will show that the box de-composition principle fails already in (Ik)2 = I2k. Consider m = µ2

1µk−12 µk−13 ∈ I2k.

Then the largest old box containing m is B2,k−1,k−1. Therefore, by Proposition 10.4,the largest new box containing m is Bk

b 2kc,b k−1

kc,b k−1

kc = Bk

0,0,0 and 0 + 0 + 0 < 1. By

Proposition 3.5, Ik is a bad ideal.

• Assume that I has at least 4 generators and let m be any minimal generator of Iwhich is not a corner. Consider Ik, k ≥ 2. We will show that the box decompositionprinciple fails already in (Ik)2 = I2k. Consider m1 = mµ1µ

k−12 µk−13 ∈ I2k. This

monomial belongs to B1,k−1,k−1 (note that this is the unique old box containing m1).By Proposition 10.4, the unique new box containing m1 is Bk

b 1kc,b k−1

kc,b k−1

kc = Bk

0,0,0

and 0 + 0 + 0 < 1. By Proposition 3.5, Ik is a bad ideal.

Remark 10.7. Let I = 〈µ1, . . . , µn〉. Then, summarizing the propositions above, weobtain that Ik is good if and only if k = 1 or n ≤ 2 or (n, k) = (3, 2). We will use thislater when we discuss the connection to Freiman ideals in Section 11.

Another natural question to ask is whether all powers of good ideals are Ratliff–Rush(even though powers of good ideals are bad in most cases). We will state a sufficientcondition that will help us to construct a family of examples of ideals whose all powersare Ratliff–Rush.

19

Lemma 10.8. Let I be a good ideal and let a1, . . . , an, a, t be nonnegative integers suchthat a1 + . . .+ an = a and t ≥ 1. Then

Ia+t : 〈µa11 · · ·µann 〉 =∑

t1,...,tn≥0t1+...+tn=t−1

µt11 · · ·µtnn It1+a1,...,tn+an .

In particular, if a = 0, the equality above is the box decomposition of I t and if t = 1, thisis just Proposition 4.4.

Proof. ⊇ Let m ∈ µt11 · · ·µtnn It1+a1,...,tn+an for some t1, . . . , tn. Then m = m1µt11 · · ·µtnn ,

where m1 ∈ It1+a1,...,tn+an , that is, m1µt1+a11 · · ·µtn+ann ∈ Ia+t. Therefore,

mµa11 · · ·µann = m1µt1+a11 · · ·µtn+ann ∈ Ia+t,

which implies m ∈ Ia+t : 〈µa11 · · ·µann 〉.⊆ Ia+t : 〈µa11 · · ·µann 〉 =

⟨m

gcd(m,µa11 ···µ

ann )| m ∈ G(Ia+t)

⟩. Fix m ∈ G(Ia+t). Then

m ∈ Bs1,...,sn , s1 + . . .+ sn = a+ t− 1, that is, m = xs1d1+α11 · · ·xsndn+αnn , 0 ≤ αi ≤ di. We

distinguish two cases considering the differences si − ai.

1. Assume that for all i we have si ≥ ai. Put ti := si − ai. Then t1 + . . . + tn = t− 1and

m

gcd(m,µa11 · · ·µann )=

m

µa11 · · ·µann= x

(s1−a1)d1+α1

1 · · ·x(sn−an)dn+αnn =

= xt1d1+α11 · · ·xtndn+αnn = µt11 · · ·µtnn · x

α11 · · ·xαnn ∈∈ µt11 · · ·µtnn It1+a1,...,tn+an ,

since xα11 · · ·xαnn ∈ Ia+t : 〈µs11 · · ·µsnn 〉 = Is1,...,sn = It1+a1,...,tn+an .

2. Assume that among differences si − ai we have r nonnegative and n − r negativeones, n − r ≥ 1. Without loss of generality, s1 − a1 ≥ 0, . . . , sr − ar ≥ 0, sr+1 −ar+1 < 0, . . . , sn − an < 0. Then m

gcd(m,µa11 ···µ

ann )

= x(s1−a1)d1+α1

1 · · · x(sr−ar)dr+αrr =

µs1−a11 · · ·µsr−arn · xα11 · · · xαrr . We know that (s1 − a1) + . . . + (sn − an) = a + t −

1 − a = t − 1, but since at least one difference is negative, we obtain (s1 − a1) +. . . + (sr − ar) ≥ t. At least one of these differences is positive, say, s1 − a1 ≥ 1.Therefore, µs1−a11 · · ·µsr−arr · xα1

1 · · ·xαrr is divisible by some µt1+11 µt22 · · ·µtrr , where

t1 + . . . + tr = t − 1. We put tr+1 = tr+2 = . . . = tn = 0. Then µt1+11 µt22 · · ·µtrr =

µ1 · µt11 µt22 · · ·µtrr µtr+1

r+1 · · ·µtnn ∈ µt11 · · ·µtnn Ia1+t1,...,an+tn , where t1 + . . . + tn = t − 1,since µ1 ∈ Ia1+t1,...,an+tn . This finishes the proof.

Recall that given a good ideal I, there exists a coloring as in Theorem 5.8, given by afinite disjoint union of cones. Let L be the number defined in the end of Section 5.

Proposition 10.9. Let I be a good ideal. Then Ik is Ratliff–Rush for all k ≥ L+ 1.

