Monomial Ideals: Course Notes

Monomial Ideals: Course Notes Mark Rogers Sean Sather-Wagstaff Mathematics Department, Missouri State University, Cheek Hall Room 45M, 901 South National Avenue, Springfield, MO 65897, USA E-mail address: [email protected] URL: http://math.missouristate.edu/43628.htm Mathematics Department, North Dakota State University Department # 2750, PO Box 6050, Fargo, ND 58108-6050, USA E-mail address: [email protected] URL: http://www.ndsu.edu/pubweb/~ssatherw/

April 8, 2010.

Contents Introduction Acknowledgments

v vi

Notation

ix

Chapter 1. Foundational Concepts 1.1. Rings 1.2. Polynomial Rings 1.3. Ideals and Generators 1.4. Constructions with Ideals 1.5. More Constructions with Ideals 1.6. Relations

1 1 5 7 10 14 17

Chapter 2. Basic Properties of Monomial Ideals 2.1. Monomial Ideals 2.2. Integral Domains (optional) 2.3. Generators of Monomial Ideals 2.4. Noetherian Rings (optional)

21 21 26 28 35

Chapter 3. Operations on Monomial Ideals 3.1. Intersections of Monomial Ideals 3.2. Unique Factorization Domains (optional) 3.3. Colons of Monomial Ideals 3.4. Bracket Powers of Monomial Ideals 3.5. Monomial Radicals 3.6. Square-Free Monomial Ideals

39 39 45 50 56 59 65

Chapter 4. M-Irreducible Ideals and Decompositions 4.1. M-Irreducible Monomial Ideals 4.2. Irreducible Ideals (optional) 4.3. M-Irreducible Decompositions 4.4. Irreducible Decompositions (optional) 4.5. Operations on M-Irreducible Decompositions

67 67 70 76 78 81

Chapter 5. Parametric Decompositions of Monomial Ideals 5.1. Parameter Ideals 5.2. An Example 5.3. Corner Elements 5.4. Finding Corner Elements in Two Variables 5.5. Finding Corner Elements in General 5.6. Existence and Uniqueness of Parametric Decompositions iii

87 87 91 94 99 104 108

iv

CONTENTS

Chapter 6. Square-free Monomial Ideals 6.1. Decompositions of Edge Ideals 6.2. Decompositions of Face Ideals

115 115 115

Chapter 7. Parametric Decomposition of Monomial Ideals 7.1. Parametric Decompositions and Operations on Ideals 7.2. Generating Sequences of Intersections

117 117 127

Chapter 8. Chains of Monomial Ideals 8.1. Monomial Covers 8.2. Saturated Chains of Monomial Ideals 8.3. Parameter Ideals and Containments

129 129 135 141

Bibliography

145

Index

147

Introduction Commutative algebra (or commutative ring theory) is the subdiscipline of abstract algebra devoted to the study of commutative rings with identity. There are many ways to study a commutative ring, and there are many aspects of a commutative ring that one can study. In these notes, we study monomial ideals in a polynomial ring A[X1 , . . . , Xd ], that is, ideals that are generated by monomials X1n1 · · · Xdnd . In many senses, these ideals are the simplest since each generator has only one summand. Accordingly, this makes monomial ideals optimal objects of study for students with little background in abstract algebra. Indeed, students begin studying polynomials in middle school, and they have seen polynomials in several variables in calculus, so they are familiar objects that are not as intimidating as arbitrary elements of an arbitrary commutative ring. In other words, students feel more comfortable with the process of formally manipulating polynomials because they feel more concrete. When one restricts to ideals generated by monomials, the ideas become even more concrete. In the process of teaching the students about abstract algebra via monomial ideals, we find that the concreteness of these objects makes it easier for students to grasp more general concepts like ideals and ideal quotients. They get used to formally manipulating these objects in this context and, after some time, we find that they are better prepared to deal with rings, ideals and quotients in greater generality and with increased proficiency. Another advantage of studying monomial ideals can be found in their connections to other areas of mathematics. For instance, one can study a monomial ideal using ideas from combinatorics, geometry and/or topology. Reciprocally, one can use monomial ideals to study certain objects in combinatorics, geometry and topology. In these notes, we focus on understanding the structure of monomial ideals in terms of their so-called “parametric decompositions”. (In certain cases, these are primary or irreducible decompositions.) The first main point of the notes is to give a proof of the following theorem that will be accessible to any mathematics student who has successfully completed a basic course in abstract algebra.

Theorem A (Heinzer, Ratliff and Shah [5, (4.1)]). Let A be a commutative ring with identity and set R = A[X1 , . . . , Xd ], the polynomial ring in d variables over A. Given integers n1 , . . . , nd > 1, set PR (X1n1 −1 · · · Xdnd −1 ) = (X1n1 , . . . , Xdnd )R. Set (X) = (X1 , . . . , Xd )R = PR (1) and let I ⊆ R be a monomial ideal such that rad (I) = rad ((X)). If z1 , . . . , zm are the distinct monomials in the compliment (I :R (X)) r I, then I = ∩m i=1 PR (zi ). This is an irredundant decomposition, and it is the unique such decomposition. v

vi

INTRODUCTION

This is a lovely result for several reasons. First, it expresses a general object (the given monomial ideal I) as a combination of simpler objects (the parameter ideals PR (zi )). Second, it is an existence and uniqueness result. Third, the proof is not terribly difficult, but the student will learn a non-trivial amount of mathematics in the process of understanding the proof (and the statement). It should be noted that we have only stated a special case of [5, (4.1)]. Chapter 1 serves as a review of (or introduction to) the fundamentals of Commutative Algebra. Much of the material therein may have been covered in a course on Abstract Algebra. Accordingly, much of this material may be skipped, though it cannot be ignored entirely as it contains many of the definitions and notations used in the rest of the text. One might make the argument that it is silly to start the text with an abstract treatment of the theory of rings and ideals when the point of the text is to make students more familiar with the abstract notions by having them work with monomial ideals. This is a bit of the old chicken/egg problem. The most obvious way for us to retort is to say that, in order for students to think about monomial ideals, they need to have been exposed to the basics of ring theory and ideal theory. Furthermore, we don’t necessarily expect students to internalize 100% of the material from Chapter 1 right away. This will happen over time as they get more practice with the monomial ideals. Hence, our approach. Chapter 2 deals with the fundamental properties of monomial ideals. Chapter 7 contains the proof of Theorem A, along with computations showing how parametric decompositions of combinations of monomial ideals can be obtained from parametric decompositions of the original monomial ideals. The computations in this chapter are the second main point of this text, after the proof of Theorem A. Chapter 8 deals chains of monomial ideals. Other chapters deal with other things, like Stanley-Reisner ideals associated to simplicial complexes, and edge ideals associated to graphs. (Or maybe these ideas will be mixed into the other chapters.) We will also have portions of each section devoted to using a computer algebra system to perform relevant computations and experiments. We use Macaulay 2 [3], but other readers may prefer to use CoCoA [1] or Singular [4]. Many of the exercises ask the reader to verify facts that were stated in the preceding text without proof. In these cases, one should only use results preceding the given fact in the proof of the exercise. For instance, Exercise 1.1.28 asks the student to verify the facts from Fact 1.1.12; the student should only use results preceding Fact 1.1.12 in the proof of this exercise. Acknowledgments We would like to express our thanks to the Mathematics Department at the University of Nebraska-Lincoln for its hospitality and generosity during the 2006 IMMERSE program (Intensive Mathematics: a Mentoring, Education and Research Summer Experience) when we taught the pilot version of this course. Specific thanks are due here to the program’s staff Allan Donsig and Marilyn Johnson, and to the graduate student mentors Jen Everson, Martha Gregg, Kris Lund, Lori McDonnell, Frank Moore, Jeremy Parrott, Emily Price, and Mark Stigge. We are grateful to the students who have worked so enthusiastically through the various versions of these notes: Ben Anderson, Katie Ansaldi, Pye Aung, Cheryl

ACKNOWLEDGMENTS

vii

Balm, Debbie Berg, Sam Cash, Jacqueline Dresch, Sarah Gilles, Chad Haugen, Colleen Hughes, Erik Insko, Nick Johnson, Samuel Kolins, Alexis Krier, Bethany Kubik, Jeff Langford, David Lovit, Trevor Potter, Becky Price, Tyler Seacrest, Kelly Smith, Sarah Tekansik, and Rich Wicklein. We are also pleased to acknowledge the financial support of the National Science Foundation (NSF grant NSF 0354281) and the National Security Agency.

Notation N is the set of nonnegative integers. N+ is the set of positive integers. Z is the set of all integers. Q is the set of rational numbers. R is the set of real numbers. C is the set of complex numbers. A × B is the Cartesian product of two sets A and B. Given a set A and a positive integer d, we set Ad = A × A × · · · × A {z } | d factors

the d-fold Cartesian product of A with itself.

ix

CHAPTER 1

Foundational Concepts This chapter contains a review of certain fundamental concepts in abstract algebra for use throughout the text. Section 1.1 deals with the basic properties of commutative rings with identity, and Section 1.2 focuses on polynomial rings. Section 1.3 introduces ideals, while Sections 1.4 and 1.5 investigate four methods for building new ideals from old ones. The chapter ends with Section 1.6, dealing with relations on sets. 1.1. Rings The following term was coined by David Hilbert. The axiomatic description was formalized by Emmy Noether. Definition 1.1.1. A commutative ring with identity is a set R equipped with two binary operations (addition and multiplication) satisfying the following axioms: (1) (2) (3) (4) (5) (6) (7) (8) (9) (10)

(closure under addition) for all r, s ∈ R we have r + s ∈ R; (associativity of addition) for all r, s, t ∈ R we have (r + s) + t = r + (s + t); (commutativity of addition) for all r, s ∈ R we have r + s = s + r; (additive identity) there exists an element a ∈ R such that for all r ∈ R, we have a + r = r; (additive inverse) for each r ∈ R there exists an element s ∈ R such that r + s = a; (closure under multiplication) for all r, s ∈ R we have rs ∈ R; (associativity of multiplication) for all r, s, t ∈ R we have (rs)t = r(st); (commutativity of multiplication) for all r, s ∈ R we have rs = sr; (multiplicative identity) there exists an element m ∈ R such that for all r ∈ R, we have mr = r; (distributivity) for all r, s, t ∈ R, we have r(s + t) = rs + rt.

We include the following examples for the sake of thoroughness, and in order to specify some notation. Example 1.1.2. Here are some examples. (a) The set of integers Z with the usual addition and multiplication, and with a = 0 and m = 1, is a commutative ring with identity. (b) The set of rational numbers Q with the usual addition and multiplication, and with a = 0 and m = 1, is a commutative ring with identity. (c) The set of real numbers R with the usual addition and multiplication, and with a = 0 and m = 1, is a commutative ring with identity. (d) The set of complex numbers C with the usual addition and multiplication, and with a = 0 and m = 1, is a commutative ring with identity. 1

2

1. FOUNDATIONAL CONCEPTS

(e) The set of natural numbers N = {n ∈ Z | n > 0} with the usual addition and multiplication is not a commutative ring with identity because it does not have additive inverses. (f) The set of even integers 2Z = {2n | n ∈ Z} with the usual addition and multiplication is not a commutative ring with identity because it does not have a multiplicative identity. (g) The set M2 (R) of 2 × 2 matrices with entries in R, with the usual addition and multiplication is not a commutative ring with identity because multiplication is not commutative. (h) Fix an integer n > 2 and let Zn = {m ∈ Z | 0 6 m < n}. For r, s ∈ Zn we define operations on Zn by the following formulas: r ⊕ s := the remainder after dividing r + s by n r s := the remainder after dividing rs by n. With a = 0 and m = 1, this makes Zn into a commutative ring with identity. Note that when 0 < m < n, the additive inverse of m in Zn is n − m. (i) The set C(R) of continuous functions f : R → R, with pointwise addition and multiplication, and with the constant functions a = 0 and m = 1, is a commutative ring with identity. (j) The set D(R) of differentiable functions f : R → R, with pointwise addition and multiplication, and with the constant functions a = 0 and m = 1, is a commutative ring with identity. Fact 1.1.3. Let R be a commutative ring with identity. (a) The additive and multiplicative identities for R are unique; we denote them 0R and 1R , respectively, or 0 and 1 when the context is clear. (b) For each r ∈ R, the additive inverse of r in R is unique; we denote it −r. (c) (cancellation) Let r, s, t ∈ R. If r + s = r + t, then s = t. If r + s = t + s, then r = t. Fact 1.1.4. Let R be a commutative ring with identity. (a) For all r ∈ R, we have 0R r = 0R . (b) For all r, s ∈ R, we have (−r)s = −(rs) = r(−s). (c) For all r ∈ R, we have −r = (−1R )r. This implies that −0R = 0R . (d) For all r ∈ R, we have −(−r) = r. This implies that for all r, s ∈ R, we have (−r)(−s) = rs. Definition 1.1.5. Let R be a commutative ring with identity For all r, s ∈ R we set r − s = r + (−s). Fact 1.1.6. Let R be a commutative ring with identity. (a) (closure under subtraction) For all r, s ∈ R we have r − s ∈ R. (b) For all r, s, t ∈ R we have r(s − t) = rs − rt. (c) For all r, s ∈ R we have −(r − s) = s − r. (d) For all r ∈ R we have r − r = 0R . Here is a formal definition of nr and rn where n ∈ N. Definition 1.1.7. Let R be a commutative ring with identity and let r ∈ R. (a) Set 0r = 0R and 1r = r. Inductively, for each n ∈ N, define (n + 1)r = r + (nr). (b) Set (−1)r = −r. Inductively, for each n ∈ N with, define (−n−1)r = (−n)r−r.

1.1. RINGS

3

(c) Set r0 = 1R and r1 = r. Inductively, for each n ∈ N with, define rn+1 = rn r. Fact 1.1.8. Let R be a commutative ring with identity. (a) For all r, s ∈ R and n ∈ Z we have (nr)s = n(rs) = r(ns). (b) For all r ∈ R and m, n ∈ Z we have (m + n)r = mr + nr and (mn)r = m(nr). (c) For all r, s ∈ R and n ∈ N we have (rs)n = rn sn . (d) For all r ∈ R and m, n ∈ N we have rm rn = rm+n and (rm )n = rmn . Definition 1.1.9. Let R be a commutative ring with identity. Let n > 1 be an integer, and let r1 , . . . , rP n ∈ R. n We define the sum i=1 ri = r1 + · · · + rn inductively. For n = 1, 2 we have P1 P2 Pn Pn−1 r1 +r2 . For n > 3, we define i=1 ri = ( i=1 ri )+rn . i=1 ri = r1 and i=1 ri = Q n We define the product i=1 ri = r1 · · · rn inductively. For n = 1, 2 we have Q1 Q2 Qn Qn−1 i=1 ri = r1 and i=1 ri = r1 r2 . For n > 3, we define i=1 ri = ( i=1 ri )rn . Fact 1.1.10. Let R be a commutative ring with identity. Let n > 1 be an integer, and let r1 , . . . , rn , s1 , . . . , sn ∈ R. Pn Qn (a) (generalized closure laws) We have i=1 ri ∈ R and i=1 ri ∈ R. (b) (generalized associative laws) Any two “meaningful sums” of the elements r1 , . . . , rn in this order are equal. For instance, we have ((r + s) + t) + u = (r + s) + (t + u) = r + ((s + t) + u) for all r, s, t, u ∈ R. Any two “meaningful products” of the elements r1 , . . . , rn in this order are equal. (c) (generalized commutative law) Given any permutation i1 , . . . , in of the numbers i, . . . , n we have r1 + · · · + rn = ri1 + · · · + rin and r1 · · · rn = ri1 · · · rin . (d) (generalized law) ForPall sequences r1 , . . . , rm , s1 , . . . , sn ∈ R we Pm distributive Pn Pm n have ( i=1 ri )( j=1 sj ) = i=1 j=1 ri sj . (There are more general generalized distributive laws for products of more than two sums. One verifies them by induction on the number of sums.) Definition 1.1.11. Let R be a commutative ring with identity. A monomial in the elements X1 , . . . , Xd ∈ R is an element of the form X1n1 · · · Xdnd ∈ R where n1 , . . . , nd ∈ N. For short, we write n = (n1 , . . . , nd ) ∈ Nd and Xn = X1n1 · · · Xdnd . Define addition and scalar multiplication in Nd coordinate-wise: for m = (m1 , . . . , md ) ∈ Nd and p ∈ N, set m + n = (m1 + n1 , . . . , md + nd ) ∈ Nd and pn = (pn1 , . . . , pnd ) ∈ Nd . Fact 1.1.12. Let R be a commutative ring with identity, and let X1 , . . . , Xd ∈ R. For all m, n ∈ Nd and p ∈ N we have Xm Xn = Xm+n and (Xm )p = Xpm . Definition 1.1.13. Let R be a commutative ring with identity. An element r ∈ R is a unit in R if there exists an element s ∈ R such that sr = 1R ; such an element s is a multiplicative inverse for r. Example 1.1.14. Here is what happens for integers. (a) An integer m is a unit in Z if and only if m = ±1. (b) Let m, n ∈ Z with 0 6 m < n and n > 2. Then m is a unit in Zn if and only if gcd(m, n) = 1. Fact 1.1.15. Let R be a commutative ring with identity. (a) If r is a unit in R then it has a unique multiplicative inverse; we denote the multiplicative inverse of r by r−1 . (b) If 1R 6= 0R and r is a unit in R, then r 6= 0.

4

1. FOUNDATIONAL CONCEPTS

Definition 1.1.16. Let R be a commutative ring with identity. Then R is a field if 1R 6= 0R and every non-zero element of R is a unit in R. Example 1.1.17. Here is what happens for the standard rings. (a) The rings Q, R and C are fields. (b) The ring Z is not a field. (c) Let n ∈ Z with n > 1. The ring Zn is a field if and only if n is prime. Definition 1.1.18. Let R be a commutative ring with identity. A subset S ⊆ R is a subring of R provided that it is a commutative ring under the operations of R with identity 1S = 1R . Example 1.1.19. The ring Z is a subring of Q, and Q is a subring of R. Fact 1.1.20. Let R be a commutative ring with identity, and let S ⊆ R be a subring. Then 0S = 0R . Example 1.1.21. We show that the condition 1S = 1R is not automatic. Let R = {( a0 0b ) | a, b ∈ R} and S = {( a0 00 ) | a ∈ R}. Then R and S are commutative rings with identity, under the standard addition and multiplication of matrices. However, we have 1R = ( 10 01 ) and 1S = ( 10 00 ), so we have 1R 6= 1S . Definition 1.1.22. Let R be a commutative ring with identity, and let r, s ∈ R. We say that r divides s when there is an element t ∈ R such that s = rt. When r divides s, we write r s. Exercises. Exercise 1.1.23. Verify the properties in Fact 1.1.3. Exercise 1.1.24. Verify the properties in Fact 1.1.4. Exercise 1.1.25. Verify the properties in Fact 1.1.6. Exercise 1.1.26. Verify the properties in Fact 1.1.8. Exercise 1.1.27. Verify the properties in Fact 1.1.10. (If you find parts (b) and (c) to be difficult, consult [6, I.1.6–7].) Exercise 1.1.28. Verify the properties in Fact 1.1.12. Exercise 1.1.29. Let R be a commutative ring with identity. Define FR : Z → R by the formula FR (n) = n1R . Show that FR (0) = 0R and FR (1) = 1R , and for all m, n ∈ Z we have FR (m + n) = FR (m) + FR (n) and FR (mn) = FR (m)FR (n). Exercise 1.1.30. Verify the properties in Example 1.1.14. Exercise 1.1.31. Verify the properties in Fact 1.1.15. Exercise 1.1.32. Verify the properties in Example 1.1.17. Exercise 1.1.33. (binomial theorem) Let R be a commutative ring with identity and R. Prove that for each positive integer n, there is an equality (r+s)n = Pn let nr,
si ∈n−i . i=0 i r s Exercise 1.1.34. Let f, g, h be elements of a ring R. Prove or disprove the following: If h 6= 0R and f h = gh, then f = g.

1.2. POLYNOMIAL RINGS

5

1.2. Polynomial Rings This section introduces polynomial rings, the primary objects of study in this text. Theorem 1.2.1. Let A be a commutative ring with identity. Then there exists a commutative ring with identity A[X] with the following properties: (1) The ring A is a subring of A[X]. (2) There is an element X ∈ A[X] r A such that for every f ∈ A[X], there exist n ∈ N and a0 , a1 . . . , an ∈ A such that f = a0 + a1 X + · · · + an X n ; (3) The elements 1A , X, X 2 , X 3 , . . . are linearly independent over A, that is, we have a0 + a1 X + · · · + an X n = 0A if and only if a0 = a1 = · · · = an = 0A . Sketch of proof. Set A[X] equal to the set of sequences in A which are eventually zero: A[X] = {(a0 , a1 , a2 , . . .) | a0 , a1 , a2 , . . . ∈ A and ai = 0A for all i  0}. We define addition and multiplication as follows: (a0 , a1 , a2 , . . .) + (b0 , b1 , b2 , . . .) = (a0 + b0 , a1 + b1 , a2 + b2 , . . .) (a0 , a1 , a2 , . . .)(b0 , b1 , b2 , . . .) = (c0 , c1 , c2 , . . .) Pn where cn = i=0 ai bn−i . It is straightforward to show that the axioms for a commutative ring with identity are satisfied with 0A[X] = (0A , 0A , 0A , . . .) and 1A[X] = (1A , 0A , 0A , . . .). For each a ∈ A, we identify a with the sequence (a, 0A , 0A , . . .). This identification makes A into a subset of A. It is straightforward to show that the addition and multiplication are compatible under this identification. Set X = (0A , 1A , 0A , 0A , . . .) and note that X n = (0A , 0A , . . . , 0A , 1A , 0A , 0A , . . .). {z } | n terms

It is straightforward to show that properties (2) ands (3) are satisfied.



Definition 1.2.2. Let A be a commutative ring with identity. An element f ∈ A[X] is a polynomial in one variable with coefficients in A. The (unique) elements a0 , a1 . . . , an ∈ A such that f = a0 + a1 X + · · · + an X n are the coefficients of f . The element X ∈ A[X] is the variable or indeterminate. If 0 6= f ∈ A[X], then the smallest n ∈ N such that f can be written in the form f = a0 + a1 X + · · · + an X n is the degree of f ; the corresponding coefficient an is the leading coefficient of f . If the leading coefficient of f is 1A , then f is monic. The coefficient a0 is the constant term of f . Elements of A are sometimes called constant polynomials. We often Pfinite employ the notation f = i∈N ai X i for elements f ∈ A[X]. Remark 1.2.3. Let A be a commutative ring with identity. (a) The constant polynomial 0A ∈ A[X] does not have a well-defined degree. (b) Theorem 1.2.1(3) implies that if a0 +a1 X+· · ·+am X m = b0 +b1 X+· · ·+bn X n in A[X] and m 6 n, then ai = bi for i = 0, . . . , m and bi = 0A for i = m + 1, . . . , n. (c) The ring A is identified with the set of constant polynomials in A[X]: A = {a + 0A X + 0A X 2 + · · · | a ∈ A}.

6

1. FOUNDATIONAL CONCEPTS

Remark 1.2.4. Let A be a commutative ring with identity. A polynomial f = a0 + a1 X + · · · + an X n ∈ A[X] gives rise to a well-defined function f : A → A by the rule f (a) = a0 + a1 a + · · · + an an . Definition 1.2.5. Let A be a commutative ring with identity. (a) Inductively, for d > 2, the ring of polynomials in d variables X1 , . . . , Xd with coefficients in A is A[X1 , . . . , Xd ] = A[X1 , . . . , Xd−1 ][Xd ]. We sometimes set X = X1 , . . . , Xd and write A[X] = A[X1 , . . . , Xd ]. When the number of variables is small, we sometimes write A[X, Y ] and A[X, Y, Z]. (b) The ring of polynomials in infinitely many variables X1 , X2 , X3 , . . . with coefficients in A is A[X1 , X2 , X3 , . . .] = ∪∞ d=1 A[X1 , . . . , Xd ]. (c) The (total) degree of a monomial f = Xn is deg(f ) = |n| = n1 + . . . + nd . Corollary 1.2.6. Let A be a commutative ring with identity. Set X = X1 , . . . , Xd and A[X] = A[X1 , . . . , Xd ]. Then A[X] is a commutative ring with identity satisfying the following properties: (a) The ring A is a subring of A[X]. (b) For every f ∈ A[X], there is a finite collection of indices n ∈ Nd and elements an ∈ A such that f=

finite X n∈Nd

an Xn =

finite X

an1 ,...,nd X1n1 · · · Xdnd .

n1 ,...,nd ∈N

(c) The set of monomials {Xn | n ∈ Nd } is linearly independent over A, that is, Pfinite we have n∈Nd an Xn = 0A if and only if each an = 0A . Proof. By induction on d. The base case d = 1 is in Theorem 1.2.1.



Definition 1.2.7. Let A be a commutative ring with identity. For a polynomial f ∈ A[X1 , . . . , Xd ], we say that a monomial Xn occurs in f if the corresponding coefficient an is non-zero. Remark 1.2.8. Let A be a commutative ring with identity. A polynomial f = Pfinite n d n∈Nd an X ∈ A[X] gives rise to a well-defined function f : A → A by the rule Pfinite f (b) = n∈Nd an bn . Exercises. Exercise 1.2.9. Perform the following computations, showing your steps and simplifying your answers. (a) In Q[X, Y, Z]: (3XY + 7Z 2 − XY 2 Z + 5)(X + Z − Y 2 + X 3 Y 2 Z) (b) In Z4 [X]: (2X 2 + 2)2 . This shows that one can have f n = 0 even when f 6= 0. (c) In Z4 [X]: (2X 2 + 1)2 . This shows that deg(f n ) can be strictly smaller than n deg(f ). Exercise 1.2.10. Let A be a commutative ring with identity. Under what conditions can A[X] be a field? Justify your answer with a proof. Exercise 1.2.11. Find a commutative ring with identity A and a non-zero polynomial f ∈ A[X] such that the induced function f : A → A is the zero function. Justify your answer.

1.3. IDEALS AND GENERATORS

7

Exercise 1.2.12. Let A be a commutative ring with identity and consider polynomials f1 , . . . , fn ∈ A[X]. Assume that each polynomial has degree at most N , that ai,0 + ai,1 X + · · · + ai,N X n for i = 1, . . . , n.PProve that if Pn is, we have fi = P n n i=1 ai,N 6= 0A , then i=1 fi has degree N and leading coefficient i=1 ai,N . Computer Exercises. Have the students work with Macaulay for declaring rings and doing computations in the rings. For instance, products of polynomials in polynomial rings. Make sure to explain how to interpret the output. 1.3. Ideals and Generators The following definition was first made by Julius Wilhelm Richard Dedekind, who called them “ideal numbers”. The definition extends the well-known properties of even integers under addition and multiplication, namely that the sum of two even integers is even, and the product of an integer and an even integer is again even. Definition 1.3.1. Let R be a commutative ring with identity. An ideal of R is a subset I ⊆ R satisfying the following axioms: (1) I 6= ∅; (2) (closure under addition) for all a, b ∈ I we have a + b ∈ I; (3) (closure under external multiplication) for all r ∈ R and a ∈ I we have ra ∈ I. Fact 1.3.2. Let R be a commutative ring with identity, and let I ⊆ R be an ideal. (a) We have 0R ∈ I. (b) (closure under additive inverses) If a ∈ I, then −a ∈ I. (c) (closure under subtraction) For all a, b ∈ I we have a − b ∈ I. (d) For all r ∈ R and all a ∈ I we have r + a ∈ I if and onlyPif r ∈ I. n (e) (generalized closure law) For all r1 , . . . , rn ∈ I we have i=1 ri ∈ I. (f) We have I = R if and only if 1R ∈ I. Example 1.3.3. Here are some examples. (a) For each n ∈ Z, the set nZ = {nm | m ∈ Z} is an ideal of Z. For instance, if n = 2, then we have the set of even integers 2Z, which is an ideal of Z. On the other hand, the set of odd integers is not an ideal of Z because it is not closed under addition. (b) If R is a commutative ring with identity, then the sets {0R } and R are ideals of R. Moreover {0R } is the unique smallest ideal of R and R is the unique largest ideal of R. For obvious reasons, we often write 0 for the ideal {0R }. (c) The ring Q has precisely two ideals: 0 and Q. Similar statements hold for R and C. (d) The set {f ∈ C(R) | f (2) = 0} is an ideal of C(R). On the other hand, the set {f ∈ C(R) | f (2) = 1} is not an ideal of C(R) because it is not closed under addition. Fact 1.3.4. Let R be a commutative ring with identity. (a) If {Iλ }λ∈Λ is a set of ideals of R, then ∩λ∈Λ Iλ is an ideal of R. (b) If {Iλ }λ∈Λ is a set of ideals of R, then ∪λ∈Λ Iλ need not be an ideal of R. For instance, if I and J are ideals of R, then I ∪ J is an ideal of R if and only if either I ⊆ J or J ⊆ I. (c) If {Iλ }λ∈Λ is a chain of ideals of R, then ∪λ∈Λ Iλ is an ideal of R. (Recall that the set of ideals {Iλ }λ∈Λ is a chain if, for every λ, µ ∈ Λ we have either Iλ ⊆ Iµ or Iµ ⊆ Iλ .)

8

1. FOUNDATIONAL CONCEPTS

In the following definition, part (a) is a special case of part (b), and part (b) is a special case of part (c). Definition 1.3.5. Let R be a commutative ring with identity. (a) Given an element s ∈ R, the ideal generated by s is the set (s)R = sR = {sr ∈ R | r ∈ R}. An ideal is principal if it can be generated by a single element. (b) For each sequence s1 , . . . , sn ∈ R, the ideal generated by s1 , . . . , sn is the set Pn (s1 , . . . , sn )R = { i=1 si ri ∈ R | r1 , . . . , rn ∈ R}. An ideal is finitely generated if it can be generated by a finite list of elements. (c) For each subset S ⊆ R, the ideal generated by S is the set Pn (S)R = { i=1 sn rn ∈ R | n > 0 and r1 , . . . , rn ∈ R}. The subset S ⊆ R is a generating set for an ideal I ⊆ R when I = (S)R. Remark 1.3.6. Let R be a commutative ring with identity. Pn For each sequence s1 , . . . , sn ∈ R we have 0R ∈ (s1 , . . . , sn )R because 0R = i=1 si 0R . Tacitly, we are assuming that n > 1. In the case n = 0 (that is, when we are considering the P0 empty sequence) we have 0R ∈ (∅)R because the empty sum i=1 si ri is 0R by Q0 convention. Similarly, the empty product i=1 ri is 1R by convention. Remark 1.3.7. Let R be a commutative ring with identity, and let S ⊆ R be a Pfinite subset. We will often find it convenient to use the notation s∈S srs for elements of (S)R. Here, the elements rs are in R, and the superscript “finite” signifies that we are assuming that all but finitely many of the rs are equal to 0R . Remark 1.3.8. Let R be a commutative ring with identity, and let r, s ∈ R. We have r ∈ (s)R if and only if s r. Example 1.3.9. If R is a commutative ring with identity, then R = (1R )R and (0R )R = {0R } = (∅)R. Remark 1.3.10. Let R be a commutative ring with identity, and let S, T ⊆ R be subsets. If S ⊆ T , then (S)R ⊆ (T )R. Note that in part (c) of the next result, we are not claiming that every ideal has a finite generating set. Proposition 1.3.11. Let R be a commutative ring with identity, and let S ⊆ R be a subset. (a) The set (S)R is the unique smallest ideal of R such that S ⊆ (S)R. In particular, the set (S)R is an ideal of R and, for each ideal I ⊆ R, we have S ⊆ I if and only if (S)R ⊆ I. (b) The ideal (S)R is the intersection of all the ideals of R containing S. (c) If I is an ideal, then (I)R = I. In particular, every ideal has a generating set. (d) If I = (S)R is finitely generated, then there exist elements s1 , . . . , sn ∈ S such that I = (s1 , . . . , sn )R. Proof. We verify part (a); parts (b) and (c) are left as exercises. We first show that (S)R is an ideal of R. To this end, we assume that S 6= ∅, since this case is covered in Example 1.3.9. Using the reasoning of Remark 1.3.6,

1.3. IDEALS AND GENERATORS

9

we conclude that 0R ∈ (S)R, and so (S)R 6= ∅. To show that (S)R is closed under addition and external multiplication, fix elements a, b ∈ (S)R and r ∈ R. Write Pfinite Pfinite a = s∈S sts and b = s∈S sus for elements ts , us ∈ R. Then, we have Pfinite Pfinite Pfinite a + b = ( s∈S sts ) + ( s∈S sus ) = s∈S s(ts + us ) ∈ (S)R and ra = r(

Pfinite s∈S

sts ) =

Pfinite s∈S

s(rts ) ∈ (S)R.

Hence the set (S)R is an ideal of R. Next, we observe that S ⊆ (S)R. For this, fix an element s0 ∈ S and set ( 1R when s = s0 rs = 0R when s 6= s0 . Pfinite With these choices, it follows directly that s0 = s∈S srs ∈ (S)R, as desired. Next, let I ⊆ R be an ideal; we show that S ⊂ I if and only if (S)R ⊆ I. One implication is straightforward: if (S)R ⊆ I, then the previous paragraph implies that S ⊆ (S)R ⊆ I. For the converse, assume that S ⊆ I. Since I is closed under external multiplication, we conclude that srs ∈ I for all s ∈ S and all rs ∈ R. Since Pfinite I is closed under finite sums, it follows that s∈S srs ∈ I for all s ∈ S and all rs ∈ R. That is, every element of (S)R is in I, as desired. It now follows directly that (S)R is the unique smallest ideal of R such that S ⊆ (S)R.  Exercises. Exercise 1.3.12. Verify the properties in Fact 1.3.2. Exercise 1.3.13. Verify the properties in Example 1.3.3. Exercise 1.3.14. Let R = Z[X]. Prove or disprove: (a) The set K of all constant polynomials in R is an ideal of R. (b) The set I of all polynomials in R with even constant terms is an ideal of R. Exercise 1.3.15. Verify the properties in Fact 1.3.4. Exercise 1.3.16. Verify the properties in Example 1.3.9. Exercise 1.3.17. Verify the following equalities for ideals in the polynomial ring R = Q[X, Y ]: (a) (X + Y, X − Y )R = (X, Y )R. (b) (X + XY, Y + XY, X 2 , Y 2 )R = (X, Y )R. (c) (2X 2 + 3Y 2 − 11, X 2 − Y 2 − 3)R = (X 2 − 4, Y 2 − 1)R. This shows that the same ideal can have many different generating sets and that different generating sets may have different numbers of elements. Exercise 1.3.18. For each of the sets in Exercise 1.3.14 that is an ideal, find a finite generating set. Prove that the set actually generates the ideal. Exercise 1.3.19. Verify the properties in Remark 1.3.10. Exercise 1.3.20. Let A be a commutative ring with identity. Let R = A[X, Y ] and show that the ideal (X, Y )R is not principal.

10

1. FOUNDATIONAL CONCEPTS

Exercise 1.3.21. Prove parts (b) and (c) of Proposition 1.3.11. Exercise 1.3.22. If I ⊆ Z is an ideal, then there exists an integer m ∈ Z such that I = mZ. (Hint: If I = 0 then m = 0, so you may assume that I 6= 0. In this case, show that I has a smallest positive element m; then use the Division Algorithm to show that I = mZ. ) Exercise 1.3.23. Given integers m, n that are not both zero, prove that the ideal (m, n)Z is generated by gcd(m, n). Exercise 1.3.24. Let R be a commutative ring with identity. Prove the following: (a) An element r ∈ R is a unit in R if and only if rR = R. (b) Let I ⊆ R be an ideal. Then I = R if and only if I contains a unit. (For this reason, the ideal R = (1R )R is often called the unit ideal.) (c) If 1R 6= 0R , then R is a field if and only if the only ideals of R are 0 and R. Exercise 1.3.25. Let R be a commutative ring with identity. Let f, g, f1 , . . . , fn ∈ R. Prove or disprove each of the following. (a) If f ∈ (f1 , . . . , fn )R, then f ∈ (fi )R for some i = 1, . . . , n. (b) If f ∈ (fi )R for some i = 1, . . . , n, then f ∈ (f1 , . . . , fn )R (c) If f fj = gfi and fi 6= fj , then f ∈ (fi )R and g ∈ (fj )R. Exercise 1.3.26. Prove or disprove the following: Let R be a commutative ring with identity. Fix elements f1 , . . . , fm , g1 , . . . , gn ∈ R and a positive integer k. If k )R = (g1k , . . . , gnk )R. (f1 , . . . , fm )R = (g1 , . . . , gn )R, then (f1k , . . . , fm Computer Exercises. Outline how to declare ideals in polynomial rings via generators. Have them, for example, test equality of ideals and ideal membership. Make sure to explain how to interpret the output. 1.4. Constructions with Ideals This section deals with some methods of combining ideals to form new ideals. Following the thought that the notion of an ideal generalizes the notion of a number, these constructions generalize familiar constructions for numbers, though not necessarily in the way one might expect. For instance, the sum of two ideals, introduced next, generalizes the greatest common divisor of two integers. Sums of ideals. In the following definition, part (a) is a special case of part (b), and part (b) is a special case of part (c). Definition 1.4.1. Let R be a commutative ring with identity. (a) Let I and J be ideals of R. The sum of I and J is I + J = {a + b | a ∈ I, b ∈ J} . (b) Let n be a positive integer and let I1 , I2 , . . . , In be ideals of R. The sum of the ideals I1 , I2 , . . . , In is Pn Pn i=1 ai | ai ∈ Ii for i = 1, 2, . . . , n} . j=1 Ij = I1 + I2 + · · · + In = { (c) Let {Iλ }λ∈Λ be a (possibly uncountably infinite)Pcollection of ideals of R. The sum of the ideals P in this collection to be the set λ∈Λ Iλ consisting of all sums of the form λ∈Λ aλ , where aλ ∈ Iλ for all λ ∈ Λ and all but a finite number of the aλ are zero (in other words, finite sums). In symbols, we write P Pfinite λ∈Λ Iλ = { λ∈Λ aλ | aλ ∈ Iλ }.

1.4. CONSTRUCTIONS WITH IDEALS

11

Example 1.4.2. If m and n are integers that are not both zero, then mZ + nZ = gcd(m, n)Z. Theorem 1.4.3. Let R be a commutative ring with identity, and let {Iλ }λ∈Λ be a collection of ideals of R. P P (a) The P set λ∈Λ Iλ is an ideal; more specifically, we have λ∈Λ Iλ = (∪λ∈Λ Iλ )R. (b) Pλ∈Λ Iλ is the unique smallest ideal of R containing ∪λ∈Λ Iλ . (c) λ∈Λ Iλ is the intersection of all Pthe ideals of R containing ∪λ∈Λ Iλ . (d) For each ideal K ⊆ R, we have λ∈Λ Iλ ⊆ K if and only if ∪λ∈Λ Iλ ⊆ K. P Proof. (a) The proof that λ∈Λ Iλ is an ideal of R is similar to the proof that (S)R is P an ideal in Proposition 1.3.11(a); we leave this as an exercise. The containment λ∈Λ Iλ ⊆ (∪λ∈Λ Iλ )R is a direct consequence of the definitions, using the fact that R has a multiplicative identity. For the reverse P containment, use the fact that 0P conclude that Iλ ⊆ λ∈Λ Iλ . It follows that R ∈ Iλ for each λ ∈ Λ toP ∪λ∈Λ Iλ ⊆ λ∈Λ Iλ , so (∪λ∈Λ Iλ ) ⊆ λ∈Λ Iλ by Proposition 1.3.11(a). Parts (b)–(d) follow from Proposition 1.3.11.  The next two results are special cases of Theorem 1.4.3. Corollary 1.4.4. Let R be a commutative ring with identity. Let n be a positive integer and let I1 , I2 , . . . , In be ideals of R. Pn Pn (a) The set j=1 Ij is an ideal; more specifically, we have j=1 Ij = (∪nj=1 Ij )R. Pn (b) I is the unique smallest ideal of R containing ∪nj=1 Ij . Pnj=1 j (c) the ideals of R containing ∪nj=1 Ij . j=1 Ij is the intersection of all Pn (d) For each ideal K ⊆ R, we have j=1 Ij ⊆ K if and only if ∪nj=1 Ij ⊆ K.  Corollary 1.4.5. Let R be a commutative ring with identity, and let I and J be ideals of R. (a) The set I + J is an ideal; more specifically, we have I + J = (I ∪ J)R. (b) I + J is the unique smallest ideal of R containing I ∪ J. (c) I + J is the intersection of all the ideals of R containing I ∪ J. (d) For each ideal K ⊆ R, we have I + J ⊆ K if and only if I ∪ J ⊆ K.  Fact 1.4.6. Let R be a commutative ring with identity (a) Let I and J be ideals of R. Then I + J = J if and only if I ⊆ J. (b) P Let n be a positive integer and let I1 ⊆ I2 ⊆ . . . ⊆ In be ideals of R. Then n j=1 Ij = In . P (c) Let {Iλ }λ∈Λ be a chain of ideals of R. Then λ∈Λ Iλ = ∪λ∈Λ Iλ . Fact 1.4.7. Let R be a commutative ring with identity (a) (0 and 1) If I is an ideal in R, then 0 + I = I and R + I = R. (b) (commutative law) If I and J are ideals of R, then I + J = J + I. More generally, P if {Iλ }λ∈Λ P is a collection of ideals of R and f : Λ → Λ is a bijection, then λ∈Λ Iλ = λ∈Λ If (λ) . (c) (associative law) If I, J and K are ideals of R, then (I + J) + K = I + J + K = I + (J + K). (More general associative laws, using more than three ideals, hold by induction on the number of ideals.) Fact 1.4.8. Let R be a commutative ring with identity. (a) Let I = (f1 , . . . , fn )R and let J = (g1 , . . . , gm )R. Then I + J is generated by the set {f1 , . . . , fn , g1 , . . . , gm }.

12

1. FOUNDATIONAL CONCEPTS

(b) P Let n be a positive integer and let S1 , S2 , . . . , Sn be subsets of R. Then n n j=1 (Sj )R = (∪j=1 Sj )R. P (c) If {Sλ }λ∈Λ is a collection of subsets of R, then λ∈Λ (Sλ )R = (∪λ∈Λ Sλ )R. Products of ideals. Definition 1.4.9. Let R be a commutative ring with identity and let I and J be ideals of R. Define the product of I and J to be the ideal IJ generated by all products xy, where x ∈ I and y ∈ J: IJ = ({xy | x ∈ I, y ∈ J})R. Similarly, we define the product of any finite family of ideals I1 , . . . , In of R to be the ideal I1 I2 · · · In generated by all products x1 x2 · · · xn , where xi ∈ Ii for i = 1, 2, . . . , n: I1 I2 · · · In = ({x1 x2 · · · xn | xi ∈ Ii for i = 1, 2, . . . , n})R. Given an ideal I of R and a positive integer n, we define I n = II · · · I} | {z

n factors

that is, I n is the ideal I1 I2 · · · In , where Ii = I for each i = 1, 2, . . . , n. When I 6= 0, we define I 0 = R. We leave 00 undefined. Example 1.4.10. If m and n are integers, then (mZ)(nZ) = (mn)Z. Fact 1.4.11. Let R be a commutative ring with identity, and let I, J, I1 , . . . , In be ideals of R. (a) The product ideal IJ is the unique smallest ideal of R that contains the set {ab | a ∈ I and b ∈ J}. In particular, an ideal K contains the product IJ if and only if it contains each product of the form ab where a ∈ I and b ∈ J. (b) The product I1 · · · In is the unique smallest ideal of R containing the set {a1 · · · an | aj ∈ Ij for j = 1, . . . , n}. In particular, an ideal K contains the product I1 · · · In if and only if it contains each product of the form a1 · · · an where aj ∈ Ij for j = 1, . . . , n. Pfinite (c) There are equalities IJ = { k ak bk | ak ∈ I and bk ∈ J for each k} and Pfinite I1 · · · In = { k a1,k · · · an,k | aj,k ∈ Ij for each j and each k}. (d) There are containments IJ ⊆ I ∩ J and I1 · · · In ⊆ I1 ∩ · · · ∩ In and I n ⊆ I. These containments may be proper or not. Even though the notation for the nth power of an ideal is the same as the notation for the n-fold Cartesian product of I with itself, these objects are not the same. We hope that it will be clear from the context which construction we mean at a given location. Fact 1.4.12. Let R be a commutative ring with identity. (a) (0 and 1) If I is an ideal of R, then RI = I and 0I = 0. (b) (commutative law) If I and J are ideals of R, then IJ = JI. Moreover, if I1 , . . . , In are ideals of R and i1 , . . . , in is a permutation of the numbers 1, . . . , n, then I1 · · · In = Ii1 · · · Iin . (c) (associative law) If I, J and K are ideals of R, then (IJ)K = IJK = I(JK). (More general associative laws hold by induction on the number of ideals.)

1.4. CONSTRUCTIONS WITH IDEALS

13

(d) (distributive law) If I, J and K are ideals of R, then (I + J)K = IK + JK = K(I + J). (More general distributive laws hold by induction on the number of ideals involved.) Fact 1.4.13. Let R be a commutative ring with identity. (a) If I = (f1 , . . . , fn )R and J = (g1 , . . . , gm )R, then IJ = ({fi gj | 1 6 i 6 n, 1 6 j 6 m})R. (b) If Ij = (Sj )R for j = 1, . . . , n then I1 · · · In = ({s1 · · · sn | si ∈ Si for i = 1, . . . , n})R. (c) If n is a positive integer and I = (f1 , . . . , fm )R, then I n = ({fi1 · · · fin | 1 6 ij 6 m for j = 1, . . . , n})R. Remark 1.4.14. Let R be a commutative ring with identity, and let I be an ideal of R. In general, the ideal I n is not generated by the nth powers of elements of I, that is I n ) ({f n | f ∈ I})R. For instance, if R = Z2 [X, Y ] and I = (X, Y )R, then I 2 = (X 2 , XY, Y 2 )R ) (X 2 , Y 2 )R = ({f 2 | f ∈ I})R. The inequalities are straightforward, as is the containment. For the inequality (X 2 , XY, Y 2 )R 6= (X 2 , Y 2 )R, note that XY ∈ (X 2 , XY, Y 2 )R r (X 2 , Y 2 )R. Definition 1.4.15. Let R be a commutative ring with identity and let J be an ideal of R. For each r ∈ R, we set rJ = {ra ∈ R | b ∈ J}. Fact 1.4.16. Let R be a commutative ring with identity. If J is an ideal of R and r is an element of R, then rJ = (rR)J. In particular, the set rJ is an ideal of R such that rJ ⊆ J. Note that we can have rJ ( J. Exercises. Exercise 1.4.17. Verify the properties in Example 1.4.2. Exercise 1.4.18. Fill in the missing details from the proof of Theorem 1.4.3. Exercise 1.4.19. Verify the properties in Fact 1.4.6. Exercise 1.4.20. Verify the properties in Fact 1.4.7. Exercise 1.4.21. Verify the properties in Fact 1.4.8. Exercise 1.4.22. Verify the properties in Example 1.4.10. Exercise 1.4.23. Verify the properties in Fact 1.4.11. Exercise 1.4.24. Verify the properties in Fact 1.4.12. Exercise 1.4.25. Verify the properties in Fact 1.4.13. Exercise 1.4.26. Verify the properties in Fact 1.4.16. Exercise 1.4.27. Let R be a commutative ring with identity, and let I and J be ideals of R. Set K = {ab | a ∈ I, b ∈ J}. Prove or disprove: K is an ideal of R. Exercise 1.4.28. Let A denote a commutative ring with identity, and set R = A[X1 , . . . , Xd ] and (X) = (X1 , . . . , Xd )R. Prove that if f ∈ R, then f ∈ (X)n if and only if each monomial occurring in f has degree at least n. Prove that this implies that (X) 6= R.

14

1. FOUNDATIONAL CONCEPTS

Computer Exercises. Outline how to perform each of the constructions from this section. Have the students perform each one for ideals in polynomial rings. Make sure to explain how to interpret the output. 1.5. More Constructions with Ideals This section deals with constructions of ideals that may be less familiar, namely colon ideals and radicals. Colon ideals. Definition 1.5.1. Let R be a commutative ring with identity. Let S be a subset of R, and let I be an ideal of R. For each element r ∈ R, set rS = {rs | s ∈ S}. The colon ideal of I with S is defined as (I :R S) = {r ∈ R | rS ⊆ I} = {r ∈ R | rs ∈ I for all s ∈ S}. For each s ∈ R, we set (I :R s) = (I :R {s}). Example 1.5.2. We have (6Z :Z 15) = (6Z :Z 15Z) = 2Z. The basic properties of colon ideals are contained in the next two results. Proposition 1.5.3. Let R be a commutative ring with identity. Let I, J, and K be ideals of R, and let S and T be subsets of R. (a) The set (I :R S) is an ideal of R. (b) We have (I :R (S)R) = (I :R S) = ∩s∈S (I :R s). (c) We have (I :R S) = R if and only if S ⊆ I. (d) There are containments J(I :R J) ⊆ I ⊆ (I :R S). (e) If I ⊆ J, then (I :R S) ⊆ (J :R S). (f) If S ⊆ T , then and (I :R S) ⊇ (I :R T ). Proof. (a) For every s ∈ S, we have 0s = 0 ∈ I; it follows that 0 ∈ (I :R S) and so (I :R S) 6= ∅. To show that (I :R S) is closed under addition, let r, r0 ∈ (I :R S). For all s ∈ S we then have rs, r0 s ∈ I. The distributive law combines with the fact that I is closed under addition to show that (r+r0 )s = rs+r0 s ∈ I, and so r+r0 ∈ (I :R S), as desired. To show that (I :R S) is closed under external multiplication, let r ∈ (I :R S) and t ∈ R. For all s ∈ S we then have rs ∈ I. The associative law combines with the fact that I is closed under external multiplication to show that (tr)s = t(rs) ∈ I, and so tr ∈ (I :R S), as desired. (c) For the forward implication, assume that (I :R S) = R. It follows that 1R ∈ (I :R S), and so S = 1R S ⊆ I. For the converse, assume that S ⊆ I. It follows that 1R ∈ (I :R S). Since (I :R S) is an ideal in R, it follows from Fact 1.3.2(f) that (I :R S) = R. The proofs of the remaining statements are left as exercises.  Proposition 1.5.4. Let R be a commutative ring with identity. Let I, J, and K be ideals of R, and let S be a subset of R. Let {Iλ }λ∈Λ be a collection of ideals of R, and let {Sλ }λ∈Λ be a collection of subsets of R. (a) There are equalities ((I :R J) :R K) = (I :R JK) = ((I :R K) :R J). (b) There is an equality (∩λ∈Λ Iλ :R S) = ∩λ∈Λ (Iλ :R S). (c) There is an equality (J :R ∪λ∈Λ Sλ ) = ∩λ∈Λ (J :R Sλ ).

1.5. MORE CONSTRUCTIONS WITH IDEALS

(d) There is an equality (J :R

P

λ∈Λ Iλ )

15

= ∩λ∈Λ (J :R Iλ ).

Proof. (a) We first show that ((I :R K) :R J) ⊆ (I :R JK). Fix elements r ∈ ((I :R K) :R J) and x ∈ JK. P It follows that there are elements b1 , . . . , bn ∈ J n and c1 , . . . , cn ∈ K such that x = i=1 bi ci . Since r ∈ ((I :R K) :R J) and bi ∈ J, we have rbi ∈ (I :R K). The fact that ci ∈ K implies that rbi ci ∈ I. It follows that Pn Pn rx = r i=1 bi ci = i=1 rbi ci ∈ I because I is an ideal, so rx ∈ (I :R JK), as desired. We next show that ((I :R K) :R J) ⊇ (I :R JK). Let s ∈ (I :R JK) and b ∈ J. To show that s ∈ ((I :R K) :R J), it suffices to show that sb ∈ (I :R K). Let c ∈ K. One needs to show that (sb)c ∈ I. The conditions b ∈ J and c ∈ K imply bc ∈ JK. Hence, the assumption s ∈ (I :R JK) implies (sb)c = s(bc) ∈ I, as desired. The other equality follows form the next sequence ((I :R K) :R J) = (I :R JK) = (I :R KJ) = ((I :R J) :R K). The proofs of the remaining statements are left as exercises.



The radical of an ideal. Definition 1.5.5. Let R be a commutative ring with identity and let I be an ideal of R. The radical of I is the set rad (I) = {x ∈ R | xn ∈ I for some n > 1} . √ Other common notations include I and r(I). Example 1.5.6. In the ring Z we have rad (12Z) = 6Z. In the ring Z8 we have rad (0Z8 ) = rad (4Z8 ) = 2Z8 . Proposition 1.5.7. Let R be a commutative ring with identity, and let I be an ideal of R. (a) The set rad (I) is an ideal of R. (b) There is a containment I ⊆ rad (I). (c) There is an equality rad (I) = rad (rad (I)). (d) We have rad (I) = R if and only if I = R. (e) For each integer n > 1, there is an equality rad (I n ) = rad (I). Proof. (a) We have 01 = 0 ∈ I; it follows that 0 ∈ rad (I), so rad (I) 6= ∅. To show that rad (I) is closed under addition, let r, s ∈ rad (I). There are integers m, n > 1 such that rm , sn ∈ I. The binomial theorem 1.1.33 implies that
i m+n−i Pm+n . (r + s)m+n = i=0 m+n rs i Note that for each i = 0, . . .
, m + n we have either i > m or m + n − i > n. It follows that each term m+n ri sm+n−i is in I. This implies that (r + s)m+n ∈ I, i so r + s ∈ rad (I). To show that rad (I) is closed under external multiplication, let r ∈ rad (I) and t ∈ R, and let n be a positive integer such that rn ∈ I. It follows that (rt)n = rn tn ∈ I, so rt ∈ rad (I). (d) For the forward implication, assume that rad (I) = R. It follows that there exists a positive integer n such that 1nR ∈ I. The equality 1nR = 1R implies that 1R ∈ I, so I = R.

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For the reverse implication, assume that I = R. Part (b) explains the second step in the sequence R = I ⊆ rad (I) ⊆ R, so rad (I) = R. The proofs of the remaining statements are left as exercises.  The next example shows that, without the commutative hypothesis, the set rad (I) need not be an ideal. Example 1.5.8. Let M2 (R) denote the set of all 2 × 2 matrices with entries in R. This is a non-commutative ring with identity using the standard definitions of matrix addition and matrix multiplication. The set rad (0) is not an ideal because it is not closed under addition. Indeed, we have M = ( 00 10 ) ∈ rad (0) because M 2 = 0. Similarly, we have N = ( 01 00 ) ∈ rad (0), but M + N = ( 01 10 ) ∈ / rad (0) because M + N is invertible. Also, we have M N = ( 10 00 ) ∈ / rad (0) so rad (0) is not closed under external multiplication. (Moreover, it is not even closed under internal multiplication.) Proposition 1.5.9. Let R be a commutative ring with identity. Let n be a positive integer, and let I, J, I1 , I2 , . . . , In be ideals of R. (a) If I ⊆ J, then rad (I) ⊆ rad (J). (b) There are equalities rad (IJ) = rad (I ∩ J) = rad (I) ∩ rad (J). (c) There are equalities rad (I1 I2 · · · In ) = rad (I1 ∩ I2 ∩ · · · ∩ In ) = rad (I1 ) ∩ rad (I2 ) ∩ · · · ∩ rad (In ) . (d) rad (I + J) = rad (rad (I) + rad (J)). (e) rad (I1 + I2 + · · · + In ) = rad (rad (I1 ) + rad (I2 ) + · · · + rad (In )). Proof. (a) Assume that I ⊆ J and let r ∈ rad (I). There is an integer m > 1 such that rm ∈ I ⊆ J, so r ∈ rad (J). (b) We show that rad (IJ) ⊆ rad (I ∩ J) ⊆ rad (I) ∩ rad (J) ⊆ rad (IJ). The containment rad (IJ) ⊆ rad (I ∩ J) follows from part (a) since IJ ⊆ I ∩ J. The containment rad (I ∩ J) ⊆ rad (I) ∩ rad (J) also follows from part (a): the containments I ∩ J ⊆ I and I ∩ J ⊆ J imply that rad (I) ∩ rad (J) ⊆ rad (I) and rad (I) ∩ rad (J) ⊆ rad (J) For the containment rad (I) ∩ rad (J) ⊆ rad (IJ), let r ∈ rad (I) ∩ rad (J). There are integers l, m > 1 such that rl ∈ I and rm ∈ J, so rl+m = rl rm ∈ IJ. It follows that r ∈ rad (IJ), as desired. (c) The case n = 2 follows from part (b); the remaining cases are proved by induction on n. The proofs of the remaining statements are left as exercises.  The following example shows that we may have rad (I + J) 6= rad (I) + rad (J); compare to Proposition 1.5.9(d). Example 1.5.10. Let R be the polynomial ring R = C[X, Y ] in two variables. Let I = (X 2 + Y 2 )R and J = (X)R. Then rad (I) = I and rad (J) = J, so 2 rad (I) + rad (J)
= I + J = (X, Y )R. On the other hand, we have rad (I + J) = 2 rad (X, Y )R = (X, Y )R, so X ∈ rad (I + J) r rad (I) + rad (J). Fact 1.5.11. Let R be a commutative ring with identity, and let I and J be ideals of R. (a) Assume that I = (f1 , . . . , fs )R with s > 1. Then rad (I) ⊆ rad (J) if and only if for each i = 1, 2, . . . , s there exists a positive integer ni such that fini ∈ J.

1.6. RELATIONS

17

(b) Assume that I = (f1 , . . . , fs )R and J = (g1 , . . . , gt )R with s, t > 1. Then rad (I) = rad (J) if and only if for each i = 1, 2, . . . , s there exists a positive integer ni such that fini ∈ J, and for each j = 1, 2, . . . , t there exists a positive m integer mj such that gj j ∈ I. (c) Suppose I ⊆ J and that J = (g1 , . . . , gt )R. Then rad (I) = rad (J) if and only m if for each j = 1, 2, . . . , t there exists an integer mj such that gj j ∈ I. Exercises. Exercise 1.5.12. Verify the properties in Example 1.5.2. Exercise 1.5.13. Complete the proof of Proposition 1.5.3. Exercise 1.5.14. Complete the proof of Proposition 1.5.4. Exercise 1.5.15. Let R be a commutative ring with identity. Let I be an ideal of R, and let S and T be subsets of R. Prove that ((I :R S) :R T ) = ((I :R T ) :R S). Exercise 1.5.16. Verify the properties in Example 1.5.6. Exercise 1.5.17. Complete the proof of Proposition 1.5.7. Exercise 1.5.18. Complete the proof of Proposition 1.5.9. Exercise 1.5.19. Verify the properties in Fact 1.5.11. Exercise 1.5.20. Let m, n ∈ Z and compute (mZ :Z nZ) and rad (nZ); justify your answer. Exercise 1.5.21. Let A be a commutative ring with identity. Consider the polynomial ring R = A[X1 , . . . , Xd ] and the ideal (X) = (X1 , . . . , Xd )R. (a) Prove that ((X)n : (X)) = (X)n−1 . (b) List the monomials in (X)n−1 r (X)n ; justify your answer. Exercise 1.5.22. Let R be a commutative ring with identity, and let I and J be ideals of R. Prove or disprove: If rad (I) ⊆ rad (J), then I ⊆ J. Exercise 1.5.23. Let A be a commutative ring with identity. Consider the polynomial ring R = A[X, Y ] and the ideals I = (X 3 , Y 4 )R and J = (XY 2 , X 2 Y )R. (a) Use Fact 1.5.11 to prove that rad (I) ) rad (J). (b) Assume that A is a field. Prove that rad (I) = (X, Y )R and rad (J) = (XY )R. Use this to give another proof that rad (I) ) rad (J). Computer Exercises. Outline how to perform each of the constructions from this section. Have the students perform each one for ideals in polynomial rings. Make sure to explain how to interpret the output. 1.6. Relations Definition 1.6.1. Let A be a set. A relation on A is a subset ∼ ⊆ A × A. If ∼ ⊆ A × A is a relation, then we write “a ∼ b” instead of “(a, b) ∈ ∼”. Remark 1.6.2. More generally, one can define a relation on the cartesian product A × B of two sets A and B as a subset ∼ ⊆ A × A. Our definition is the special case A = B. We will not need the more general version. Example 1.6.3. Every equivalence relation on a set A is a relation on A.

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Definition 1.6.4. Let A be a non-empty set and let > be a relation on A. The relation > is a partial order on A if it satisfies the following properties: (a) (reflexivity) For all a ∈ A we have a > a; (b) (transitivity) For all a, b, c ∈ A, if a > b and b > c, then a > c; (c) (antisymmetry) For all a, b ∈ A, if a > b and b > a, then a = b. The negation of “a > b” is written “a 6> b”. Example 1.6.5. The usual orders > on the sets N, Z, Q and R are partial orders. The “divisibility relation” on the set of positive integers N+ , given by n >div m when m n, is a partial order on N. The obvious divisibility relation on Z is not a partial order on Z because it is not antisymmetric: We have 1 − 1 and −1 1, but 1 6= −1. Definition 1.6.6. Let A be a set with a partial order >. Two elements a, b ∈ A are comparable if either a > b or b > a. The partial order > is a total order if every a, b ∈ A are comparable. Remark 1.6.7. Sets with partial orders are often called “partially ordered sets”, or “posets” for short. The adjective “partial” is used for these relations because there is no guarantee it is a total order. For instance, the divisibility order on N+ is not a total order because 2 - 3 and 3 - 2. Here is the main example for this text. Definition 1.6.8. Let d be a positive integer and define a relation < on Nd as follows: (a1 , . . . , ad ) < (b1 , . . . , bd ) when, for i = 1, . . . , d we have ai > bi in the usual order on N. Example 1.6.9. When d = 2, we graph the set {n ∈ N2 | N < (1, 2)} as follows. O

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This set can be described as the “translate” (1, 2) + N2 . Definition 1.6.10. Let d be a positive integer. For each n ∈ Nd , set [n] = {m ∈ Nd | m < n} = n + Nd .

1.6. RELATIONS

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Example 1.6.11. When d = 2, the graph of [(1, 2)] ∪ [(3, 1)] is the following. O

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Exercises. Exercise 1.6.12. Prove that the relation from Definition 1.6.8 is a partial order. Prove that it is a total order if and only if d = 1. Exercise 1.6.13. Let R be a commutative ring with identity. Prove that the divisibility order on R is reflexive and transitive. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output.

CHAPTER 2

Basic Properties of Monomial Ideals In this chapter, we develop the fundamental tools for working with monomial ideals. Section 2.1 introduces the main players. Motivated by some properties from this section, Section 2.2 contains a brief discussion of integral domains. Section 2.3 deals with some aspects of generating sets for monomial ideals. The concluding optional Section 2.4 on noetherian rings is a natural follow-up. 2.1. Monomial Ideals Here is the main object of study in this text. Definition 2.1.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. A monomial ideal in R is an ideal of R that can be generated by monomials in X1 , . . . , Xd ; see Definition 1.1.11. Example 2.1.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in 2 variables. (a) The ideal I = (X 2 , Y 3 )R is a monomial ideal. Note that I contains the polynomial X 2 − Y 3 , so monomial ideals may contain more than monomials. (b) The ideal J = (Y 2 − X 3 , X 3 )R is a monomial ideal because J = (Y 2 , X 3 )R. (c) The trivial ideals 0 and R are monomial ideals since 0 = (∅)R and R = 1R R = X10 · · · Xd0 R. Definition 2.1.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. For each monomial ideal I ⊆ R, let [I] denote the set of all monomials contained in I. Remark 2.1.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. For each non-zero monomial ideal I ⊆ R, the set [I] ⊂ R is an infinite set that is not an ideal. By definition, we have [I] = I ∩ [R]. Lemma 2.1.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. For each monomial ideal I ⊆ R, we have I = ([I])R. Proof. Let S be a set of monomials generating I. It follows that S ⊆ [I] ⊆ I, so I = (S)R ⊆ ([I])R ⊆ I, implying the desired equality.  Proposition 2.1.6. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I and J be monomial ideals of R. (a) We have I ⊆ J if and only if [I] ⊆ [J]. (b) We have I = J if and only if [I] = [J]. 21

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Proof. (a) If I ⊆ J, then [I] = I ∩[R] ⊆ J ∩[R] = [J]. Conversely, if [I] ⊆ [J], then Lemma 2.1.5 implies that I = ([I])R ⊆ ([J])R = J. Part (b) follows directly from (a).  Definition 2.1.7. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. (a) Let f and g be monomials in R. Then f is a monomial multiple of g if there is a monomial h ∈ R such that f = gh. (b) For a monomial f = Xn ∈ R, the d-tuple n ∈ Nd is the exponent vector of f . Remark 2.1.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Because the monomials in X1 , . . . , Xd are linearly independent over A, the exponent vector of each monomial f ∈ R is well-defined. Thus, for d-tuples m, n ∈ Nd , we have Xm = Xn if and only if m = n. Lemma 2.1.9. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f = Xm and g = Xn be monomials in R. If h is a polynomial in R such that f = gh, then mi > ni for i = 1, . . . , d and h is the monomial h = Xp where pi = mi − ni . P Proof. Assume that f = gh where h = p∈Λ ap Xp ∈ R and Λ ⊂ Nd is a finite subset such that ap 6= 0A for each p ∈ Λ. The equation f = gh then reads X X Xm = Xn ap Xp = ap Xn+p . p∈Λ

p∈Λ

The linear independence of the monomials in R implies that ( 0A when n + p 6= m ap = 1A when n + p = m. Since we have assumed that each of the coefficients ap is non-zero, this implies that the only possible non-zero term in h is ap Xp when n + p = m; in other words, the set Λ consists of a single element Λ = {p}. The coefficient ap = 1A then implies that h = Xp , so the equation f = gh implies that f is a monomial multiple of g by definition, as desired.  Example 2.1.10. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in 2 variables. Then XY 7 is not a multiple of X 2 Y , but X 2 Y 7 is a multiple of X 2 Y . Lemma 2.1.11. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f = Xm and g = Xn be monomials in R. The following conditions are equivalent: (i) we have f ∈ gR; (ii) the element f is a multiple of g; (iii) the element f is a monomial multiple of g; (iv) we have m < n; and (v) we have m ∈ [n]. Proof. The equivalences (i) ⇐⇒ (ii) and (iv) ⇐⇒ (v) hold by definition, and the implication (iii) =⇒ (ii) is straightforward to verify. The implication (ii) =⇒ (iii) follows from Lemma 2.1.9.

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(iii) =⇒ (iv): Assume that f = gh where h = Xp . It follows that Xm = Xn Xp = Xn+p . The well-definedness of the exponent vector for a monomial implies that m = n+p, i.e., that mi = ni +pi for i = 1, . . . , d. Since each pi > 0, this implies mi = ni +pi > ni for each i, so m < n by definition, as desired. (iv) =⇒ (iii): Assume that m < n. By definition, this implies that mi − ni > 0 for each i. Setting pi = mi − ni , we have p ∈ Nd and, as above, it follows that f = Xm = Xn+p = Xn Xp = gh where h = Xp . This says that f is a monomial multiple of g, as desired.



Definition 2.1.12. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. The divisibility order on the monomial set [R] is the following order: Xm < Xn when Xm is a multiple of Xn . Lemma 2.1.13. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. The divisibility order on [R] is a partial order. Proof. We have already seen that the order < on Nd is a partial order. The result now follows from Lemma 2.1.11 because (a) the monomials in [R] are in bijection with the elements of Nd , and (b) Lemma 2.1.11 implies that Xm < Xn if and only if m < n.  Lemma 2.1.14. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f, f1 , . . . , fm be monomials in R. Then f ∈ (f1 , . . . , fm )R if and only if f ∈ fi R for some i. Proof. ⇐= : Since fi ∈ {f1 , . . . , fm }, we have fi R =P(fi )R ⊆ (f1 , . . . , fm )R. m =⇒ : Assume that f ∈ (f1 , . . . , fm )R and write f = i=1 fi gi with elements gi ∈ R. By assumption, we have f = Xn and fi = Xni for some n, n1 , . . . , nm ∈ Nd . Pfinite By definition, we can write each gi = p∈Nd ai,p Xp where each ai,p ∈ A, so   finite m finite m m X X X X X ai,p Xni +p . ai,p Xp  = Xn = f = fi gi = Xni  i=1

i=1

p∈Nd

i=1 p∈Nd

Since the monomials in R are linearly independent over A, the monomial Xn must occur in the right-most sum in this display. (Note that we are not suggesting that the monomials occurring in the right-most sum are distinct. Hence, the monomial Xn it may occur more than once in this expression.) In other words, we have Xn = Xni +p for some i and some p. It follows that f = Xn = Xni +p = Xni Xp = fi Xp ∈ fi R for some i, as desired.



Remark 2.1.15. The non-trivial implication of Lemma 2.1.14 can fail if the fi are not monomials; see Exercise 1.3.25(a). Definition 2.1.16. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. The graph of a monomial ideal I is the set Γ(I) = {n ∈ Nd | Xn ∈ I}.

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Theorem 2.1.17. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. If I = (Xn1 , . . . , Xnm )R, then Γ(I) = [n1 ] ∪ · · · ∪ [nm ]. Proof. ⊇: Let m ∈ [n1 ] ∪ · · · ∪ [nm ]. Then we have m ∈ [ni ] for some i, and so m < ni . Lemma 2.1.11 implies that

Xm ∈ Xni R ⊆ (Xn1 , . . . , Xnm )R = I

and it follows by definition that m ∈ Γ(I). ⊆: Assume that p ∈ Γ(I). Then Xp ∈ I = (Xn1 , . . . , Xnm )R, so Lemma 2.1.14 implies that Xp ∈ Xnj R for some j. From Lemma 2.1.11 we conclude that p ∈ [nj ] ⊆ [n1 ] ∪ · · · ∪ [nm ], as desired. 

Example 2.1.18. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in 2 variables. The graph of the monomial ideal I = (X 4 , X 3 Y, Y 2 )R is the set Γ(I) = [(4, 0)] ∪ [(3, 1)] ∪ [(0, 2)] ⊆ N2 , represented graphically in the next diagram.

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Example 2.1.19. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set I = (X 2 )R and J = (Y 3 )R. Then I + J = (X 2 , Y 3 )R. Theorem 2.1.17 shows that Γ(I) = [(2, 0)] and Γ(J) = [(0, 3)] and

Γ(I + J) = [(2, 0)] ∪ [(0, 3)] = Γ(I) ∪ Γ(J).

2.1. MONOMIAL IDEALS

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Graphically, we have the following: O

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More generally, in Exercise Pn2.3.20(e) we see that if I1 , . . . , In are monomial ideals in A[X1 , . . . , Xd ], then Γ( j=1 Ij ) = ∪nj=1 Γ(Ij ). Exercises. Exercise 2.1.20. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let I be a monomial ideal. Prove that I 6= R if and only if I ⊆ (X). Exercise 2.1.21. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Prove that if I is a monomial ideal such that I 6= R, then rad (I) = rad ((X)) if and only if for each i = 1, . . . , d there exists an integer ni > 0 such that Xini ∈ I. Exercise 2.1.22. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let I be a monomial ideal. Prove that if rad (I) ⊆ rad ((X)), then I ⊆ (X) ( R.

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Exercise 2.1.23. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables with d > 2. Let I be a monomial ideal in R and let f be a monomial in R. (a) Prove that if Xi | f , then rad (f I) ⊆ rad ((Xi )R). (b) Prove that if f 6= 1, then rad (f I) ( rad ((X)). Exercise 2.1.24. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Prove that for monomials f, g ∈ [R], one has (f )R = (g)R if and only if f = g. Exercise 2.1.25. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be an ideal of R. Prove that the following conditions are equivalent. (i) I is a monomial ideal; and (ii) for each f ∈ I each monomial occurring in f is in I. Exercise 2.1.26. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I and J be monomial ideals in R. Prove that I = J if and only if Γ(I) = Γ(J). Exercise 2.1.27. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Prove that if I is a monomial ideal in R such that I 6= R, then I ∩ A = 0. Exercise 2.1.28. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f , g, h be monomials in R. (a) Prove that if f h = gh, then f = g. (b) Prove that if f Xi = gXj for some i 6= j, then f ∈ (Xj )R and g ∈ (Xi )R. (c) Prove that deg(f g) = deg(f ) + deg(g). Exercise 2.1.29. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in 2 variables. Find Γ(I) for the following ideals and justify your answers: (a) I = (X 5 , Y 4 )R. (b) I = (X 5 , XY 2 , Y 4 )R. (c) I = (X 5 Y, X 2 Y 2 , XY 4 )R. Exercise 2.1.30. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let (X) = (X1 , . . . , Xd )R. Let I be a monomial ideal in R, and let S denote the set of monomials in R r I. Prove that S is a finite set if and only if rad (I) ⊇ rad ((X)). Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 2.2. Integral Domains (optional) Given polynomials f, g ∈ A[X] one may have deg(f g) = deg(f ) + deg(g) or deg(f g) < deg(f ) + deg(g). If f and g are monomials, then deg(f g) = deg(f ) + deg(g). This is, in many regards, the nicest situation. This section focuses on rings for which this equality always holds. Definition 2.2.1. An integral domain is a commutative ring R with identity such that 1R 6= 0R and such that for all r, s ∈ R if r, s 6= 0R , then rs 6= 0R .

2.2. INTEGRAL DOMAINS (OPTIONAL)

27

Example 2.2.2. The rings Z, Q, R, and Z are integral domains. The ring Zn is an integral domain if and only if n is prime. For instance, the ring Z6 is not an integral domain because 2, 3 6= 0 in Z6 , but 2 · 3 = 0 in Z6 . The ring R = Z is an integral domain that is not a field. Hence, the converse of the next result fails. Proposition 2.2.3. If R is a field, then R is an integral domain. Proof. Assume that R is a field, and let r, s ∈ R such that rs = 0R . We need to show that r = 0R or s = 0R . Assume that r 6= 0R . Since R is a field, we have s = 1R s = (r−1 r)s = r−1 (rs) = r−1 0R = 0R as desired.



Compare the next result with Exercise 2.1.28(a). See also Exercise 2.2.10. Proposition 2.2.4 (cancellation). Let R be an integral domain. If r, s, t ∈ R such that r 6= 0R and rs = rt, then s = t. Proof. Given that rs = rt, we have r(s − t) = 0R . Since R is an integral domain and r 6= 0R , it follows that s − t = 0R , that is, that s = t.  See Exercise 2.2.11 for a compliment to the next result. Proposition 2.2.5. Let A be an integral domain, and let R be the polynomials ring R = A[X] in 1 variable. (a) Given non-zero polynomials f, g ∈ A[X], one has deg(f g) = deg(f ) + deg(g), and the leading coefficient of f g is the product of the leading coefficients of f and g. (b) The ring A[X] is an integral domain. Pm Pn Proof. (a) Write f = i=0 ai X i and g = i=0 bi X i with m = deg(f ) and n = deg(g). It follows that we have f g = a0 b0 + (a1 b0 + a0 b1 )X + · · · + (am bn−1 + am−1 bn )X m+n−1 + am bn X m+n . Since R is an integral domain and am , bn 6= 0R , it follows that am bn 6= 0R , and hence the desired properties. (b) Part (a) shows that the product of two non-zero polynomials in A[X] is non-zero. Since 1A[X] = 0A 6= 1A = 1A[X] , it follows that A[X] is an integral domain.  Corollary 2.2.6. Let A be an integral domain. Then the polynomial ring R = A[X1 , . . . , Xd ] in d variables is an integral domain. Proof. Induction on d, using Proposition 2.2.5(b).



Exercises. Exercise 2.2.7. Prove that Zn is an integral domain if and only if n is prime. Exercise 2.2.8. Prove Corollary 2.2.6. Exercise 2.2.9. Let A be a commutative ring with identity. (a) Prove that if A is a subring of an integral domain, then A is an integral domain. (b) Find a counter-example to the converse of part (a); justify your answer.

28

2. BASIC PROPERTIES OF MONOMIAL IDEALS

(c) Prove that if the polynomial ring A[X1 , . . . , Xd ] in d variables is an integral domain for some d > 1, then A is an integral domain. Exercise 2.2.10. Let R be a commutative ring with identity such that 1R 6= 0R . Assume that for all r, s, t ∈ R if r 6= 0R and rs = rt, then s = t Prove that R is an integral domain. Exercise 2.2.11. Let A be a commutative ring with identity such that 1A 6= 0A . Assume that for all non-zero polynomials f, g ∈ A[X] in one variable, one has deg(f g) = deg(f ) + deg(g). Prove that A is an integral domain. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 2.3. Generators of Monomial Ideals Theorem 2.1.17 is only stated for a monomial ideal I that is generated by a finite number of monomials. The next result shows that this condition is automatic for all monomial ideals. Theorem 2.3.1 (Monomial Hilbert Basis Theorem). Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Then every monomial ideal or R is finitely generated; moreover, it is generated by a finite set of monomials. Proof. Let I ⊆ R be a monomial ideal, and assume without loss of generality that I 6= 0. We proceed by induction on the number of variables d. Base case: d = 1. For this case, we write R = A[X]. Let r = min{e > 0 | X e ∈ I}. Then X r ∈ I and so X r R ⊆ I. We will be done with this case once we show that X r R ⊇ I. Since I is generated by its monomials, it suffices to show that X r R ⊇ [I]. For this, note that if X s ∈ I, then s > r and so X s ∈ X r R by Lemma 2.1.11. Induction step: Assume that d > 2 and assume that every monomial ideal of the ring R0 = A[X1 , . . . , Xd−1 ] is finitely generated. Given a monomial ideal I, set S = {monomials z ∈ R0 | zXde ∈ I for some e > 0} and J = (S)R0 . By definition J is a monomial ideal in R0 , so our induction hypothesis implies that it is finitely generated, say J = (z1 , . . . , zn )R0 where z1 , . . . , zn ∈ S; see Exercise 2.4.10. For i = 1, . . . , n there exists an integer ei > 0 such that zi Xdei ∈ I. With e = max{e1 , . . . , en }, it follows that zi Xde ∈ I for i = 1, . . . , n. For m = 0, . . . , e − 1 we set Sm = {monomials w ∈ R0 | wXdm ∈ I} and Jm = (Sm )R0 . By definition Jm is a monomial ideal in R0 , so our induction hypothesis implies that it is finitely generated, say Jm = (wm,1 , . . . , wm,nm )R0 where wm,1 , . . . , wm,nm ∈ Sm . Let I 0 be the ideal of R generated by the zi Xde and the wm,i Xdm : I 0 = ({zi Xde | i = 1, . . . , n} ∪ {wm,i Xdm | m = 0, . . . , e − 1; i = 1, . . . , nm })R. By construction I 0 is a finitely generated monomial ideal of R. By definition, each zi Xde , wm,i Xdm ∈ I, so we have I 0 ⊆ I.

2.3. GENERATORS OF MONOMIAL IDEALS

29

Claim: I 0 = I. (Once the claim is established, we conclude that I is generated by a finite set of its monomials, completing the proof.) It suffices to show that I 0 ⊇ I. Since I is generated by its monomials, it suffices to show that I 0 ⊇ [I], so pd−1 pd let Xp = X1p1 · · · Xd−1 Xd ∈ [I]. pd−1 Case 1: pd > e. We have X1p1 · · · Xd−1 ∈ S ⊆ J = (z1 , . . . , zn )R0 . Thus, we pd−1 pd−1 p1 have X1 · · · Xd−1 ∈ zi R0 for some i, by Lemma 2.1.14. Writing X1p1 · · · Xd−1 = zi z for some z ∈ R0 we have p

d−1 Xp = X1p1 · · · Xd−1 Xdpd = zi zXde Xdpd −e = (zi Xde )(zXdpd −e ) ∈ (zi X e )R ⊆ I 0

as desired. pd−1 Case 2: pd < e. The monomial X1p1 · · · Xd−1 is in the set Spd ⊆ Jpd = pd−1 0 (wpd ,1 , . . . , wpd ,npd )R . Lemma 2.1.14 implies that X1p1 · · · Xd−1 ∈ wpd ,i R0 for some pd−1 p1 i. Writing X1 · · · Xd−1 = wpd ,i w for some w ∈ R0 we have p

d−1 Xdpd = wpd ,i wXdpd = (wpd ,i Xdpd )(w) ∈ (wpd ,i Xdpd )R ⊆ I 0 Xp = X1p1 · · · Xd−1

as desired.