Proof. Let k ≥ L+ 1. We want to show that Ikl+k : Ikl = Ik for all l ≥ 0. It is clear thatIkl+k : Ikl ⊇ Ik. For the other inclusion note that Ikl+k : Ikl ⊆ Ikl+k : 〈µ1, . . . , µn〉kl. Thusit is sufficient to show that Ikl+k : 〈µ1, . . . , µn〉kl ⊆ Ik. We know that Ikl+k : 〈µ1, . . . , µn〉kl

20

equals the intersection of ideals of the type Ikl+k : 〈µk11 · · ·µknn 〉, where k1 + . . .+ kn = kl.From Lemma 10.8 we know that

Ikl+k : 〈µk11 · · ·µknn 〉 =∑

t1,...,tn≥0t1+...+tn=k−1

µt11 · · ·µtnn It1+k1,...,tn+kn .

Therefore,

Ikl+k : 〈µ1, . . . , µn〉kl =⋂

k1,...,kn≥0k1+...+kn=kl

∑t1,...,tn≥0

t1+...+tn=k−1

µt11 · · ·µtnn It1+k1,...,tn+kn

.

Let t1, . . . , tn, t′1, . . . , t

′n be nonnegative integers such that t1 + . . .+ tn = t′1 + . . .+ t′n =

k−1. Every minimal generator of µt11 · · ·µtnn It1+k1,...,tn+kn is of the form µt11 · · ·µtnn m, where

m ∈ G(It1+k1,...,tn+kn) and similarly every minimal generator of µt′11 · · ·µ

t′nn It′1+k′1,...,t′n+k′n is

of the form µt′11 · · ·µ

t′nn m′, where m′ ∈ G(It′1+k′1,...,t′n+k′n). Thus any minimal generator of

(µt11 · · ·µtnn It1+k1,...,tn+kn)⋂⋂

(µt′11 · · ·µ

t′nn It′1+k′1,...,t′n+k′n) is of the form lcm(µt11 · · ·µtnn m,µ

t′11 · · ·µ

t′nn m′), which is divisible

by

lcm(µt11 · · ·µtnn , µt′11 · · ·µt

′nn ) = µ

max(t1,t′1)1 · · ·µmax(tn,t′n)

n .

Note that if (t1, . . . , tn) 6= (t′1, . . . , t′n), then max(t1, t

′1) + . . .+ max(tn, t

′n) ≥ k and so

µt11 · · ·µtnn It1+k1,...,tn+kn⋂

µt′11 · · ·µt

′nn It′1+k′1,...,t′n+k′n ⊆ Ik.

Therefore, it remains to consider the sum of intersections which share a common(t1, . . . , tn), that is, we are left with

∑t1,...,tn≥0

t1+...+tn=k−1

⋂k1,...,kn≥0

k1+...+kn=kl

µt11 · · ·µtnn It1+k1,...,tn+kn

.

Fix t1, . . . , tn. Then (t1, . . . , tn) belongs to some cone C of our coloring. This cone cannot have dimension 0. Indeed, if we assume the opposite, then the vertex of this cone is(t1, . . . , tn) itself, but t1 + . . . + tn = k − 1 ≥ L and, given the way L was defined, theonly case when this could potentially be possible is t1 + . . .+ tn = L. But, as mentionedin Remark 5.10, L can not be attained at a zero dimensional cone, therefore, we get acontradiction. Thus the dimension of C is at least 1. Then at least one coordinate of thevertex of C, say, the first one, is not underlined. Then (t1 + kl, t2, . . . , tn) ∈ C as well as(t1, . . . , tn) and thus It1+kl,t2,...,tn = It1,...,tn . Then taking k1 = kl, k2 = . . . = kn = 0 weobtain⋂

k1,...,kn≥0k1+...+kn=kl

µt11 · · ·µtnn It1+k1,...,tn+kn ⊆

⊆ µt11 · · ·µtnn It1+kl,t2,...,tn = µt11 · · ·µtnn It1,...,tn .

21

Summing over all (t1, . . . , tn), we obtain

∑t1,...,tn≥0

t1+...+tn=k−1

⋂k1,...,kn≥0

k1+...+kn=kl

µt11 · · ·µtnn It1+k1,...,tn+kn

⊆⊆

∑t1,...,tn≥0

t1+...+tn=k−1

µt11 · · ·µtnn It1,...,tn = Ik,

which finishes the proof.

Example 10.10. Let I be an ideal of K[x, y] generated by all the possible monomialsof degree 2d except xdyd, d ≥ 2. Then I is not Ratliff–Rush, but all higher powers are.Indeed, I is a good ideal by the sufficient condition and Ia1,a2 = I ′ for all (a1, a2) 6= (0, 0),where I ′ = I + 〈xdyd〉. Then we can choose the following coloring: {C1,0, C0,1, C0,0}.Therefore, L = 1 and thus all powers of I except I itself are Ratliff–Rush. I is notRatliff–Rush since I = I ′.

So now we know that if we have a good ideal and its coloring with the maximal sumof coordiates of vertices of cones equal to L, then all Ik with k ≥ L+ 1 are Ratliff–Rush.It is natural to mention ideals for which L = 0, or, in other words, ideals for which thereis a coloring consisting of a single cone C0,0,...,0.

Definition 10.11. Let I be a good ideal such that for all (a1, . . . , an) we have Ia1,...,an = I.Then we call I a very good ideal.

Clearly, any ideal in K[x] is a very good one. If I is a very good ideal, then fromProposition 10.9 all its powers are Ratliff–Rush. It is also clear that if I is a very goodideal, then in particular I2 = IJ , where J = 〈µ1, . . . , µn〉. We will now show that theseconditions are in fact equivalent.