Corollary 2.3.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let S ⊆ [R] and set I = (S)R. Then there is a finite sequence s1 , . . . , sn ∈ S such that I = (s1 , . . . , sn )R. Proof. Theorem 2.3.1 implies that I is finitely generated, so the desired conclusion follows from Proposition 1.3.11(d).  Part (a) of the next result says that the polynomial ring R = A[X1 , . . . , Xd ] satisfies the ascending chain condition for monomial ideals. Part (b) says that every nonempty set Σ of monomial ideals in R has a maximal element, and that every element of Σ is contained in a maximal element of Σ. Theorem 2.3.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. (a) Given a chain I1 ⊆ I2 ⊆ · · · of monomial ideals in R, there is an integer N > 1 such that IN = IN +1 = · · · . (b) Given a nonempty set Σ of monomial ideals in R, there is an ideal I ∈ Σ such that for all J ∈ Σ, if I ⊆ J, then I = J. Moreover, for each K ∈ Σ, there is an ideal I ∈ Σ such that K ⊆ I and such that for all J ∈ Σ, if I ⊆ J, then I = J. Proof. (a) a chain I1 ⊆ I2 ⊆ · · · of monomial ideals in R. Then PConsider ∞ the ideal J = j=1 Ij = ∪∞ j=1 Ij is a monomial ideal in R; see Fact 1.4.6(c) and Exercise 2.3.20(a). Theorem 2.3.1 implies that J is generated by a finite list of monomials f1 , . . . , fm ∈ [J]. Since J = ∪∞ j=1 Ij , for i = 1, . . . , m there is an index ji such that fi ∈ Iji . The condition I1 ⊆ I2 ⊆ · · · implies that there is an index N such that fi ∈ IN for i = 1, . . . , m. It follows that J = (f1 , . . . , fm )R ⊆ IN ⊆ IN +1 ⊆ IN +2 ⊆ · · · ⊆ J. Thus, we have equality at each step, as desired. (b) Let Σ be a nonempty set of monomial ideals in R, and let K ∈ Σ. Suppose by way of contradiction that K is not contained in a maximal element of Σ. In particular, K is not a maximal element of Σ, so there is an element I1 ∈ Σ such that K ( I1 . Since K is not contained in a maximal element of Σ, it follows that I1 is not a maximal element of Σ. Thus, there is an element I2 ∈ Σ such that K ( I1 ( I2 .

30

2. BASIC PROPERTIES OF MONOMIAL IDEALS

Continue this process inductively to construct a chain K ( I1 ( I2 ( I3 ( · · · of elements of Σ. The existence of this chain contradicts part (a).  Definition 2.3.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal of R. Let z1 , . . . , zm ∈ [I] such that I = (z1 , . . . , zm )R. The list z1 , . . . , zm is an irredundant generating sequence for I if zi is not a monomial multiple of zj whenever i 6= j. The list is a redundant generating sequence for I if it is not irredundant. Example 2.3.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. The sequence X 3 , X 2 Y, X 2 Y 2 , Y 5 is a redundant generating sequence for the ideal (X 3 , X 2 Y, X 2 Y 2 , Y 5 )R because 2 2 2 X Y X Y . The sequence X 3 , X 2 Y, XY 2 , Y 3 is an irredundant generating sequence for (X 3 , X 2 Y, XY 2 , Y 3 )R because none of the monomials X 3 , X 2 Y, XY 2 , Y 3 is a multiple of the other. Proposition 2.3.6. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal of R, and let z1 , . . . , zm ∈ [I] such that I = (z1 , . . . , zm )R. The following conditions are equivalent: (i) the generating sequence z1 , . . . , zm is irredundant; (ii) for i = 1, . . . , m we have zi ∈ / (z1 , . . . , zi−1 , zi+1 , . . . , zm )R; and (iii) for i = 1, . . . , m we have (z1 , . . . , zi−1 , zi+1 , . . . , zm )R ( I. Proof. (i) =⇒ (ii): Assume that the generating sequence z1 , . . . , zm is irredundant. If zi ∈ (z1 , . . . , zi−1 , zi+1 , . . . , zm )R, then Lemma 2.1.14 implies that zi ∈ zj R for some j 6= i; Lemma 2.1.11 then implies that zi is a monomial multiple of zj , contradicting the assumption the irredundancy of the generating sequence. (ii) =⇒ (iii): If zi ∈ / (z1 , . . . , zi−1 , zi+1 , . . . , zm )R, then zi is in I and is not in (z1 , . . . , zi−1 , zi+1 , . . . , zm )R so (z1 , . . . , zi−1 , zi+1 , . . . , zm )R ( I. (iii) =⇒ (i): Assume that (z1 , . . . , zi−1 , zi+1 , . . . , zm )R ( I for i = 1, . . . , m. Suppose that the sequence z1 , . . . , zm is redundant and fix indices i, j such that i 6= j and zi is a monomial multiple of zj . Then zi ∈ zj R ⊆ (z1 , . . . , zi−1 , zi+1 , . . . , zm )R. It follows that I = (z1 , . . . , zi−1 , zi , zi+1 , . . . , zm )R ⊆ (z1 , . . . , zi−1 , zi+1 , . . . , zm )R contradicting our assumption.



Theorem 2.3.7. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal of R. (a) Every monomial generating set for I contains an irredundant monomial generating sequence for I. (b) The ideal I has an irredundant monomial generating sequence. (c) Irredundant monomial generating sequences are unique up to re-ordering. Proof. (a) Assume without loss of generality that S 6= 0. Corollary 2.3.2 implies that I can be generated by a finite list of monomials z1 , . . . , zm ∈ S. If the sequence is redundant, then Proposition 2.3.6 provides an index i such that (z1 , . . . , zi−1 , zi+1 , . . . , zm )R = I; hence, the shorter list z1 , . . . , zi−1 , zi+1 , . . . , zm of monomials generates I. Repeat this process with the new list, removing elements from the list until the remaining elements form an irredundant generating sequence for I. The process will terminate in finitely many steps as the original list is finite.

2.3. GENERATORS OF MONOMIAL IDEALS

31

(b) The ideal I has a monomial generating set by definition, so the desired conclusion follows from part (a). (c) Assume that z1 , . . . , zm and w1 , . . . , wn are two irredundant generating sequences. We show that m = n and that there is a permutation σ ∈ Sn such that zi = wσ(i) for i = 1, . . . , n. Fix an index i. Since zi is in I = (w1 , . . . , wn )R, Lemma 2.1.14 implies that zi is a monomial multiple of wj for some index j. Similarly, there is an index k such that wj is a monomial multiple of zk . The transitivity of the divisibility order on the monomial set [R] implies that zi is a multiple of zk ; that is, zi is a monomial multiple of zk . Since the generating sequence z1 , . . . , zm is irredundant, it follows that i = k. It follows that zi wj and wj zi . The fact that the divisibility order on [R] is antisymmetric implies zi = wj . In summary, we see that for each index i = 1, . . . , m there exists an index j = σ(i) such that zi = wj = wσ(i) . Since the zi are distinct and the wj are distinct, we conclude that the function σ : {1, . . . , m} → {1, . . . , n} is 1-1. By symmetry, there is a 1-1 function δ : {1, . . . , n} → {1, . . . , m} such that wi = zδ(i) for i = 1, . . . , n. It follows that m 6 n 6 m and so m = n. Furthermore, since σ is 1-1 and m = n, the pigeon-hole principle implies that σ is also onto. This is the desired conclusion.  Here is an algorithm for finding an irredundant generating sequence. Algorithm 2.3.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix monomials z1 , . . . , zm ∈ [R] and set J = (z1 , . . . , zm )R. We assume that m > 1. Step 1. Check whether the generating sequence z1 , . . . , zm is irredundant using the definition. Step 1a. If, for all indices i and j such that i 6= j, we have zj ∈ / (zi )R, then the generating sequence is irredundant; in this case, the algorithm terminates. Step 1b. If there exist indices i and j such that i 6= j and zj ∈ (zi )R, then the generating sequence is redundant; in this case, continue to Step 2. Step 2. Reduce the generating sequence by removing the generator which causes the redundancy in the generating sequence. By assumption, there exist indices i and j such that i 6= j and zj ∈ (zi )R. Reorder the indices to assume without loss of generality that j = m. Thus, we have i < m and zm ∈ (zi )R. It follows that J = (z1 , . . . , zm )R = (z1 , . . . , zm−1 )R. Now apply Step 1 to the new list of monomials z1 , . . . , zm−1 . The algorithm will terminate in at most m − 1 steps because one can remove at most m − 1 monomials from the list and still form a non-zero ideal. Example 2.3.9. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Using Algorithm 2.3.8, one finds that the sequence X 3 , X 2 Y, Y 5 is an irredundant generating sequence for the ideal (X 3 , X 2 Y, X 2 Y 2 , Y 5 )R. Here is a proposition that shows you how to this in one shot. Part (b) says something about the monoid Nd being noetherian. Proposition 2.3.10. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix a non-empty set of monomials S ⊆ [R] and set J = (S)R. For each z ∈ S, write z = Xnz with

32

2. BASIC PROPERTIES OF MONOMIAL IDEALS

nz ∈ Nd . Set ∆ = {nz | z ∈ S} ⊆ Nd and consider the order < on Nd from Definition 1.6.8. Let ∆0 denote the set of minimal elements of ∆ under this order. (a) Then S 0 = {z | nz ∈ ∆0 } is an irredundant monomial generating set for J. (b) The set ∆0 is finite. Proof. Note: the set ∆ has minimal elements by the Well-Ordering Axiom. (a) The minimality of the elements of ∆0 implies that for each nz ∈ ∆, there is an element nw ∈ ∆0 such that nz < nw ; it follows that z ∈ (w)R. From this, we conclude that J = (S)R ⊆ ({w ∈ S | nw ∈ ∆0 })R ⊆ (S)R = J so J = ({w ∈ S | nw ∈ ∆0 })R = (S 0 )R. For distinct elements w, z ∈ S 0 we have nw 6< nz since nw and nz are both minimal among the elements of ∆ and they are distinct. It follows that w ∈ / (z)R. Theorem 2.3.7(a) implies that S 0 contains an irredundant monomial generating sequence s1 , . . . , sn ∈ S 0 for J. We claim that S 0 = {s1 , . . . , sn }. (From this, it follows that S 0 is an irredundant monomial generating set for J. We know that {s1 , . . . , sn } ⊆ S 0 , so suppose by way of contradiction that {s1 , . . . , sn } ( S 0 , and let s ∈ S 0 r {s1 , . . . , sn }. It follows that s ∈ J = (s1 , . . . , sn )R so s ∈ (si )R for some i by Lemma 2.1.14. The previous paragraph implies that s = si ∈ {s1 , . . . , sn }, a contradiction. (b) The set ∆0 is in bijection with S 0 , which is finite.  Definition 2.3.11. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R, and let νR (J) denote the number of elements in the irredundant monomial generating sequence for J. Example 2.3.12. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Example 2.3.9 shows that the sequence X 3 , X 2 Y, Y 5 is an irredundant generating sequence for the ideal J = (X 3 , X 2 Y, X 2 Y 2 , Y 5 )R, so we have νR (J) = 3. Definition 2.3.13. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. We define the lexicographical order on the set of monomials [R] as follows: For monomials f = X a Y b and g = X c Y d write f
2.3. GENERATORS OF MONOMIAL IDEALS

33

by points that are to the right of f or directly above f .

O

.. .O

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4

−

•O

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In the graph, we have represented the order with sequential vertical arrows. Fact 2.3.16. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. The lexicographical order 6lex on [R] is a total order; see Definition 1.6.6. Moreover, it is a well-order, that is, a total order such that every non-empty subset of [R] has a unique minimal element. Also, given monomials f, g ∈ [R] we have f 2. For i = 1, . . . , n we write fi = X ai Y bi . If fi bj . Proof. By definition, the inequality fi aj . The previous paragraph shows that this implies that ai = aj and bi < bj . In other words, in the order on N2 we have (aj , bj ) < (ai , bi ). It follows that fj is a monomial multiple of fi , contradicting the irredundancy of the generating sequence. Suppose next that bi 6 bj . The condition ai < aj that we have already established then implies (aj , bj ) < (ai , bi ), and this again contradicts the irredundancy of the generating sequence. 

Example 2.3.18. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Let J be a monomial ideal in R. Graphically, the previous lemma says that, when the monomials of the irredundant generating sequence for J are arranged in lexicographical order, they form a strictly

34

2. BASIC PROPERTIES OF MONOMIAL IDEALS

descending sequence O

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3

4

Exercises. Exercise 2.3.19. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I1 , . . . , In be monomial ideals in R and set (X) = (X1 , . . . , Xd )R. (a) Prove that the product I1 · · · In is a monomial ideal. (b) Prove that if rad (Ij ) = rad ((X)) for j = 1, . . . , n, then rad (I1 · · · In ) = rad ((X)). (c) Prove that if Ij 6= R for j = 1, . . . , n and rad (I1 · · · In ) = rad ((X)), then rad (Ij ) = rad ((X)) for j = 1, . . . , n. (d) Prove that [I1 · · · In ] = {z1 · · · zn | z1 ∈ [I1 ], . . . , zn ∈ [In ]}. Exercise 2.3.20. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I1 , . . . , In be monomial ideals in R and set (X) = (X1 , . . . , Xd )R. (a) Prove that the sum I1 + · · · + In is a monomial ideal. (b) Prove that if rad (Ij ) = rad ((X)) for j = 1, . . . , n, then rad (I1 + · · · + In ) = rad ((X)).

2.4. NOETHERIAN RINGS (OPTIONAL)

35

(c) Prove or give a counter-example for the following: if rad (I1 + · · · + In ) = rad ((X)), then rad (Ij ) = rad ((X)) for j = 1, . . . , n. (d) Prove that [I1 + · · · + In ] = [I1 ] ∪ · · · ∪ [In ]. (e) Prove that Γ(I1 + · · · + In ) = Γ(I1 ) ∪ · · · ∪ Γ(In ). Exercise 2.3.21. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. (a) Let I1 , . . . , Ik , and J be monomial ideals in R. Prove that (I1 + · · · + Ik ) ∩ J = (I1 ∩ J) + · · · + (Ik ∩ J). (b) Give an example (where d = 2) to show that this is not true without the assumption that each of the ideals I1 , I2 , and J are monomial ideals; justify your answer. Exercise 2.3.22. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Find irredundant generating sequences and compute νR (I) for the following monomial ideals and justify your answers: (a) I = (X15 , X1 X2 , X12 X23 , X23 )R (b) I = (X1 X22 X33 , X1 X3 , X2 X4 , X13 X22 X4 X5 )R Exercise 2.3.23. Let A be a commutative ring with identity. (a) Verify the properties in Fact 2.3.16. (b) Define a lexicographical order on the monomials [R] in the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Prove that this order is a total order. Is it a well-order? Justify your answer. Exercise 2.3.24. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal of R, and set (X) = (X1 , . . . , Xd )R. Prove that rad (I) = rad ((X)) if and only if the irredundant monomial generating sequence for I contains a positive power of each variable. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 2.4. Noetherian Rings (optional) In the previous section, we saw that every monomial ideal in the polynomial ring A[X1 , . . . , Xd ] is finitely generated. This condition can fail for more general ideals. Specifically, there exist rings with ideals that are not finitely generated; see Example 2.4.4(d). This motivates the study of noetherian rings; see Theorem 2.4.2. Definition 2.4.1. Let R be a commutative ring with identity. We say that an ascending chain I1 ⊆ I2 ⊆ I3 ⊆ · · · of ideals in R stabilizes if there is a positive integer n such that In = In+1 = In+2 = · · · . The ring R satisfies the ascending chain condition (ACC) for ideals if every ascending chain of ideals in R stabilizes. Compare the next result with Theorems 2.3.1 and 2.3.3. Theorem 2.4.2. Let R be a commutative ring with identity. The following conditions are equivalent: (i) every ideal of R is finitely generated; (ii) every non-empty set of ideals of R contains a maximal element; and

36

2. BASIC PROPERTIES OF MONOMIAL IDEALS

(iii) R satisfies the ascending chain condition for ideals. Proof. (iii) =⇒ (ii): We argue by contradiction. Assume that R satisfies the ascending chain condition for ideals, and suppose that R admits a non-empty set Σ of ideals with no maximal element. Let I1 ∈ Σ. Since Σ has no maximal element, there is an element I2 ∈ Σ such that I1 ( I2 . Since Σ has no maximal element, there is an element I3 ∈ Σ such that I2 ( I3 . Inductively, this process yields an ascending chain I1 ( I2 ( I3 ( · · · . By construction, this chain does not stabilize, contradicting the assumption that R satisfies ACC for ideals. (ii) =⇒ (i): Assume that every non-empty set of ideals of R contains a maximal element. Let I be an ideal of R, and let Σ = {finitely generated ideals J ⊆ R | J ⊆ I}. Note that Σ 6= ∅ since 0 = (0R )R ∈ Σ. By assumption, the set Σ has a maximal element J. Then J ⊆ I and J is finitely generated. We claim that J = I. By way of contradiction, suppose that J ( I and let a ∈ I r J. Since J is finitely generated, so is J + aR. Furthermore, since a ∈ I, we have aR ⊆ I, so the assumption J ⊆ I implies that J + aR ⊆ I. Furthermore, we have a ∈ aR ⊆ J + aR, so the condition a ∈ / J implies that J + aR is an ideal in Σ that properly contains J, contradicting the maximality of J. Hence, we have J = I and so I is finitely generated. (i) =⇒ (iii): Assume that every ideal of R is finitely generated. Consider an ascending chain I1 ⊆ I2 ⊆ I3 ⊆ · · · of ideals in R, and set I = ∪∞ j=1 Ij . Fact 1.3.4(c) implies that I is an ideal of R. Our assumption implies that I = (r1 , . . . , rk )R for some elements ri ∈ R. For i = 1, . . . , k we have ri ∈ I = ∪∞ j=1 Ij , and so there exists a positive integer ni such that ri ∈ Ini . With n = max{n1 , . . . , nk } we have ri ∈ Ini ⊆ In and so I = (r1 , . . . , rk )R ⊆ In . Hence, for each m > 0, we have In+m ⊆ I ⊆ In ⊆ In+m and so In+m = In . That is, the chain stabilizes, as desired.



Definition 2.4.3. Let R be a commutative ring with identity. Then R is noetherian if it satisfies the equivalent conditions of Theorem 2.4.2. Example 2.4.4. (a) The ring Z is noetherian because every ideal is principal; see Exercise 1.3.22. (b) Every finite ring is noetherian. In particular, the ring Zn is noetherian. (c) Every field k is noetherian since its only ideals are 0 = 0k and k = 1k k. In particular, the rings Q, R and C are noetherian. (d) Given a non-zero commutative ring A with identity, the polynomial ring R = A[X1 , X2 , X3 , . . .] in infinitely many variables is not noetherian, since the ideal (X1 , X2 , X3 , . . .)R is not finitely generated. Also, the chain of ideals (X1 )R ( (X1 , X2 )R ( (X1 , X2 , X3 )R ( · · · never stabilizes. (e) The ring R = C(R) of continuous functions is not noetherian. Indeed, for each n ∈ N define In = {f ∈ R | f (m) = 0 for all k ∈ N such that k > n}. Then the chain of ideals I0 ( I1 ( I2 ( · · · never stabilizes. See Exercise 2.4.11. Here is the Hilbert Basis Theorem. Compare it with Theorem 2.3.1.

2.4. NOETHERIAN RINGS (OPTIONAL)

37

Theorem 2.4.5 (Hilbert Basis Theorem). Let A be a commutative ring with identity. If A is noetherian, then the polynomial ring A[X] in one variable is noetherian. Proof. Let I be an ideal of A[X]. We need to show that I is finitely generated. Assume without loss of generality that I 6= 0. Let J = {a ∈ A | a is the leading coefficient of some f ∈ I} ∪ {0A }. The set J is an ideal of A; see Exercise 2.4.9. The fact that A is noetherian implies that there exist a1 , . . . , an ∈ J such that J = (a1 , . . . , an )A. Assume without loss of generality that each ai is non-zero. By definition, for each i = 1, . . . , n there is a polynomial fi ∈ I such that ai is the leading coefficient of fi . By multiplying each polynomial fi by a power of X, we may assume that the polynomials f1 , . . . , fn have the same degree; let N denote this common degree. For d = 0, 1, . . . , N − 1 let Jd = {a ∈ A | a is the leading coefficient of some f ∈ I of degree d} ∪ {0A }. The set Jd is an ideal of A; see Exercise 2.4.9. The fact that A is noetherian implies that there exist bd,1 , . . . , bd,nd ∈ Jd such that J = (bd,1 , . . . , bd,nd )A. For each d such that Jd 6= 0, we assume without loss of generality that each bd,i is non-zero. By definition, for each d such that Jd 6= 0 and for each i = 1, . . . , nd there is a polynomial gd,i ∈ I such that bd,i is the leading coefficient of gd,i . For each d such that Jd = 0, we set nd = 1 and bd,1 = 0 = gd,1 . Let I 0 be the ideal of A[X] generated by the polynomials fi and gd,i : I 0 = ({fi | i = 1, . . . , n} ∪ {gd,i | d = 0, . . . , N − 1; i = 1, . . . , nd }) A[X]. By construction I 0 is a finitely generated ideal of A[X]. Since each fi , gd,i ∈ I, we have I 0 ⊆ I. Claim: I 0 = I. (Once the claim is established, we conclude that I is finitely generated, completing the proof.) Suppose by way of contradiction that I 0 ( I Let f be a polynomial of minimal degree in the compliment I r I 0 , Let e denote the degree of f and let a be the leading coefficient of f . Suppose that e > N . Since f ∈PI, we have a ∈ J, and so there exist elen ments r1 , . . . , rn ∈ A such that a = i=1 ri ai . For i = 1, . . . , n the polynomial e−N 0 ri X fi is an element of I with degree at most e. Also, the coefficient of X e e−N 0 in thePpolynomial ri X e−N fi is ri ai . Since each polynomial Pn ri X e−Nfi is in I , we n e−N 0 have i=1 ri X fi ∈ I P . Exercise 1.2.12 implies that i=1 ri XP fi has degree n n e and leading coefficient i=1 ri ai . Fact 1.3.2 that f − i=1 ri X e−N fi is Pn impliese−N 0 fi has degree strictly less in I r I . Furthermore, the polynomial f − i=1 ri X than e, contradicting the minimality of e. Hence, we have e < P N . It follows that a ∈ Je and so exist elements Pnthere ne e s1 , . . . , sne such that a = i=1 si be,i . The polynomial f − i=1 si ge,i is in I r I 0 and has degree strictly less than e. Again, this contradicts the minimality of e. It  follows that I 0 = I, as claimed. Remark 2.4.6. “This is not mathematics, it is theology.” This quote is attributed to the mathematician Paul Gordan, circa 1890, in response to Hilbert’s proof of Theorem 2.4.5. Gordan was apparently dismayed by the non-constructive nature of the proof. (The authors of this text do not necessarily agree with Gordan’s assessment.)

38

2. BASIC PROPERTIES OF MONOMIAL IDEALS

Remark 2.4.7. In the language of the 1890’s, the conclusion of Theorem 2.4.5 says that every ideal of A[X] has a finite basis, hence the name Hilbert Basis Theorem. In today’s language, we say “finite generating set” instead of “finite basis”. Corollary 2.4.8. Let A be a commutative ring with identity. If A is noetherian, then the polynomial ring A[X1 , . . . , Xd ] in d variables is noetherian. Proof. By induction on d using Theorem 2.4.5.



Exercises. Exercise 2.4.9. Verify that the sets J and Jd from the proof of Theorem 2.4.5 are ideals of A. Exercise 2.4.10. Let R be a commutative ring with identity. Let I be a finitely generated ideal of R and S ⊂ R a generating set for I. (a) Prove that there exist s1 , . . . , sn ∈ S such that I = (s1 , . . . , sn )R. (b) Prove that if R is noetherian and I = (S)R, then I is generated by a finite subset of elements of S. Exercise 2.4.11. Prove that the ring of continuous functions C(R) is not noetherian. Exercise 2.4.12. Prove Corollary 2.4.8. Exercise 2.4.13. Let A be a commutative ring with identity. Prove that the following conditions are equivalent: (i) the ring A is noetherian; (ii) for each integer d > 1 the polynomial ring A[X1 , . . . , Xd ] in d variables is noetherian; and (iii) there is an integer d > 1 such that the polynomial ring A[X1 , . . . , Xd ] in d variables is noetherian. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output.

CHAPTER 3

Operations on Monomial Ideals In this chapter we look at situations where the operations of Section 1.4, applied to monomial ideals, yield monomial ideals. We have already seen this a little in Exercises 2.3.19 and 2.3.20. We also consider the effect of some of these operations on the graphs of the monomial ideals. 3.1. Intersections of Monomial Ideals Theorem 3.1.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. If I1 , . . . , In are monomial ideals of R, then the intersection I1 ∩ · · · ∩ In is generated by the set of monomials in I1 ∩ · · · ∩ In . In particular, the ideal I1 ∩ · · · ∩ In is a monomial ideal of R and [I1 ∩ · · · ∩ In ] = [I1 ] ∩ · · · ∩ [In ]. Proof. Let S denote the set of monomials ∩nj=1 [Ij ] and set J = (S)R. By construction J is a monomial ideal such that J ⊆ ∩nj=1 Ij , since S ⊆ ∩nj=1 Ij . We claim that J = ∩nj=1 Ij . To show this, fix an element f ∈ ∩nj=1 Ij and write Pfinite f = n∈Nd an Xn . For j = 1, . . . , n we have f ∈ Ij . Hence, Exercise 2.1.25 implies that if an 6= 0, then Xn ∈ [Ij ] for each j, that is, if an 6= 0, then Xn ∈ ∩nj=1 [Ij ] = S. Hence, we have f ∈ (S)R = J, as desired. The previous paragraph shows that I1 ∩ · · · ∩ In is a monomial ideal of R and is generated by the monomial set ∩nj=1 [Ij ]. We complete the proof with the following computation:
[∩nj=1 Ij ] = ∩nj=1 Ij ∩ [R] = ∩nj=1 (Ij ∩ [R]) = ∩nj=1 [Ij ].  Remark 3.1.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I1 , . . . , In be monomial ideals of R. The fact that the intersection I1 ∩ · · · ∩ In is a monomial ideal such that [I1 ]∩· · ·∩[In ] = [I1 ]∩· · ·∩[In ] implies that Γ(I1 ∩· · ·∩In ) = Γ(I1 )∩· · ·∩Γ(In ). Next we show how to find a monomial generating sequence for an intersection of monomial ideals. Definition 3.1.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f = Xm and g = Xn for some m, n ∈ Nd . For i = 1, . . . , d set pi = max{mi , ni }. Define the least common multiple of f and g to be the monomial lcm(f, g) = Xp . Example 3.1.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables. We compute the least common multiple of f = XY 4 Z 8 and g = X 3 Z 5 . In the notation of Definition 3.1.3, we have m = (1, 4, 8) and n = (3, 0, 5), and thus p = (3, 4, 8). This yields lcm(XY 4 Z 8 , X 3 Z 5 ) = X 3 Y 4 Z 8 . 39

40

3. OPERATIONS ON MONOMIAL IDEALS

The next example motivates the connection between intersections of monomial ideals and least common multiples. Example 3.1.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. We compute the intersection (X, Y 2 )R ∩ (X 2 Y )R. Because of Theorem 3.1.1 and Remark 3.1.2, we need to compute Γ((X, Y 2 )R ∩ (X 2 Y )R) = Γ((X, Y 2 )R) ∩ Γ((X 2 Y )R).

KEY

O

.. .

.. .

.. .

.. .

4

−

◦

~

~

~

···

Γ((X, Y 2 )R)

◦

3

−

◦

~

~

~

···

Γ((X 2 Y )R)

∗

2

−

◦

~

~

~

···

Γ((X, Y 2 )R ∩ (X 2 Y )R)

~

1

−

∗

∗

∗

···

|

|

|

|

/

1

2

3

4

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.. .

.. .

.. .

4

−

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···

3

−

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···

2

−

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~

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···

1

− |

|

|

|

/

1

2

3

4

0 0

0 0

From this, we see that (X, Y 2 )R ∩ (X 2 Y )R = (X 2 Y 2 )R = (lcm(XY 2 , X 2 Y ))R. In words, the intersection of the principal ideals generated by XY 2 and X 2 Y is principal and is generated by lcm(XY 2 , X 2 Y ). The next result shows that this is true for any two principal monomial ideals. Lemma 3.1.6. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. For monomials f, g ∈ [R], one has (f )R ∩ (g)R = (lcm(f, g))R. Proof. Exercise 3.1.12.



Proposition 3.1.7. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Suppose I is generated by the set of monomials {f1 , . . . , fm } and J is generated by the set of monomials {g1 , . . . , gn }.

3.1. INTERSECTIONS OF MONOMIAL IDEALS

41

Then I ∩ J is generated by the set of monomials

{lcm(fi , gj ) | 1 6 i 6 m, 1 6 j 6 n}. Proof. We begin by setting

K = ({lcm(fi , gj ) | 1 6 i 6 m, 1 6 j 6 n})R.

This is a monomial ideal in R since each element lcm(fi , gj ) is a monomial in R. For the containment I ∩ J ⊆ K, it suffices to show that every monomial z ∈ [I ∩ J] is in K. The element z is a monomial in I = (f1 , . . . , fm )R so Lemma 2.1.14 implies that z ∈ (fi )R for some index i. Similarly, the condition z ∈ J = (g1 , . . . , gn )R implies that z ∈ (gj )R for some index j. Hence, Lemma 3.1.6 yields z ∈ (fi )R ∩ (gj )R = (lcm(fi , gj ))R ⊆ K as desired. For the containment I ∩J ⊇ K, it suffices to show that each monomial generator lcm(fi , gj ) of K is in I ∩ J. Lemma 2.1.14 implies the equality in the next sequence

lcm(fi , gj ) ∈ (lcm(fi , gj ))R = (fi )R ∩ (gj )R ⊆ I ∩ J

while the rest of the sequence are standard. This gives the desired conclusion.



Example 3.1.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. We compute a generating sequence for the ideal I = (X 2 , Y 3 )R ∩ (X 3 , Y )R:

.. .O

.. .

.. .

.. .

.. .

4

•

•

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···

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1

−

• | 1

0 0 |

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.. .

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4

•

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···

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···

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···

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···

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0

|

|

•

•

/

2 {z

3

4

1

2 {z

3

4

Γ((X 2 ,Y 3 )R)

\

0 }

|

Γ((X 3 ,Y )R.)

}

42

3. OPERATIONS ON MONOMIAL IDEALS

Theorem 3.1.1 implies that [I] = [(X 2 , Y 3 )R] ∩ [(X 3 , Y )R], so the graph of I is the following.

.. .O

.. .

.. .

.. .

.. .

4

•

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···

3

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···

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···

1

−

•

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···

|

|

•

•

/ ···

1

2

3

4

0 0

Using Proposition 2.3.10 and a visual inspection of the graph, we conclude that the irredundant monomial generating sequence for I is Y 3 , X 2 Y, X 3 . We now use Proposition 3.1.7 and Algorithm 2.3.8 to find the irredundant monomial generating sequence for I. In the notation of Proposition 3.1.7, we have f1 = X 2 , f2 = Y 3 , g1 = X 3 and g2 = Y . We compute the relevant LCM’s:

lcm(f1 , g1 ) = X 3

lcm(f2 , g1 ) = X 3 Y 3

lcm(f1 , g2 ) = X 2 Y

lcm(f2 , g2 ) = Y 3 .

Proposition 3.1.7 implies that the sequence X 3 , X 3 Y 3 , X 2 Y, Y 3 generates I. Now we use Algorithm 2.3.8 to find the irredundant monomial generating sequence for this ideal. The list of zi ’s to consider is X 3 , X 3 Y 3 , X 2 Y, Y 3 . The monomial X 3 Y 3 is a multiple of X 3 , so we remove X 3 Y 3 from the list. The new list of zi ’s to consider is X 3 , X 2 Y, Y 3 . No monomial in this list is a multiple of another since the exponent vectors (3, 0), (2, 1), and (0, 3) are incomparable. Hence, the list X 3 , X 2 Y, Y 3 is an irredundant monomial generating sequence for J. Let I be a monomial ideal of R. The goal of the paper is the following: given a monomial ideal I, to find simpler monomial ideals I1 , . . . , In ⊆ R such that I = I1 ∩ · · · ∩ In . A hint as to how this might be done is found in the previous example, as we discuss next. Example 3.1.9. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set I = (X 3 , X 2 Y, Y 3 )R. The

3.1. INTERSECTIONS OF MONOMIAL IDEALS

43

graph Γ(I) has the following form. .. .O

.. .

.. .

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.. .

4

•

•

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···

3

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···

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···

1

−

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···

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/ ···

1

2

3

4

0 0

The two corners of the form q break the “negative space” into two pieces .. .O

.. .

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.. .

4

•

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···

3

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···

1

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···

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and suggest the following decomposition. \ .. .. .. .. .. .O . . . .

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2 {z

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2 {z

3

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Γ((X 2 ,Y 3 )R)

0 }

|

Γ((X 3 ,Y )R)

}

Remark 3.1.10. The the methods from Proposition 3.1.7 and Example 3.1.8 can be extended to intersections of three or more monomial ideals inductively, simply

44

3. OPERATIONS ON MONOMIAL IDEALS

by repeating the process. For instance, to find a monomial generating sequence for I ∩ J ∩ K, first find a monomial generating sequence for I ∩ J, then find one for (I ∩ J) ∩ K. See also Exercise 3.1.16. Exercises. Exercise 3.1.11. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Prove that if {Iλ }λ∈Λ is a (possibly infinite) set of monomial ideals in R, then ∩λ∈Λ Iλ is a monomial ideal. Exercise 3.1.12. Prove Lemma 3.1.6. Exercise 3.1.13. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Assume that I is generated by the set of monomials S and that J is generated by the set of monomials T . Prove or disprove the following: The ideal I ∩ J is generated by the set of monomials L = {lcm(f, g) | f ∈ S and g ∈ T }. Exercise 3.1.14. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f = Xm , g = Xn , and w = Xp be monomials in R. Prove that the following conditions are equivalent. (i) w = lcm(f, g); (ii) w is a common multiple of f and g, and if h ∈ R is a common multiple of f and g, then h is a multiple of w (Note that we have not assumed that h is a monomial.); and (iii) we have m 6 p and n 6 p, and if e ∈ Nd is such that m 6 e and n 6 e, then p 6 e. Exercise 3.1.15. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f = Xm and g = Xn for some m, n ∈ Nd . For i = 1, . . . , d set qi = min{mi , ni }. Define the greatest common divisor of f and g to be the monomial gcd(f, g) = Xq . Prove that lcm(f, g) gcd(f, g) = f g. Exercise 3.1.16. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f = Xm and g = Xn and h = Xp for some m, n, p ∈ Nd . For i = 1, . . . , d set qi = max{mi , ni , pi }. Define the least common multiple of f , g, and h to be the monomial lcm(f, g, h) = Xq . (a) Prove that lcm(lcm(f, g), h) = lcm(f, g, h) = lcm(f, lcm(g, h)). (b) Prove that (f )R ∩ (g)R ∩ (h)R = (lcm(f, g, h))R. (c) Assume that I is generated by the set of monomials {f1 , . . . , fm } and that J is generated by the set of monomials {g1 , . . . , gn } and K is generated by the set of monomials {h1 , . . . , hp }. Prove that I ∩ J ∩ K is generated by the set of monomials L = {lcm(fi , gj , hk ) | 1 6 i 6 m, 1 6 j 6 n, 1 6 k 6 p}. (d) For a sequence z1 , . . . , za ∈ [R], propose a definition of the term “least common multiple of z1 , . . . , za . State and prove the versions of parts (a)–(c) for your definition. Exercise 3.1.17. Let K be a field, and let I be a monomial ideal in the polynomial ring R = K[X1 , . . . , Xd ] in d variables. (a) Prove that rad (I) is a monomial ideal. (b) Describe [rad (I)] and Γ(rad (I)) in terms of [I] and Γ(I).

3.2. UNIQUE FACTORIZATION DOMAINS (OPTIONAL)

45

Exercise 3.1.18. Give an example of a commtuative ring A with identity that satisfies the following property: there exists a monomial ideal I in the polynomial ring R = A[x, y] in two variables such that rad (I) is not a monomial ideal; justify your answer. Exercise 3.1.19. Let A be a commutative ring with identity. An element a ∈ A is nilpotent if there exists n > 0 such that an = 0A . The nilradical of A is N(A) = {nilpotent elements of A}. The ring A is reduced if N(A) = 0. (a) Prove that N(A) is an ideal of A. (b) Prove that if a ∈ A and n > 0 such that an ∈ N(A), then a ∈ N(A). (c) Prove that if R is an integral domain, then R is reduced. Exercise 3.1.20. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Prove that the following conditions are equivalent: (i) the ring A is reduced; (ii) for every monomial ideal I ⊂ R, the ideal rad (I) is a monomial ideal; and (iii) there exists a monomial ideal I ⊂ R such that rad (I) is a monomial ideal. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 3.2. Unique Factorization Domains (optional) We have seen that in a polynomial ring A[X], each monomial Xn has a unique factorization as a product of powers of the variables Xi , up to re-ordering the factors. This is similar to the Fundamental Theorem of Arithmetic which states that every positive integer has a unique factorization as a product of powers of prime numbers. This section deals with classes of rings with similar properties. Definition 3.2.1. Let R be an integral domain. Two elements r, s ∈ R are associates if (r)R = (s)R. A non-zero non-unit t ∈ R is irreducible if it admits no non-trivial factorization, i.e., if for every a, b ∈ R such that t = ab either a is a unit or b is a unit. non-unit p ∈ R is prime if for every a, b ∈ R such that p ab either A non-zero p a or p b. Example 3.2.2. Two integers m and n are associates in Z if and only if m = ±n. A non-zero positive integer is irreducible in Z if and only if it is a prime number, that is, if and only if it is a prime element in the terminology of Definition 3.2.1. Fact 3.2.3. Let R be an integral domain, and let r, s ∈ R. The following conditions are equivalent: (i) r and s are associates; (ii) r s and s r; (iii) r ∈ (s)R and s ∈ (r)R; and (iv) there is a unit u ∈ R such that r = us. Fact 3.2.4. Let R be an integral domain, and let r, r0 , s, u ∈ R. (a) If u ∈ R is a unit and r ∈ R, then r u if and only if r is a unit.

46

3. OPERATIONS ON MONOMIAL IDEALS

(b) If r and r0 are associates, then r s if and only if r0 s. Fact 3.2.5. Let R be an integral domain. Let p and q be non-zero non-units in R, and let a1 , . . . , an ∈ R. (a) If p is prime and p a1 · · · an , then there is an index i such that p ai . (b) If p is prime, then p is irreducible. (c) The converse to part (b) can fail in general. See however Lemma 3.2.10. (d) If p and q are prime and p q, then p and q are associates. Definition 3.2.6. An integral domain R is a unique factorization domain provided that every non-zero non-unit of R can be factored as a product of irreducible elements in an essentially unique way, that is: (1) for every non-zero non-unit r ∈ R there are (not necessarily distinct) irreducible elements s1 , . . . , sm ∈ R such that r = s1 · · · sm ; and (2) for all irreducible elements s1 , . . . , sm , t1 , . . . , tn if s1 · · · sm = t1 · · · tn , then m = n and there is a permutation σ of the integers 1, . . . , m such that si and tσ(i) are associates for i = 1, . . . , m. Remark 3.2.7. The term “unique factorization domain” is frequently abbreviated as UFD. Example 3.2.8. The Fundamental Theorem of Arithmetic states that the ring Z is a unique factorization domain. Every field is (vacuously) a unique factorization domain. The ring of Gaussian integers Z[i] = {a + bi ∈ C | a, b ∈ Z} is a unique factorization domain. The ring √ √ Z[ 10] = {a + b 10 ∈ R | a, b ∈ Z} is not a unique factorization domain; see [6, Sec. III.3]. Remark 3.2.9. If A is a unique factorization domain, then so is the polynomial ring A[X1 , . . . , Xd ] in d variables for each d. In particular, if k is a field, then k[X1 , . . . , Xd ] is a unique factorization domain. See, e.g., [6, Thm. III.6.12]. Lemma 3.2.10. Let R be a unique factorization domain. An element p ∈ R is irreducible if and only if it is prime. Proof. One implication is in Fact 3.2.5(b). For the converse, assume that p is irreducible in R. To show that p is prime, a or p b. Then there let a, b ∈ R and assume that p ab. We need to prove that p is an element c ∈ R such that ab = pc. If a = 0, then p a because a = 0 = 0 · p, and we are done. If a is a unit, then b = a−1 pc, so we have p b and we are done. Similarly, if b = 0 or if b is a unit, then we are done. So, we may assume that a and b are non-zero non-units. In particular, the equation pc = ab implies that c 6= 0. Assume that c is not a unit. (The case where c is a unit is handled similarly; see Exercise 3.2.24.) Since R is a unique factorization domain, there are irreducible elements a1 , . . . , ak , b1 , . . . , bm , c1 , . . . cn ∈ R such that a = a1 · · · ak and b = b1 · · · bm and c = c1 · · · cn . The equation pc = ab then reads pc1 · · · cn = a1 · · · ak b1 · · · bm . The uniqueness of factorizations in R implies that p is associate to one of the factors on the right-hand side. (Here is where we use the fact that p is irreducible.) If p

3.2. UNIQUE FACTORIZATION DOMAINS (OPTIONAL)

47

and ai are associates, then p a 1 , so p divides a1 · · · ak = a, as desired. Similarly, if p and bj are associates, then p b. Thus p is prime.  The proofs of the next two lemmas are tedious exercises in bookkeeping. Lemma 3.2.11. Let R be a unique factorization domain, and let r ∈ R be a nonzero non-unit. There exist irreducible elements p1 , . . . , pm ∈ R, a unit u ∈ R, and integers e1 , . . . , em > 1 such that (1) r = upe11 · · · pemm , and (2) for all i, j ∈ {1, . . . , m} such that i 6= j, the elements pi and pj are not associates. Proof. By definition there are (not necessarily distinct) irreducible elements s1 , . . . , sk ∈ R such that r = s1 · · · sk . Re-order the si so that they are grouped by associates. That is, re-order the si to assume that there are integers i0 = 0 < 1 = i1 < i2 < · · · < im < im+1 = k + 1 such that (1) for j = 1, . . . , m and l = ij , . . . , ij+1 −1 the elements sl and sij are associates, and (2) for j = 1, . . . , m if 1 6 l < ij 6 h the sl and sh are not associates. For j = 1, . . . , n set pj = sij and ej = ij − ij−1 > 1. For l = ij , . . . , ij+1 − 1 fix units ul ∈ R such that sl = ul sij = ul pj . Note that uij = 1R for j = 1, . . . , n. For j = 1, . . . , n set vj = uij · · · uij+1 −1 . This yields e

sij sij +1 · · · sij+1 −1 = pj (uij +1 pj ) · · · (uij+1 −1 pj ) = vj pj j . Qn Qk With u = j=1 vj = l=1 ul , it follows that we have r = s1 · · · sk = [si1 · · · si2 −1 ][si2 · · · si3 −1 ] · · · [sim · · · sim+1 −1 ] = [v1 pe11 ][v2 pe22 ] · · · [vn penn ] = upe11 pe22 · · · pemm . This is the desired factorization.



Lemma 3.2.12. Let R be a unique factorization domain, and let r, s ∈ R be nonzero non-units. There exist irreducible elements p1 , . . . , pn ∈ R, units u, v ∈ R, and integers e1 , . . . , en , f1 , . . . , fn > 0 such that (1) r = upe11 · · · penn and s = vpf11 · · · pfnn , and (2) for all i, j ∈ {1, . . . , n} such that i 6= j, the elements pi and pj are not associates. Proof. Lemma 3.2.11 yields irreducible elements p1 , . . . , pm , p01 , . . . , p0m0 ∈ R, integers e1 , . . . , em , e01 , . . . , e0m0 > 1, and units u, u0 ∈ R such that 0

0

(1) r = upe11 · · · pemm and s = u0 (p01 )e1 · · · (p0m0 )em , (2) for all i, j ∈ {1, . . . , m} such that i 6= j, the elements pi and pj are not associates, and (3) for all i, j ∈ {1, . . . , m0 } such that i 6= j, the elements p0i and p0j are not associates. Re-order the p0i if necessary to assume that there is an integer µ > 1 such that (4) for 1 6 i < µ the elements pi and p0i are associates, and (5) for µ 6 i 6 m and µ 6 i0 6 m0 the elements pi and p0i0 are not associates.

48

3. OPERATIONS ON MONOMIAL IDEALS

Set n = m + (m0 − µ + 1). For i = m + 1, . . . , n set pi = p0µ−m−1+i and ei = 0. Then one has r = upe11 · · · pemm = upe11 · · · pemm p0m+1 · · · p0n = upe11 · · · penn For i = 1, . . . , µ−1 set fi = e0i and fix a unit xi such that p0i = xi pi . For i = µ, . . . , m set fi = 0. For i = m + 1, . . . , n set fi = e0µ−m−1+i . Then one has 0

0

s = u0 (p01 )e1 · · · (p0m0 )em 0

0

0

0

= u0 (x1 p1 )e1 · · · (xµ pµ )eµ p0µ+1 · · · p0m (xm+1 pm+1 )eµ+1 · · · (xn pn )en = vpf11 · · · pfnn e0

e0

where v = u0 x11 · · · xmm0 .



Lemma 3.2.13. Let R be a unique factorization domain, and let r ∈ R be a nonzero non-unit. Fix irreducible elements p1 , . . . , pn ∈ R, integers e1 , . . . , en > 0, and a unit u ∈ R such that r = upe11 · · · penn , and for all i, j ∈ {1, . . . , n} such that i 6= j, the elements pi and pj are not associates. Given a prime element p ∈ R, one has p r if and only if there is an index i such that p and pi are associates and ei > 1. Proof. First, assume that there is an index i such that ei > 1 and the elements p and pi are associates. Then p pi , and hence p upe11 · · · pei i · · · penn = r. assume that p r = upe11 · · · penn . Fact 3.2.5(a) implies that either Conversely, e p u or p pi i for some index i. Since p is not a unit and u is a unit, Fact 3.2.4(a) implies that p - u. Thus, we have p pei i for some index i. If ei = 0 then p p0i = 1, which is impossible, again because p is not a unit. It follows that ei > 1 and p pi · · · pi . Another application of Fact 3.2.5(a) shows that p pi . We conclude from Fact 3.2.5(d) that p and pi are associates.  The next result compares to Lemma 2.1.11. Lemma 3.2.14. Let R be a unique factorization domain, and let r, s ∈ R be nonzero non-units. Given irreducible elements p1 , . . . , pn ∈ R, units u, v ∈ R, and integers e1 , . . . , en , f1 , . . . , fn > 0 as in Lemma 3.2.12, one has r s if and only if ei 6 fi for i = 1, . . . , n. Proof. Assume first that ei 6 fi for i = 1, . . . , n and consider the element x = vu−1 pf11 −e1 · · · pfnn −e n . It is straightforward to show that rx = s, so r s. Assume now that r s, and fix an element t ∈ R such that s = rt. Our assumptions imply that (3.2.14.1)

vpf11 · · · pfnn = upe11 · · · penn t.

We prove that fi > ei by induction on e = e1 + · · · + en . Base case: e = 0. In this case, each ei = 0, so we have fi > 0 = ei for each i. Induction step: Assume that e > 1 and that the result holds for elements of e0 e0 the form r0 = up11 · · · pnn where e01 + · · · e0n = e − 1. Since e > 1, we have ei > 1 for some index i. It follows that pi upe11 · · · penn t = vpf11 · · · pfnn . Lemma 3.2.13 implies that there is an index j such that pi and pj are associates and fj > 1. It follows that i = j, so we have fi > 1. Equation (3.2.14.1) now reads as pi (vpf11 · · · pifi −1 · · · pfnn ) = pi (upe11 · · · pei i −1 · · · penn t).

3.2. UNIQUE FACTORIZATION DOMAINS (OPTIONAL)

49

Since R is an integral domain, the cancellation property 2.2.4 implies that vpf11 · · · pifi −1 · · · pfnn = upe11 · · · piei −1 · · · penn t. The sum of the exponents on the right-hand side of this equation is e − 1, so the induction hypothesis implies that fj > ej for each j 6= i, and fi − 1 > ei − 1 so fi > ei , as desired.  Lemma 3.2.15. Let R be a unique factorization domain, and let r ∈ R be a nonzero non-unit. Fix irreducible elements p1 , . . . , pn ∈ R, integers e1 , . . . , en > 0, and a unit u ∈ R such that r = upe11 · · · penn , and for all i, j ∈ {1, . . . , n} such that i 6= j, the elements pi and pj are not associates. Given a non-zero element t ∈ R, one has t r if and only if there exist integers l1 , . . . , ln , and a unit w ∈ R such that t = wpl11 · · · plnn and 0 6 li 6 ei for i = 1, . . . , n. Proof. One implication follows from Lemma 3.2.14. For the converse, assume that t r. If t is a unit, then the integers l1 = · · · = ln = 0 and the unit w = t satisfy the desired conclusions. Assume that t is not a unit. Since t is also non-zero, it has a prime factor, say p. Since t r, we have p r, so Lemma 3.2.13 provides an index i such that p and pi are associates and ei > 1. Thus, Fact 3.2.4(b) implies that pi is a prime factor of t. In other words, in a prime factorization of t, the element p can be replaced with pi . This implies that there exist integers l1 , . . . , ln > 0, and a unit w ∈ R such that t = wpl11 · · · plnn . Since t r, Lemma 3.2.14 implies that li 6 ei for i = 1, . . . , n.  Definition 3.2.16. Let R be a commutative ring with identity, and let r, s ∈ R. (a) An element g ∈ R is a greatest common divisor for r and s if (1) one has g r and g s; and (2) for all h ∈ R such that h r and h s, one has h g. (b) An element l ∈ R is a least common multiple for r and s if (1) one has r l and s l; and (2) for all m ∈ R such that r m and s m, one has l m. Fact 3.2.17. Let R be an integral domain, and let r, s ∈ R. If g ∈ R is a greatest common divisor for r and s, then g 0 ∈ R is a greatest common divisor for r and s if and only if g and g 0 are associates. If l ∈ R is a least common multiple for r and s, then l0 ∈ R is a least common multiple for r and s if and only if l and l0 are associates. The next result compares to Exercise 3.1.14. Proposition 3.2.18. Let R be a unique factorization domain, and let r, s ∈ R be non-zero non-units. Then R contains a greatest common divisor and a least common multiple for r and s. Specifically, fix irreducible elements p1 , . . . , pn ∈ R, integers e1 , . . . , en , f1 , . . . , fn > 0, and units u, v ∈ R as in Lemma 3.2.12. For mn 1 i = 1, . . . , n let mi = min{ei , fi } and Mi = max{ei , fi }. Then g = pm 1 · · · pn M1 Mn is a greatest common divisor for r and s, and l = p1 · · · pn is a least common multiple for r and s. Proof. We prove that g is a greatest common divisor for r and s. The proof that l is a least common multiple for r and s is left as an exercise. Lemma 3.2.14 shows that g r and g s, since mi 6 ei and mi 6 fi . Now, assume that h ∈ R such that h r and h s; we need to show that h g. Since

50

3. OPERATIONS ON MONOMIAL IDEALS

h r, Lemma 3.2.15 provides integers l1 , . . . , ln , and a unit w ∈ R such that h = wpl11 · · · plnn and 0 6 li 6 ei for i = 1, . . . , n. Since h s, Lemma 3.2.14 implies that li 6 fi for i = 1, . . . , n so we have li 6 min{e i , fi } = mi for i = 1, . . . , n. Another application of Lemma 3.2.14 implies that h g, as desired.  The next result compares to Exercise 3.1.15. Corollary 3.2.19. Let R be a unique factorization domain, and let r, s ∈ R be non-zero non-units. Let g ∈ R be a greatest common divisor for r and s, and let l ∈ R be a least common multiple for r and s. Then there is a unit w ∈ R such that gl = wrs. Proof. See Exercise 3.2.27.



Exercises. Exercise 3.2.20. Verify the conclusions of Example 3.2.2. Exercise 3.2.21. Verify the conclusion of Fact 3.2.3. Exercise 3.2.22. Verify the conclusions of Fact 3.2.4. Exercise 3.2.23. Verify the conclusions of Fact 3.2.5. Exercise 3.2.24. Complete the proof of Lemma 3.2.10 by proving the desired conclusion when c is a unit. Exercise 3.2.25. Verify the conclusions of Fact 3.2.17. Exercise 3.2.26. Complete the proof of Proposition 3.2.18 by proving that l is a least common multiple for r and s. Exercise 3.2.27. Prove Corollary 3.2.19. Exercise 3.2.28. Let R be a unique factorization domain, and let r, s ∈ R. (a) Prove that (r)R ∩ (s)R = (lcm(r, s))R. (b) Prove or disprove the following: (r)R + (s)R = (gcd(r, s))R. (Hint: Feel free to use Remark 3.2.9.) Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 3.3. Colons of Monomial Ideals The next construction to analyze is the colon ideal. Theorem 3.3.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. If I and J are monomial ideals of R, then the colon ideal (J :R I) is a monomial ideal of R. Proof. Case 1: I = zR for some monomial z = Xm ∈ R. Let S denote the set of monomials in (J :R I) = (J :R zR) and set K = (S)R. By construction K is a monomial ideal such that K ⊆ (J :R I), since S ⊆ (J :R I). We claim that K = (J :R I). To show this, fix an element f ∈ (J :R I) and write f = Pfinite Pfinite n n+m ∈ J. By Exercise 2.1.25 we know that n∈Nd an X . Then f z = n∈Nd an X n+m if an 6= 0, then X ∈ J, since J is a monomial ideal. So, if an 6= 0, then

3.3. COLONS OF MONOMIAL IDEALS

51

zXn = Xn+m ∈ J. In other words, if an 6= 0, then Xn ∈ (J :R zR) = (J :R I), and Pfinite so Xn ∈ S ⊆ (S)R = K. It follows that f = n∈Nd an Xn ∈ K, as desired. Case 2: The general case. The ideal I is generated by a finite list of monomials z1 , . . . , zn . It follows that

(J :R I) = (J :R (z1 , . . . , zn )R) = ∩ni=1 (J :R zi R).

Thus, the ideal (J :R I) is a finite intersection of monomial ideals. By Theorem 3.1.1, it follows that (J :R I) is a monomial ideal. 

Remark 3.3.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I and J be monomial ideals of R. In general it is difficult to identify the monomial set [(J :R I)] in terms of [I] and [J]. Of course, we have J ⊆ (J :R I), so [J] ⊆ [(J :R I)]. Sometimes these containments are proper, and sometimes they are not proper. We will be interested in the special case I = (X) = (X1 , . . . , Xd )R. A hint as to how we might find monomials in (J :R (X)) r J is found in the graph Γ(I), as we see in the next examples. Example 3.3.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Let I be a monomial ideal in R and set (X) = (X, Y )R. A monomial f ∈ R is in (I :R (X)) if and only if f X, f Y ∈ I; see Proposition 1.5.3(b). The elements f , f X and f Y relate to each graphically as follows. O −

f YO

−

f

/ fX

|

|

.. . ···

/

Thus, the point (a, b) ∈ N2 represents a point in (I :R (X)) if and only if the ordered pairs (a + 1, b) and (a, b + 1) are in the graph Γ(I). Example 3.3.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set I = (X 3 , X 2 Y, Y 3 )R. The

52

3. OPERATIONS ON MONOMIAL IDEALS

graph Γ(I) has the following form.

.. .O

.. .

.. .

.. .

.. .

4

•

•

•

•

•

···

3

•

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···

2

−

•

•

•

···

1

−

•

•

•

···

•

•

/ ···

3

4

|

0 0

1

2

The two corners of the form q show us where to find elements of (I :R (X)) r I.

.. .O

.. .

.. .

.. .

.. .

4

•

•

•

•

•

···

3

•

•

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···

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···

1

−

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•

···

•

•

/ ···

3

4

0 0

'&%$ !"# q

|

'&%$ !"# q

1

2

It is not difficult to show that the monomials X 2 and XY 2 are precisely the monomials in (I :R (X)) r I; see Exercise 3.3.14. Note that these “corners” correspond to the “corners” in the ideals (X 2 , Y 3 )R and (X 3 , Y )R in the decomposition

3.3. COLONS OF MONOMIAL IDEALS

53

I = (X 2 , Y 3 )R ∩ (X 3 , Y )R; see Examples 3.1.8 and 3.1.9. .. .O

.. .

.. .

.. .

.. .

4

•

•

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2 {z

3

4

3

4

Γ((X 2 ,Y 3 )R)

| 0

}

|

'&%$ !"# q

1

2 {z

Γ((X 3 ,Y )R)

}

The point of the paper is that the “corner elements” of certain monomial ideals J ⊆ R give rise to the decomposition of J as an intersection of monomial ideals of the form (X a , Y b )R. Our next concern is the question of when (I :R (X)) ) I. Definition 3.3.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal of R. The monomial ideal I is non-degenerate if its irredundant generating sequence z1 , . . . , zm satisfies the following property: for each i = 1, . . . , d there exists an index j such that zj is a monomial multiple of Xi ; in other words, if each variable Xi is a factor of some generator zj . The monomial ideal I is degenerate if it is not non-degenerate, i.e., if there is an index i between 1 and d such that for each j = 1, . . . , m the monomial zj is not a multiple of Xi ; in other words, there is a variable Xi that is not a factor of any zj . Example 3.3.6. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. The ideal (X 3 )R is degenerate because X 3 is the irredundant generating sequence, and the variable Y is not a factor of X 3 . The ideal (X 2 Y 2 )R is non-degenerate because X and Y both occur in the irredundant generating sequence X 2 Y 2 . Remark 3.3.7. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R, and let I be a non-degenerate monomial ideal of R. Then I ⊆ (X). Indeed, if not, then we have I = R by Exercise 2.1.20, so the irredundant generating sequence for I is 1, which is not a multiple of any of the variables. The next lemma says that each monomial ideal I such that rad (I) = rad ((X)) is non-degenerate. Lemma 3.3.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R, and let I be a monomial ideal of R such that rad (I) = rad ((X)). Let z1 , . . . , zm ∈ [I] be the irredundant generating sequence for I. Then for i = 1, . . . , d there exists ni > 1 and there exists j such that zj = Xini . In particular, the ideal I is non-degenerate.

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Proof. First, note that the condition rad (I) = rad ((X)) implies that I ⊆ (X) ( R by Exercise 2.1.22. For j = 1, . . . , m write zi = Xni where ni = (ni,1 , . . . , ni,d ) ∈ Nd . Fix an index i such that 1 6 i 6 d. The condition rad (I) = rad ((X)) implies that Ximi ∈ I = (z1 , . . . , zm )R for some mi > 1, and so Ximi is a monomial multiple of zj = Xnj for some index j. Lemma 2.1.11 implies that mi > nj,i and, for k 6= i we n have 0 > nj,k > 0. It follows that nj,k = 0 whenever k 6= i and so zj = Xi j,i . It nj,i ∈ I; this also follows that nj,i > 1 since, otherwise we have nj,i = 0 so 1 = Xi contradicts the condition I ( R. From this we conclude that I is non-degenerate because the variable Xi is a n factor of the generator zj = Xi j,i .  Fact 3.3.9. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J ⊆ R be a monomial ideal, and let z1 , . . . , zn ∈ [J] be an irredundant generating sequence for J. Let f, g ∈ R be monomials such that f 6= 1A and z1 = f g. Then g ∈ / J; see Exercise 3.3.16. Theorem 3.3.10. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I and J be monomial ideals of R such that I ( R. Assume that J is non-degenerate and that (X2 , . . . , Xd )R ⊆ rad (J). Then (J :R I) ) J. Proof. We know that (J :R I) ⊇ J, so we need to show that (J :R I) 6⊆ J. Let z1 , . . . , zn ∈ [J] be the irredundant monomial generating sequence for J. Since J is non-degenerate, for each index i there exists an index j such that zj is a monomial multiple of Xi . Reorder the generators z1 , . . . , zn to assume that z1 is a monomial multiple of X1 , and fix a monomial w1 ∈ [R] such that z1 = X1 w1 . Fact 3.3.9 implies that w1 ∈ / J. Claim: For j = 1, . . . , d there exists a monomial wj ∈ [R] such that wj ∈ / J and X1 wj , . . . , Xj wj ∈ J. We prove the claim by induction on j. The base case j = 1 is established in the previous paragraph. Induction step, assume that j > 1 and that there is a monomial wj−1 ∈ [R] such that wj−1 ∈ / J and X1 wj−1 , . . . , Xj−1 wj−1 ∈ J. The assumption (X2 , . . . , Xd )R ⊆ m rad (J) implies that there is an integer mj > 1 such that Xj j ∈ J. It follows that mj Xj wj−1 ∈ J, so the set Kj = {m > 1 | Xjm wj−1 ∈ J} is a non-empty set of positive integers. The Well-Ordering Axiom implies that k −1 Kj has a unique minimal element kj = min(Kj ). Set wj = Xj j wj−1 . By the definition of kj we have wj ∈ / J and Xj wj ∈ J. Furthermore, for i = 1, . . . , j − 1 k −1 we have Xi wj = (Xi wj−1 )Xj j ∈ J. This establishes the claim. The monomial wd is not in J and satisfies X1 wd , . . . , Xd wd ∈ J. It follows that wd is in (J :R (X)) ⊆ (J :R I), so (J :R I) 6⊆ J.  Corollary 3.3.11. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R, and let J be a monomial ideal of R. If rad (J) = rad ((X)), then (J :R (X)) ) J. Proof. The assumption rad (J) = rad ((X)) implies that (X2 , . . . , Xd )R ⊆ rad ((X)) = rad (J). Lemma 3.3.8 implies that J is non-degenerate. Hence, the hypotheses of Theorem 3.3.10 are satisfied with the ideal I = (X). 

3.3. COLONS OF MONOMIAL IDEALS

55

We end this section with two helpful lemmas for later use. Lemma 3.3.12. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal, and consider monomials f, z ∈ [R]. Then z ∈ J if and only if f z ∈ f J. Proof. The forward implication follows from the definition. For the reverse implication, we assume that f z ∈ f J and show that z ∈ J. Let z1 , . . . , zk ∈ [J] be a monomial generating sequence for J. Fact 1.4.13(a) and Exercise 1.4.16 combine to show that f z1 , . . . , f zm ∈ [f J] is a monomial generating sequence for f J. The condition f z ∈ f J implies that f z is a monomial multiple of f zi for some i, so there is a monomial w ∈ [R] such that f z = f zi w. Monomial cancellation (Exercise 2.1.28(a)) implies that z = zi w ∈ (zi )R ⊆ J, as desired.  Lemma 3.3.13. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal, and consider monomials f, h ∈ [R]. If h ∈ f J, then there is a monomial z ∈ [J] such that h = f z. Thus, we have [f J] = {f z | z ∈ [J]}. Proof. Assume first that h ∈ f J. We have h ∈ f J ⊆ (f )R, so h is a monomial multiple of f . Thus, there is a monomial z ∈ [R] such that h = f z. Since f z = h ∈ f J, Lemma 3.3.12 implies that z ∈ J, as desired. The containment [f J] ⊆ {f z | z ∈ [J]} follows from the previous paragraph. The containment [f J] ⊇ {f z | z ∈ [J]} is by definition, so [f J] = {f z | z ∈ [J]}.  Exercises. Exercise 3.3.14. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set J = (X 3 , X 2 Y, Y 3 )R and (X) = (X, Y )R. Verify that the monomials in (J :R (X)) r J are XY 2 and X 2 . Exercise 3.3.15. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set (X) = (X, Y )R, and find a monomial ideal I in R such that (I :R (X)) = I; justify your answer. Exercise 3.3.16. Verify Fact 3.3.9. Exercise 3.3.17. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. In this exercise you are asked to work through the proof of Theorem 3.3.10 with I = (X, Y )R and J = (X 3 , X 2 Y 2 , Y 4 )R. (a) Prove that the hypotheses of Theorem 3.3.10 are satisfied for this ideal J. (b) Start with z1 = X 3 and follow the proof to find an element w2 ∈ (J :R I) r J; show your steps. Graph z1 , w2 , and J on the same set of coordinate axes. (c) Start with z1 = X 2 Y 2 and follow the proof to find an element w2 ∈ (J :R I)\J; show your steps. Graph z1 , w2 , and J on the same set of coordinate axes. Challenge Exercise 3.3.18. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set (X) = (X, Y )R, and let I be a monomial ideal in R. Give a necessary and sufficient condition for the strictness of the containment (I :R (X)) ⊇ I; prove this result. Do the same for an arbitrary monomial ideal in A[X1 , . . . , Xd ]. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output.

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3.4. Bracket Powers of Monomial Ideals Here is [5, (4.13)] for our situation. Definition 3.4.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal of R. For k = 1, 2, . . . we set J [k] = (Tk )R where Tk = {f k | f ∈ [J]}. Remark 3.4.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal of R. By definition, the ideal J [k] is a monomial ideal for each k = 1, 2, . . .. The next lemma will help us to find a generating set for J [k] . Lemma 3.4.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Consider a set of monomials S ⊆ [R] and an integer k > 1. Set J = (S)R and I = ({f k | f ∈ S})R. For each monomial g ∈ [R] we have g ∈ J if and only if g k ∈ I. Proof. For the forward implication, we assume that g ∈ J. The ideal J is a monomial ideal, Theorem 2.3.1 implies that there is a finite subset S 0 ⊆ S such that J = (S 0 )R. Lemma 2.1.14 implies that g ∈ (f )R for some f ∈ S 0 , and it follows that g k ∈ (f k )R ⊆ I. For the converse, assume that g k ∈ I. The set Sk = {f k | f ∈ S} is a monomial generating set for I. Hence, there is a finite subset Sk0 ⊆ Sk such that I = (Sk0 )R. Lemma 2.1.14 implies that g k ∈ (f k )R for some f k ∈ Sk0 . Note that f ∈ S by definition. Write f = Xm and g = Xn with m, n ∈ Nd . Then f k = Xkm and g k = Xkn , so Lemma 2.1.11 implies that km < kn. It follows readily that m < n, so g = Xn ∈ (Xm )R = (f )R ⊆ J.  Proposition 3.4.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R. (a) If S is a monomial generating set for J, then the set Sk = {f k | f ∈ S} is a monomial generating set for J [k] . (b) If f1 , . . . , fn ∈ [J] is a monomial generating sequence for J, then J [k] = (f1k , . . . , fnk )R. Proof. (a) Let Tk = {f k | f ∈ [J]}. By definition, we have J [k] = (Tk )R, so we need to show that (Sk )R = (Tk )R. To verify the containment (Sk )R ⊆ (Tk )R, we need to show that Sk ⊆ (Tk )R. An arbitrary element of Sk has the form f k for some f ∈ S. By definition, we have f k ∈ Tk ⊆ (Tk )R, so Sk ⊆ (Tk )R, as desired. To verify the containment (Sk )R ⊇ (Tk )R, we need to show that Tk ⊆ (Sk )R. An arbitrary element of Tk has the form f k for some f ∈ [J]. Lemma 3.4.3 implies that f k ∈ (Sk )R, so Tk ⊆ (Sk )R, as desired. (b) This is the special case of part (a) with S = {f1 , . . . , fn }.  Example 3.4.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. We consider the monomial ideal

3.4. BRACKET POWERS OF MONOMIAL IDEALS

57

J = (X 3 , X 2 Y, Y 2 )R which has the following graph. .. .O

.. .

.. .

.. .

.. .

.. .

.. .

4

•

•

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···

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···

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/ ···

1

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0 0 We have J

[2]

6

4

2

4

= (X , X Y , Y )R which has the following graph. .. .O

.. .

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4

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3

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··· / ···

0 0

|

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1

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5

6

Notice that the graph of J [2] is essentially a scale model of the graph of J. Lemma 3.4.6. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal of R, and let g ∈ [R] be a monomial in R. For k = 1, 2, . . . we have g ∈ J if and only if g k ∈ J [k] . Proof. This is a restatement of Lemma 3.4.3. Remark 3.4.7. Note that the previous lemma does not imply that we have [J {hk | h ∈ [J]}; see Exercise 3.4.12.

 [k]

]=

Proposition 3.4.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables.
Let J be a monomial ideal of R. For k = 1, 2, . . . we have J [k] ⊆ J and rad J [k] = rad (J). Proof. Exercise 3.4.13.



Proposition 3.4.9. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal of R and let f1 , . . . , fn ∈ [J] be an irredundant monomial generating sequence for J. For k = 1, 2, . . . the irredundant monomial generating sequence for J [k] is f1k , . . . , fnk .

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Proof. By Proposition 3.4.4(b), the sequence f1k , . . . , fnk is a monomial generating sequence for J [k] , so it suffices to show irredundancy. Suppose that the sequence is redundant. Then there are indices i, j such that i 6= j and fik ∈ [k] (fjk )R = ((fj )R) . Then Lemma 3.4.6 implies that fi ∈ (fj )R, contradicting the irredundancy of the original generating sequence.  Lemma 3.4.10. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I and J be monomial ideals in R and fix an integer k > 1. (a) I ⊆ J if and only if I [k] ⊆ J [k] . (b) I = J if and only if I [k] = J [k] . Proof. (a) Let f1 , . . . , fm ∈ [I] be a monomial generating sequence for I, and let g1 , . . . , gn ∈ [J] be a monomial generating sequence for J. For the forward implication, we assume that I ⊆ J, and we show that I [k] ⊆ [k] J . It suffices to show that each generator fik ∈ I [k] is in J [k] . By assumption, we have fi ∈ I ⊆ J. Lemma 3.4.6 implies that fik ∈ J [k] . For the converse, we assume that I [k] ⊆ J [k] , and we show that I ⊆ J. For i = 1, . . . , m we have fik ∈ I [k] ⊆ J [k] so Lemma 3.4.6 implies that fi ∈ J. It follows that I = (f1 , . . . , fm )R ⊆ J, as desired. (b) We have I = J if and only if I ⊆ J and J ⊆ I. By part (a), we have (I ⊆ J and J ⊆ I) if and only if (I [k] ⊆ J [k] and J [k] ⊆ J [k] ), that is, if and only if I [k] = J [k] .  Recall that the intersection of monomial ideals is a monomial ideal by Theo[k] rem 3.1.1. Hence, the ideal (∩ni=1 Ji ) in the next lemma is defined, and it is a monomial ideal. Similarly, the ideal ∩ni=1 Ji [k] is also a monomial ideal. Lemma 3.4.11. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J1 , . . . , Jn be monomial ideals [k] in R. For each integer k > 1, we have (∩ni=1 Ji ) = ∩ni=1 Ji [k] . Proof. We proceed by induction on n, the number of ideals. Base case: n = 2. Let f1 , . . . , fm ∈ [J1 ] be a monomial generating sequence for J1 . Let g1 , . . . , gn ∈ [J2 ] be a monomial generating sequence for J2 . Since J1 and J2 are monomial ideals, the same is true of J1 ∩ J2 . As we noted before the [k] statement of the result, the ideals (J1 ∩ J2 ) and J1 [k] ∩ J2 [k] are both monomial ideals. [k] For the containment (J1 ∩ J2 ) ⊆ J1 [k] ∩ J2 [k] , observe that J1 ∩ J2 ⊆ J1 , so [k] [k] Lemma 3.4.10(a) implies that (J1 ∩ J2 ) ⊆ J1 [k] . Similarly, we have (J1 ∩ J2 ) ⊆ [k] J2 [k] , and hence (J1 ∩ J2 ) ⊆ J1 [k] ∩ J2 [k] . [k] For the containment (J1 ∩ J2 ) ⊇ J1 [k] ∩ J2 [k] , we need only show that every monomial z ∈ [J1 [k] ∩ J2 [k] ] = [J1 [k] ] ∩ [J2 [k] ] [k]

k is in (J1 ∩ J2 ) . The condition z ∈ [J1 [k] ] = [(f1k , . . . , fm )R] implies that z ∈ (fik )R for some index i. Similarly, the condition z ∈ [J2 [k] ] = [(g1k , . . . , gnk )R] implies that z ∈ (gjk )R for some index j. Write fi = Xm and gj = Xn . Then we have fik = Xkm and gjk = Xkn . For l = 1, . . . , d set pl = max{ml , nl }. It is straightforward to show

3.5. MONOMIAL RADICALS

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that kpl = max{kml , knl }, so Lemma 3.1.6 yields the first and third equalities in the next sequence z ∈ (fik )R ∩ (gjk )R = (Xkp )R = ((Xp )R)

[k]

= ((fi )R ∩ (gj )R)

[k]

[k]

⊆ (J1 ∩ J2 ) .

The second equality is by definition. The containment at the end of the sequence follows from Lemma 3.4.10(a) since (fi )R ∩ (gj )R ⊆ J1 ∩ J2 . Induction step: Exercise 3.4.15.  Exercises. Exercise 3.4.12. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a non-zero monomial ideal in R. Prove that for each integer k > 2, we have [J [k] ] ) {hk | h ∈ [J]}; see Remark 3.4.7. Exercise 3.4.13. Prove Proposition 3.4.8. Exercise 3.4.14. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R with monomial generating sequence f1 , . . . , fn . Fix an integer k > 1, and show that f1 , . . . , fn is an irredundant monomial generating sequence for J if and only if f1k , . . . , fnk is an irredundant monomial generating sequence for J [k] . Exercise 3.4.15. Complete the proof of Lemma 3.4.11. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 3.5. Monomial Radicals This section focuses on a notion of the radical for monomial ideals. To motivate it, we give an example showing that the radical of a monomial ideal need not be a monomial ideal. See also the exercises in Section 3.1 for more details about this phenomenon. Example 3.5.1. In the polynomial ring R = Z4 [X] in one variable, the ideal J = (X)R is a monomial ideal. However, the ideal rad (J) = (2, X)R is not a monomial ideal. Definition 3.5.2. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R. The monomial radical of J is the monomial ideal m-rad (J) = (S)R where S = {z ∈ [R] | z n ∈ J for some n > 1}. Remark 3.5.3. In the notation of Definition 3.5.2, we have S = rad (J) ∩ [R]. Example 3.5.4. Let A be a commutative ring with identity and let R be
the polynomial ring R = A[X, Y ]
in two variables. We have m-rad (X 3 , Y 2 )R = (X, Y )R and m-rad (X 3 Y 2 )R = (XY )R. (This can be verified directly, or using Proposition 3.5.14.) The next result explains the relation between rad (J) and m-rad (J). Note that the equality m-rad (J) = rad (J) holds more generally when R is reduced or an integral domain; see Exercise 3.5.20.

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Proposition 3.5.5. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R. (a) One has m-rad (J) ⊆ rad (J). (b) One has m-rad (J) = rad (J) if and only if rad (J) is a monomial ideal. (c) If A is a field, then m-rad (J) = rad (J). Proof. (a) The ideal m-rad (J) is generated by the set S = rad (J) ∩ [R] ⊆ rad (J), so we have m-rad (J) ⊆ rad (J). (b) If rad (J) is a monomial ideal, then rad (J) = ([rad (J)])R = (S)R = m-rad (J) . Conversely, if rad (J) = m-rad (J), then the fact that m rad (J) is a monomial ideal implies that rad (J) is a monomial ideal. (c) Assuming that R is a field, Exercise 3.1.17 shows that rad (J) is a monomial ideal, part (b) implies that m-rad (J) = rad (J).  The next result compares to Proposition 1.5.7. Proposition 3.5.6. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R. (a) There is a containment J ⊆ m-rad (J). (b) One has [m-rad (J)] = rad (J) ∩ [R]. (c) There is an equality m-rad (J) = m-rad (m-rad (J)). (d) One has m-rad (J) = R if and only if J = R. (e) One has m-rad (J) = 0 if and only if J = 0.
(f) For each integer n > 1, one has m-rad (J) = m-rad (J n ) = m-rad J [k] . Proof. (a) The set S = rad (J) ∩ [R] ⊇ J ∩ [R] = [J] generates m-rad (J), so we have m-rad (J) = (S)R ⊇ ([J])R = J. (b) Since S = rad (J) ∩ [R] is a monomial generating set for m-rad (J), we have [m-rad (J)] ⊇ rad (J) ∩ [R]. For the reverse containment, fix a monomial f ∈ [m-rad (J)]. Proposition 3.5.5 implies that f ∈ rad (J), so f ∈ rad (J) ∩ [R], as desired. (c) The containment m-rad (J) ⊆ m-rad (m-rad (J)) follows from part (a). For the reverse containment, is suffices to show that [m-rad (J)] ⊇ [m-rad (m-rad (J))]. Let f ∈ [m-rad (m-rad (J))] = rad (m-rad (J)) ∩ [R], and fix an exponent n > 1 such that f n ∈ m-rad (J). Then f n ∈ m-rad (J) ∩ [R] = [m-rad (J)] = rad (J) ∩ [R], so there is an exponenet m > 1 such that f mn = (f n )m ∈ J. This shows that f ∈ rad (J) ∩ [R] = [m-rad (J)], as desired. The proofs of parts (d)–(f) are left as exercises.  Example 3.5.7. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. With (X) = (X1 , . . . , Xd )R, we have m-rad ((X)) = (X). Indeed, the containment m-rad ((X)) ⊇ (X) is from Proposition 3.5.6(a). For the reverse containment, observe that, since (X) 6= R, the ideal m-rad ((X)) is a monomial ideal of R such that m-rad ((X)) 6= R by Proposition 3.5.6(d) Thus, we have m-rad ((X)) ⊆ (X) by Exercise 2.1.20. (Alternately, this can be verified using Proposition 3.5.14.) The next result compares to Proposition 1.5.9. Note however the difference in parts (d) and (e); see Example 1.5.10.

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Proposition 3.5.8. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let n be a positive integer, and let I, J, I1 , I2 , . . . , In be monomial ideals of R. (a) If I ⊆ J, then m-rad (I) ⊆ m-rad (J). (b) There are equalities m-rad (IJ) = m-rad (I ∩ J) = m-rad (I) ∩ m-rad (J). (c) There are equalities
m-rad (I1 I2 · · · In ) = m-rad ∩nj=1 Ij = ∩nj=1 m-rad (Ij ) . (d) There is an equality m-rad  (I + J) =m-rad (I) + m-rad (J). Pn Pn (e) There is an equality m-rad j=1 Ij = j=1 m-rad (Ij ). Proof. (a) Assume that I ⊆ J. The containment rad (I) ⊆ rad (J) is from Proposition 1.5.9(a), so we have rad (I) ∩ [R] ⊆ rad (J) ∩ [R] and m-rad (I) = (rad (I) ∩ [R])R ⊆ (rad (J) ∩ [R])R = m-rad (J) . (b) As in the proof of Proposition 1.5.9(b), the containments m-rad (IJ) ⊆ m-rad (I ∩ J) ⊆ m-rad (I) ∩ m-rad (J) follow from part (a). For the containment m-rad (I)∩m-rad (J) ⊆ m-rad (IJ), it suffices to show that [m-rad (I)∩m-rad (J)] = [m-rad (IJ)]. We compute: [m-rad (I) ∩ m-rad (J)] = [m-rad (I)] ∩ [m-rad (J)] = (rad (I) ∩ [R]) ∩ (rad (J) ∩ [R]) = (rad (I) ∩ rad (J)) ∩ [R] = rad (IJ) ∩ [R] = [m-rad (IJ)]. The first step in this sequence is from Theorem 3.1.1. The second and fifth steps are from Proposition 3.5.6(b). The third step is routine, and the fourth step is from Proposition 1.5.9(b). (d) We first show that (3.5.8.1)

rad (I + J) ∩ [R] = (rad (I) ∩ [R]) ∪ (rad (J) ∩ [R]).

For the containment “⊆”, let f ∈ rad (I + J) ∩ [R] and fix an integer n > 1 such that f n ∈ I + J. Since f n is a monomial, Exercise 2.3.20(b) implies that f n ∈ I ∪ J. If f n ∈ I, then f ∈ rad (I) ∩ [R]. If f n ∈ J, then f ∈ rad (J) ∩ [R]. So, we conclude that f ∈ (rad (I) ∩ [R]) ∪ (rad (J) ∩ [R]). For the reverse containment, we compute: (rad (I) ∩ [R]) ∪ (rad (J) ∩ [R]) = (rad (I) ∪ rad (J)) ∩ [R] ⊆ (rad (I) + rad (J)) ∩ [R] ⊆ rad (rad (I) + rad (J)) ∩ [R] = rad (I + J) ∩ [R]. The first step is routine, and the second step is from the containment rad (I) ∪ rad (J) ⊆ rad (I) + rad (J). The remaining steps are from Propositions 1.5.7(b) and 1.5.9(d).