Proposition 10.12. Let I be an m-primary monomial ideal in K[x1, . . . , xn]. Then I isvery good if and only if I2 = IJ , where J = 〈µ1, . . . , µn〉.

Proof. As mentioned before, one implication is trivial. For the other implication, assumethat I2 = IJ , where J = 〈µ1, . . . , µn〉. Then for any k ≥ 1 we have

Ik = Jk−1I =∑

k1+...+kn=k−1

µk11 · · ·µknn I.

Thus the box decomposition principle holds and moreover Ia1,...,an = I for all (a1, . . . , an).

Example 10.13. Let I be an m-primary monomial ideal in K[x1, . . . , xn], n ≥ 2 such thatthe following holds: there is a pair of indices i 6= j such that for each minimal generatorof I, except the corners µ1, . . . , µn, its xi-exponent is greater or equal to di/2 and itsxj-exponent is greater or equal to dj/2. Then products of non-corners are not needed asminimal generators of I2, thus

I2 =∑

k1+...+kn=1

µk11 · · ·µknn I = IJ,

where, as before, J = 〈µ1, . . . , µn〉. Therefore, I is a very good ideal and thus all itspowers are Ratliff–Rush. This example generalizes the example constructed in [2].

22

Proposition 10.14. Let I = 〈µ1, . . . , µn〉. Then Ik is good if and only if it is very good.

Proof. One way to prove this statement is to consider all the cases for which Ik is good(Remark 10.7) and directly check that I2k = IkJ in each case. Note that here J =〈µk1, . . . , µkn〉.

Another way is the following. Since generators of Ik are monomials in variablesµ1, . . . , µn, we can, without loss of generality, assume µ1 = x1, . . . , µn = xn, even if Iis not equigenerated. Let Ik be a good ideal and let Bk

a1,...,andenote the boxes, associated

to Ik. It is enough to show that Ik1,0,...,0 = Ik, all other Ika1,...,an with a1 + . . . + an = 1are analogous (by Ika1,...,an we mean (Ik)a1,...,an and not (Ia1,...,an)k). Ik is minimally gen-erated by all monomials of degree k and I2k is minimally generated by all monomials ofdegree 2k. By definition, Ik1,0,...,0 is generated by monomials in Bk

1,0...,0 ∩ G(I2k), dividedby µk1 = xk1. In other words, we take all monomials of degree 2k, which are divisible byxk1, and divide them by xk1. Clearly, we obtain all monomials of degree k (and only them).Therefore, Ik1,0,...,0 = Ik and thus Ik is a very good ideal.

We remark that this proposition means the following: for Ik = 〈µ1, . . . , µn〉k there isnothing ”extra” in boxes Bk

1,0...,0, . . . , Bk0,...,0,1 besides the translations of Ik and the only

reason Ik might fail to be a very good ideal is that it fails to be a good ideal, that is,there exists a monomial in G(I2k) which only belongs to Bk

0,0,...,0. Therefore, ideals of suchtype are either good (and thus very good), or the box decomposition principle should failalready in (Ik)2 = I2k. If we look back at the proofs of Proposition 10.2, Proposition 10.5and Proposition 10.6, we see that it is indeed the case.

11 Connection to Freiman ideals

Let I be an equigenerated monomial ideal with analytic spread l(I). It has been shownin [8] (Theorem 1.9) that |G(I2)| ≥ l(I)|G(I)| −

(l(I)2

). Note that this bound is no longer

valid if I is not equigenerated. Indeed, for each m ≥ 6 there exists a monomial ideal intwo variables such that |G(I)| = m and |G(I2)| = 9, see [4].

Definition 11.1. An equigenerated monomial ideal is called Freiman if

|G(I2)| = l(I)|G(I)| −(l(I)

2

).

The next proposition is exactly Theorem 3.1 in [9], but we give an alternative proof.

Theorem 11.2. Let I ⊂ K[x1, . . . , xn] be an m-primary monomial ideal, equigeneratedin degree d. Then I is Freiman if and only if I satisfies the equivalent conditions fromProposition 10.12. In other words, for an m-primary d-equigenerated monomial ideal Ithe following are equivalent:

(1) I is Freiman;

(2) I is very good;

(3) I2 = IJ ,

where, as before, J = 〈µ1, . . . , µn〉 and µi = xdi .

23

Proof. Since I is m-primary, we have l(I) = n. The ideal I is equigenerated in degreed, thus I2 is equigenerated in degree 2d. Note that any product of two elements of G(I)gives us an element of G(I2). Indeed, if such a product is not in G(I2), then there is anelement in G(I2) strictly dividing it. But this is impossible since these monomials areboth of degree 2d. Note that different products of two elements from G(I) might of coursegive us the same monomial from G(I2). The idea is to list all the different elements fromG(I2) and count them. Clearly, µ1G(I) := {µ1m | m ∈ G(I)} is a set of |G(I)| differentelements of G(I2). We can similarly define sets µ2G(I), . . . , µnG(I). Monomials insideeach such set are different. Every pair of different sets, say µiG(I) and µjG(I), has aunique monomial in common, which is µiµj. Every triple of different sets has an empty

intersection. Thus in all these sets we have n|G(I)| −(n2

)= l(I)|G(I)| −

(l(I)2

)different

elements from G(I2). Now it is clear that the Freiman equality holds if and only if thereare no other minimal generators of I2 except those in µ1G(I) ∪ . . . ∪ µnG(I) if and onlyif I2 = 〈µ1〉I + . . .+ 〈µn〉I = IJ .