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In the next sequence, the second step is from equation (3.5.8.1): [m-rad (I + J)] = rad (I + J) ∩ [R] = (rad (I) ∩ [R]) ∪ (rad (J) ∩ [R]) = [m-rad (I)] ∪ [m-rad (J)] = [m-rad (I) + m-rad (J)]. The first and third steps are from Proposition 3.5.6(b), and the fourth step is from Exercise 2.3.20(b). The proofs of the remaining statements follow by induction and are left as exercises.  Definition 3.5.9. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. The support of a monomial f = Xn ∈ [R] is the set Supp(f ) = {i ∈ N | 1 6 i 6 d and ni 6= 0}. The reduction of f is the monomial red(f ) =

Y

Xi .

i∈Supp(f )

Remark 3.5.10. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. The support of a monomial f ∈ [R] is the set of indices i such that Xi f . The reduction of f is the product of the variables dividing f : Y Xi . red(f ) = Xi f Example 3.5.11. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , X2 , X3 ] in three variables. Then Supp(X12 X35 ) = {1, 3} and red(X12 X35 ) = X1 X3 . Fact 3.5.12. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. For each integer n > 1 and each monon n mial f ∈ [R], we have Supp(f ) = Supp(f ) = Supp(red(f )) and red(f ) = red(f ) and red(f f, g ∈ [R], one has Supp(f ) ⊆ Supp(g) if and only if ) f . For monomials red(f ) red(g). Also, if f g, then Supp(f ) ⊆ Supp(g) and red(f ) red(g). Lemma 3.5.13. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R, and let f ∈ [R]. (a) There is an integer n > 1 such that red(f )n ∈ (f )R. (b) If f ∈ J, then red(f ) ∈ m-rad (J). Proof. (a) Let Supp(f ) = {i1 , . . . , ik } with 1 6 i1 < · · · < ik 6 d. It follows that red(f ) = Xi1 · · · Xik . Write f = Xm and let n = max{m1 , . . . , md }. Then mi m f = Xi1 i1 · · · Xik k . Since n > mi for i = 1, . . . , d we have mi m f = Xi1 i1 · · · Xik k Xin1 · · · Xink = red(f )n so red(f )n ∈ (f )R, as desired. (b) Assume that f ∈ J. Part (a) yields an integer n > 1 such that red(f )n ∈ (f )R ⊆ J. Hence red(f ) ∈ m-rad (J). 

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Proposition 3.5.14. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let S ⊆ [R] and set J = (S)R. Then one has m-rad (J) = ({red(f ) | f ∈ S})R. Proof. Set T = {red(f ) | f ∈ S} and K = ({red(f ) | f ∈ S})R. For each f ∈ S ⊆ [J], we have red(f ) ∈ m-rad (J) by Lemma 3.5.13(b). This explains the containment m-rad (J) ⊇ T , hence m-rad (J) ⊇ K. For the reverse containment, let g ∈ [m-rad (J)] = rad (J) ∩ [R]. Then there is an integer n > 1 such that g n ∈ [J]. It follows that there is a monomial f ∈S such that f g n . Fact 3.5.12 implies that red(f ) red(g n ) = red(g) g, so we have g ∈ (red(f ))R ⊆ (T )R = K. It follows that m-rad (J) = (m-rad (J))R ⊆ K.  Remark 3.5.15. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal of R with irredundant monomial generating sequence f1 , . . . , fn . Proposition 3.5.14 shows that the sequence red(f1 ), . . . , red(fn ) is a monomial generating sequence for m-rad (J). However, the next example shows that the sequence red(f1 ), . . . , red(fn ) may be redundant. Example 3.5.16. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X, Y ] in two variables. The sequence X 2 Y, XY 2 is an irredundant monomial generating sequence for the ideal (X 2 Y, XY 2 )R. We have red(X 2 Y ) = XY = red(XY 2 ), so Proposition 3.5.14 shows that the sequence
XY, XY is a monomial generating sequence for the ideal m-rad (X 2 Y, XY 2 )R ; however, this sequence is redundant. Corollary 3.5.17. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set J = (Xte11 , . . . , Xtekk )R where k, t1 , . . . , tk , e1 , . . . , ek > 1 are such that 1 6 t1 < · · · < tk 6 d. Then m-rad (J) = (Xt1 , . . . , Xtk )R Proof. For i = 1, . . . , k we have red(Xteii ) = Xti , so the result follows from Proposition 3.5.14.  Exercises. Exercise 3.5.18. Verify the conclusions of Example 3.5.1. Exercise 3.5.19. Verify the conclusions of Example 3.5.4. Exercise 3.5.20. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R. (a) Prove that if A is an integral domain, then m-rad (J) = rad (J); see Exercises 3.1.17–3.1.20. (b) Prove that if A is reduced, then m-rad (J) = rad (J). (c) Prove that if J 6= R and m-rad (J) = rad (J), then A is reduced. Exercise 3.5.21. Complete the proof of Proposition 3.5.6. Exercise 3.5.22. Complete the proof of Proposition 3.5.8. Exercise 3.5.23. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I and J be monomial ideals of R. Prove or disprove: If m-rad (I) ⊆ m-rad (J), then I ⊆ J.

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Exercise 3.5.24. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f1 , . . . , fs , g1 , . . . , gt ∈ [R], and set I = (f1 , . . . , fs )R and J = (g1 , . . . , gt )R. (a) Prove that m-rad (I) ⊆ m-rad (J) if and only if for each i = 1, 2, . . . , s there exists a positive integer ni such that fini ∈ J. (b) Prove that m-rad (I) = m-rad (J) if and only if for each i = 1, 2, . . . , s there exists a positive integer ni such that fini ∈ J, and for each j = 1, 2, . . . , t there m exists a positive integer mj such that gj j ∈ I. (c) Assume that I ⊆ J. Prove that m-rad (I) = m-rad (J) if and only if for each m j = 1, 2, . . . , t there exists an integer mj such that gj j ∈ I. Compare these properties with Fact 1.5.11. Exercise 3.5.25. Let A be a commutative ring with identity. Consider the polynomial ring R = A[X, Y ] and the ideals I = (X 3 , Y 4 )R and J = (XY 2 , X 2 Y )R. (a) Use Exercise 3.5.24 to prove that m-rad (I) ) m-rad (J). (b) Prove that m-rad (I) = (X, Y )R and m-rad (J) = (XY )R. Use this to give another proof that m-rad (I) ) m-rad (J). Exercise 3.5.26. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I and J be monomial ideals in R. (a) Prove that m-rad (I) ⊆ m-rad (J) if and only if rad (I) ⊆ rad (J). (b) Prove that m-rad (I) = m-rad (J) if and only if rad (I) = rad (J). Exercise 3.5.27. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Prove that if I is a monomial ideal, then m-rad (I) = (X) if and only if rad (I) = rad ((X)). Exercise 3.5.28. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Prove that if I is a monomial ideal such that I 6= R, then m-rad (I) = (X) if and only if for each i = 1, . . . , d there exists an integer ni > 0 such that Xini ∈ I. Exercise 3.5.29. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let I be a monomial ideal. Prove that if rad (I) ⊆ rad ((X)), then I ⊆ (X) ( R. Exercise 3.5.30. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R and let f be a monomial in R. (a) Prove that if Xi | f , then m-rad (f I) ⊆ m-rad ((Xi )R). (b) Prove that if f 6= 1 and d > 2, then m-rad (f I) ( m-rad ((X)). Exercise 3.5.31. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let (X) = (X1 , . . . , Xd )R. Let I be a monomial ideal in R, and let S denote the set of monomials in R r I. Prove that S is a finite set if and only if m-rad (I) ⊇ (X). Exercise 3.5.32. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I1 , . . . , In be monomial ideals in R and set (X) = (X1 , . . . , Xd )R. (a) Prove that if m-rad (Ij ) = (X) for j = 1, . . . , n, then m-rad (I1 · · · In ) = (X).

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(b) Prove that if Ij 6= R for j = 1, . . . , n and m-rad (I1 · · · In ) = (X), then m-rad (Ij ) = (X) for j = 1, . . . , n. Exercise 3.5.33. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I1 , . . . , In be monomial ideals in R and set (X) = (X1 , . . . , Xd )R. (a) Prove that if m-rad (Ij ) = (X) for j = 1, . . . , n, then m-rad (I1 + · · · + In ) = (X). (b) Prove or give a counter-example for the following: if m-rad (I1 + · · · + In ) = (X), then m-rad (Ij ) = (X) for j = 1, . . . , n. Exercise 3.5.34. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal of R, and set (X) = (X1 , . . . , Xd )R. Prove that rad (I) = (X) if and only if the irredundant monomial generating sequence for I contains a power of each variable. Exercise 3.5.35. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R. Describe [m-rad (I)] and Γ(m-rad (I)) in terms of [I] and Γ(I); justify your answer. Exercise 3.5.36. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R, and let I and J be monomial ideals of R. Prove that if I ⊆ (X) and m-rad (J) = (X), then (J :R I) ) J. Exercise 3.5.37. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J
be a monomial ideal of R. Prove that for each integer k > 1 we have m-rad J [k] = m-rad (J). Exercise 3.5.38. Verify the conclusions of Example 3.5.11. Exercise 3.5.39. Verify the conclusions of Fact 3.5.12. Exercise 3.5.40. Verify the conclusions of Example 3.5.16. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 3.6. Square-Free Monomial Ideals Definition 3.6.1. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. A monomial Xn ∈ [R] is squarefree if, for i = 1, . . . , d one has ni ∈ {0, 1}. A monomial ideal J ⊆ R is square-free if it is generated by square-free monomials. Example 3.6.2. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X, Y, Z] in three variables. The square-free monomials in R are 1, X, Y, Z, XY, XZ, Y Z, XY Z. The ideal (XY, Y Z)R is square-free. The ideal (X 2 Y, Y Z 2 )R is not square-free. Remark 3.6.3. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. A monomial f ∈ [R] is squarefree if and only if it has no factor of the form Xi2 , i.e., if and only if f = red(f ). (In particular, the monomial red(f ) is square-free.) Thus, the term “square-free” refers to the fact that it is free of square factors.

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Remark 3.6.4. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. A monomial ideal in R is square-free if and only if the monomials in its irredundant monomial generating sequence are square-free. Proposition 3.6.5. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. A monomial ideal J ⊆ R is square-free if and only if m-rad (J) = J. In particular, the monomial radical of J is square-free. Proof. Let f1 , . . . , fn ∈ [J] be the irredundant monomial generating sequence for J. Assume first that J is square-free; we show that m-rad (J) = J. Remark 3.6.4 implies that each monomial fi is square-free, so fi = red(fi ) for i = 1, . . . , n. Thus, Proposition 3.5.14 implies that m-rad (J) = (red(f1 ), . . . , red(fn ))R = (f1 , . . . , fn )R = J as desired. Assume next that m-rad (J) = J. To show that J is square-free, we need to show that each fi is square-free, that is, that fi = red(fi ) for i = 1, . . . , n. Proposition 3.5.14 implies that J = m-rad (J) = (red(f1 ), . . . , red(fn ))R so the irredundant monomial generating sequence for J is a sequence of the form red(fi1 ), . . . , red(fik ). The uniqueness of irredundant monomial generating sequences implies that k = n, so the irredundant monomial generating sequence for J is red(f1 ), . . . , red(fn ), and further that {f1 , . . . , fn } = {red(f1 ), . . . , red(fn )}. Thus, for i = 1, . . . , n there is an index ji such that fi = red(fji ). Fact 3.5.12 implies that red(fji ) fji , so we have fi fji . The irredundancy of the the sequence f1 , . . . , fn implies that ji = i, so fi = red(fi ). Finally, if J is an arbitrary monomial ideal, then we have m-rad (m-rad (J)) = m-rad (J) by Proposition 3.5.6(c), so the previous paragraph shows that m-rad (J) is square-free.  Exercises. Exercise 3.6.6. Verify the conclusions of Remark 3.6.3. Exercise 3.6.7. Verify the conclusions of Remark 3.6.4. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output.

CHAPTER 4

M-Irreducible Ideals and Decompositions 4.1. M-Irreducible Monomial Ideals M-irreducible ideals are, in a sense, the simplest monomial ideals, in that they cannot be written as non-trivial intersections. One of the main points of this book is (1) to show that every monomial ideal can be written as a finite intersection of m-irreducible monomial ideals, and (2) to show how, given a monomial ideal J to find m-irreducible ideals J1 , . . . , Jn such that J = ∩ni=1 Ji . The first of these goals is accomplished in Theorem 4.3.1. Definition 4.1.1. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. A monomial ideal J ( R is m-reducible if there are monomial ideals J1 , J2 6= J such that J = J1 ∩ J2 . A monomial ideal J ( R is m-irreducible if it is not m-reducible. Remark 4.1.2. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. A monomial ideal J ⊆ R is m-irreducible if and only if it satisfies the following: (1) J 6= R, and (2) given two monomial ideals J1 , J2 such that J = J1 ∩J2 , either J1 = J or J2 = J. If J is m-irreducible and J1 , . . . , Jn are monomial ideals (with n > 2) such that J = ∩ni=1 Ji , then J = Ji for some index i. In general, if J = J1 ∩ J2 , then J ⊆ J1 and J ⊆ J2 . Example 4.1.3. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X, Y ] in 2 variables. The monomial ideal J = (X 3 , X 2 Y, Y 3 )R is m-reducible. Indeed, we have J = (X 2 , Y 3 )R ∩ (X 3 , Y )R by Example 3.1.8. Also, we have X 2 ∈ (X 2 , Y 3 )R r J so J 6= (X 2 , Y 3 )R. Also Y ∈ (X 3 , Y )R r J, so J 6= (X 3 , Y )R. On the other hand, the ideals (X 2 , Y 3 )R and (X 3 , Y )R are m-irreducible. This can be verified directly, or by appealing to Proposition 4.1.4. The next two results characterize the non-zero m-irreducible monomial ideals. Proposition 4.1.4. Let A be a commutative ring with identity, and let R denote the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix positive integers k, t1 , . . . , tk , e1 , . . . , ek such that 1 6 t1 < · · · < tk 6 d. Then the monomial ideal J = (Xte11 , . . . , Xtekk )R is m-irreducible. Proof. Note first that J ⊆ (Xti , . . . , Xtk )R ⊆ (X1 , . . . , Xd )R, so J 6= R. Also, since k > 1, we have J 6= 0. 67

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Fix monomial ideals J1 , J2 in R such that J = J1 ∩ J2 . Suppose that J ( Ji for i = 1, 2 and fix a monomial fi ∈ [Ji ] r [J]. Write f1 = Xm and f2 = Xn . For i = 1, . . . , d set pi = max{mi , ni }. For i = 1, . . . , k we have mti < ei : otherwise, we have mi > ei for some i, so a comparison of exponent vectors shows that f1 ∈ (Xteii )R ⊆ J, a contradiction. Similarly, for i = 1, . . . , k we have ni < ei , and hence pi = max{mi , ni } < ei . A similar argument shows that lcm(f1 , f2 ) = Xp ∈ / J. However, we have lcm(f1 , f2 ) ∈ J1 ∩ J2 = J, a contradiction.  The next result shows that the only non-zero m-irreducible ideals are the ones from the previous result. Proposition 4.1.5. Let A be a commutative ring with identity, and let R denote the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a non-zero mirreducible monomial ideal of R. There are positive integers k, t1 , . . . , tk , e1 , . . . , ek such that 1 6 t1 < · · · < tk 6 d such that J = (Xte11 , . . . , Xtekk )R. Proof. Let f1 , . . . , fk be an irredundant generating sequence for J. It suffices to show that each fi is of the form Xteii . Suppose by way of contradiction that one of the fi is not of this form. Re-order the fj if necessary to assume that fk is not of the form Xteii . This means that we can write fk = Xteii g where ei > 1 and Xti - g and g 6= 1. Re-order the variables if necessary to assume that fk = X1e g where e > 1 and X1 - g and g 6= 1. Set I = (f1 , . . . , fk−1 , X1e )R and I 0 = (f1 , . . . , fk−1 , g)R. Claim: J = I ∩ I 0 . For this, we use Proposition 3.1.7 to conclude that the following sequence generated I ∩ I 0 : f1 , . . . , fn−1 , lcm(f1 , X1e ), lcm(f1 , g), . . . , lcm(fn−1 1, X1e ), lcm(fn−1 , g), lcm(X1e , g) . {z } {z } | {z } | | ∈(f1 )R

∈(fn−1 )R

=fn

Removing the redundancies from this list (using Algorithm 2.3.8) we see that I ∩ I 0 is generated by f, . . . , fn−1 , fn . Since this sequence generates J, we have the desired equality. Claim: J ( I. To show that J ⊆ I, it suffices to show that f1 , . . . , fn ∈ I. The elements f1 , . . . , fn−1 are generators for I, by definition. Also, the element X1e is a generator for I, so we have fn = X1e g ∈ I. To show that J 6= I, we need to e e show that X1e ∈ / J. Suppose by way of contradiction that X1 ∈ J. Then fi X1 e for some index i. Since X1 fk , this implies that fi fk . The sequence f1 , . . . , fk is irredundant, so we have fi = fk . Thus, we have fk = X1e g X1e . By comparing exponent vectors, we conclude that g = 1, a contradiction. Similarly, we have J ( I 0 . In summary, we have J = I ∩ I 0 and J ( I and J ( I 0 . This contradicts the assumption that J is m-irreducible, completing the proof.  The next result is useful. It is kind of parallel to the definition of m-irreducible. Lemma 4.1.6. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I, J1 , . . . , Jn be monomial ideals in R such that I is m-irreducible. If ∩ni=1 Ji ⊆ I, then there is an index j such that Jj ⊆ I. Proof. If I = 0, then the condition ∩ni=1 Ji ⊆ I = 0 implies that ∩ni=1 Ji = 0; it is straightforward to show that this implies that Ji = 0 = I for some index i.

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69

Thus, we assume that I 6= 0. Also, the case n = 1 is routine, so we assume that n > 2. Proposition 4.1.5 provides positive integers k, t1 , . . . , tk , e1 , . . . , ek such that 1 6 t1 < · · · < tk 6 d such that I = (Xte11 , . . . , Xtekk )R. We proceed by induction on n. Base case: n = 2. Assume that J1 ∩ J2 ⊆ I. Suppose by way of contradiction that J1 6⊆ I and J2 6⊆ I. This implies that [J1 ] 6⊆ [I] and [J2 ] 6⊆ [I], so there are monomials f1 ∈ J1 r I and f2 ∈ J2 r I. Write f1 = Xm and f2 = Xn . For i = 1, . . . , d set pi = max{mi , ni } so we have Xp = lcm(f1 , f2 ) ∈ J1 ∩ J2 ⊆ I = (Xte11 , . . . , Xtekk )R. e It follows that there is an index j such that Xtjj Xp . Comparing exponent vectors, we find that ej 6 ptj = max{mtj , ntj }. It follows that either ej 6 mtj or e j 6 ntj . If e ej 6 mtj , then another comparison of exponent vectors implies that Xtjj Xm = f1 , ej so f1 ∈ (Xtj )R ⊆ I, a contradiction. Similarly, if ej 6 ntj , we conclude that f2 ∈ I, a contradiction. This concludes the proof of the base case. Induction step: Exercise 4.1.10.  Exercises. Exercise 4.1.7. Verify the conclusions from Remark 4.1.2. Exercise 4.1.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Prove that the ideal 0 is mirreducible. Exercise 4.1.9. Complete the proof of Proposition 4.1.5. Exercise 4.1.10. Complete the proof of Lemma 4.1.6. Exercise 4.1.11. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a non-zero monomial ideal in R. (a) Prove that if J is m-irreducible, then m-rad (J) is m-irreducible. (b) Prove or disprove the following: if m-rad (J) is m-irreducible, then J is mirreducible. Exercise 4.1.12. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix integers k, t1 , . . . , tk , e1 , . . . , ek > 1 such that 1 6 t1 < · · · < tk 6 d, and set J = (Xte11 , . . . , Xtekk )R. Prove that the monomial Xte11 −1 · · · Xtekk −1 is not in J. Exercise 4.1.13. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a non-zero m-irreducible monomial ideal in R. Prove that there are positive integers n, t1 , . . . , tn such that m-rad (J) = (Xt1 , . . . , Xtn )R. Exercise 4.1.14. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a non-zero m-irreducible monomial ideal in R. Prove or disprove the following: If {Iλ }λ∈Λ is a (possibly infinite) set of monomial ideals in R such that J = ∩λ∈Λ Iλ , then there is an index λ ∈ Λ such that J = Iλ .

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Exercise 4.1.15. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R such that J 6= R. Prove that the following conditions are equivalent: (i) J is m-irreducible; (ii) for all monomial ideal J1 , J2 if J1 ∩ J2 ⊆ J, then either J1 ⊆ J or J2 ⊆ J; and (iii) for all monomials f, g ∈ [R] if lcm(f, g) ∈ J, then either f ∈ J or g ∈ J. Exercise 4.1.16. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. We say that a monomial ideal J in R is m-prime if it satisfies the following condition: for all monomials f, g ∈ [R] if f g ∈ J, then either f ∈ J or g ∈ J. (a) Prove that 0 is m-prime. (b) Prove that the following conditions are equivalent when J 6= 0: (i) J is m-prime; (ii) for all non-zero monomial ideals I and K if IK ⊆ J, then either I ⊆ J or K ⊆ J; (iii) J is m-irreducible and square-free; and (iv) there are positive integers k, t1 , . . . , tk such that J = (Xt1 , . . . , Xtk )R. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 4.2. Irreducible Ideals (optional) Definition 4.2.1. Let A be a commutative ring with identity. An ideal J ( A is reducible if there are ideals J1 , J2 6= J such that J = J1 ∩ J2 . An ideal J ( A is irreducible if it is not reducible. Remark 4.2.2. Let A be a commutative ring with identity. An ideal J ⊆ A is irreducible if and only if it satisfies the following: (1) J 6= A, and (2) given two ideals J1 , J2 such that J = J1 ∩ J2 , either J1 = J or J2 = J. If J is irreducible and J1 , . . . , Jn are ideals (with n > 2) such that J = ∩ni=1 Ji , then J = Ji for some index i. Example 4.2.3. The ideal 0 ⊆ Z is irreducible. If p is a prime number and n is a positive integer, then the ideal pn Z ⊆ Z is irreducible. These are the only irreducible ideals of Z; for instance, we have 6Z = 3Z ∩ 2Z. Let k be a field and let R be the polynomial ring R = k[X] in one variable. For each a ∈ k and each positive integer n, the ideal (X − a)n R ⊆ R is irreducible. Example 4.2.4. In the ring Z6 , the ideal 0 is reducible since 2Z6 ∩ 3Z6 = 0. In the ring Z4 , the ideal 0 is irreducible since the only ideals in Z4 are 0, 2Z4 and Z4 . Example 4.2.5. In light of Example 4.4.3 one might be tempted to ask whether every power of an irreducible ideal is again irreducible. While it does happen sometimes, the answer is “no” in general, even for powers of prime ideal. Indeed, in the ring R = C[X1 , . . . , Xn ] let P be the ideal generated by the size two minors of the matrix   X1 X2 X3 M = X4 X5 X6  X7 X8 X9

4.2. IRREDUCIBLE IDEALS (OPTIONAL)

71

that is,we set P = (X5 X9 − X6 X8 , X4 X9 − X6 X7 , X4 X8 − X5 X7 , X2 X9 − X3 X8 , X1 X9 − X3 X7 , X1 X8 − X2 X7 , X2 X6 − X3 X5 , X1 X6 − X3 X4 , X1 X5 − X2 X4 )R. One can show that this ideal is prime, hence irreducible, but that P 2 is not irreducible; see [2, (3.9.1)]. Remark 4.2.6. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal of R. If J is irreducible, then it is m-irreducible. The converse of this statement can fail. Definition 4.2.7. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Given a polynomial f = Pfinite n n∈Nd an X ∈ R, the support of f is the set γ(f ) = {n ∈ Nd | an 6= 0}. Example 4.2.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables. The support of the polynomial f = X 2 + XY + X 2 Z 3 − XY 2 Z 3 is the set γ(f ) = {(2, 0, 0), (1, 1, 0), (2, 0, 3), (1, 2, 3)} ⊆ N3 . Remark 4.2.9. Let A be a commutative ring with identity and let R be the polynomial ring R = PA[X1 , . . . , Xd ] in d variables. Let f ∈ R. Then γ(f ) is a finite set such that f = n∈γ(f ) an Xn . Furthermore, we have γ(f ) = ∅ if and only if f = 0. Lemma 4.2.10. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix positive integers k, e1 , . . . , ek and set J = (X1e1 , . . . , Xkek )R. Let I be an ideal of R such that J ( I. Then there is a polynomial hk = zb h(Xk+1 , . . . , Xd ) in I r J where z = X1e1 −1 · · · Xkek −1 . Proof. Fix a polynomial h ∈ I r J. Then we have X X X h= an Xn = an Xn + an Xn = f + g n∈γ(h)

n∈γ(h) n∈Γ(J) /

n∈γ(h) n∈Γ(J)

where f=

X an Xn

n∈γ(h) n∈Γ(J) /

and

g=

X an Xn .

n∈γ(h) n∈Γ(J)

By construction, every monomial occuring in g is in J, so g ∈ J. On the other hand, since h ∈ / J and g ∈ J, we have f = h − g ∈ / J. In particular, we have f 6= 0. Also, we have h ∈ I by assumption, so the condition g ∈ J ⊆ I implies that f = h − g ∈ I. Furthermore, we have γ(f ) = {n ∈ γ(h) | n ∈ / Γ(J)}

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so for each n ∈ γ(f ) and for i = 1, . . . , k, we have ni < ei . Indeed, if ni > ei , then / Γ(J). Xn ∈ (Xiei )R ⊆ J, contradicting the condition n ∈ (In a sense, the existence of f is stronger than the existence of h. Not only is f in I and not in J, but also no monomial occurring in f is in P . For this reason, we turn our attention from h to f .) Claim: For j = 1, . . . , k there exists a polynomial hj ∈ I r J such that for each n ∈ γ(hj ), we have ni = ei − 1 when 1 6 i 6 j and we have ni 6 ei when j + 1 6 i 6 k. We prove the claim by induction on j. Base case: j = 1. Consider the powers of X1 appearing in the monomials of f , and set r1 = min{n1 ∈ N | n ∈ γ(f )} which is the smallest of these powers. It follows that r1 < e1 since, if not, then every monomial occurring in f would be in (X1e1 )R ⊆ J; this would imply that f ∈ J, a contradiction. This implies that e1 − r1 > 0, that is, that e1 − r1 > 1. Write X X an Xn = f1 + g1 an Xn + f= n∈γ(f ) n1 >r1

n∈γ(f ) n1 =r1

where f1 =

X

an Xn

and

n∈γ(f ) n1 =r1

g=

X

an Xn

n∈γ(f ) n1 >r1

and note that γ(f1 ) = {n ∈ γ(f ) | n1 = r1 } = 6 ∅ γ(g1 ) = {n ∈ γ(f ) | n1 > r1 + 1}. We set h1 = X1e1 −r1 −1 f1 6= 0. To show that h1 has the desired properties, we first show that X1e1 −r1 −1 g1 ∈ J ⊆ I. For each d-tuple n ∈ γ(g1 ) we have n1 > r1 + 1. By construction, we have X X X1e1 −r1 −1 g1 = X1e1 −r1 −1 an Xn = an X1e1 −r1 −1 Xn . n∈γ(g1 )

It follows that every d-tuple m ∈

γ(X1p1 −r1 g1 )

n∈γ(g1 )

satisfies

m1 = (e1 − r1 − 1) + n1 > (e1 − r1 − 1) + r1 + 1 = e1 so

X1e1 −r1 −1 g1

∈ (X1e1 )R ⊆ J. Since f ∈ I, it follows that h1 = X1e1 −r1 −1 f1 = X1e1 −r1 −1 f − X1e1 −r1 −1 g1 ∈ I.

Each monomial occurring in h1 has the form Xm = X1e1 −r1 −1 Xn for some n ∈ γ(f1 ). The condition n ∈ γ(f1 ) implies that n1 = r1 so m1 = e1 − r1 − 1 + n1 = e1 − r1 + r1 − 1 = e1 − 1. Furthermore, for i > 2, the condition n ∈ γ(f1 ) ⊂ γ(f ) implies that ni < ei , that is ni 6 ei − 1, so mi = ni 6 ei − 1. To complete the proof of the base case, we need to show that h1 ∈ / J. Since J is a monomial ideal, it suffices to show that no monomial Xm occurring in h1 is in J. Again write Xm = X1p1 −r1 Xn for some n ∈ γ(f1 ), and suppose that Xm ∈ J = (X1e1 , . . . , Xkek )R. Lemma 2.1.14 implies Xm ∈ (Xiei )R for some i 6 k,

4.2. IRREDUCIBLE IDEALS (OPTIONAL)

73

so mi > ei by Lemma 2.1.11. This contradicts the condition mi 6 ei − 1, which has already been shown for each i. This completes the proof of the base case of our induction. The induction step is left as an exercise; see Exercise 4.2.20. This step is quite similar to the base case. Here are some hints. Assume that 1 6 j 6 d − 1 and that hj has been constructed. We want to construct hj+1 . Set rj+1 = min{nj+1 ∈ N | n ∈ γ(hj )} and show that rj+1 6 pj+1 . Then write X X an Xn + hj = n∈γ(hn ) nj+1 =rj+1 p

an Xn = fj+1 + gj+1 .

n∈γ(hn ) nj+1 =rj+1

−r

j+1 j+1 fj+1 and show that this polynomial satisfies the desired Set hj+1 = Xj+1 properties. It follows that there is a polynomial hk ∈ I rJ such that for each n ∈ γ(hk ), we have ni = ei − 1 when 1 6 i 6 k. In other words, every monomial occurring in hk has the form zw where w is a monomial in Xk+1 , . . . , Xd . This implies that there is a polynomial b h(Xk+1 , . . . , Xd ) such that hk = zb h(Xk+1 , . . . , Xd ), as desired. 

Theorem 4.2.11. Let A be an integral domain and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. A non-zero monomial ideal J ⊆ R is irreducible if and only if it is m-irreducible. Proof. One implication is from Remark 4.2.6. For the converse, assume that J is m-irreducible. Proposition 4.1.5, yields positive integers k, t1 , . . . , tk , e1 , . . . , ek such that 1 6 t1 < · · · < tk 6 d such that J = (Xte11 , . . . , Xtekk )R. By re-ordering the variables if necessary, we may assume without loss of generality that J = (X1e1 , . . . , Xkek )R. Set z = X1e1 −1 · · · Xnen −1 , and note that Exercise 4.1.12 implies that z ∈ / J. By way of contradiction, suppose that there are ideals I, K ⊆ R such that J = I ∩ K and J ( I and J ( K. Lemma 4.2.10 provides polynomials fk = z fb(Xk+1 , . . . , Xd ) ∈ I r J gk = zb g (Xk+1 , . . . , Xd ) ∈ K r J. Write fb = fb(Xk+1 , . . . , Xd ) and gb = gb(Xk+1 , . . . , Xd ). Since fk ∈ I, we have z fbgb = fk gb ∈ I. Similarly, the condition gk ∈ K implies that z fbgb ∈ K, hence z fbgb ∈ I ∩ K = J. Because of the conditions fb = fb(Xk+1 , . . . , Xd ) and gb = gb(Xk+1 , . . . , Xd ), every monomial occurring in z fbgb has the form m

n+1 w = zv = X1e1 −1 · · · Xnen −1 Xn+1 · · · Xdmd .

Since J is a monomial ideal, every monomial occurring in z fbgb is in J; see Exercise 2.1.25. The condition w ∈ J implies that there is an index j such that 1 6 j 6 k e and Xj j w. By comparing exponent vectors, we deduce that ej 6 ej − 1, which is impossible. We conclude that the polynomial z fbgb does not have any monomials, that is, we have z fbgb = 0. Since R is an integral domain and z is a monomial, it follows

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4. M-IRREDUCIBLE IDEALS AND DECOMPOSITIONS

that either fb = 0 or gb = 0. If fb = 0, then 0 = z fb = fk ∈ / J, which is impossible. A similar contradiction arises if gb = 0. Thus, the ideal J is irreducible, as desired.  The next example shows that not every irreducible ideal in a polynomial ring over an integral domain is a monomial ideal. Similarly, Exercise 4.2.21 shows that, if A is not an integral domain, then m-irreducible monomial ideals need not be irreducible. Example 4.2.12. Let k be a field and let R denote the polynomial ring R = k[X] in one variable. Then the ideal (X + 1)R is irreducible. Exercises. Exercise 4.2.13. Verify the statements in Remark 4.2.2. Exercise 4.2.14. Prove that if A is an integral domain, then 0 is irreducible in A. Exercise 4.2.15. Verify the properties in Example 4.2.3. Exercise 4.2.16. Verify the statements in Remark 4.2.9. Exercise 4.2.17. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. (a) Prove that the ideal 0 is irreducible in R if and only if it is irreducible in A. (b) Prove that if A is a field, then 0 is irreducible in R. (c) Prove that if A is an integral domain, then 0 is irreducible in R. Exercise 4.2.18. Let A be a commutative ring with identity and let I ⊆ A be an ideal. (a) Assume that I has the following property: there exists an element f ∈ R such that f is not in I, but f ∈ J for every ideal J of R that properly contains I. Prove that I is irreducible. (b) Does the converse of part (a) hold? That is, if I is irreducible, must there exist an element f ∈ R such that f is not in I, but f ∈ J for every ideal J of R that properly contains I? Exercise 4.2.19. Let R denote the polynomial ring R = Q[X, Y, Z] in three variables. Set J = (X 4 , Y 5 , Z 3 )R and h = 2X 2 Y +3X 2 Y Z 2 +5X 2 Y 2 +2X 2 Y 4 Z+X 3 Y 2 +9X 4 Y 2 +6XY 4 Z 4 +11Y 3 Z 5 +Y 6 and suppose that I is an ideal of R that contains h. Work through the proof of Lemma 4.2.10 in this special case by completing the following steps. (a) Prove that h ∈ / J. (b) List the elements in the set γ(h). (c) What is a(4,2,0) ? (d) Find f and g. (e) Find r1 . Is r1 6 3? (f) Find f1 and g1 . (g) Find h1 . Does h1 have the property that X appears raised to the power 3 in each monomial of h1 , while Y appears raised to a power of at most 4 and Z appears raised to a power of at most 2? (h) Find r2 . Is r2 6 4? (i) Find f2 and g2 .

4.2. IRREDUCIBLE IDEALS (OPTIONAL)

(j) Find h2 . Does h2 have the property that X appears raised to the power 3 each monomial of h2 , while Y appears raised to a power of at most 4 and appears raised to a power of at most 2? (k) Find r3 . Is r3 6 2? (l) Find f3 and g3 . (m) Find h3 . Does h3 have the property that X appears raised to the power 3 each monomial of h3 , while Y appears raised to a power of at most 4 and appears raised to a power of at most 2? (n) What is b h? What is z? Is it true that h3 = zb h?

75

in Z

in Z

Exercise 4.2.20. Prove the induction step in Lemma 4.2.10. Exercise 4.2.21. Find an example of a commutative ring A (with identity) such that the ideal (X1 , . . . , Xd )R in the polynomial ring R = A[X1 , . . . , Xd ] is reducible. Exercise 4.2.22. Verify the claim in Example 4.2.12. Exercise 4.2.23. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Prove that the following conditions are equivalent: (a) the ideal 0 is irreducible in A; (b) R has an irreducible monomial ideal; and (c) every non-zero monomial ideal J ( R is irreducible. Exercise 4.2.24. Let A be a commutative ring with identity, and let J be a nonzero irreducible ideal in A. Prove or disprove the following: If {Iλ }λ∈Λ is a (possibly infinite) set of ideals in A such that J = ∩λ∈Λ Iλ , then there is an index λ ∈ Λ such that J = Iλ . Exercise 4.2.25. Let A be a commutative ring with identity. We say that an ideal J in R is prime if it satisfies the following condition: for all f, g ∈ R if f g ∈ J, then either f ∈ J or g ∈ J. (a) Prove that 0 is prime if and only if A is an integral domain. (b) Prove that the non-zero prime ideals of Z are the ideals of the form pZ where p is a prime number. (This is where the name “prime ideal” comes from.) (c) Let n be an integer such that n > 2. Prove that the prime ideals of Zn are the ideals pZn such that p is a prime integer such that p n. (d) Let k be a field and let R be the polynomial ring R = k[X1 , . . . , Xd ] with d > 1. Prove that each ideal of the form (Xi1 , . . . , Xin )R is prime. (It can be shown that R has infinitely many prime ideals, once one knows Gauss’ Theorem which states that R is a unique factorization domain.) (e) Prove that the following conditions are equivalent: (i) J is prime; (ii) for all ideals I and K if IK ⊆ J, then either I ⊆ J or K ⊆ J; and (iii) J is irreducible and rad (J) = J. (Here is an outline for the proof of (iii) =⇒ (i). Assume that J is irreducible and rad (J) = J, and suppose that J is not prime. Then there exist f, g ∈ R r J such that f g ∈ J. Set I = J + f R and K = J + gR. The containments IK ⊆ J ⊆ I ∩ K imply that J = rad (J) = rad (I ∩ K) = rad (I) ∩ rad (K). Since J is irreducible, we have J = rad (I) or J = rad (K). The fact that f ∈ I ⊆ rad (I) and g ∈ K ⊆ rad (K) yield a contradiction.)

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4. M-IRREDUCIBLE IDEALS AND DECOMPOSITIONS

Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 4.3. M-Irreducible Decompositions The next proof is essentially due to Emmy Noether. Theorem 4.3.1. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. If J ( R is a monomial ideal, then there are m-irreducible monomial ideals J1 , . . . , Jn of R such that J = ∩ni=1 Ji . Proof. If J = 0, then J is m-irreducible by Exercise 4.1.8, so it is an intersection of one m-irreducible monomial ideal. Suppose that there is a non-zero monomial ideal J ( R that is not an intersection of finitely many m-irreducible monomial ideals of R. Then the set Σ = {non-zero monomial ideals J ( R | J is not an intersection of finitely many m-irreducible monomial ideals of R} is a non-empty set of monomial ideals of R. Theorem 2.3.3(b) implies that Σ has a maximal element J. In particular J is not m-irreducible, so there exist monomial ideals I, K ⊆ R such that J = I ∩K and J ( I, K. In particular, we have 0 6= I 6= R and 0 6= K 6= R. Since J is maximal in Σ, we have I, K ∈ / Σ. Hence, there are m-irreducible monomial ideals I1 , . . . , Im , K1 , . . . , Kn such that I = ∩m j=1 Ij and K = ∩ni=1 Ki . It follows that
n J = I ∩ K = ∩m j=1 Ij ∩ (∩i=1 Ki ) so J is a finite intersection of m-irreducible monomial ideals, a contradiction.



Definition 4.3.2. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J ( R be a monomial ideal. An m-irreducible decomposition of J is an expression J = ∩ni=1 Ji where each Ji is m-irreducible. Example 4.3.3. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X, Y ] in 2 variables. An m-irreducible decomposition of the the monomial ideal J = (X 3 , X 2 Y, Y 3 )R is J = (X 2 , Y 3 )R ∩ (X 3 , Y )R. See Example 4.1.3. Definition 4.3.4. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J ( R be a monomial ideal. An m-irreducible decomposition J = ∩ni=1 Ji is redundant if there exist indices i 6= i0 such that Ji ⊆ Ji0 . An m-irreducible decomposition J = ∩ni=1 Ji is irredundant if if it is not redundant, that is if for all indices i 6= i0 one has Ji 6⊆ Ji0 . Example 4.3.5. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X, Y ] in 2 variables. Set J = (X 3 , X 2 Y, Y 3 )R. The mirreducible decomposition of J from Example 4.3.3 J = (X 2 , Y 3 )R ∩ (X 3 , Y )R is irredundant. Indeed, we have X 2 ∈ (X 2 , Y 3 )R r (X 3 , Y )R so (X 2 , Y 3 )R 6⊆ (X 3 , Y )R. Also, we have Y ∈ (X 3 , Y )R r (X 2 , Y 3 )R so (X 3 , Y )R 6⊆ (X 2 , Y 3 )R.