Remark 11.3. Note that if I is a very good, but not an equigenerated ideal, the Freimanequality |G(I2)| = l(I)|G(I)| −

(l(I)2

)still holds, but it is not of any particular interest

and thus such ideals are not called Freiman. If I satisfies the Freiman equality, but isnot an equigenerated ideal, it is not necessarily very good and not even necessarily good.Consider, for instance, I = 〈x3, y3, xy〉.

Remark 11.4. From Theorem 11.2 we know that an equigenerated m-primary monomialideal is Freiman if and only if it is very good. From Proposition 10.14 we also knowthat Ik = 〈µ1, . . . , µn〉k, µi = xdii , is very good if and only if it is good. Finally, fromRemark 10.7 we know that Ik = 〈µ1, . . . , µn〉k, µi = xdii , is good if and only if k = 1,n ≤ 2, or (n, k) = (3, 2). Therefore, we conclude the following: Ik = 〈µ1, . . . , µn〉k isFreiman (note that here µi = xdi since we want the ideal to be equigenerated) if and onlyif k = 1, n ≤ 2, or (n, k) = (3, 2). This can be seen as an alternative proof of Theorem 2.3in [9].

References

[1] I. Al-Ayyoub, An algorithm for computing the Ratliff–Rush closure, Journal ofAlgebra and its Applications, 8 (2009), pp. 521–532.

[2] I. Al-Ayyoub and O. Solaiman, Infinite families of Ratliff–Rush ideals, Interna-tional Journal of Algebra, 6 (2012), pp. 815–824.

[3] W. Decker, G.-M. Greuel, G. Pfister, and H. Schonemann, Singu-lar 4-1-2 — A computer algebra system for polynomial computations. http:

//www.singular.uni-kl.de, 2019.

[4] S. Eliahou, J. Herzog, and M. M. Saem, Monomial ideals with tiny squares,Journal of Algebra, 514 (2018), pp. 99–112.

[5] J. Elias, On the computation of the Ratliff–Rush closure, Journal of Symbolic Com-putation, 37 (2004), pp. 717–725.

24

[6] W. Heinzer, B. Johnston, D. Lantz, and K. Shah, Coefficient ideals inand blowups of a commutative Noetherian domain, Journal of Algebra, 162 (1993),pp. 355–391.

[7] W. Heinzer, D. Lantz, and K. Shah, The Ratliff–Rush ideals in a Noetherianring, Communications in algebra, 20 (1992), pp. 591–622.

[8] J. Herzog, M. M. Saem, and N. Zamani, On the number of generators of powersof an ideal, arXiv preprint arXiv:1707.07302, (2017).

[9] J. Herzog and G. Zhu, Freiman ideals, Communications in Algebra, (2019), pp. 1–17.

[10] R. Kallstrom, Liftable derivations for generically separably algebraic morphismsof schemes, Transactions of the American Mathematical Society, 361 (2009), pp. 495–523.

[11] T. J. Puthenpurakal, Bockstein cohomology of associated graded rings, ActaMathematica Vietnamica, (2013), pp. 1–22.

[12] V. C. Quinonez, Ratliff–Rush monomial ideals, algebraic and geometric combina-torics, Contemp. Math., 423 (2007), pp. 43–50.

[13] L. J. Ratliff and D. E. Rush, Two notes on reductions of ideals, Indiana Univ.Math. J, 27 (1978), pp. 929–934.

[14] M. E. Rossi and I. Swanson, Notes on the behavior of the Ratliff–Rush filtration,Comm. Algebra (Grenoble/Lyon, 2001), Contemp. Math., 331 (2003), pp. 313–328.

[15] M. E. Rossi, D. T. Trung, and N. V. Trung, Castelnuovo–Mumford regularityand Ratliff–Rush closure, Journal of Algebra, 504 (2018), pp. 568–586.

[16] D. T. Trung, On the computation of Castelnuovo–Mumford regularity of the Reesalgebra and of the fiber ring, arXiv preprint arXiv:1904.07484v1, (2019).

25

Monomial ideals with arbitrarily high tiny powers inany number of variables

Oleksandra Gasanova

Abstract

Powers of (monomial) ideals is a subject that still calls attraction in various ways.Let I ⊂ K[x1, . . . , xn] be a monomial ideal and let G(I) denote the (unique) minimalmonomial generating set of I. How small can |G(Ii)| be in terms of |G(I)|? We ex-pect that the inequality |G(I2)| > |G(I)| should hold and that |G(Ii)|, i ≥ 2, growsfurther whenever |G(I)| ≥ 2. In this paper we will disprove this expectation andshow that for any n and d there is an m-primary monomial ideal I ⊂ K[x1, . . . , xn]such that |G(I)| > |G(Ii)| for all i ≤ d.

1 Introduction

Let I ⊂ K[x1, . . . , xn] be a monomial ideal and let G(I) denote its minimal monomialgenerating set. It is known (see for example [2]) that the function f(i) = |G(I i)|, forlarge i, is a polynomial in i of degree l(I)−1 with a positive leading coefficient. Here l(I)denotes the analytic spread of I, that is, the Krull dimension of the fiber ring F (I) ofI. In particular, for all i large enough we have |G(I i+1)| > |G(I i)| unless I is a principalideal.

But what kind of pathologies can occur if i is small? How small can |G(I i)| be interms of |G(I)|? This question has been explored in [1] and [3]. We intuitively expectthat the inequality |G(I2)| > |G(I)| should hold and that |G(I i)|, i ≥ 2, grows furtherwhenever |G(I)| ≥ 2. This expectation has been disproven in [1]: the authors constructa family of ideals in K[x, y] for which |G(I)| > |G(I2)|.

In Section 2 we will generalize the above result and show that for any n ≥ 2 and d ≥ 2there is an m-primary ideal I ⊂ K[x1, . . . , xn] such that |G(I)| > |G(I i)| for all i ≤ d.This section contains an explicit construction and several examples.

In Section 3 we will discuss Theorem 3.1 of [1]. This theorem says that if a monomialideal I = 〈u1, . . . , um〉 ⊂ K[x, y] satisfies certain conditions, then |G(I2)| = 9. We willrelax the conditions of this theorem and give a more intuitive proof.

2 Ideals with arbitrarily high tiny powers in any num-

ber of variables

Let n and d be positive integers with n, d ≥ 2. We will construct an m-primary monomialideal I ⊂ K[x1, . . . , xn] such that |G(I)| > |G(I2)|, |G(I)| > |G(I3)|, . . . , |G(I)| > |G(Id)|.We will briefly describe the idea, after which we will give all the necessary proofs.

1

Let µ := xt1 · · ·xtn, where t is yet to be determined. We start with the ideal J =〈x4t1 , x4t2 , . . . , x4tn , x2t1 µ, x2t2 µ, . . . , x2tn µ〉. Note that the number of generators of J i only de-pends on i and n and not on t. Let A(n, d) := max

i∈{1,...,d}|G(J i)|. For fixed n and d it is a

constant which can be found using computer algebra (for simplicity, one may set t = 1here).

So far we only have 2n generators in J . Our goal is to find at least A(n, d) − 2n + 1monomials q1, . . . , qs such that the set G(J)∪{q1, . . . , qs} contains no monomials dividingeach other and (J + 〈q1, . . . , qs〉)i = J i for any i ≥ 2. In other words, we would like toadd generators to J without changing any higher powers of J . The resulting ideal will bedenoted by I, and J will be called its skeleton. Clearly, the more monomials we can addthe better it is and this is where t comes into play.

Lemma 2.1. Let J = 〈x4t1 , x4t2 , . . . , x4tn , x2t1 µ, x2t2 µ, . . . , x2tn µ〉 with µ = xt1 · · ·xtn, as above.Let q := µ2 = x2t1 x

2t2 · · ·x2tn and let Q := 〈q〉. Then JQ ⊆ J2 and Q2 ⊆ J2.

Proof. In order to prove that JQ ⊆ J2, it is enough to show that x4t1 q ∈ J2 and x2t1 µq ∈J2. Indeed, x4t1 q = x4t1 µ

2 = (x2t1 µ)2 and x2t1 µq = x5t1 x3t2 · · ·x3tn is divisible by x4t1 x

2t2 µ =

x5t1 x3t2 x

t3 · · ·xtn.

In order to show that Q2 ⊆ J2, it is enough to show that q2 ∈ J2, which is true sinceq2 = x4t1 · · ·x4tn is divisible by x4t1 x

4t2 .

Corollary 2.2. Let J and Q be as above and let Q′ ⊆ Q. Then for any i ≥ 2 we have(J +Q′)i = J i.

Proof. (J +Q′)i ⊆ (J +Q)i = J i + J i−1Q+ . . .+ JQi−1 +Qi.

If j is even, then J i−jQj = J i−j(Q2)j2 ⊆ J i−j(J2)

j2 = J i.

If j is odd, then J i−jQj = J i−j(Q2)j−12 Q ⊆ J i−j(J2)

j−12 Q = J i−1Q = J i−2(JQ) ⊆

J i−2J2 = J i.Therefore, (J + Q′)i ⊆ J i for any i ≥ 2. The other inclusion is trivial, which finishes

the proof.

Now we know that monomials from Q are those that could potentially add moregenerators to J , but do not change the higher powers of J . We want to choose somemonomials q1, . . . , qs ∈ Q such that the set G(J) ∪ {q1, . . . , qs} contains no monomialsdividing each other. In other words, we need to find monomials q1, . . . , qs ∈ Q that satisfythe following three conditions:

1. no monomial from {q1, . . . , qs} is divisible by any monomial in G(J);

2. no monomial from G(J) is divisible by any monomial in {q1, . . . , qs};

3. monomials in {q1, . . . , qs} do not divide each other.

An obvious way to have the first condition satisfied is to consider only monomials fromQ\J . This set has a nice description.

Lemma 2.3. Let J and Q be as above. Then

Q\J = {xα11 · · · xαn

n : (α1, . . . , αn) ∈ [2t, 3t− 1]n ∩ Nn}

2

Proof. ⊇: All monomials within the given hypercube are divisible by q and none of themis divisible by any of the minimal generators of J since every minimal generator of J hasan exponent greater than or equal to 3t. This is the only inclusion we will use in thefuture construction, but we can show the other inclusion as well.⊆: Let xα1

1 · · ·xαnn ∈ Q\J . Then αi ≥ 2t for all i. But xα1

1 · · ·xαnn 6∈ J , that is, in

particular, it is not divisible by x2t1 µ = x3t1 xt2 · · ·xtn. This implies α1 ≤ 3t−1. Analogously,

αi ≤ 3t− 1 for all i.

00

t

t

2t

2t

3t

3t

4t

4t

Figure 1: monomials in Q\J , n = 2.