4.3. M-IRREDUCIBLE DECOMPOSITIONS

77

On the other hand, the m-irreducible decomposition J = (X 2 , Y 3 )R ∩ (X 3 , Y )R ∩ (X, Y )R is redundant because (X 2 , Y 3 )R ⊆ (X, Y )R. The previous example shows that m-irreducible decompositions are not unique in general. However, we show below that irredundant m-irreducible decompositions are unique. First, we show that every m-irreducible decomposition can be transformed into an irredundant one by removing redundancies. Algorithm 4.3.6. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal with m-irreducible decomposition J = ∩ni=1 Ji . Note that n > 1. Step 1. Check whether the intersection J = ∩ni=1 Ji is irredundant. Step 1a. If, for all indices j and j 0 such that j 6= j 0 , we have Jj 6⊆ Jj 0 , then the intersection is irredundant; in this case, the algorithm terminates. Step 1b. If there exist indices j and j 0 such that j 6= j 0 and Jj ⊆ Jj 0 , then the intersection is redundant; in this case, continue to Step 2. Step 2. Reduce the intersection by removing the ideal that causes the redundancy in the intersection. By assumption, there exist indices j and j 0 such that j 6= j 0 and Jj ⊆ Jj 0 . Reorder the indices to assume without loss of generality that j 0 = n. Thus, we have j < n and Jj ⊆ Jn . It follows that J = ∩ni=1 Ji = ∩n−1 i=1 Ji . Step 3: Apply Step 1 to the new decomposition J = ∩n−1 J . i i=1 The algorithm will terminate in at most n − 1 steps because one can remove at most n − 1 monomials from the list and still form an ideal that is a non-empty intersection of parameter ideals. Corollary 4.3.7. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Every monomial ideal J ( R has an irredundant m-irreducible decomposition. Proof. Theorem 4.3.1 and Algorithm 4.3.6.



The next result explains the our use of the term “irredundant” for m-irreducible decompositions. Proposition 4.3.8. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R with m-irreducible decomposition J = ∩ni=1 Ji . Then the following conditions are equivalent: (i) the decomposition J = ∩ni=1 Ji is redundant; and (ii) there is an index j such that J = ∩i6=j Ji . Proof. (i) =⇒ (ii) Assume that the decomposition J = ∩ni=1 Ji is redundant. This implies that there are indices j 6= j 0 such that Jj ⊆ Jj 0 , and it follows that J = ∩i6=j 0 Ji . (ii) =⇒ (i) Assume that there is an index j such that J = ∩i6=j Ji . (This implies that n > 2.) Then we have ∩i6=j Ji = J = ∩ni=1 Ji ⊆ Jj . Lemma 4.1.6 implies that there is an index j 0 such that Jj 0 ⊆ Jj , so the intersection is redundant.  The next result shows that irredundant m-irreducible decompositions are unique up to re-ordering the terms.

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Theorem 4.3.9. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R with irredundant m-irreducible decompositions J = ∩ni=1 Ji = ∩m h=1 Ih . Then m = n and there is a permutation σ ∈ Sn such that Jt = Iσ(t) for t = 1, . . . , n. Proof. Claim 1: For t = 1, . . . , n there is an index u such that Iu = Jt . To show this, we compute: n ∩m h=1 Ih = J = ∩i=1 Ji ⊆ Jt .

Lemma 4.1.6 implies that there is an index u such that Iu ⊆ Jt . Similarly, we have ∩ni=1 Ji = J = ∩m h=1 Ih ⊆ Iu so Lemma 4.1.6 implies that there is an index v such that Jv ⊆ Iu ⊆ Jt . Since the decomposition ∩ni=1 Ji is irredundant, the containment Jv ⊆ Jt implies that v = t, so we have Jt ⊆ Iu ⊆ Jt , that is Iu = Jt . Claim 2: For t = 1, . . . , n there is a unique index u such that Iu = Jt . Indeed, if Iu = Jt = Iu0 , then the irredundancy of the intersection ∩m h=1 Ih implies that u = u0 . Define the function σ : {1, . . . , n} → {1, . . . , m} by letting σ(t) be the unique index u such that Iu = Jt . The same reasoning shows the following: For u = 1, . . . , m there is a unique index t such that Iu = Jt . Define the function ω : {1, . . . , m} → {1, . . . , n} by letting ω(u) be the unique index t such that Iu = Jt . By construction, the function ω is a two-sided inverse function for σ, hence the desired conclusions.  Exercises. Exercise 4.3.10. Verify the equality J = (X 2 , Y 3 )R ∩ (X 3 , Y )R ∩ (X, Y )R from Example 4.3.5. Exercise 4.3.11. Verify the equality ∩ni=1 Ji = ∩n−1 i=1 Ji from Algorithm 4.3.6. Exercise 4.3.12. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R with irredundant m-irreducible decomposition J = ∩ni=1 Ji . (a) Assume that for i = 1, . . . , n the ideal Ji is square-free. Prove that J is squarefree. (b) Prove or disprove the following: If J is square-free, then each ideal Ji is squarefree. Exercise 4.3.13. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a non-zero monomial ideal in R with irredundant monomial generating sequence f1 , . . . , fk . Use the ideas in the proof of Proposition 4.1.5 to explicitly construct an m-irreducible decomposition J = ∩ni=1 Ji where the generators of each Ji are determined by factorizations of the fj . Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output.

4.4. IRREDUCIBLE DECOMPOSITIONS (OPTIONAL)

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4.4. Irreducible Decompositions (optional) The next proof is essentially due to Emmy Noether. Proposition 4.4.1. Let A be a commutative noetherian ring with identity. If J ( A is an ideal, then there are irreducible ideals J1 , . . . , Jn of A such that J = ∩ni=1 Ji . Proof. Suppose that there is an ideal J ( A that is not an intersection of finitely many irreducible monomial ideals of A. Then the set Σ = {ideals J ( A | A is not an intersection of finitely many irreducible ideals of A} is a non-empty set of ideals of A. Since A is noetherian, the set Σ has a maximal element J. In particular J is not irreducible, so there exist ideals I, K ⊆ A such that J = I ∩ K and J ( I, K. In particular, we have I 6= A and K 6= A. Since J is maximal in Σ, we have I, K ∈ / Σ. Hence, there are irreducible ideals n I1 , . . . , Im , K1 , . . . , Kn such that I = ∩m j=1 Ij and K = ∩i=1 Ki . It follows that
n J = I ∩ K = ∩m j=1 Ij ∩ (∩i=1 Ki ) so J is a finite intersection of irreducible ideals, a contradiction.



Definition 4.4.2. Let A be a commutative ring with identity, and let J ( A be an ideal. An irreducible decomposition of J is an expression J = ∩ni=1 Ji where each Ji is irreducible. Example 4.4.3. Let n be an integer such that n > 2. The Fundamental Theorem of Arithmetic says that n has a factorization n = pe11 · · · pemm where the pi are distinct prime numbers and each ei is a positive integer. It follows that nZ = pe11 Z ∩ · · · ∩ pemm Z is an irredundant irreducible decomposition. Let R be the polynomial ring R = C[X] in one variable, and let f be a nonconstant polynomial in R. The Fundamental Theorem of Algebra says that f has a factorization f = c(X − a1 )e1 · · · (X − am )em where the ai are distinct complex numbers, each ei is a positive integer, and c is the leading coefficient of f . It follows that f R = (X − a1 )e1 R ∩ · · · ∩ (X − am )em R is an irredundant irreducible decomposition. In the ring C(R) of continuous functions, the ideal I = {f ∈ C(R) | f (n) = 0 for all n ∈ N} does not have an irreducible decomposition. For an indication of why this is true, note that I = ∩n∈Z In where In = {f ∈ C(R) | f (n) = 0}. Each ideal is irreducible since it is prime (see Exercise 4.2.25) but there is no finite sequence n1 , . . . , nt such that I = In1 ∩ · · · ∩ Int . Example 4.4.4. Let k be an integral domain, and let A be the polynomial ring A = k[X, Y ] in 2 variables. An irreducible decomposition of the the ideal J = (X 3 , X 2 Y, Y 3 )A is J = (X 2 , Y 3 )A ∩ (X 3 , Y )A. See Example 4.1.3 and Theorem 4.2.11.

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Definition 4.4.5. Let A be a commutative ring with identity, and let J ( A be an ideal. An irreducible decomposition J = ∩ni=1 Ji is redundant if if there exists an index i0 such that J = ∩i6=i0 Ji . An irreducible decomposition J = ∩ni=1 Ji is irredundant if if it is not redundant, that is if for all indices i0 one has J 6= ∩i6=i0 Ji . Example 4.4.6. Let k be an integral domain, and let A be the polynomial ring A = k[X, Y ] in 2 variables. Set J = (X 3 , X 2 Y, Y 3 )A. The irreducible decomposition of J from Example 4.4.4 J = (X 2 , Y 3 )A ∩ (X 3 , Y )A is irredundant. Indeed, we have X 2 ∈ (X 2 , Y 3 )A r J so J 6= (X 3 , Y )A. Also, we have Y ∈ (X 3 , Y )A r J so J 6= (X 2 , Y 3 )A. On the other hand, the irreducible decomposition J = (X 2 , Y 3 )A ∩ (X 3 , Y )A ∩ (X, Y )A is redundant because J = (X 2 , Y 3 )A ∩ (X 3 , Y )A. The previous example shows that irreducible decompositions are not unique in general. The situation is even worse, however, as the next example shows that even irredundant irreducible decompositions are not unique in general. Contrast this with the situation of m-irreducible decompositions. See Exercise 4.4.19 for a weak uniqueness statement. Example 4.4.7. Let k be an integral domain, and let A be the polynomial ring A = k[X, Y ] in 2 variables. The following ideals are irreducible and pair-wise distinct: (X, Y 2 )A and (X 2 , Y )A and (X 2 , X + Y )A and (X, (X + Y )2 )A. Furthermore, one has (X, Y 2 )A ∩ (X 2 , Y )A = (X 2 , XY, Y 2 )A = (X 2 , X + Y )A ∩ (X, (X + Y )2 )A. Algorithm 4.4.8. Let A be a commutative ring with identity, and let J be an ideal with irreducible decomposition J = ∩ni=1 Ji . Note that n > 1. Step 1. Check whether the intersection J = ∩ni=1 Ji is irredundant. Step 1a. If, for all indices j and j 0 such that j 6= j 0 , we have Jj 6⊆ Jj 0 , then the intersection is irredundant; in this case, the algorithm terminates. Step 1b. If there exist indices j and j 0 such that j 6= j 0 and Jj ⊆ Jj 0 , then the intersection is redundant; in this case, continue to Step 2. Step 2. Reduce the intersection by removing the ideal that causes the redundancy in the intersection. By assumption, there exist indices j and j 0 such that j 6= j 0 and Jj ⊆ Jj 0 . Reorder the indices to assume without loss of generality that j 0 = n. Thus, we have j < n and Jj ⊆ Jn . It follows that J = ∩ni=1 Ji = ∩n−1 i=1 Ji . n−1 Step 3: Apply Step 1 to the new decomposition J = ∩i=1 Ji . The algorithm will terminate in at most n − 1 steps because one can remove at most n − 1 monomials from the list and still form an ideal that is a non-empty intersection of parameter ideals. Corollary 4.4.9. Let A be a commutative noetherian ring with identity. Every ideal in A has an irredundant irreducible decomposition. Proof. Proposition 4.4.1 and Algorithm 4.4.8.



Proposition 4.4.10. Let A be a commutative ring with identity, and let J be an ideal in A with irreducible decomposition J = ∩ni=1 Ji . If there are indices j 6= j 0 such that Jj ⊆ Jj 0 , then the decomposition J = ∩ni=1 Ji is redundant.

4.4. IRREDUCIBLE DECOMPOSITIONS (OPTIONAL)

Proof. If there are indices j 6= j 0 such that Jj ⊆ Jj 0 , then J = ∩i6=j 0 Ji .

81



The next example shows that the converse of the previous result fails in general. Contrast this with the situation for monomial ideals from Theorem 4.3.9. The example also shows that the version of Lemma 4.1.6 fails in this setting. Example 4.4.11. Let k be an integral domain, and let A be the polynomial ring A = k[X, Y ] in 2 variables. We consider the irreducible decompositions (X, Y 2 )A ∩ (X 2 , Y )A = (X 2 , XY, Y 2 )A = (X 2 , X + Y )A ∩ (X, (X + Y )2 )A from Example 4.4.7. It follows that the decomposition (X 2 , XY, Y 2 )A = (X, Y 2 )A ∩ (X 2 , Y )A ∩ (X 2 , X + Y )A is redundant; however, there are no containment relations between the three ideals in this decomposition. Thus, the converse of Proposition 4.4.10 fails in general. Also, we have (X, Y 2 )A ∩ (X 2 , Y )A = (X 2 , XY, Y 2 )A ⊆ (X 2 , X + Y )A even though (X, Y 2 )A 6⊆ (X 2 , X +Y )A and (X 2 , Y )A 6⊆ (X 2 , X +Y )A. This shows that the version of Lemma 4.1.6 fails in this setting. Corollary 4.4.12. Let A be an integral domain, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Every non-zero monomial ideal in R has an irredundant irreducible decomposition. Proof. Corollary 4.3.7 and Theorem 4.2.11.



Exercises. Exercise 4.4.13. Verify the properties in Example 4.4.3. Exercise 4.4.14. Verify the equality J = (X 2 , Y 3 )R ∩ (X 3 , Y )R ∩ (X, Y )R from Example 4.4.4. Exercise 4.4.15. Verify the equality ∩ni=1 Ji = ∩n−1 i=1 Ji from Algorithm 4.4.8. Exercise 4.4.16. Let A be a commutative ring with identity, and let J be an ideal in A with irredundant irreducible decomposition J = ∩ni=1 Ji . (a) Assume that for i = 1, . . . , n one has rad (Ji ) = Ji . Prove that J = rad (J). (b) Prove or disprove the following: If J = rad (J), then rad (Ji ) = Ji for i = 1, . . . , n. Exercise 4.4.17. Verify the details from Example 4.4.7. Exercise 4.4.18. Verify the details from Example 4.4.11. Exercise 4.4.19. Let A be a commutative ring with identity, and let J be an ideal in A with irredundant irreducible decompositions J = ∩ni=1 Ji = ∩m h=1 Ih . Then
m = n and there is a permutation σ ∈ Sn such that rad (Jt ) = rad Iσ(t) for t = 1, . . . , n. (Hint: Show that for i = 1, . . . , m there is an index j such that J = J1 ∩ · · · ∩ Ji−1 ∩ Ij ∩ Ji+1 ∩ · · · ∩ Jn .) Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output.

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4.5. Operations on M-Irreducible Decompositions The theme for this section is the following: Given monomial ideals I and J, use m-irreducible decompositions I = ∩nj=1 Ij and J = ∩m i=1 Ji to find m-irreducible decompositions of other ideals obtained from I and J. Our first example of this is particularly straightforward. Remark 4.5.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I and J be monomial ideals with m-irreducible decompositions I = ∩nj=1 Ij and J = ∩m i=1 Ji . Then the monomial ideal I ∩ J has m-irreducible decomposition I ∩ J = (∩nj=1 Ij ) ∩ (∩m i=1 Ji ). (Of course, this also works for intersections of more than two ideals.) Note that this decomposition need not be irredundant, even when the original decompositions are irredundant. M-Irreducible Decompositions of Bracket Powers. Now, we consider bracket powers of monomial ideals; see Section 3.4. Proposition 4.5.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal with m-irreducible decomposition I = ∩nj=1 Ij , and let k be a positive integer. (a) An m-irreducible decomposition of I [k] is I [k] = ∩nj=1 Ij [k] . (b) The decomposition I = ∩nj=1 Ij is irredundant if and only if the decomposition I [k] = ∩nj=1 Ij [k] is irredundant. Proof. Claim: if J is an m-irreducible monomial ideal of R, then J [k] is mirreducible. Indeed, if J = 0, then J [k] = 0, which is m-irreducible. Assume that J 6= 0. Proposition 4.1.5 provides positive integers m, t1 , . . . , tm , e1 , . . . , em such that 1 6 t1 < · · · < tm 6 d such that J = (Xte11 , . . . , Xtemm )R. From Proposition 3.4.9, we have m 1 )R , . . . , Xtke J [k] = (Xtke m 1 so Proposition 4.1.4 implies that J [k] is m-irreducible. (a) Lemma 3.4.11 shows that I [k] = ∩nj=1 Ij [k] , and the first paragraph of this proof shows that each ideal Ij [k] is m-irreducible. (b) “ ⇐= ”: If the decomposition I = ∩nj=1 Ij is redundant, then there are indices i 6= i0 such that Ii ⊆ Ii0 . Lemma 3.4.10(a) implies that Ii [k] ⊆ Ii0 [k] , so the decomposition I [k] = ∩nj=1 Ij [k] is redundant. “ =⇒ ”: If the decomposition I [k] = ∩nj=1 Ij [k] is redundant, then there are indices i 6= i0 such that Ii [k] ⊆ Ii0 [k] . Lemma 3.4.10(a) implies that Ii ⊆ Ii0 , so the decomposition I = ∩nj=1 Ij is redundant.  M-Irreducible Decompositions of Sums. Next, we look at sums of monomial ideals; see Exercise 2.3.20. Proposition 4.5.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. If J1 , . . . , Jn are m-irreducible monomial ideals, then the sum J1 + · · · + Jn is an m-irreducible monomial ideal. Proof. We prove the result by induction on n. The case n = 1 is straightforward.

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Base case: n = 2. Assume that I and J are m-irreducible monomial ideals; we show that I + J is m-irreducible. If I = 0, then I + J = 0 + J = J which is m-irreducible. Similarly, if J = 0, then we are done, so we assume that I and J are both non-zero. Proposition 4.1.5 shows that there are positive integers j, k, s1 , . . . , sj , t1 , . . . , tk , d1 , . . . , dj , e1 , . . . , ek such that 1 6 s1 < · · · < sj 6 d and d 1 6 t1 < · · · < tk 6 d and I = (Xsd11 , . . . , Xsjj )R and J = (Xte11 , . . . , Xtekk )R. Fact 1.4.8(a) implies that I + J = (Xsd11 , . . . , Xsdjj , Xte11 , . . . , Xtekk )R. It is straightforward but tedious to show that it follows that I+J = (Xuf11 , . . . , Xufll )R for appropriate positive integers l, u1 , . . . , ul , f1 , . . . , fl . Proposition 4.1.4 implies that this ideal is m-irreducible.  Proposition 4.5.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Given monomial ideals I1 , . . . , In and J1 , . . . , Jm in R, one has n m (∩nj=1 Ij ) + (∩m i=1 Ji ) = ∩j=1 ∩i=1 (Ij + Ji ). n m Proof. The ideals (∩nj=1 Ij ) + (∩m i=1 Ji ) and ∩j=1 ∩i=1 (Ij + Ji ) are monomial ideals, so we need only show that the sets of monomials in each ideals are the same. We compute: n m [(∩nj=1 Ij ) + (∩m i=1 Ji )] = [∩j=1 Ij ] ∪ [∩i=1 Ji ]

= (∩nj=1 [Ij ]) ∪ (∩m i=1 [Ji ]) = ∩nj=1 ∩m i=1 ([Ij ] ∪ [Ji ]) = ∩nj=1 ∩m i=1 [Ij + Ji ] = [∩nj=1 ∩m i=1 (Ij + Ji )]. The first and fourth steps are from Exercise 2.3.20(c). The second and fifth steps are from Theorem 3.1.1. The third step is from the distributive laws for intersections and unions.  Proposition 4.5.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I and J be monomial ideals with m-irreducible decompositions I = ∩nj=1 Ij and J = ∩m i=1 Ji . Then an m-irreducible decomposition of I + J is I + J = ∩nj=1 ∩m i=1 (Ij + Ji ). Proof. From Proposition 4.5.4 we have n m I + J = (∩nj=1 Ij ) + (∩m i=1 Ji ) = ∩j=1 ∩i=1 (Ij + Ji )

and Proposition 4.5.3 shows that each ideal Ij + Ji is m-irreducible.



M-Irreducible Decompositions of Colon Ideals. Next, we look at colon ideals of monomial ideals; see Section 3.3. Proposition 4.5.6. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let k, t1 , . . . , tk , et1 , . . . , etk be

84

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e

positive integers, and set J = (Xt1t1 , . . . , Xtk k )R. Given a monomial f = Xn ∈ [R], we have ( R if there is an index i such that nti > eti (J :R f ) = etk −ntk et1 −nt1 (Xt1 , . . . , Xtk )R if for i = 1, . . . , k we have nti > eti . Proof. We know that f ∈ J if and only if if there is an index i such that et f ∈ (Xti i )R. By comparing exponent vectors, this says that f ∈ J if and only if there is an index i such that nti > eti . If there is an index i such that nti > eti , then f ∈ J, so (J :R f ) = R by Proposition 1.5.3(c). Assume now that for i = 1, . . . , k we have nti > eti . For i = 1, . . . , k the et −nti monomial Xti i is in (J :R f ) because et −nti

Xti i

et −nti +nti

f = X1n1 · · · Xti i

et

· · · Xdnd ∈ (Xti i )R ⊆ J.

To complete the proof, we need to fix a monomial g ∈ (J :R f ) and show that et −nti )R for some index i. (This uses the fact that (J :R f ) is a monomial g ∈ (Xti i ideal; see Theorem 3.3.1.) Let g = Xm ∈ [(J :R f )]. Then f g ∈ J, so there is an et index i such that 1 6 i 6 n and f g ∈ (Xti i )R. A comparison of exponent vectors shows that this implies that nti +mti > eti , so mti > eti −nti . Another comparison et −nti )R as desired.  of exponent vectors implies that g ∈ (Xti i Proposition 4.5.7. Let A be a commutative ring with identity, and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a non-zero monomial ideal of R with monomial generating sequence f1 , . . . , ft and write fj = Xnj where nj = (nj,1 , . . . , nj,d ). Let J be a monomial ideal of R with m-irreducible decometi,k eti,1 i position J = ∩m i=1 Ji , and write Ji = (Xti,1 , . . . , Xti,ki )R; in particular, we have J 6= R. Assume that I 6⊆ J. Then an m-irreducible decomposition of (J :R I) is (J :R I) = ∩(i,j)∈S (Ji :R fj ) where the intersection is taken over S = {(i, j) | for l = 1, . . . , ki we have ntj,l > etj,l }. Proof. We compute: (J :R I) = (J :R (f1 , . . . , ft )R) = ∩tj=1 (J :R fj ) = ∩tj=1 (∩m i=1 Ji :R fj ) = ∩tj=1 ∩m i=1 (Ji :R fj ) = ∩(i,j)∈S (Ji :R fj ). The first and third equalities are by assumption. The second equality is from Proposition 1.5.3(b), and the fourth equality is from Proposition 1.5.4(b). The fifth equality is from Proposition 4.5.6 which shows that for each (i, j) ∈ / S we have (Ji :R fj ) = R. Another application of Proposition 4.5.6 (with Proposition 4.1.4) shows that for each (i, j) ∈ S the ideal (Ji :R fj ) is m-irreducible. 

4.5. OPERATIONS ON M-IRREDUCIBLE DECOMPOSITIONS

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Exercises. Exercise 4.5.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in 2 variables. Find non-zero monomial ideal I, J ( R with irredundant m-irreducible decompositions I = ∩nj=1 Ij and J = ∩m i=1 Ji such n m that the decomposition I ∩ J = (∩j=1 Ij ) ∩ (∩i=1 Ji ) is redundant. Can this be done for monomial ideals in 1 variable? Justify your answers. Exercise 4.5.9. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in 2 variables. Find an irredundant m-irreducible decomposition of the ideal J [3] where J = (X 3 , X 2 Y, Y 3 )R. Justify your answer. Exercise 4.5.10. Complete the proof of Proposition 4.5.3. Exercise 4.5.11. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Prove or disprove the following: if J1 , . . . , Jn are monomial ideals such that the sum J1 + · · · + Jn is an m-irreducible monomial ideal, then Ji is m-irreducible for j = 1, . . . , n. Exercise 4.5.12. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in 2 variables. Find an irredundant m-irreducible decomposition of the ideal I +J where I = (X 3 , XY 2 , Y 3 )R and J = (X 3 , X 2 Y, Y 3 )R. Justify your answer. Exercise 4.5.13. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in 2 variables. Find non-zero monomial ideals I, J ( R with irredundant m-irreducible decompositions I = ∩nj=1 Ij and J = ∩m i=1 Ji such that the decomposition I + J = ∩nj=1 ∩m (I + J ) is redundant. Can this be done j i i=1 for monomial ideals in 1 variable? Justify your answers. Exercise 4.5.14. State and prove the version of Proposition 4.5.5 for K1 +· · ·+Kp where K1 , . . . , Kp are monomial ideals. Exercise 4.5.15. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in 2 variables. Find an irredundant m-irreducible decomposition of the ideal (J :R I) where I = (X 3 , XY 2 , Y 3 )R and J = (X 3 , X 2 Y, Y 3 )R. Justify your answer. Exercise 4.5.16. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in 2 variables. Find non-zero monomial ideals I, J ( R with (1) irredundant monomial generating sequence f1 , . . . , ft ∈ [I] and (2) irredundant m-irreducible decomposition J = ∩m i=1 Ji such that the decomposition (J :R I) = ∩(i,j)∈S (Ji :R fj ) from Proposition 4.5.7 is redundant. Can this be done for monomial ideals in 1 variable? Justify your answers. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output.

CHAPTER 5

Parametric Decompositions of Monomial Ideals 5.1. Parameter Ideals This section deals with special cases of m-irreducible monomial ideals. In the following definition, the “P” stands for “parameter”. The term “parameter ideal” comes from the idea that each power of each variable is a parameter, so these ideals are generated by a complete sequence of parameters. Definition 5.1.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. A parameter ideal in R is an ideal of the form (X1a1 , . . . , Xdad )R with a1 , . . . , ad > 1. If f = Xn with n ∈ Nd , then set PR (f ) = (X1n1 +1 , . . . , Xdnd +1 )R. Example 5.1.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Then PR (XY 2 ) = (X 2 , Y 3 )R. One can see from the graph .. .O

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that the “corner” in the graph of PR (XY 2 ) corresponds exactly to the monomial XY 2 . (We will see this more generally later.) This explains why this ideal is denoted PR (XY 2 ) instead of PR (f ) for some different monomial f . Other computations in this situation include PR (1) = (X, Y )R and PR (X) = (X 2 , Y )R and PR (Y ) = (X, Y 2 )R. The next lemma is a version of [5, (2.3)] for our situation. Lemma 5.1.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f and g be monomials in R. 87

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(a) We have f ∈ / PR (f ). (b) We have g ∈ PR (f ) if and only if f ∈ / (g)R.

Proof. Write f = Xm and g = Xn . (a) We have PR (f ) = (X1m1 +1 , . . . , Xdmd +1 )R. Suppose that f ∈ PR (f ). Lemma 2.1.14 implies that f is a monomial multiple of some Ximi +1 , and it follows from Lemma 2.1.11 that mi > mi + 1, which is impossible. (b) =⇒ : Assume that g ∈ PR (f ) and suppose that f ∈ (g)R. The condition g ∈ PR (f ) implies f ∈ (g)R ⊆ PR (f ), which contradicts part (a). ⇐= : Assume that g ∈ / PR (f ). Since PR (f ) is generated by X1m1 +1 , . . . , Xdmd +1 mi +1 )R for i = 1, . . . , d. Lemma 2.1.11 implies that either this implies that g ∈ / (Xi nj < 0 for some j 6= i or ni < mi + 1. Since we have nj > 0 for all j, this implies that ni < mi + 1, that is, that ni 6 mi . This is true for each index i, so m > n. This implies that f ∈ (g)R by Lemma 2.1.11. 

Here is a graphical explanation of the previous lemma.

Example 5.1.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Then PR (XY 2 ) = (X 2 , Y 3 )R. For part (a) one can see from the graph

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that f = XY 2 ∈ / PR (XY 2 ). (The monomial XY 2 is represented by ◦ in this graph.)

5.1. PARAMETER IDEALS

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For part (b) we look at two examples. In the first example, the monomial g1 = X 3 Y 2 ∈ PR (f ) is circled .. .O

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and we have g1 ∈ PR (f ) and f ∈ / g1 R. In the second example, the monomial g2 = XY ∈ / PR (f ) designated with an asterisk ∗ .. .O

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and we have g2 ∈ / PR (f ) and f ∈ g1 R. Definition 5.1.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal of R. A parametric decomposition of J is a decomposition of J of the form J = ∩ni=1 PR (zi ). A parametric decomposition J = ∩ni=1 PR (zi ) is redundant if if there exists indices j 6= j 0 such that PR (zj ) ⊆ PR (zj 0 ). A parametric decomposition J = ∩ni=1 PR (zi ) is irredundant if if it is not redundant, that is if for all indices j 6= j 0 one has PR (zj ) 6⊆ PR (zj 0 ). Remark 5.1.6. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal of R.

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The monomial J may or may not have a parametric decomposition. In fact, we shall see later that J has a parametric decomposition if and only if m-rad (J) = (X1 , . . . , Xd )R. Proposition 4.1.4 implies that every parameter ideal in R is m-irreducible. Hence, each parametric decomposition of J is an m-irreducible decomposition; also, such a decomposition is (ir)redundant as a parametric decomposition if and only if it is (ir)redundant as an m-irreducible decomposition. Exercises. Exercise 5.1.7. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Compute PR (1) and PR (Xi ) for i = 1, . . . , d. What are the generators for these ideals? Exercise 5.1.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f , g be monomials in [R]. (a) Prove that PR (f g) ⊆ PR (f ) ∩ PR (g). (b) Prove or disprove: PR (f g) = PR (f ) ∩ PR (g). (c) Prove or disprove: PR (f g) ⊆ PR (f )PR (g). (d) Prove or disprove: PR (f g) = PR (f )PR (g). (e) Prove that the following conditions are equivalent. (i) PR (f g) = PR (f ); (ii) f g = f ; and (iii) g = 1R . Exercise 5.1.9. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix a monomial z ∈ [R] and set (X) = (X1 , . . . , Xd )R. (a) Prove that m-rad (PR (z)) = (X). (b) Prove that if A is a field, then rad (PR (z)) = (X). Exercise 5.1.10. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let z1 , . . . , zn be monomials in [R] and (X) = (X1 , . . . , Xd )R. Prove that if I = ∩ni=1 PR (zi ), then m-rad (I) = (X). Exercise 5.1.11. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Prove that the following conditions are equivalent. (i) A is reduced; (ii) for every monomial z ∈ R, one has rad (PR (z)) = (X); and (iii) there exists a monomial z ∈ R such that rad (PR (z)) = (X). Exercise 5.1.12. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. For a monomial z ∈ [R], prove that z ∈ / I if and only if I ⊆ PR (z). Exercise 5.1.13. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let w, z be monomials in R. Prove that the following conditions are equivalent: (i) z ∈ (w)R; (ii) w ∈ / PR (z); (iii) PR (z) ⊆ PR (w); and

5.2. AN EXAMPLE

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(iv) (PR (z) :R w) 6= R. Exercise 5.1.14. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables with d > 2. Let I be a monomial ideal in R. If f and w are monomials in R, show that (PR (f w) :R f R) = PR (w). Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output.

5.2. An Example Lemma 5.2.1. Let A be a commutative ring with identity and let R be the poly-
nomial ring R = A[X1 , . . . , Xd ] in d variables. Then R contains precisely d+k−1 d−1 monomials of total degree k. Proof. Each monomial X1a1 · · · Xdad ∈ R of degree k corresponds to a binary number with exactly k ones and d − 1 zeroes: 1| 1 1{z. . . 1} 0 |1 1 1{z. . . 1} 0 · · · 0 |1 1 1{z. . . 1} . a1 ones

a2 ones

ad ones

Notice that each such binary number has exactly k + d − 1 digits and is determined
by the placement of the d − 1 zeroes. It follows that there are exactly k+d−1 such d−1 binary numbers. The correspondence between the monomials of degree k in R and the binary numbers with k + d − 1 digits and d − 1 zeroes
is a bijection, and so the number of monomials of degree k in R is exactly k+d−1  d−1 . The next result is a version of [5, (2.4)] for our situation. It is a special case of Theorem A. Theorem 5.2.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. For each n > 1, we have \ (X)n = PR (Xm ) |m|=n−1

where the intersection runs over all d-tuples m ∈ Nd such that |m| = m1 +· · ·+md = n − 1. Furthermore, this intersection is irredundant. In particular, the ideal (X)n
n+d−2 is the irredundant intersection of d−1 parameter ideals. Before proving this result, we give an example. Example 5.2.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. If (X) = (X, Y )R, then we have (X)3 = (X 3 , X 2 Y, XY 2 , Y 3 )R. The of total degree 2 in R are
monomials
3 X 2 , XY, Y 2 and there are exactly 2+2−1 = = 3 of them. The theorem says 2−1 1 that we have (X)3 = PR (X 2 ) ∩ PR (XY ) ∩ PR (Y 2 ) = (X 3 , Y )R ∩ (X 2 , Y 2 )R ∩ (X, Y 3 )R.

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Graphically, we are decomposing the graph Γ((X)3 ) .. .O

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One can check easily that this decomposition is irredundant. For instance, the monomial XY is in PR (X 2 ) ∩ PR (Y 2 ) and not in (X)3 . Proof of Theorem 5.2.2. In the case n = 1, the theorem states that (X) = PR (1). This follows from Exercise 5.1.7. This expression
is clearly
irredundant and the d−1 number of ideals in the decomposition is 1+d−2 = d−1 d−1 = 1. Thus, we assume without loss T of generality that n > 2. Set J = |m|=n−1 PR (Xm ) where the intersection runs over all d-tuples m ∈ Nd such that |m| = m1 + · · · + md = n − 1. We show that J = (X)n . Since each ideal PR (Xm ) is a monomial ideal, Theorem 3.1.1 implies that J is a monomial ideal. Exercise 2.3.19(d) shows that (X)n is the monomial ideal generated by all the monomials in R of total degree n; see also Exercise 1.4.28. Thus, in order to show that J = (X)n , it suffices to show that [J] = [(X)n ]; see Proposition 2.1.6(b). To this end, let g be a monomial in [R]. We show that g ∈ / J if and only if g ∈ / (X)n . We have g ∈ / J if and only if there exists a monomial f ∈ [R] of total degree n − 1 such that g ∈ / PR (f ), by the definition of J, that is, if and only if there exists a monomial f ∈ [R] of total degree n − 1 such that f ∈ (g)R; see Lemma 5.1.3(b). Exercise 5.2.8 shows that this condition occurs if and only if deg(g) < n, and this is so if and only if g ∈ / (X)n . It follows that J = (X)n , as desired. To see that the intersection is irredundant, let f = Xp ∈ [R] be a monomial with deg(f ) = n − 1 and set \ J= PR (Xm ) |m|=n−1 m6=p

where the intersection runs over all d-tuples m ∈ Nd such that |m| = m1 +· · ·+md = n − 1 and m 6= p. To show that the intersection is irredundant, we show that J 0 6⊆ PR (f ). To this end, we show that f ∈ J 0 and f ∈ / PR (f ). The condition f ∈ / PR (f ) follows from Lemma 5.1.3(a). Exercise 5.2.7(d) shows that for each monomial h ∈ [R] such that deg(h) = n − 1, if h 6= f , then f ∈ PR (h). This shows that f ∈ J 0 , as desired. Finally, the number of ideals occurring in this decomposition is precisely the number of monomials in [R] of degree n − 1. Lemma 5.2.1 implies that this number

is d+(n−1)−1 = d+n−2  d−1 d−1 , as desired. Here is a souped up version of the previous result. Theorem 5.2.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix positive integers n, t1 , . . . , tn and set I = (Xt1 , . . . , Xtn )R. For each k > 1, we have \ Ik = (Xte11 , . . . , Xtenn )R e1 +···+en =k+n−1

where the intersection runs over all sequences e1 , . . . , en ∈ Z+ such that e1 + · · · + en = k + n − 1. Furthermore, this intersection is
irredundant. In particular, the ideal I k is the irredundant intersection of k+n−2 m-irreducible monomial ideals. n−1 Proof.



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Exercises. Exercise 5.2.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in 3 variables. Use Theorem 5.2.2 to find a the irredundant parametric decomposition of ((X, Y, Z)R)4 . Exercise 5.2.6. Find another proof of Lemma 5.2.1. Exercise 5.2.7. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f and g be monomials in R. (a) (b) (c) (d)

Prove Prove Prove Prove

that if f ∈ (g)R, then deg(f ) > deg(g). or disprove: If deg(f ) > deg(g), then f ∈ (g)R. that if deg(f ) = deg(g) and g ∈ (f )R, then g = f . that if deg(f ) = deg(g) and f 6= g, then f ∈ PR (g).

Exercise 5.2.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f be a monomial in R and let n be an integer such that n > 1. Prove that deg(f ) < n if and only if there exists a monomial g of degree n − 1 such that g ∈ (f )R. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 5.3. Corner Elements Here is [5, (3.1)] for our situation. Definition 5.3.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R. A monomial z ∈ [R] is a J-corner element if z ∈ / J and X1 z, . . . , Xd z ∈ J. The set of J-corner elements of J in [R] is denoted CR (J). Here is [5, (3.2)] for our situation. Proposition 5.3.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let J be a monomial ideal in R. (a) The J-corner elements are precisely the monomials in (J :R (X)) r J, in other words, we have CR (J) = [(J :R (X))] r [J]. (b) The following conditions are equivalent: (i) 1 ∈ CR (J); (ii) J = (X); and (iii) CR (J) = {1}. (c) If z and z 0 are distinct J-corner elements, then z ∈ / (z 0 )R and z 0 ∈ / (z)R. (d) There are only finitely many J-corner elements, in other words, the set CR (J) is finite. Proof. (a) This follows from Proposition 1.5.3(b). (b) (i) =⇒ (ii): Assume that 1 ∈ CR (J). By definition, this implies that 1 ∈ /J and so J 6= R. Exercise 2.1.20 implies that J ⊆ (X). On the other hand, we have 1 ∈ CR (J) ⊆ (J :R (X)) by part (a), so (J :R (X)) = R. Proposition 1.5.3(c) implies that (X) ⊆ J, so (X) = J.

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(ii) =⇒ (iii): Assume that J = (X). Then J 6= R, so 1 ∈ / J. Furthermore, we have (J :R (X)) = ((X) :R (X)) = R by Proposition 1.5.3(c), so 1 ∈ CR (J) by definition. On the other hand, we have {1} ⊆ CR ((X)) ⊆ [R] r [(X)] = {1} and hence CR (J) = {1}. (iii) =⇒ (i): This is straightforward. (c) Assume that z and z 0 are distinct J-corner elements and suppose that z ∈ (z 0 )R. It follows that there is a monomial f ∈ [R] such that z = f z 0 . Since z 6= z 0 , we conclude that f 6= 1. Since f is a monomial, it follows that f ∈ (X). By part (a), we have z 0 ∈ (J :R (X)), so the condition f ∈ (X) implies that z = f z 0 ∈ J. This contradicts the condition z ∈ CR (J), and so the condition z ∈ (z 0 )R must be false. Similarly, we conclude that z 0 ∈ / (z)R as desired. (d) The ideal K = (CR (J))R is a monomial ideal, so Theorem 2.3.1 implies that K is generated by a finite list of monomials z1 , . . . , zn ∈ CR (J). We claim that {z1 , . . . , zn } = CR (J). The containment {z1 , . . . , zn } ⊆ CR (J) holds by assumption. For the reverse containment, let z 0 ∈ CR (J). Then we have z 0 ∈ CR (J) ⊆ K = (z1 , . . . , zn )R and so z 0 is a monomial multiple of zj for some index j. Part (c) implies z 0 = zj , as desired.  Here is [5, (3.3)] for our situation. Corollary 5.3.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let J be a monomial ideal in R. Let z1 , . . . , zm be the distinct J-corner elements and set J 0 = PR (z1 ) ∩ · · · ∩ PR (zm ). (a) For i = 1, . . . , m we have (z1 , . . . , zi−1 , zi+1 , . . . , zm )R ⊆ PR (zi ) and zi ∈ / PR (zi ) and zi ∈ / J 0. (b) The intersection J 0 = ∩m i=1 PR (zi ) is irredundant. (c) There is a containment J ⊆ J 0 . Proof. (a) The condition zi ∈ / PR (zi ) is from Lemma 5.1.3(b). From this, the special case m = 1 is straightforward, so we assume that m > 2. For indices i and j such that j 6= i, we have zi 6= zj and so zi ∈ / (zj )R by Proposition 5.3.2(c); it follows that zj ∈ PR (zi ) by Lemma 5.1.3(b). We conclude that z1 , . . . , zi−1 , zi+1 , . . . , zm ∈ PR (zi ) so (z1 , . . . , zi−1 , zi+1 , . . . , zm )R ⊆ PR (zi ). The condition zi ∈ / J 0 follows because J 0 ⊆ PR (zi ). (b) This follows from part (a) because zi ∈ ∩j6=i PR (zj ) and zi ∈ / PR (zi ). (c) Exercise 5.3.18(c).  Remark 5.3.4. Continue with the notation of Corollary 5.3.3. The theorem we are working toward proving (Theorem A) says that if rad (J) = rad ((X)), then J = J 0 . In other words, if rad (J) = rad ((X)), then the J-corner elements yield an irreduandant decomposition of J as an intersection of parameter ideals. Definition 5.3.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R. Let cR (J) denote the number of J-corner elements, that is cR (J) = | CR (J)|. Remark 5.3.6. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R. Proposition 5.3.2(d) shows that cR (J) < ∞.