Now we know that any subset of monomials from [2t, 3t−1]n satisfies the first condition.It is also quite obvious that any subset of monomials from [2t, 3t−1]n satisfies the secondcondition. The only thing to be taken care of is that the chosen monomials from [2t, 3t−1]n

do not divide each other. The most natural way to do so is to choose monomials of thesame degree. To get as many of them as possible, we should choose monomials on acentral integer cross-section of this hypercube, that is, monomials of the form{

xα11 · · ·xαn

n : (α1, . . . , αn) ∈ [2t, 3t− 1]n ∩ Nn, α1 + . . .+ αn =

⌊n(5t− 1)

2

⌋}.

Note that if n(t − 1) is even, we have a unique central integer cross-section, otherwisethere are two of them giving the same number of integer points; the other central integercross-section can be obtained by replacing b.c by d.e in the expression above. The numberof integer points on every central integer cross-section of [2t, 3t − 1]n equals the numberof integer points on every central integer cross-section of [0, t− 1]n and equals the central(and largest) coefficient(s) in the expansion of (1 + x + . . . + xt−1)n. For a fixed n thenumber of these monomials only depends on t and can be made arbitrarily large for tlarge enough.

To summarize all of the above, we need to perform the following steps:

1. Fix n and d.

3

2. Let J = 〈x4t1 , x4t2 , . . . , x4tn , x2t1 µ, x2t2 µ, . . . , x2tn µ〉 with µ = xt1 · · ·xtn, as above. Computethe number of generators in J, J2, J3, . . . , Jd using computer algebra. These numbersare independent of t, so we may set t = 1 in this step. Take the maximum of thesenumbers and call it A(n, d).

3. We already have 2n generators in J ; we would like to have at least A(n, d) + 1, thatis, we need at least A(n, d)− 2n + 1 more. There exists t such that the number ofinteger points on each central integer cross-section of [2t, 3t− 1]n is greater or equalto A(n, d) − 2n + 1. Choose any such t and add all the appropriate monomials toour skeleton J . This is our ideal I.

Let us discuss a few examples demonstrating the algorithm above.

Example 2.4. Let us first consider an easy case with a small number of variables.

1. We will fix n = 2 and d = 6.

2. Let J = 〈x4t, y4t, x3tyt, xty3t〉. In this step we may set t = 1. Computing the powersof this ideal up to the sixth, we obtain: |G(J2)| = 9, |G(J3)| = 13, |G(J4)| = 17,|G(J5)| = 21, |G(J6)| = 25. Thus A(2, 6) = 25.

3. We have 4 generators, but we would like to have at least 26. That is, we need to addat least 22 more. A square of the form [2t, 3t−1]2 has t integer points on the diagonal,thus we can choose t = 22. Therefore, our skeleton is J = 〈x88, y88, x66y22, x22y66〉.The monomials we want to add are on the diagonal of [2t, 3t− 1]2 = [44, 65]2, thatis, x65y44, x64y45, . . . , x45y64, x44y65.

Therefore,

I = 〈x88, x66y22, x65y44, x64y45, x63y46, x62y47, x61y48, x60y49, x59y50,x58y51, x57y52, x56y53, x55y54, x54y55, x53y56, x52y57, x51y58,

x50y59, x49y60, x48y61, x47y62, x46y63, x45y64, x44y65, x22y66, y88〉.

We see that |G(I)| = 26, |G(I2)| = 9, |G(I3)| = 13, |G(I4)| = 17, |G(I5)| = 21, |G(I6)| =25, as desired.

Example 2.5. Here is another example.

1. Fix n = 3 and d = 3.

2. Let J = 〈x4t, y4t, z4t, x3tytzt, xty3tzt, xtytz3t〉. In this step we may set t = 1.Computing the powers of this ideal up to the third, we obtain: |G(J2)| = 18,|G(J3)| = 34. Thus A(3, 3) = 34.

3. We have 6 generators, but we would like to have at least 35. That is, we need toadd at least 29 more. As we already know, the number of integer points on everycentral integer cross-section of [2t, 3t − 1]3 equals the number of integer points onevery central integer cross-section of [0, t− 1]3 and equals the central (and largest)coefficient(s) in the expansion of (1+x+ . . .+xt−1)3. An explicit computation showsthat t = 7 is the smallest suitable integer and we can add 37 monomials. Thus ourskeleton is J = 〈x28, y28, z28, x21y7z7, x7y21z7, x7y7z21〉. The monomials we want toadd are those on the (unique) central integer cross-section of [2t, 3t−1]3 = [14, 20]3.

4

Up to a shift, it is the same as the central integer cross-section of [0, 6]3. The pointswe are looking for satisfy α1 + α2 + α3 = 9, αi ≤ 6. All the possible triples are:(6, 3, 0) - 6 points considering all the permutations, (6, 2, 1) - 6 points, (5, 4, 0) - 6points, (5, 3, 1) - 6 points, (5, 2, 2) - 3 points, (4, 4, 1) - 3 points, (4, 3, 2) - 6 points,(3, 3, 3) - 1 point. This gives us 37 extra monomials, as desired. Shifting them backand adding to our ideal (they are written in the same order as above) gives us

I = 〈x28, y28, z28, x21y7z7, x7y21z7, x7y7z21,x20y17z14, x20y14z17, x17y20z14, x17y14z20, x14y20z17, x14y17z20,

x20y16z15, x20y15z16, x16y20z15, x16y15z20, x15y20z16, x15y16z20,

x19y18z14, x19y14z18, x18y19z14, x18y14z19, x14y19z18, x14y18z19,

x19y17z15, x19y15z17, x17y19z15, x17y15z19, x15y19z17, x15y17z19,

x19y16z16, x16y19z16, x16y16z19, x18y18z15, x18y15z18, x15y18z18,

x18y17z16, x18y16z17, x17y18z16, x17y16z18, x16y18z17, x16y17z18,

x17y17z17〉.