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Here is [5, (3.4)] for our situation. Proposition 5.3.7. Let A be a commutative ring with identity and let R be the polynomial ring R =
A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Then cR ((X)n ) = n+d−2 for all n > 1. d−1 Proof. Exercise 5.3.19.



Remark 5.3.8. Continue with the notation of Proposition 5.3.7. The theorem we are working toward proving (Theorem A) says that if J is a monomial ideal in R such that rad (J) = rad ((X)), then J-corner elements yield an irreduandant decomposition of J as an intersection of parameter ideals. Theorem 5.2.2 expresses
the ideal J = (X)n as an irredundant intersection of n+d−2 parameter ideals. d−1 Proposition 5.3.7 shows that the number of corner elements for this ideal is also
n+d−2 . So, it all ties together. d−1 Here is a digression about Hilbert functions and Hilbert polynomials. Remark 5.3.9. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R, and let I and J be monomial ideals in R. The number cR (J) is an invariant of J. This means that if cR (I) 6= cR (J), then I 6= J. One theme in mathematics is the search for patterns in such invariants. For instance, one can look for patterns in the sequence cR (J), cR (J 2 ), cR (J 3 ), . . .. For instance, we have the following.   n+d−2 n cR ((X) ) = d−1 (n + d − 2)(n + d − 2 − 1) · · · (2)(1) = (d − 1)!(n + d − 2 − (d − 1))! (n + d − 2)(n + d − 2 − 1) · · · (2)(1) = (d − 1)!(n − 1)! (n + d − 2)(n + d − 2 − 1) · · · (n − d − 2 − (d − 2))[(n − 1)!] = (d − 1)!(n − 1)! (n + d − 2)(n + d − 2 − 1) · · · (n − d − 2 − (d − 2)) = (d − 1)! Pd−2 1 j=0 j d−2 = nd−1 + n + ··· (d − 1)! (d − 1)! In particular, the number cR ((X)n ) is described by a polynomial in Q[n] of degree d − 1. This is an amazing fact: the function that counts a certain number of polynomials is itself a polynomial! We’ll see some more on this subject later. Here is [5, (3.5)] for our situation. Proposition 5.3.10. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f, g ∈ [R] be monomials. If there exist distinct integers i and j such that Xi f, Xj f ∈ (g)R, then f ∈ (g)R. Proof. The condition Xi f, Xj f ∈ (g)R implies that there are monomials hi , hj ∈ [R] such that Xi f = ghi and Xj f = ghj by Lemma 2.1.14. It follows that g(hi Xj ) = f Xi Xj = g(hj Xi )

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and so Exercise 2.1.28(a) implies that hi Xj = hj Xi . We conclude from Exercise 2.1.28(b) that hi ∈ (Xi )R and so there is a monomial k ∈ [R] such that hi = Xi k. This implies f Xi Xj = ghi Xj = gkXi Xj and so f = gk ∈ (g)R, as desired.  Corollary 5.3.11. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R that has a corner element z. If d > 2, then the monomials X1 z, . . . , Xd z are members of distinct principal ideals generated by monomials in J. Proof. Exercise 5.3.22.



Here is [5, (3.6)] for our situation. Corollary 5.3.12. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix monomials f1 , . . . , fn ∈ [R] and set J = (f1 , . . . , fn )R. If J has a corner element, then n > d. Proof. Exercise 5.3.23.



Here is [5, (3.7)] for our situation. Theorem 5.3.13. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let J be a monomial ideal in R such that J ⊆ (X). Let z1 , . . . , zm be the distinct J-corner elements. (a) There is an equality (J :R (X)) = (z1 , . . . , zm )R + J. (b) If m > 1, then we have (J :R (X)) ) (z1 , . . . , zi−1 , zi+1 , . . . , zm )R + J for each i = 1, . . . , m. Proof. (a) Exercise 5.3.16. (b) Set Ki = (z1 , . . . , zi−1 , zi+1 , . . . , zm )R + J. The containment (J :R (X)) ⊇ Ki follows from part (a). To show that the containment is proper, it suffices to show that zi ∈ / Ki since zi ∈ (J :R (X)) by definition. Theorem 2.3.1 shows that there is a finite list of monomials f1 , . . . , fn ∈ [J] such that J = (f1 , . . . , fn )R. It follows that Ki = (z1 , . . . , zi−1 , zi+1 , . . . , zm )R + J = (z1 , . . . , zi−1 , zi+1 , . . . , zm )R + (f1 , . . . , fn )R = (z1 , . . . , zi−1 , zi+1 , . . . , zm , f1 , . . . , fn )R. Suppose by way of contradiction that zi ∈ Ki . Lemma 2.1.14 implies that zi is a monomial multiple of one of the monomial generators of Ki . If zi ∈ (fj )R for some index j, then zi ∈ (fj )R ⊆ J, contradicting the fact that zi is a J-corner element. It follows that zi ∈ (zj )R for some index j 6= i, but this contradicts Proposition 5.3.2(c), so we must have zi ∈ / Ki , as desired.  Exercises. Exercise 5.3.14. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d > 1 variables. Let f be a monomial in [R] and prove that CR ((f )R) = ∅.

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Exercise 5.3.15. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let J be a monomial ideal in R. Prove that if m-rad (J) = (X), then J has a corner element. Exercise 5.3.16. Prove Theorem 5.3.13(a). Exercise 5.3.17. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R, and fix a monomial f = Xn ∈ [R]. For i = 1, . . . , d set ei = (0, . . . , 0, 1, 0, . . . , 0) ∈ Nd where the 1 occurs in the ith position. Prove that f is a J-corner element if and only if n ∈ / Γ(J) and n + ei ∈ Γ(J) for each i = 1, . . . , d. Exercise 5.3.18. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R, and fix a monomial w ∈ [R]. (a) Prove that if w ∈ / J, then J ⊆ PR (w). (b) Prove that if w is a J-corner element, then J ⊆ PR (w). (c) In the notation of Corollary 5.3.3, prove that J ⊆ J 0 . Exercise 5.3.19. Prove Proposition 5.3.7. Exercise 5.3.20. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let J be a monomial ideal in R such that m-rad (J) = (X). (a) Prove that for each n > 0, the set of monomials [R] r [J n ] is finite. (b) Define the function F : N → N by the formula F (n) = [the number of monomials in R − (X)n ] = |[R] r [(X)n ]|. Prove that there is a polynomial f ∈ Q[x] of degree d such that F (n) = f (n) for all n ∈ N. The function F is the Hilbert function of R associated to the ideal (X), and f is the associated Hilbert polynomial. A deeper result (one that we do not expect you to prove) says the following. Let FJ : N → N be the function F (n) = [the number of monomials in R − J n ] = |[R] r [J n ]|. There exists n0 ∈ N and a polynomial fJ ∈ Q[x] of degree d such that FJ (n) = fJ (n) for all n > n0 . Exercise 5.3.21. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R and consider the function GI : N → N given by the formula GI (n) = cR (I n ). (a) Let z be a monomial in R and set I = PR (z). Show that there exists a polynomial gI ∈ Q[x] and an integer n0 > 1 such that gI (n) = GI (n) for all n > n0 . (b) (Challenge) Does the conclusion of part (a) hold if I is an arbitrary monomial ideal in R? Exercise 5.3.22. Prove Corollary 5.3.11. Exercise 5.3.23. Prove Corollary 5.3.12.

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99

Challenge Exercise 5.3.24. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R. Give a necessary and sufficient for J to have a corner element. Prove that your condition is necessary and sufficient. Exercise 5.3.25. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R such that m-rad (I) = (X). Let f be a monomial in [R] and let z ∈ CR (I). Prove that f z ∈ CR (f I). Exercise 5.3.26. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R and let z1 , . . . , zm be the distinct corner elements for I. Prove that z1 , . . . , zm is an irredundant generating sequence for the ideal J = (z1 , . . . , zm )R. Exercise 5.3.27. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R. Prove that if I 6= R, then CR (I) ⊆ m-rad (I). Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 5.4. Finding Corner Elements in Two Variables Here is [5, (3.8)] for our situation. Note that in loc. cit. it is assumed that J is “open”, that is, a monomial ideal such that (X, Y )R ⊆ rad (J). However, as we shall see, this hypothesis is not necessary in our situation. Theorem 5.4.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Let J be a non-zero monomial ideal in R and let f1 , . . . , fn ∈ [J] be the irredundant monomial generating sequence for J. Then cR (J) = n − 1. Proof. If n = 1, then the result follows from Exercise 5.3.14. Thus, we assume for the remainder of the proof that n > 2. Since the generating sequence f1 , . . . , fn is irredundant, we have fi 6= fj whenever i 6= j. It follows from Fact 2.3.16 that we have either fi
ai < aj

and

bi > bj

when i < j.

In particular, the inequalities 0 6 a1 < aj for j > 2 imply that 1 6 aj when j > 2, and the inequalities 0 6 bn < bi for i < n imply that 1 6 bi when i < n. For i = 1, . . . , n−1 we set zi = X ai+1 −1 Y bi −1 . We will show that the monomials z1 , . . . , zn−1 are precisely the distinct J-corner elements. Claim 1: For i = 1, . . . , n − 1 we have zi ∈ / J. Suppose by way of contradiction that zi ∈ J. Lemma 2.1.14 then implies that zi is a monomial multiple of fj for some j. Comparing exponents, we have ai+1 − 1 > aj and bi − 1 > bj . If j 6 i, then this implies the first inequality in the following sequence bi − 1 > bj > bi

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5. PARAMETRIC DECOMPOSITIONS OF MONOMIAL IDEALS

while the second inequality is from (5.4.1.1); this is impossible. If j > i, then we have j > i + 1 and so similar reasoning explains the sequence ai+1 − 1 > aj > ai+1 which is also impossible. This establishes the claim. Claim 2: For i = 1, . . . , n − 1 we have zi ∈ CR (J). Since we have already seen that zi ∈ / J, it suffices to show that Xzi , Y zi ∈ J. By construction, we have Xzi = XX ai+1 −1 Y bi −1 = X ai+1 Y bi −1 = X ai+1 Y bi+1 Y bi −1−bi+1 = fi+1 Y bi −1−bi+1 ∈ (fi+1 )R ⊆ J. Note that the element Y bi −1−bi+1 is a bona fide element of R because the condition bi > bi+1 implies that bi − 1 − bi+1 > 0. Similarly, we have Y zi = Y X ai+1 −1 Y bi −1 = X ai+1 −1 Y bi = X ai Y bi X ai+1 −1−ai = fi X ai+1 −1−ai ∈ (fi )R ⊆ J. This establishes the claim. Claim 3: For indices i, j such that i 6= j we have zi 6= zj . This comes from a direct comparison of X-exponents when i < j: the inequailty ai < aj implies that ai − 1 < aj − 1 and so zi 2. It follows that Xz is not a monomial multiple of fi for all i > 2. Hence, the condition Xz ∈ J implies that Xz is a monomial multiple of f1 . This implies that a1 6 a + 1 and b1 6 b. However, since z 6∈ J, we know that z is not a monomial multiple of f1 , and so either a1 > a or b1 > b. The condition b1 6 b implies that b1 6> b, so we must have a1 > a. Hence, we have a < a1 6 a + 1, and so a1 = a + 1. This implies that a = a1 − 1 < a1 6 ai for all i > 1. Comparing X-exponents, we conclude that Y z is not a monomial multiple of any fi , and so Y z ∈ / J, a contradiction. It follows that we must have a = a2 − 1 and b < b1 − 1. In this case, the Y -exponent of Y z is b + 1 < b1 which shows that Y z is not a monomial multiple of f1 . The X-exponent of Y z is a2 − 1 < ai for all i > 2, and this shows that Y z is not a monomial multiple of fi for i = 2, . . . , n. This shows that Y z ∈ / J, a contradiction. This shows that the condition z
5.4. FINDING CORNER ELEMENTS IN TWO VARIABLES (n−1)+2−2 2−1




=

n−1 1




101

= n − 1 which agrees with the computation from Theo-

rem 5.4.1. Remark 5.4.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. For each monomial ideal J ⊆ R such that 0 6= J ⊆ (X, Y ), Theorem 5.4.1 says cR (J) = νR (J) − 1. Here is some of [5, (3.11)] for our situation. Proposition 5.4.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. For each monomial ideal J ⊆ R such that 0 6= J ⊆ (X, Y )R we have νR (J) − 1 6 νR ((J :R (X, Y )R)) 6 2νR (J) − 1. Proof. Let n = νR (J) and consider the irredundant monomial generating sequence f1 , . . . , fn ∈ [J] for J. Theorem 5.4.1 implies that cR (J) = n − 1, so we have CR (J) = {z1 , . . . , zn−1 }. Theorem 5.3.13(a) implies that (J :R (X, Y )R) = (f1 , . . . , fn , z1 , . . . , zn−1 )R The list f1 , . . . , fn , z1 , . . . , zn−1 is a monomial generating sequence for the ideal (J :R (X, Y )R). Hence, this list contains the irredundant monomial generating sequence for (J :R (X, Y )R) and so νR ((J :R (X, Y )R)) 6 n + n − 1 = 2n − 1. Theorem 5.3.13(b) shows that (J :R (X, Y )R) 6= (f1 , . . . , fn , z1 , . . . , zi−1 , zi+1 , . . . , zn−1 )R for i = 1, . . . , n − 1. It follows that each zi occurs in the irredundant monomial generating sequence for (J :R (X, Y )R). This implies that νR ((J :R (X, Y )R)) > n − 1 as desired.  Here is the rest of [5, (3.11)] for our situation. Proposition 5.4.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. For all integers n and t such that n > 1 and n − 1 6 t 6 2n − 1, there is a monomial ideal J such that νR (J) = n and νR ((J :R (X, Y )R)) = t. Proof. Set s = t − n + 1 and note that the conditions n − 1 6 t 6 2n − 1 imply that 0 6 s 6 n. For i = 1, . . . , s we set fi = X 2(i−1) Y n+s−2i , which has total degree n + s − 2. For i = s + 1, . . . , n we set fi = X s+i−1 Y n−i , which has total degree n + s − 1. Consider the monomial ideal J = (f1 , . . . , fn )R. Claim 1: The sequence f1 , . . . , fn is an irredundant monomial generating sequence for J. Since the sequence generates J by definition, it suffices to show that fi is not a monomial multiple of fj when i 6= j. To this end, note that the sequence of fi ’s has the following form: (5.4.5.1)

Y n+s−2 , X 2 Y n+s−4 , X 4 Y n+s−6 , . . . , X 2s−4 Y n−s+2 , X 2s−2 Y n−s , X 2s Y n−s−1 , X 2s+1 Y n−s−2 , . . . , X s+n−2 Y, X s+n−1 .

In particular, the sequence of X-exponents is strictly increasing, while the sequence of Y -exponents is strictly decreasing. It follows that we have fi ∈ / (fj )R whenever i 6= j. This establishes the claim. Note that the monomials in the list (5.4.5.1) are listed in strictly increasing order with respect to the lexicographical order. Graphically, they are arranged as

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5. PARAMETRIC DECOMPOSITIONS OF MONOMIAL IDEALS

follows:

.. .O • − •

− −

•

− −

•

−

•

−

•

− |

|

|

|

|

|

|

|

|

•

/ ···

From the proof of Theorem 5.4.1 we see that the corner elements of J are the following:

(5.4.5.2)

XY n+s−3 , X 3 Y n+s−5 , X 5 Y n+s−7 , . . . , X 2s−3 Y n−s−1 , X 2s−1 Y n−s−1 , X 2s Y n−s−2 , . . . , X s+n−3 Y, X s+n−2 .

5.4. FINDING CORNER ELEMENTS IN TWO VARIABLES

103

They are marked by ∗’s in the next graph. .. .O • −

∗ •

−

∗

−

•

−

∗

−

•

−

∗

− − |

|

|

|

|

|

|

• ∗

•

|

∗

•

/ ···

Formally, they are given in lexicographical order as: zi = X 2i−1 Y n+s−2i−1 for i = 1, . . . , s and zi = X s+i−1 Y n−2−1 for i = s + 1, . . . , n − 1. The elements f1 , . . . , fn , z1 , . . . , zn−1 generate (J :R (X, Y )R). From the graph, we can see that some of the fi ’s are part of the irredundant generating sequence for (J :R (X, Y )R) (the fi in the upper left are not monomial multiples of the zj ) and some of the fi are not part of the irredundant generating sequence for (J :R (X, Y )R) (the fi in the lower right are monomial multiples of some zj ). We show that νR ((J :R (X, Y )R)) = t by considering two cases. (Then we are done.) Case 1: s = 0. This is the case where t = n − 1. Graphically, this case looks like the following. .. .O • ∗

•

−

∗

•

|

∗

•

/ ···

In this case, one checks readily that f1 ∈ (z1 )R and fi ∈ (zi−1 )R ∩ (zi )R for i = 2, . . . , n − 1 and fn ∈ (zn )R; see Exercise 5.4.7. This shows that (J :R (X, Y )R) = (z1 , . . . , zn−1 )R. Since the zi are distinct J-corner elements, it follows from Proposition 5.3.2(c) that we have zi ∈ / (zj )R when i 6= j. It follows that z1 , . . . , zn−1 is an irredundant monomial generating sequence for (J :R (X, Y )R), and so νR ((J :R (X, Y )R)) = n − 1 = t, as desired.

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Case 2: s > 1. This is the case where t > n. Graphically, this case looks like our larger graphs. One checks readily that fi ∈ (zi−1 )R for i = s + 1, . . . n and fi ∈ / (z1 , . . . , zn−1 )R for i = 1, . . . , s; see Exercise 5.4.7. Since the fi form an irredundant monomial generating sequence for J, it follows that the sequence f1 , . . . , fs , z1 , . . . , zn−1 is an irredundant monomial generating sequence for (J :R (X, Y )R), and so νR ((J :R (X, Y )R)) = s + n − 1 = t, as desired.  Exercises. Exercise 5.4.6. Complete the proof of Theorem 5.4.1. Exercise 5.4.7. Complete the proof of Proposition 5.4.5. Exercise 5.4.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in 3 variables. Show by example that there is a monomial ideal I in R such that m-rad (I) = (X, Y, Z)R and cR (I) > νR (I). Exercise 5.4.9. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set J = (Y 9 , XY 7 , X 3 Y 4 , X 5 Y 2 , X 11 )R. (a) Find the corner elements of J without drawing the graph of J; instead, use the proof of Theorem 5.4.1 from the class notes. (b) Now draw the graph of J and check that your answers above agree with the graph. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 5.5. Finding Corner Elements in General Here is [5, (3.14)] for our situation. Proposition 5.5.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R, and let I be a monomial ideal in R such that m-rad (I) = (X). Set S = [R] r [I], the set of monomials in R that are not in I, and set w = max{deg(f ) | f ∈ S}. For j = 0, . . . , w set Dj = {f ∈ S | deg(f ) = j}. For j = 0, . . . , w − 1 set Cj = {f ∈ Dj | for i = 1, . . . , d we have Xi f 6∈ Dj+1 } and set Cw = Dw . Then CR (I) is the disjoint union CR (I) = ∪w j=0 Cj . Proof. Note that CR (I) 6= ∅ by Exercise 5.3.15. Also, the set S is finite by Exercise 2.1.30. In particular, the number w is well-defined as it is the maximum element of a finite set of natural numbers. Note also that for i 6= j we have Ci ∩ Cj = ∅ because the elements of Ci have different degrees from the elements of Cj . Claim 1: ∪w j=0 Cj ⊆ CR (I). To verify this containment, we fix a monomial f ∈ Cj for some j and show that f ∈ CR (I). The containments Cj ⊆ Dj ⊆ S ⊆ R r I show that f ∈ / I. Since deg(f ) = j by definition, we have deg(Xi f ) = j + 1 for i = 1, . . . , d. If j = w, then f ∈ Cw = Dw , and so the condition deg(Xi f ) = j + 1 = w + 1 > w = max{deg(g) | g ∈ S} implies that Xi f ∈ / S for i = 1, . . . , d. On the other hand, if j < w, then Xi f 6∈ Dj+1 by definition of Cj , and so the definition of Dj+1

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implies that Xi f ∈ / S. In either case, the elements X1 f, . . . , Xd f are monomials of R that are not in S = [R] r [I]. It follows that X1 f, . . . , Xd f ∈ I and so f ∈ CR (I). This establishes the claim. Claim 2: ∪w j=0 Cj ⊇ CR (I). To verify this containment, we fix a monomial g ∈ CR (I) and set j = deg(g); we show that g ∈ Cj . The assumption g ∈ CR (I) implies that g is a monomial in R that is not in I, and so g ∈ S. By definition, we have g ∈ Dj . When j = w, this implies g ∈ Cj , so we assume that j < w. For i = 1, . . . , d we have Xi g ∈ I and so Xi g ∈ / S. It follows that Xi g cannot be in Dj+1 . This shows that g ∈ Cj , thus completing the proof of the claim and the proof of the proposition. 

Example 5.5.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in 2 variables. Let J = (X 6 , X 4 Y, X 3 Y 2 , Y 6 )R which has the following graph.

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The monomials in the set S are designated with ∗’s, and the elements of Dj are the represented by the ∗’s on the diagonal line of slope −1 and Y -intercept j. .. .O

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D5 = {X 5 , X 2 Y 3 , XY 4 , Y 5 }

D6 = {X 2 Y 4 , XY 5 }

D7 = {X 2 Y 5 }.

It follows that C7 = D7 = {X 2 Y 5 }. For j < w = 7, the elements of Cj are the monomials of degree j represented by ∗’s such that (a) the point one unit to the right is a •, and (b) the point one unit up is a •. Such points are designated in the next graph as ~’s .. .O

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and so we have C0 = ∅

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While this gives us a longer algorithm for finding corner elements in the case of two variables, it also works in more than two variables. Here is [5, (3.15)] for our situation. Proposition 5.5.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R, and let I be a monomial ideal in R such that m-rad (I) = (X). If f is a monomial in [R] r [I], then there is a monomial g ∈ [R] such that f g ∈ CR (I). Proof. Set S = [R] r [I] which is a finite set by Exercise 2.1.30. Set T = {g ∈ [R] | f g ∈ / I}. If g ∈ T , then g 6∈ I since f g 6∈ I and I is an ideal. It follows that T ⊆ S and so T is a finite set. Let g be a monomial in T with maximal degree. By the maximality of deg(g), we have Xi g ∈ / T for i = 1, . . . , d. In other words, we have Xi f g ∈ I for i = 1, . . . , d. This says that f g ∈ (I :R (X)), so the condition f g ∈ / I implies that f g ∈ CR (I), as desired.  Here is [5, (3.15)] for our situation. Proposition 5.5.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R, and let I and J be monomial ideals in R such that I ( J and m-rad (I) = (X). Then CR (I) ∩ J 6= ∅. Proof. Fix an element f ∈ [J] r [I]. Proposition 5.5.3 implies that there is a monomial g ∈ [R] such that f g ∈ CR (I). Since J is an ideal and f ∈ J, we have f g ∈ J, and so f g ∈ CR (I) ∩ J.  Exercises. Exercise 5.5.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables. Set J = (Z 4 , Y 2 Z 3 , Y 3 , XY Z, XY 2 , X 2 )R. Use Proposition 5.5.1 to find the J-corner-elements. State the value of w and list the elements in each Ci and Di . Exercise 5.5.6. Let A be a commutative ring with identity and let R be the polynomial ring R = A[U, X, Y, Z] in four variables. Set J = (Z 5 , Y Z 4 , Y 2 Z 2 , Y 3 , XZ 2 , XY Z, X 3 Z, X 3 Y 2 , X 4 , U )R. Use Proposition 5.5.1 to find the J-corner-elements. State the value of w and list the elements in each Ci and Di . Exercise 5.5.7. Let A be a commutative ring with identity and let R be the polynomial ring R = A[U, X, Y, Z] in four variables. Set J = (Z d , Y c , X b , U XY Z, U a )R

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where a, b, c, and d are integers that are greater than 1. Use Proposition 5.5.1 to find the J-corner-elements. State the value of w and list the elements in each Ci and Di . Exercise 5.5.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables. Consider the monomial ideal J = (Z 4 , Y 2 Z 3 , Y 3 , XY Z, XY 2 , X 2 )R and set f = X. Show that f 6∈ J and find a monomial g such that f g ∈ CR (I). (See Proposition 5.5.4.) Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 5.6. Existence and Uniqueness of Parametric Decompositions Here is [5, (4.1)] for our situation. It contains part of Theorem A from the introduction. Theorem 5.6.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let J be a monomial ideal in R such that m-rad (J) = (X). If the distinct J-corner elements are z1 , . . . , zm then J = ∩m j=1 PR (zj ) is an irredundant parametric decomposition of J. Proof. Note first that Exercise 5.3.15 shows that J has a corner element. Set J 0 = ∩m j=1 PR (zj ). Corollary 5.3.3 implies that this intersection is irredundant and that J 0 ⊆ J. Thus, it remains to show that J 0 ⊇ J. Since each PR (zj ) is a monomial ideal, Theorem 3.1.1 implies that J 0 is a monomial ideal. Suppose by way of contradiction that J 0 ( J. Proposition 5.5.4 implies that J 0 contains an J-corner element, say zi ∈ J 0 . This implies that zi ∈ PR (zi ), which contradicts Lemma 5.1.3(a).  Example 5.6.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. We compute an irredundant parametric decomposition of the monomial ideal J = (X 6 , X 4 Y, X 3 Y 2 , Y 6 )R. Example 5.5.2 shows that the distinct J-corner elements are X 2 Y 5 , X 3 Y, X 5 . Hence, Theorem 5.6.1 yields J = PR (X 2 Y 5 ) ∩ PR (X 3 Y ) ∩ PR (X 5 ) = (X 3 , Y 6 )R ∩ (X 4 , Y 2 )R ∩ (X 6 , Y )R. Here is [5, (4.2)] for our situation. It says that a monomial ideal J such that rad (J) = rad ((X)) is uniquely determined by its corner elements. Corollary 5.6.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let I and J be monomial ideals in R such that m-rad (J) = (X) = m-rad (I). Then I = J if and only if CR (I) = CR (J). Proof. It is straightforward to show that if I = J, then CR (I) = CR (J). For the converse, assume that CR (I) = CR (J). Then Theorem 5.6.1 implies that I = ∩z∈CR (I) PR (z) = ∩z∈CR (J) PR (z) = J as desired.



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Remark 5.6.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let I and J be monomial ideals in R such that rad (J) = rad ((X)) = rad (I). The operator CR (−) does not respect containments: If I ⊆ J, then we may have CR (I) 6⊆ CR (J) and CR (I) 6⊇ CR (J); see Exercise 5.6.19. On the other hand, one can use Theorem 5.6.1 to show that following: if CR (I) ⊆ CR (J), then I ⊇ J; see Exercise 5.6.20. Theorem 5.6.1 shows that the J-corner elements determine an irredundant parametric decomposition of J. The next result shows that the converse holds, that an irredundant parametric decomposition of J determine the J-corner elements. It is [5, (4.4)] for our situation. Proposition 5.6.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix monomials z1 , . . . , zm ∈ [R] and assume that J = ∩m j=1 PR (zj ) is an irredundant parametric decomposition of J. Then the distinct J-corner elements are z1 , . . . , zm . Proof. Claim 1: Each zi is a J-corner element. First, note that zi ∈ / PR (zi ) by Lemma 5.1.3(a); it follows that zi ∈ / J since J = ∩m j=1 PR (zj ) ⊆ PR (zi ). To complete the proof of the claim, we need to show that Xj zi ∈ J for each j. By way of contradiction, suppose that Xj zi ∈ / J. It follows that Xj zi ∈ / PR (zk ) for some k. Lemma 5.1.3(b) implies that zk ∈ (Xj zi )R ⊆ (zi )R. Exercise 5.1.13 implies that PR (zk ) ⊆ PR (zi ). The irredundancy of the intersection implies that PR (zk ) = PR (zi ) and so i = k. The condition zk ∈ (Xj zi )R then reads as zk ∈ (Xj zk )R. A degree argument shows that this is impossible. Thus we must have Xj zi ∈ J for each i and j, and so each zi ∈ CR (J), as claimed. Claim 2: The elements z1 , . . . , zm are distinct. Indeed, if zi = zj , then PR (zi ) = PR (zj ), so the irreduncancy of the intersection implies that i = j. Claim 3: CR (J) ⊆ {z1 , . . . , zm }. (Once this claim is established, the proof is complete.) Let z ∈ CR (J). This implies that z ∈ / J, and so there is an index k such that z ∈ / PR (zk ). Lemma 5.1.3(b) implies that zk ∈ (z)R, and so Proposition 5.3.2(c) implies that z = zk . This establishes the claim.  Example 5.6.6. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set J = (X 3 , Y 6 )R ∩ (X 4 , Y 4 )R ∩ (X 5 , Y 2 )R = PR (X 2 Y 5 ) ∩ PR (X 3 Y 3 ) ∩ PR (X 4 Y ). The ideal J is an intersection of parameter ideals, so it has a corner element by Exercise 5.1.10. More specifically, we have CR (J) = {X 2 Y 5 , X 3 Y 3 , X 4 Y }. To show this, we first note that by Proposition 5.6.5, using this parametric decomposition, it suffices to show that the intersection is irredundant. To this end, we have the following: X 4 ∈ [(X 3 , Y 6 )R ∩ (X 4 , Y 4 )R] r J X 3 Y 2 ∈ [(X 3 , Y 6 )R ∩ (X 5 , Y 2 )R] r J Y 4 ∈ [(X 4 , Y 4 )R ∩ (X 5 , Y 2 )R] r J. These elements can be found by inspecting the graph of J.

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Corollary 5.6.7. Let A be a commtuative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. For each monomial z ∈ [R] we have CR (PR (z)) = {z}. Proof. Then the trivial intersection J = PR (z) is an irredundant parametric decomposition. Hence, Proposition 5.6.5 implies that CR (PR (z)) = CR (J) = {z}.  Here is [5, (4.3)] for our situation. It says that when A is a field, the only irreducible monomial ideals J such that rad (J) = rad ((X)) are the parameter ideals. Corollary 5.6.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let J be a monomial ideal in R such that m-rad (J) = (X). Then the following conditions are equivalent: (i) the ideal J is m-irreducible; (ii) the ideal J is a parameter ideal; and (iii) we have cR (J) = 1. Proof. Theorem 5.6.1 implies that we have an irredundant parametric decomposition J = ∩z∈CR (J) PR (z). (i) =⇒ (ii): Assume that J is irreducible. Using the given decomposition of J, Theorem 4.3.9 shows that J = PR (z) for some z ∈ CR (J). (ii) =⇒ (iii): If J is a parameter ideal, say J = PR (w) for some monomial w ∈ [R], then Corollary 5.6.7 implies that CR (J) = {w}, so we conclude that cR (J) = 1. (iii) =⇒ (ii): If cR (J) = 1 then we have CR (J) = {w} for some w ∈ [R]. Thus, the given decomposition has the form J = PR (w) which is a parameter ideal, hence m-irreducible by Proposition 4.1.4.  Here is [5, (4.9)] for our situation. Corollary 5.6.9. Let A be a commtuative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let J be a monomial ideal in R such that m-rad (J) = (X). Let z1 , . . . , zm be the irredundant monomial generating set for J and set I = ∩m j=1 PR (zj ). Then the distinct I-corner elements are z1 , . . . , zm and we have (I :R (X)) = I + J. Proof. Claim: The intersection I = ∩m j=1 PR (zj ) is irredundant. For each i and each j 6= i, we have zj ∈ / (zi )R since the generating sequence is irredundant. It follows from Lemma 5.1.3(b) that zi ∈ PR (zj ), and so zi ∈ ∩j6=i PR (zj ). Since zi ∈ / PR (zi ) by Lemma 5.1.3(a), it follows that ∩j6=i PR (zj ) 6⊆ PR (zi ), and so the intersection is irredundant. This establishes the claim. The equality CR (I) = {z1 , . . . , zm } now follows from Proposition 5.6.5. The equality (I :R (X)) = I + J then follows from Exercise 5.3.16.  Here is [5, (4.10)] for our situation. It contains the uniqueness statement of Theorem A from the introduction. Compare with Theorem 4.3.9. Theorem 5.6.10. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let z1 , . . . , zm , w1 , . . . , wn ∈ [R] n be monomials such that ∩m j=1 PR (zj ) = ∩j=1 PR (wj ). If each of these intersections is irredundant, then n = m and {z1 , . . . , zm } = {w1 , . . . , wn }.

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n Proof. Set I = ∩m j=1 PR (zj ) = ∩j=1 PR (wj ). Proposition 5.6.5 implies that {z1 , . . . , zm } = CR (I) = {w1 , . . . , wn }. Since the monomials in the list z1 , . . . , zm are distinct, and the monomials in the list w1 , . . . , wn are distinct, it follows that m = n. 

Checking that a given parametric decomposition (as in Example 5.6.6) is irredundant can be tedious. Imagine how it goes in three or more variables when there is no good visual to help guide you. The next proposition makes it a lot easier. Proposition 5.6.11. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix monomials z1 , . . . , zm ∈ [R] and set I = ∩m j=1 PR (zj ) and J = (z1 , . . . , zm )R. The following conditions are equivalent: (i) the intersection ∩m j=1 PR (zj ) is irredundant; (ii) for all indices i and j, if i 6= j, then PR (zi ) 6⊆ PR (zj ); (iii) for all indices i and j, if i 6= j, then zi ∈ / (zj )R; and (iv) the sequence z1 , . . . , zm is an irredundant generating sequence for J. Proof. (iv) =⇒ (i): Assume that the sequence z1 , . . . , zm is an irredundant generating sequence for J. Corollary 5.6.9 implies that the distinct I-corner elements are z1 , . . . , zm . Theorem 5.6.1 implies that the intersection ∩m j=1 PR (zj ) is irredundant. (i) =⇒ (ii): We prove the contrapositive. Assume that there are indices i and j such that i 6= j and PR (zi ) ⊆ PR (zj ). Then ∩k6=j PR (zk ) ⊆ PR (zi ) ⊆ PR (zj ) so the intersection is redundant. (ii) =⇒ (iii): Exercise 5.1.13. (iii) =⇒ (iv): Lemma 2.1.14.  Example 5.6.12. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set J = (X 3 , Y 6 )R ∩ (X 4 , Y 4 )R ∩ (X 5 , Y 2 )R = PR (X 2 Y 5 ) ∩ PR (X 3 Y 3 ) ∩ PR (X 4 Y ) as in Example 5.6.6. To show that the intersection is irredundant, it suffices (by Proposition 5.6.11) to observe that no monomial in the list X 2 Y 5 , X 3 Y 3 , X 4 Y is a monomial multiple of any other monomial in this list. This is much easier than the method used in Example 5.6.6, as we did not have to find the elements wi ∈ ∩j6=i PR (zj ) r PR (zi ). Furthermore, this method is much easier for computations in three or more variables as the elements wi are even harder to find in that case. Proposition 5.6.11 and Exercise 5.1.13 combine to give the following algorithm for transforming redundant parametric intersections into irredundant parametric intersections. Algorithm 5.6.13. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix monomials z1 , . . . , zm ∈ [R] and set I = ∩m j=1 PR (zj ). We assume that m > 1. Step 1. Check whether the intersection ∩m j=1 PR (zj ) is irredundant using Proposition 5.6.11. Step 1a. If, for all indices i and j such that i 6= j, we have zj ∈ / (zi )R, then the intersection is irredundant; in this case, the algorithm terminates.

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Step 1b. If there exist indices i and j such that i 6= j and zj ∈ (zi )R, then the intersection is redundant; in this case, continue to Step 2. Step 2. Reduce the intersection by removing the parameter ideal which causes the redundancy in the intersection. By assumption, there exist indices i and j such that i 6= j and zj ∈ (zi )R. Reorder the indices to assume without loss of generality that i = m. Thus, we have j < m and zj ∈ (zm )R. Exercise 5.1.13 implies that m−1 PR (zj ) ⊆ PR (zm ), and it follows that I = ∩m j=1 PR (zj ) = ∩j=1 PR (zj ). Now apply Step 1 to the new list of monomials z1 , . . . , zm−1 . The algorithm will terminate in at most m − 1 steps because one can remove at most m − 1 monomials from the list and still form an ideal that is a non-empty intersection of parameter ideals. Example 5.6.14. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set J = (X, Y 5 )R ∩ (X 2 , Y 3 )R ∩ (X 4 , Y 4 )R ∩ (X 3 , Y )R ∩ (X 5 , Y 2 )R = PR (Y 4 ) ∩ PR (XY 2 ) ∩ PR (X 3 Y 3 ) ∩ PR (X 2 ) ∩ PR (X 4 Y ). The list of zi ’s to consider is Y 4 , XY 2 , X 3 Y 3 , X 2 , X 4 Y . The monomial X 3 Y 3 is a multiple of XY 2 , so we remove XY 2 from the list. The new list of zi ’s to consider is Y 4 , X 3 Y 3 , X 2 , X 4 Y . The monomial X 4 Y is a multiple of X 2 , so we remove X 2 from the list. The new list of zi ’s to consider is Y 4 , X 3 Y 3 , X 4 Y . No monomial in the list is a multiple of another since the exponent vectors (0, 4), (3, 3) and (4, 1) are incomparable. Hence, the intersection J = PR (Y 4 ) ∩ PR (X 3 Y 3 ) ∩ PR (X 4 Y ) = (X, Y 5 )R ∩ (X 4 , Y 4 )R ∩ (X 5 , Y 2 )R. is an irredundant parametric decomposition of J. Here is a one-step procedure for transforming redundant parametric intersections into irredundant parametric intersections. Proposition 5.6.15. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix distinct monomials z1 , . . . , zm ∈ [R] with m > 1 and set I = ∩m j=1 PR (zj ). For j = 1, . . . , m write zj = Xnj with nj ∈ Nd . Set ∆ = {n1 , . . . , nm } ⊆ Nd and consider the order < on Nd from Definition 1.6.8. Let ∆0 denote the set of maximal elements of ∆ under this order. Then I = ∩nj ∈∆0 PR (zj ) is an irredundant parametric decomposition of I and CR (I) = {zj | nj ∈ ∆0 } ⊆ {z1 , . . . , zn }. Proof. Note that the set ∆ has maximal elements since ∆ is finite; see Proposition 5.3.2(d). The maximality of the elements of ∆0 implies that for each ni ∈ ∆, there is an element nj ∈ ∆0 such that nj < ni . It follows that zj ∈ (zi )R and so Exercise 5.1.13 implies that PR (zj ) ⊆ PR (zi ). From this, we conclude that I = ∩nj ∈∆0 PR (zj ). (Note that we are using the following straightforward fact from set-theory: If S1 , . . . , Sn are subsets of a fixed set T , and i, j are indices such that i 6= j and Si ⊆ Sj , then ∩nk=1 Sk = ∩k6=j Sk .) For each nj , nk ∈ ∆0 such that j 6= k, we have nj 6< nk since nj and nk are both maximal among the elements of ∆ and they are distinct. It follows that zj ∈ / (zk )R and so PR (zk ) 6⊆ PR (zj ). Proposition 5.6.11 then implies that the intersection

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I = ∩nj ∈∆0 PR (zj ) is irredundant, and the equality CR (I) = {zj | nj ∈ ∆0 } comes from Proposition 5.6.5.  Example 5.6.16. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set J = (X, Y 5 )R ∩ (X 2 , Y 3 )R ∩ (X 4 , Y 4 )R ∩ (X 3 , Y )R ∩ (X 5 , Y 2 )R = PR (Y 4 ) ∩ PR (XY 2 ) ∩ PR (X 3 Y 3 ) ∩ PR (X 2 ) ∩ PR (X 4 Y ). The list of exponent vectors is ∆ = {(0, 4), (1, 2), (3, 3), (2, 0), (4, 1)}. The list of maximal elements in ∆ is ∆0 = {(0, 4), (3, 3), (4, 1)}. Hence, the intersection J = PR (Y 4 ) ∩ PR (X 3 Y 3 ) ∩ PR (X 4 Y ) = (X, Y 5 )R ∩ (X 4 , Y 4 )R ∩ (X 5 , Y 2 )R. is an irredundant parametric decomposition of J. Compare this to Example 5.6.14. If the ground ring A is not a field, then the parameter ideals PR (z) are not necessarily irreducible. However, the next result shows that when we are working with monomial ideals, they are “irreducible enough”. Exercises. Exercise 5.6.17. Find irredundant parametric decompositions for the ideals from Exercises 5.4.9–5.5.7. Exercise 5.6.18. Show that the conclusion of Corollary 5.6.3 does not hold when we drop the assumption that rad (I) = rad ((X)). Exercise 5.6.19. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set (X) = (X, Y )R. Find monomial ideals I and J in R such that rad (I) = rad ((X)) = rad (J) and I ⊆ J and CR (I) ∩ CR (J) = ∅; in particular, you will have CR (I) 6⊆ CR (J) and CR (I) 6⊇ CR (J). Exercise 5.6.20. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I and J be monomial ideals of R such that rad (I) = rad ((X)) = rad (J). Prove that if CR (I) ⊆ CR (J), then I ⊇ J. Exercise 5.6.21. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables. Set I = (X 2 , Y, Z)R ∩ (X, Y 2 , Z)R ∩ (X 3 , Y, Z 2 )R ∩ (X, Y 2 , Z 3 )R ∩ (X 2 , Y 2 , Z 2 )R. (a) Find a finite set {z1 , . . . , zm } of monomials such that I = PR (z1 ) ∩ · · · ∩ PR (zm ). (b) Find an irredundant parametric decomposition for I and list the I-corner elements using: (1) Algorithm 5.6.13. (2) Proposition 5.6.15. In particular, this example shows that, with the notation of Proposition 5.6.15, the containment CR (I) ⊆ {z1 , . . . , zm } can be proper.