We see that |G(I)| = 43, |G(I2)| = 18, |G(I3)| = 34, as desired.

3 Improved conditions for tiny squares

Let I ⊂ K[x, y] be a monomial ideal and let G(I) denote the minimal monomial generatingset of I. Then G(I) = {u1, . . . , um}, where ui = xaiybi for all i and where the exponentsai, bi ∈ N satisfy a1 > a2 > . . . > am and b1 < b2 < . . . < bm.

Let V := {(i, j) ∈ N2 : 1 ≤ i ≤ j ≤ m} and consider the map

f : V → I2

(i, j) 7→ uiuj.

Then f(V ) generates I2, but it is not a minimal generating set in general.

Example 3.1. Let I = 〈x4, x3y2, y3〉. Then G(I2) = {x8, x7y2, x4y3, x3y5, y6}. Note thatu22 6∈ G(I2) since u1u3|u22. The picture below represents V ; (i, j) is marked with a starif f(i, j) ∈ G(I2) and (i, j) is marked with a usual dot if f(i, j) 6∈ G(I2) and the arrowsshow which monomials divide which. If several monomials are equal, we can choose onethat will be marked with a star and mark the others with dots.

1

1 ∗

∗

2

2 ∗

∗

3

3 ∗ ∗

Figure 2: generators of I2.

5

Theorem 3.2. (Theorem 3.1 from [1]) Let m ≥ 5 and let I = 〈u1, . . . , um〉 be an idealwith ui = xaiybi for all i, where a1 > . . . > am and b1 < . . . < bm. Assume that thefollowing divisibility conditions hold:

u1um|u2um−1 (2)

u1um−1|u2u3 (3.1)

u1um−1|u2m−2 (3.2)

u22|u1u3 (4.1)

u22|u1um−2 (4.2)

u2um|u3um−1 (5.1)

u2um|um−2um−1 (5.2)

u2m−1|u3um (6.1)

u2m−1|um−2um. (6.2)

Then G(I2) = {u21, u1u2, u22} ∪ {u1um−1, u1um, u2um} ∪ {u2m−1, um−1um, u2m}.

Let us look closer at the conditions above.If we multiply conditions (2) and (4.2), we will get u1um · u22|u2um−1 · u1um−2, that is,

u2um|um−2um−1, which is exactly condition (5.2).If we multiply conditions (2) and (4.1), we will get u1um · u22|u2um−1 · u1u3, that is,

u2um|u3um−1, which is exactly condition (5.1).If we multiply conditions (2) and (6.1), we will get u1um · u2m−1|u2um−1 · u3um, that is,

u1um−1|u2u3, which is exactly condition (3.1).In other words, conditions (5.1), (5.2) and (3.1) follow from the other conditions and

are redundant. We are now left with conditions (2), (3.2), (4.1), (4.2), (6.1) and (6.2).This set of conditions has a nice property of being ”almost self-dual” in the following sense.Recall that V = {(i, j) ∈ N2 : 1 ≤ i ≤ j ≤ m}. This set of points is symmetric withrespect to the line i+j = m+1. If (i, j) ∈ V , then (m+1−j,m+1−i) ∈ V is obtained byreflecting (i, j) about the line i + j = m + 1. If uiuj ∈ f(V ), then um+1−jum+1−i ∈ f(V )will be called its dual. If i + j = m + 1, the corresponding monomial will be calledself-dual.

Consider condition (2). If we dualize all monomials in this condition, we will get itback.

Consider condition (4.1). If we dualize all monomials in this condition, we will getcondition (6.2).

Consider condition (4.2). If we dualize all monomials in this condition, we will getcondition (6.1).

The only condition that has no dual is condition (3.2). Intuitively, we would like touse a self-dual set of conditions, that is, a set of conditions such that if we dualize everycondition, we get the same set of conditions. Our set of conditions is not self-dual. Atthe first glance, it can be resolved in several ways:

1. It could be the case that condition (3.2) follows from other conditions. However,this is not the case, as Remark 3.6 shows.

2. We can add the dual of condition (3.2) to our set. However, this seems unnatural,given that the theorem holds without adding any other conditions.

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3. We can remove condition (3.2) and try to prove the theorem without it. This isexactly what we will do in Theorem 3.4. We will use the following relabelling ofconditions: (2)→ (A), (4.1)→ (B), (6.2)→ (B*), (4.2)→ (C), (6.1)→ (C*). Butbefore proving Theorem 3.4, we need a preliminary lemma.

Lemma 3.3. (Lemma 2.3 from [1]) Let v, v1, v2 ∈ V . Assume that v1 ≤ v2 and that f(v)divides both f(v1), f(v2). Then f(v) divides f(v′) for all v′ ∈ V such that v1 ≤ v′ ≤ v2.

Theorem 3.4. (Improved conditions for tiny squares) Let m ≥ 5 and let I = 〈u1, . . . , um〉be an ideal with ui = xaiybi for all i, where a1 > . . . > am and b1 < . . . < bm. Assumethat the following divisibility conditions hold:

u1um|u2um−1 (A)

u22|u1u3 (B)

u2m−1|um−2um (B*)

u22|u1um−2 (C)

u2m−1|u3um. (C*)

Then G(I2) = {u21, u1u2, u22} ∪ {u1um−1, u1um, u2um} ∪ {u2m−1, um−1um, u2m}.

Proof. In order to see that these monomials generate I2, it is enough to show that eachmonomial in f(V ) is divisible by one of these nine monomials. We distinguish severalcases. Figure 3 shows which monomials are covered by which cases.

1

1

2

2

3

3

mm−1m−2· · ·

...

m−2

m−1

m

∗

∗ ∗

∗

∗ ∗

∗

∗ ∗

Figure 3: illustration of the proof of Theorem 3.4.

Case 0 (self-dual): i = 2, j = m− 1. We are done by condition (A).Case 1: (1, 3) ≤ (1, j) ≤ (1,m−2). Conditions (B) and (C), together with Lemma 3.3,

imply u22|u1uj for all 3 ≤ j ≤ m− 2.Case 1* (dual to Case 1): (3,m) ≤ (i,m) ≤ (m− 2,m). By the dual argument (that

is, using conditions (B*) and (C*)) and Lemma 3.3 we conclude that u2m−1|uium for all3 ≤ i ≤ m− 2.

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Case 2: (2, 3) ≤ (2, j) ≤ (2,m− 2). If we multiply conditions (A) and (B*), we obtain

u1um · u2m−1|u2um−1 · um−2um ⇔ u1um−1|u2um−2.

If we multiply conditions (A) and (C*), we obtain

u1um · u2m−1|u2um−1 · u3um ⇔ u1um−1|u2u3.

Combining these two divisibility conditions with Lemma 3.3, we conclude that u1um−1|u2ujfor all 3 ≤ j ≤ m− 2.

Case 2*(dual to case 2): (3,m−1) ≤ (i,m−1) ≤ (m−2,m−1). We apply argumentsdual to those from Case 2, that is, we multiply conditions (A) and (B) and we multiplyconditions (A) and (C). Combining these two divisibility conditions with Lemma 3.3, weconclude that u2um|uium−1 for all 3 ≤ i ≤ m− 2.

Case 3 (self-dual): (3, 3) ≤ (i, j) ≤ (m− 2,m− 2). If we multiply conditions (A), (B)and (C*), we obtain

u1um · u22 · u2m−1|u2um−1 · u1u3 · u3um ⇔ u2um−1|u23.

Dually, if we multiply conditions (A), (B*) and (C), we obtain u2um−1|u2m−2. Combiningthese two divisibility conditions with Lemma 3.3, we conclude that u2um−1|uiuj for all(3, 3) ≤ (i, j) ≤ (m− 2,m− 2).

So far we know that |G(I2)| ≤ 9. The proof of |G(I2)| ≥ 9 can be found in [1]. Inany case, we are not so interested in proving that |G(I2)| ≥ 9, the essential point here is|G(I2)| ≤ 9.

Example 3.5. (a three-parameter family of ideals with tiny squares) Let l, k, t be positiveintegers such that k ≥ 4t. Let I = 〈u1, . . . , utl+4〉, where ui = xaiybi with (a1, . . . atl+4) =(btl+4, . . . , b1) = (kl, (k − t)l, (k − t)l − 1, . . . , (k − 2t)l, tl, 0). Clearly, kl > (k − t)l >(k − t)l − 1 and (k − 2t)l > tl > 0 under the condition that k ≥ 4t. Also, (k − t)l − 1 ≥(k − 2t)l⇔ tl ≥ 1, that is, it is indeed a decreasing sequence (utl+2 is u3 if t = l = 1, butthere is no problem). We check the conditions for tiny squares:

Condition (A) holds trivially.Condition (B): u22|u1u3 ⇔ x2(k−t)ly2tl|xkl · x(k−t)l−1y(k−2t)l ⇔ 2(k − t)l ≤ (2k − t)l −

1 and 2tl ≤ (k − 2t)l⇔ tl ≥ 1 and k ≥ 4t.Condition (B*) holds by the symmetry of our ideal.Condition (C): u22|u1um−2 ⇔ x2(k−t)ly2tl|xkl · x(k−2t)ly(k−t)l−1 ⇔ 2(k − t)l ≤ 2(k −

t)l and 2tl ≤ (k − t)l − 1⇔ (k − 3t)l ≥ 1.Condition (C*) holds by the symmetry of our ideal.

Remark 3.6. Consider Example 3.5 again. Put l = 1 and k = 4t, then I = 〈u1, . . . , ut+4〉,where ui = xaiybi with (a1, . . . at+4) = (bt+4, . . . , b1) = (4t, 3t, 3t − 1, ..., 2t, t, 0). Moreexplicitly,

I = 〈x4t, x3tyt, x3t−1y2t, . . . , x2ty3t−1, xty3t, y4t〉.

First of all note that this ideal does not satisfy condition (3.2) which requires u1um−1|u2m−2.Also note that this is exactly the ideal obtained using the construction, described inSection 2 for n = 2: the first two and the last two monomials generate the skeleton J ofI. All other monomials are those on the central integer cross-section of the hypercube[2t, 3t− 1]2, which is simply a diagonal in this case. If we put t = 22, we will recover theideal from Example 2.4.

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References

[1] S. Eliahou, J. Herzog, and M. M. Saem, Monomial ideals with tiny squares,Journal of Algebra, 514 (2018), pp. 99–112.

[2] J. Herzog and T. Hibi, Monomial ideals, in Graduate Texts in Mathematics,vol. 260, Springer, 2011.

[3] J. Herzog, M. M. Saem, and N. Zamani, The number of generators of the powersof an ideal, International Journal of Algebra and Computation, 29 (2019), pp. 827–847.

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