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Exercise 5.6.22. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables. Set (X) = (X, Y, Z)R. Set Σ = {monomial ideals J ⊆ R | rad (J) = rad ((X))} and set s = sup {cR (J) − νR (J) | Q ∈ Σ} . In Exercise 5.4.8 it is shown that s > 0. Prove that s = ∞. Exercise 5.6.23. Continue with the notation of Exercise 5.6.22. For integers n > 3, define f (n) = inf {cR (J) | J ∈ Σ and νR (J) = n} and f (n) = sup {cR (J) | J ∈ Σ and νR (J) = n} . (a) Prove that 1 6 f (n) 6 F (n) 6 ∞ for all integer n > 3. (b) Prove that f (3) = F (3) = 1. (c) Find f (4) and F (4). Justify your answer. Challenge Exercise 5.6.24. Continue with the notation of Exercise 5.6.22. Prove or disprove the following: If n ∈ Z and n > 3, then F (n) < ∞. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output.

CHAPTER 6

Square-free Monomial Ideals 6.1. Decompositions of Edge Ideals Graphs and Edge Ideals. Geometrically, a graph consists of a set of points (called “vertices”) and a set of lines or arcs (called “edges”) connecting pairs of vertices. (For us, the term “graph” is short for “finite simple graph”.) We will take the more combinatorial approach (as opposed to the geometric approach) to the study of graphs. Definition 6.1.1. Let V = {v1 , . . . , vd } be a finite set. A graph with vertex set V is an ordered pair G = (V, E) where E is a set of un-ordered pairs vi vj with vi 6= vj . (Since the pairs are un-ordered, we have vi vj = vj vi .) The set E is the edge set sometimes denoted E = E(G). Given an edge e = vi vj , the endpoints of e are the vertices vi and vj . Two vertices vi , vj ∈ V are incident if there is an edge e ∈ E with endpoints vi and vj , that is, if vi vj ∈ E. Remark 6.1.2. Our definition implies that our graphs are finite (i.e., have finite vertex sets) and are simple (i.e., have no loops and no multiple edges). Exercises. Exercise 6.1.3. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 6.2. Decompositions of Face Ideals Exercises. Exercise 6.2.1. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output.

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Parametric Decomposition of Monomial Ideals 7.1. Parametric Decompositions and Operations on Ideals The theme of this section is the following: given a monomial ideal J such that rad (J) = rad ((X)), use the parametric decomposition of J to find interesting decompositions of other ideals. Here is [5, (4.11)] for our situation. Proposition 7.1.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Let J be a monomial ideal such that rad (J) = rad ((X)), and consider a monomial
f ∈ [R]. The ideal f J is a monomial ideal such that f J = ∩z∈CR (J) PR (f z) ∩(f )R. Proof. We have f J = ((f )R)J by Exercise 1.4.16. Since (f )R and J are both monomial ideals, so Exercise 2.3.19(a) implies that
f J is a monomial ideal. We next show that f J ⊆ ∩z∈CR (J) PR (f z) ∩ (f )R. Since f J is a monomial
ideal, it suffices to show that every monomial of f J is in ∩z∈CR (J) PR (f z) ∩ (f )R. Lemma 3.3.13 implies that [f J] = {f g | g ∈ [J]}, so an arbitrary monomial in f J has the form f g for some monomial g ∈ [J]. The containment f g ∈ (f )R is immediate. It remains to show that f g ∈ PR (f z) for each z ∈ CR (J). Theorem 5.6.1 implies that J = ∩z∈CR (J) PR (z), so the condition g ∈ J implies that g ∈ PR (z) for each z ∈ CR (J). Lemma 5.1.3(b) implies that z ∈ / (g)R. Cancellation of monomials (Exercise 2.1.28(a)) implies that f z ∈ / (f g)R: if f z ∈ (f g)R then there is a monomial w ∈ [R] such that f z = f gw, so cancellation implies z = gw ∈ (g)R, a contradiction. Thus, we have f g ∈ PR (f z). This establishes the desired containment.
We next show that
f J ⊇ ∩z∈CR (J) PR (f z) ∩ (f )R. Theorem 3.1.1 implies that ∩z∈CR (J) PR (f z) ∩ (f )R is a monomial ideal such that

[ ∩z∈CR (J) PR (f z) ∩ (f )R] = ∩z∈CR (J) [PR (f z)] ∩ [(f )R] To verify the desired
containment, it suffices to show that every monomial h ∈ [ ∩z∈CR (J) PR (f z) ∩ (f )R] is in f J. Fix such a monomial h. Since h ∈ (f )R, there is a monomial k ∈ [R] such that h = f k. For each z ∈ CR (J), we have f k ∈ PR (f z), since h ∈ ∩z∈CR (J) PR (f z). Hence, Lemma 5.1.3(b) implies that f z ∈ / (f k)R. We conclude that z ∈ / (k)R, and so k ∈ PR (z). Since this is so for every z ∈ CR (J), we have k ∈ ∩z∈CR (J) PR (z) = J, and so the element h = f k is in f J, as desired.  Example 7.1.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. We consider the monomial ideal I = (X 8 Y, X 6 Y 2 , X 5 Y 3 , X 2 Y 7 )R = X 2 Y (X 6 , X 4 Y, X 3 Y 2 , Y 6 )R. 117

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(See Fact 1.4.13(a) and Exercise 1.4.16.) It follows from Example 5.6.2 that we have I = f J where f = X 2 Y and J = (X 6 , X 4 Y, X 3 Y 2 , Y 6 )R = PR (X 2 Y 5 ) ∩ PR (X 3 Y ) ∩ PR (X 5 ). Note that we have CR (J) = {X 2 Y 5 , X 3 Y, X 5 }. Here is the graph of J where the associated parameter ideals are marked with dotted lines. .. .O

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Note that the graph of f J is obtained by translating the graph of J in the direction of f ; more technically, we translate in the direction of the exponent vector of f . The next result identifies the corner elements of f J. Note that it does not assume that rad (J) = rad ((X)). It is a reasonable result in light of the final comment in the previous example. Proposition 7.1.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal such that J 6= R, and consider a monomial f ∈ [R]. Given a monomial z ∈ [R], we have z ∈ CR (J) if and only if f z ∈ CR (f J). Also, there are equalities CR (f J) = {f w | w ∈ CR (J)} and cR (J) = cR (f J). Proof. Lemma 3.3.12 implies that z ∈ J if and only if f z ∈ f J. In other words, we have z ∈ / J if and only if f z ∈ / f J. This lemma also shows that for i = 1, . . . , d we have Xi z ∈ J if and only if Xi f z ∈ f J. From this it follows that z ∈ CR (J) if and only if f z ∈ CR (f J). Hence, we have a well-defined function α : CR (J) → CR (f J) given by α(z) = f z. To complete the proof, we need to show that α is a bijection. To this end, we note that given z, z 0 ∈ CR (J), we have z = z 0 if and only if f z = f z 0 : this follows from monomial cancellation (Exercise 2.1.28(a)). It follows that z 6= z 0 if and only if f z 6= f z 0 . Hence, the map α : CR (J) → CR (f J) given by α(z) = f z is injective. Lastly, we show that α is surjective. To this end, fix a corner element g ∈ CR (f J). We show that g ∈ (f )R. (Once this is shown, we can find a monomial w ∈ [R] such that g = f w. Our assumptions imply that f w = g ∈ CR (f J), and so the first paragraph of this proof shows that w ∈ CR (J). It follows that g = f w = α(w) and so α is surjective.) Case 1: d = 1. In this case, there are integers m, n, p > 0 such that f = X1m and g = X1n and J = (X1p )R. Then f J = (X1m+p )R. The condition J 6= R implies that p > 1. By assumption, we have X1n+1 = Xg ∈ f J = (X1p+m )R and so n + 1 > p + m. Thus, we have n>p+m−1>m and so g = X n ∈ (X1m )R = (f )R as desired. Case 2: d > 2. In this case, the desired conclusion follows from Proposition 5.3.10 since X1 g, X2 g ∈ f J ⊆ (f )R.  Example 7.1.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. We consider the monomial ideal J = (X 6 , X 4 Y, X 3 Y 2 , Y 6 )R = PR (X 2 Y 5 ) ∩ PR (X 3 Y ) ∩ PR (X 5 ) which has CR (J) = {X 2 Y 5 , X 3 Y, X 5 }; see Examples 5.6.2 and 7.1.2. With f = X 2 Y , Fact 1.4.13(a) shows that f J = (f X 6 , f X 4 Y, f X 3 Y 2 , f Y 6 )R = (X 8 Y, X 6 Y 2 , X 5 Y 3 , X 2 Y 7 )R.

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Here is the graph of J where the J-corner elements X 2 Y 5 , X 3 Y, X 5 are marked ~. .. .O

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Proposition 7.1.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Let J be a monomial ideal such that rad (J) = rad ((X)), and let z1 , . . . , zm ∈ CR (J) be the distinct J-corner elements. Fix a monomial f ∈ [R] and assume that there is an integer k such that 0 6 k 6 m and such that zi ∈ (f )R if and only if i 6 k. For i = 1, . . . , k we fix monomials wi ∈ [R] such that zi = f wi . Then (J :R (f )R) = ∩ki=1 PR (wi ) is an irredundant parametric decomposition, and CR ((J :R (f )R)) = {w1 , . . . , wk } and cR ((J :R (f )R)) = k 6 cR (J). Remark 7.1.6. Continue with the notation of Proposition 7.1.5. The condition zi ∈ (f )R if and only if i 6 k is explained by the following. z1 , . . . , zk , . . . , zm | {z } | {z } ∈(f )R

∈(f / )R

Such an arrangement can always be achieved by reorder the zi ’s if necessary. We discuss why the given parametric decomposition of (J :R (f )R) is reasonable. If f ∈ / J, then Proposition 5.5.3 implies that zi ∈ (f )R for some i; hence, in this case we have k > 1 and the intersection ∩ki=1 PR (wi ) is taken over
a non-empty set. In particular, Exercise 5.1.10 implies that rad ∩ki=1 PR (wi ) = rad ((X)). On the other hand, we have J ⊆ (J :R (f )R) ⊆ (X), and this implies that rad ((X)) = rad (J) ⊆ rad ((J :R (f )R)) ⊆ rad ((X)). It follows that rad ((J :R (f )R)) = rad ((X)), so it makes sense for the ideal (J :R (f )R) to have a parametric decomposition. On the other hand, if f ∈ J, then Proposition 1.5.3(c) implies that (J :R (f )R) = R. For the Proposition to make sense, we need to show that zj ∈ / (f )R for j = 1, . . . , m; in other words, we need k = 0. Suppose that zj ∈ (f )R for some j. It follows that f ∈ / PR (zj ), and so f ∈ / ∩m i=1 PR (zi ) = J, a contradiction. Proof of Proposition 7.1.5. First, we show that ( R if i > k (7.1.6.1) (PR (zi ) :R (f )R) = PR (wi ) if i 6 k. If i > k, then zi ∈ / (f )R and so f ∈ PR (zi ) by Lemma 5.1.3(b); Proposition 1.5.3(c) implies that (PR (zi ) :R (f )R) = R in this case. If i 6 k, then (PR (zi ) :R (f )R) = (PR (f wi ) :R (f )R) = PR (wi ) by Exercise 5.1.14. Theorem 5.6.1 implies that J = ∩m i=1 PR (zi ) and this explains the first equality in the following sequence. (J :R (f )R) = (∩m i=1 PR (zi ) :R (f )R) = ∩m i=1 (PR (zi ) :R (f )R) = ∩ki=1 PR (wi ) The second equality follows from Proposition 1.5.4(b). The third equality follows from equation (7.1.6.1). To show that this intersection is irredundant, it suffices to show that for all indices i and j, if i 6= j, then wj ∈ / (wi )R; see Proposition 5.6.11. Suppose that

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wj ∈ (wi )R. It follows that zj = f wj ∈ (f wi )R = (zi )R. This contradicts Proposition 5.3.2(c). The remaining conclusions now follows from Proposition 5.6.5. 

Example 7.1.7. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. We consider the monomial ideal

J = (X 6 , X 4 Y, X 3 Y 2 , Y 6 )R = PR (X 2 Y 5 ) ∩ PR (X 3 Y ) ∩ PR (X 5 )

which has CR (J) = {X 2 Y 5 , X 3 Y, X 5 }; see Examples 5.5.2 and 5.6.2. We compute an irredundant parametric decomposition of (J :R (f )R) where f = X 2 Y 3 , and we compute CR ((J :R (f )R)). Here is the graph of J where the J-corner elements are marked ~ and the monomial f is marked ⊗.

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(J :R (f )R) = PR (w1 ) = PR (Y 2 ) = (X, Y 3 )R

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and so CR ((J :R (f )R)) = {w1 } = {Y 2 }. Here is the graph of (J :R (f )R). .. .O

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To summarize the process: we throw away all the corner elements that are not up and to the right of f ; then we move the remaining corner elements down and to the left, corresponding to division by f . The next result shows how to find a parametric decomposition of (I :R J) when rad (I) = rad ((X)). It is [5, (4.12)] for our situation. Note that the intersection in this result may be redundant; see Exercise 7.1.20. Proposition 7.1.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Let I and J be monomial ideals such that rad (I) = rad ((X)) and such that J 6⊆ I. Let z1 , . . . , zm ∈ CR (I) be the distinct I-corner elements, and let f1 , . . . , fn ∈ [J] be an (irredundant) monomial generating sequence for J. For i = 1, . . . , n and j = 1, . . . , m set ( the monomial in [R] such that zj = wj,i fi if zj ∈ (fi )R wj,i = 1 if zj ∈ / (fi )R. Then n (I :R J) = ∩m j=1 ∩i=1 PR (wj,i ) is a parametric decomposition, and

CR ((I :R J)) ⊆ {wj,i | i = 1, . . . , n and j = 1, . . . , m} and cR ((I :R J)) 6 νR (J) cR (I). Proof. Since J 6⊆ I, we have (I :R J) 6= R by Proposition 1.5.3(c). Since I and J are monomial ideals, Theorem 3.3.1 implies that (I :R J) is a monomial ideal. Hence, Exercise 2.1.20 implies that (I :R J) ⊆ (X) = PR (1). Proposition 7.1.5 implies that for i = 1, . . . , n we have ( R if fi ∈ I (I :R (fi )R) = ∩m P (w ) if fi ∈ / I. j,i j=1 R Reorder the fi ’s if necessary to assume that there is an integer k such that 0 6 k 6 n and such that fi ∈ I if and only if i > k. In other words, we assume that we have f1 , . . . , fk , . . . , fn . | {z } | {z } ∈I /

∈I

124

7. PARAMETRIC DECOMPOSITION OF MONOMIAL IDEALS

Note that k > 1 because, if not, we would have J ⊆ I. Proposition 1.5.3(b) implies the second equality in the following sequence (I :R J) = (I :R (f1 , . . . , fn )R) = ∩ni=1 (I :R (fi )R) = ∩ki=1 (I :R (fi )R) n m = ∩ki=1 ∩m j=1 PR (wj,i ) = ∩i=1 ∩j=1 PR (wj,i ).

Proposition 5.6.15 implies that CR ((I :R J)) ⊆ {wj,i | i = 1, . . . , n and j = 1, . . . , m}. For the inequality cR ((I :R J)) 6 νR (J) cR (I) we assume without loss of generality that f1 , . . . , fn is an irredundant monomial generating sequence for J. By definition, it follows that n = νR (J) and m = cR (I), so the desired inequality follows from the displayed containment of corner elements.  The following notation is non-standard. Definition 7.1.9. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix a monomial z ∈ [R] and write z = Xn where n ∈ Nd . For k = 1, 2, . . . we set z (k) = X1kn1 +k−1 · · · Xdknd +k−1 . Example 7.1.10. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Then (4)

(X 2 Y 3 )

= X (4)(2)+4−1 Y (4)(3)+4−1 = X 11 Y 15 .

Lemma 7.1.11. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix monomials w, z ∈ [R] and an integer k > 1. (a) We have w ∈ (z)R if and only if w(k) ∈ (z (k) )R. (b) We have w = z if and only if w(k) = z (k) . Proof. Write w = Xm and z = Xn . (a) Assume first that w ∈ (z)R. It follows that mi > ni for i = 1, . . . , d and so kmi + k − 1 > kni + k − 1. Hence, we have w(k) ∈ (z (k) )R. For the converse, assume that w(k) ∈ (z (k) )R. It follows that kmi + k − 1 > kni + k − 1 for i = 1, . . . , d and it follows readily that mi > ni . Hence, we have w ∈ (z)R. (b) Proceed as in part (a) changing the inequalities to equalities.  Lemma 7.1.12. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix a monomial z ∈ [R] and an [k] integer k > 1. Then we have PR (z) = PR (z (k) ). Proof. With z = Xn we have [k]

PR (z)

= ((X1n1 +1 , . . . , Xdnd +1 )R) k(n1 +1)

= (X1 =

[k]

k(nd +1)

, . . . , Xd

)R

(kn +k−1)+1 (kn +k−1)+1 (X1 1 , . . . , Xd d )R

= PR (z (k) ) as desired.



7.1. PARAMETRIC DECOMPOSITIONS AND OPERATIONS ON IDEALS

125

Example 7.1.13. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Then PR (X 2 Y 3 ) = (X 3 , Y 4 )R, and (4) Example 7.1.10 implies that (X 2 Y 3 ) = X 11 Y 15 . We verify the conclusion of Lemma 7.1.12 directly: PR (X 2 Y 3 )

[4]

= (X 3 , Y 4 )R

[4]

(4)

= (X 12 , Y 16 )R = PR (X 11 Y 15 ) = PR ((X 2 Y 3 )

)

as claimed. Here is [5, (4.15) and (4.16)] for our situation. Theorem 7.1.14. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let J be a monomial ideal in R such that rad (J) = rad ((X)). Let z1 , . . . , zm ∈ CR (J) be the distinct J-corner elements, and write zi = Xni where ni = (ni,1 , . . . , ni,d ) ∈ Nd . Fix an integer k > 1. For i = 1, . . . , m set kni,1 +k−1

zi (k) = X1

kni,d +k−1

· · · Xd

.

Then the following is an irredundant parametric decomposition [k]

J [k] = ∩m i=1 PR (zi )

(k) = ∩m ) i=1 PR (zi

and we have CR (J [k] ) = {zm (k) , . . . , zm (k) } and cR (J [k] ) = m = cR (J). [k]

Proof. Lemma 7.1.12 implies that PR (zi ) [k]

J [k] = ∩m i=1 PR (zi )

= PR (zi (k) ), and so we have

(k) = ∩m ) i=1 PR (zi

by Lemma 3.4.11. Next, we show that this decomposition is irredundant. By Proposition 5.6.11, it suffices to show that PR (zi (k) ) 6⊆ PR (zj (k) ) for all distinct indices i and j. Theorem 5.6.1 implies that the intersection J = ∩m i=1 PR (zi ) is irredundant, so we know that PR (zi ) 6⊆ PR (zj ). Lemma 3.4.10(a) implies that [k] [k] PR (zi ) 6⊆ PR (zj ) , and so PR (zi (k) ) 6⊆ PR (zj (k) ) by Lemma 3.4.11. This shows that the intersection is irredundant. The final conclusions of the result follow from Proposition 5.6.5.  Exercises. Exercise 7.1.15. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables. Set J = (Z 4 , Y 2 Z 3 , Y 3 , XY Z, XY 2 , X 2 )R and f = X. Compute the decomposition of f J guaranteed by Proposition 7.1.1 and find the complete list of f J-corner elements. Exercise 7.1.16. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let J be a non-zero monomial ideal in R. (a) Prove that if d 6 2, then there exists a monomial f ∈ [R] and a monomial ideal I such that J = f I and rad (I) = rad ((X)). Therefore, Proposition 7.1.1 always applies when d 6 2. (b) Show by example that the conclusion of part (a) can fail when d = 3. Exercise 7.1.17. Prove that the intersection in the conclusion of Proposition 7.1.1 is irredundant.

126

7. PARAMETRIC DECOMPOSITION OF MONOMIAL IDEALS

Exercise 7.1.18. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables. Set I = (Z 4 , Y 2 Z 3 , Y 3 , XY Z, XY 2 , X 2 )R and f = X. Set J = (I :R (f )R). Compute the irredundant parametric decomposition of J and find the complete list of J-corner elements. Exercise 7.1.19. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Let J = (X 6 , X 4 Y, X 3 Y 2 , Y 6 )R = PR (X 2 Y 5 ) ∩ PR (X 3 Y ) ∩ PR (X 5 ) and I = (X 5 , X 4 Y 2 , XY 3 , Y 4 )R. Compute an irredundant parametric decomposition of (J :R I) and compute CR ((J :R I)). Exercise 7.1.20. Under the assumptions of Proposition 7.1.8: (a) Show by example that the parametric decomposition n (I :R J) = ∩m j=1 ∩i=1 PR (wj,i )

can be redundant or irredundant. (b) Prove or disprove: There is an equality cR ((I :R J)) = {wj,i | i = 1, . . . , n and j = 1, . . . , m}. Exercise 7.1.21. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let f and g be monomials in R. Show that f = g if and only if f (k) = g (k) . Exercise 7.1.22. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let z1 , . . . , zm be monomials in R. Prove that z1 , . . . , zm is an irredundant generating sequence for the ideal (k) (k) (z1 , . . . , zm )R if and only if z1 , . . . , zm is an irredundant generating sequence for (k) (k) the ideal (z1 , . . . , zm )R. Exercise 7.1.23. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables. Set I = (Z 4 , Y 2 Z 3 , Y 3 , XY Z, XY 2 , X 2 )R. Compute the irredundant parametric decomposition of J [3] and find the complete list of J [3] -corner elements. Exercise 7.1.24. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables. Show that (X 3 , Y 2 , XZ, Z 3 )R = (X 3 , Y 2 , Z)R ∩ (X, Y 2 , Z 3 )R. Use this to find the irredundant parametric decomposition of (X 6 , Y 4 , X 2 Z 2 , Z 6 )R. Exercise 7.1.25. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let J be a monomial ideal in R. Prove or disprove: CR (J [k] ) = {z (k) | z ∈ CR (J). Exercise 7.1.26. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix a d-tuple e ∈ Nd such that e1 , . . . , ed > 1. For a monomial z = Xn , set z e = (Xn )e = X1n1 e1 · · · Xdnd ed . Let f, f1 , . . . , fm , g1 , . . . , gn be monomials in R.

7.2. GENERATING SEQUENCES OF INTERSECTIONS

127

e (a) Prove that f e ∈ (f1e , . . . , fm )R if and only if f ∈ (f1 , . . . , fm )R. e (b) Prove that we have (f1 , . . . , fm )R ⊆ (g1 , . . . , gn )R if and only if (f1e , . . . , fm )R ⊆ e e (g1 , . . . , gn )R. e (c) Prove that we have (f1 , . . . , fm )R = (g1 , . . . , gn )R if and only if (f1e , . . . , fm )R = e e (g1 , . . . , gn )R. (d) If I = (f1 , . . . , fm )R, define e I [e] = (f1e , . . . , fm )R.

(e) (f) (g)

(h)

(i)

Prove that this is independent of the choice of monomial generating sequence for I.
Prove that I [e] ⊆ I and rad I [e] = rad (I). Prove that f1 , . . . , fn is an irredundant generating sequence for I if and only if f1e , . . . , fne is an irredundant generating sequence for I [e] . Let R = A[x, y] and set I = (y 3 , xy 2 , x2 )R and e = (2, 3). Write out an irredundant generating sequence for I [e] . Sketch the graphs Γ(I) and Γ(I [e] ), indicating the generators and corner elements in each case. How do the corner elements for I and I [e] compare? How do the irredundant parametric decompositions compare? Let I be an arbitrary monomial ideal in R = A[X, Y ]. Based on your observations from part (g), make a conjectural description of the irredundant parametric decomposition and corner elements of I [e] , based on the irredundant parametric decomposition and corner elements of I. Prove your conjecture. Repeat part (h) for monomial ideals in R = A[X1 , . . . , Xd ] with d > 2.

Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 7.2. Generating Sequences of Intersections Remark 7.2.1. The similarity between Algorithms 5.6.13 and 2.3.8 is striking. Similarly for Propositions 5.6.15 and 2.3.10. We indicate the connection by showing how Proposition 5.6.15 gives another proof of Proposition 2.3.10 in the case where S is finite. It uses a concept called “Alexander duality” in conjunction with Proposition 5.6.15. (This is backward. We should really use Propositions 2.3.10 to prove Proposition 5.6.15.) The details here are a bit sketchy; Exercise 7.2.5 asks the reader to fill in the missing details. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Fix a list of monomials z1 , . . . , zm ∈ [R] and set J = (z1 , . . . , zm )R. For each index i, write zi = Xni with ni ∈ Nd . Set ∆ = {n1 , . . . , nm } ⊆ Nd and consider the order < on Nd from Definition 1.6.8. Let ∆00 denote the set of minimal elements of ∆ under this order. We need to show that {zi | ni ∈ ∆00 } is an irredundant monomial generating sequence for J. Fix an element p ∈ Nd such that p < ni for i = 1, . . . , m. (Note that we are 0 not saying that p is in ∆.) For each index i, set n0i = p − n and zi0 = Xni . Set Λ = {n01 , . . . , n0m } ⊆ Nd , and let Λ0 denote the set of maximal elements in Λ. It is straight forward to show that there is a bijection α : ∆ → Λ given by α(ni ) = n0i . Furthermore, this restricts to a bijection α0 : ∆00 → Λ0 given by α(ni ) = n0i . 0 Set I = ∩m i=1 PR (zi ). Proposition 5.6.15 shows that I = ∩n0i ∈Λ0 PR (zi0 )

128

7. PARAMETRIC DECOMPOSITION OF MONOMIAL IDEALS

is an irredundant decomposition. The fact that the elements of Λ and Λ0 define the same ideal via parametric decomposition implies that the elements of ∆ and ∆00 define the same ideal by generators. Furthermore, the fact that Λ0 determines an irredundant intersection implies that ∆00 determines an irredundant monomial generating sequence of J. Example 7.2.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables. We show how to find a monomial ideal J ⊆ R with CR (J) = {XY, Z 3 }. We also give the irredundant monomial generating sequence for J. The sequence XY, Z 3 is an irredundant monomial generating sequence for the ideal (XY, Z 3 )R. Hence, Proposition 5.6.11 implies that the intersection J = PR (XY ) ∩ PR (Z 3 ) is irredundant. Proposition 5.6.5 implies that CR (J) = {XY, Z 3 }. It remains to show that J is irredundantly generated by the sequence X 2 , Y 2 , XZ, Y Z, Z 4 . Proposition 3.1.7 explains the third equality in the next sequence J = PR (XY ) ∩ PR (Z 3 ) = (X 2 , Y 2 , Z)R ∩ (X, Y, Z 4 )R = (lcm(X 2 , X), lcm(X 2 , Y ), lcm(X 2 , Z 4 ), lcm(Y 2 , X), lcm(Y 2 , Y ), lcm(Y 2 , Z 4 ), lcm(Z, X), lcm(Z, Y ), lcm(Z, Z 4 ))R = (X 2 , X 2 Y, X 2 Z 4 , XY 2 , Y 2 , Y 2 Z 4 , XZ, Y Z, Z 4 )R = (X 2 , Y 2 , XZ, Y Z, Z 4 )R while the final equality is from Algorithm 2.3.8 or Proposition 2.3.10. Exercises. Exercise 7.2.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables. Let (X) = (X, Y, Z)R. (a) Find an irredundant monomial generating sequence (in lexicographic order) for PR (X 3 Y 6 Z 5 ) ∩ PR (Y 3 Z 6 ) ∩ PR (X 4 Y ). (b) Find an irredundant monomial generating sequence (in lexicographic order) for an ideal I such that rad (I) = rad ((X)) and CR (I) = {X 2 Z, XY Z}. (c) Is your answer for part (b) unique? Justify your response. (d) Can you find a monomial ideal J such that CR (J) = {XY, Y Z, XY Z}? Justify your response. Exercise 7.2.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R. Prove that if z and w are distinct I-corner-elements, then lcm(z, w) ∈ I. Exercise 7.2.5. Fill in the details for Remark 7.2.1. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output.

CHAPTER 8

Chains of Monomial Ideals In this chapter we explore containments of monomial ideals. 8.1. Monomial Covers We begin this section with a result generalizing Propositions 5.5.3 and 5.5.4. Proposition 8.1.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R and let I be a monomial ideal in R such that rad (I) = rad ((X)). Fix a polynomial f ∈ R r I, not necessarily a monomial, and fix an ideal J ⊆ R, not necessarily a monomial ideal. (a) There there is a monomial g ∈ [R] such that f g ∈ (I :R (X)) r I. (b) If I ( J, then I ( J ∩ (I :R (X)). (c) If I ( J, then J ∩ (CR (I))R 6⊆ I, and so J ∩ (CR (I))R 6= 0. Proof. (a) Let Σ = {g ∈ [R] | f g ∈ / I}. Since f ∈ / I, we have 1 ∈ Σ; in particular, we have Σ 6= ∅. Also, we have Σ ⊆ [R] r [I]. The condition rad (I) = rad ((X)) implies that [R] r [I] is a finite set by Exercise 2.1.30, and so Σ is also finite. Let g ∈ Σ be an element of maximal degree among the elements of Σ. Claim: f g ∈ (I :R (X)) r I. Since g ∈ Σ, we know that f g ∈ / I. Also, for i = 1, . . . , d we have deg(Xi g) > deg(g), and so the maximality of deg(g) among the elements of Σ implies that Xi g ∈ / Σ. Hence Xi (gf ) = (Xi g)f ∈ I, and so f g ∈ (I :R (X)). (b) We always have I ⊆ (I :R (X)), so the assumption I ( J implies that I ⊆ J ∩ (I :R (X)). Hence, it remains to show that I 6= J ∩ (I :R (X)). Since I ( J, there is a polynomial f1 ∈ J r I. Part (a) yields a monomial g ∈ [R] such that f1 g ∈ (I :R (X)) r I. Since f1 ∈ J, we have f1 g ∈ J and so f1 g ∈ (J ∩ (I :R (X))) r I. (c) By part (b) we have I ( J ∩ (I :R (X)), so there is an element h ∈ J ∩ (I :R (X)) such that h ∈ / I. Theorem 5.3.13(a) implies that (I :R (X)) = I + (CR (I))R so we can write h = h1 + h2 for some h1 ∈ I and h2 ∈ (CR (I))R. Claim: h2 ∈ J ∩ (CR (I))R r I. Note that h2 = h − h1 . Since h ∈ / I and h1 ∈ I, we have h2 ∈ / I by Fact 1.3.2(d). By construction, we have h2 ∈ (CR (I))R. Also, we have h ∈ J and h1 ∈ I ⊆ J, and so h2 = h − h1 ∈ J. The properties of the element h2 show that J ∩ (CR (I))R 6⊆ I, and it follows that J ∩ (CR (I))R 6= 0.  Definition 8.1.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R. A monomial cover of I is a monomial ideal J such that I ( J and, for all monomial ideals K in R, if I ⊆ K ⊆ J, then either I = K or K = J. 129

130

8. CHAINS OF MONOMIAL IDEALS

Remark 8.1.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. (a) It is not clear that every monomial ideal I in R has a monomial cover. (b) The idea behind monomial covers is the following: If J is a monomial cover of I, then J is “right above” I, that is, there is no room for another monomial ideal to fit between I and J. Example 8.1.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set I = (X 4 , X 2 Y 2 , Y 4 )R. The monomial ideal K = (X 3 , XY, Y 3 ) contains I, but is not a monomial cover for I. Here are the graphs for I and K graphed together: O

.. .

.. .

.. .

.. .

4

•

•

•

•

•

···

3

◦

◦

•

•

•

···

2

−

◦

•

•

•

···

1

−

◦

◦

◦

•

···

|

|

◦

•

/

1

2

3

4

0 0

We see that K is not a monomial cover for I because, for instance, the ideal J = (X 4 , X 3 Y, X 2 Y 2 , Y 4 )R satisfies I ( J ( K. O

.. .

.. .

.. .

.. .

4

•

•

•

•

•

···

3

◦

◦

•

•

•

···

2

−

◦

•

•

•

···

1

−

◦

◦

⊗

•

···

|

|

◦

•

/

1

2

3

4

0 0

The ideal J is obtained from I by adding the generator X 3 Y . Note that, in this case, we could have added any of the monomials from [K] r [I] to I and arrived at a monomial ideal I 0 such that I ( I 0 ( K. We claim that the ideal J is a monomial cover for I. In fact, the ideal I has exactly two monomial covers, namely, the ideals J = (X 4 , X 3 Y, X 2 Y 2 , Y 4 )R and J 0 = (X 4 , X 2 Y 2 , XY 3 , Y 4 )R. This will follow directly from the next proposition.

8.1. MONOMIAL COVERS

131

Proposition 8.1.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I and J be monomial ideals in R. The following conditions are equivalent: (i) the ideal J is a monomial cover for I; (ii) we have I ( J and, for every monomial z ∈ [J] r [I], we have J = I + (z)R; (iii) there exists a monomial z ∈ CR (I) such that J = I + (z)R; (iv) there exists a monomial z ∈ CR (I) such that [J] = [I] ∪ {z}; and (v) we have I ⊆ J and |[J] r [I]| = 1. Proof. (i) =⇒ (ii): Assume that J is a monomial cover for I. In particular, we have I ( J. Since I and J are both monomial ideals, this implies that [I] ( [J], so there is a monomial z ∈ [J] r [I]. It follows that I ⊆ I + (z)R ⊆ J. Furthermore, since z ∈ / I, we have I ( I + (z)R ⊆ J. Since J is a monomial cover of I, this implies that I + (z)R = J. (ii) =⇒ (iii): Assume that I ( J and, for every monomial z ∈ [J] r [I], we have J = I + (z)R. Since I ( J, there is a monomial z ∈ [J] r [I]. It suffices to show that z ∈ CR (I). Since z ∈ / I, we need to show that Xi z ∈ I for i = 1, . . . , d. Suppose that Xi z ∈ / I for some I. Since z ∈ [J], we have Xi z ∈ [J], and so we have I + (Xi z)R = J = I + (z)R, by assumption. It follows that z ∈ [I + (Xi z)R] = [I] ∪ [(Xi z)R] by Exercise 2.3.20(d). Since z ∈ / [I], we conclude that z ∈ (Xi z)R. This is impossible, though, since deg(z) < deg(Xi z). Hence, we have Xi z ∈ I for all i, as desired. (iii) =⇒ (iv): Assume that there exists a monomial z ∈ CR (I) such that J = I + (z)R. It follows that z ∈ / [I] and [J] = [I + (z)R] = [I] ∪ [(z)R]. It suffices to show that z is the unique monomial in [J] r [I]. Fix a monomial w ∈ [J] r [I]. From the displayed equalities, we have w ∈ [(z)R], so there is a monomial v ∈ [R] such that w = zv. If v ∈ [(X)], then w = zv ∈ [I] because z ∈ CR (I); this contradicts our assumption. It follows that v ∈ [R] r [(X)] = {1}, and so v = 1 and w = zv = z, as desired. (iv) =⇒ (v): Assume that there exists a monomial z ∈ CR (I) such that [J] = [I] ∪ {z}. It follows that [I] ⊆ [I] ∪ {z} = [J] and so I = ([I])R ⊆ ([J])R = J. The inequality |[J] r [I]| > 1 follows from the proper containment [I] ( [J]. The inequality |[J] r [I]| 6 1 follows from the proper condition [J] = [I] ∪ {z}. (v) =⇒ (i): Assume that I ⊆ J and |[J] r [I]| = 1. It follows that I ( J. To show that J is a monomial cover of I, let K be a monomial ideal in R such that I ( K ⊆ J. We show that K = J. Our assumptions imply that [I] ( [K] ⊆ [J]. The condition [I] ( [K] implies that there is a monomial z ∈ [K] r [I]. The condition [K] ( [J] implies that z ∈ [J]. It follows that z is the unique monomial in J that is not in I. In other words, we have [J] = [I] ∪ {z} ⊆ [K] ⊆ [J] and so [J] = [K], which implies that J = ([J])R = ([K])R = K as desired.



Corollary 8.1.6. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R.

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(a) The number of distinct monomial covers of I is cR (I). (b) The ideal I has a monomial cover if and only if it has a corner element. Proof. Exercise 8.1.18.



Example 8.1.7. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let z = Xn ∈ [R]. The unique monomial cover of PR (z) is PR (z) + (z)R = (X1n1 +1 , . . . , Xdnd +1 , z)R. In particular, the unique monomial cover of (X) = PR (1) is R. Remark 8.1.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R, and let I be a monomial ideal in R. (a) The ideal R is a monomial cover of I if and only if I = (X). (b) If I has a monomial cover, then I ⊆ (X). See Exercise 8.1.19. Remark 8.1.9. We will be interested in constructing sequences of monomial ideals I = I0 ( I1 ( I2 ( · · · where Ij+1 is a monomial cover of Ij for each index j. (An increasing sequence of ideals is often called a chain of ideals.) To do this effectively, we need to start with a monomial ideal I = I0 that has a monomial cover, that is, that has a corner element z1 . The ideal I1 = I + (z1 )R is a monomial cover of I. Next we need to make sure that the ideal I1 has a corner element. Thus, we need to restrict our attention to a class of monomial ideals such that (1) every ideal in the class has a corner element, and (2) for every ideal I in the class and every monomial ideal J containing I, the ideal J has a corner element. A class of ideals satisfying these conditions is the class of monomial ideals I such that rad (I) = rad ((X)). So we focus, again, on this class, after some more general frumph. Proposition 8.1.10. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Let I be a monomial ideal in R and assume that I has a monomial cover K. (a) rad (I) = rad ((X)) if and only if rad (K) ⊇ rad ((X)). (b) The following conditions are equivalent: (i) rad (K) = rad ((X)); (ii) rad (I) = rad ((X)) and I ( (X); (iii) rad (I) = rad ((X)) and I 6= (X); (iv) rad (I) = rad ((X)) and J ⊆ (X); and (v) rad (I) = rad ((X)) and J 6= R. Proof. (a) If rad (I) = rad ((X)), then we have rad (K) ⊇ rad (I) = rad ((X)). Conversely, assume that rad (K) ⊇ rad ((X)). Exercise 2.1.30 implies that [R] r [K] is a finite set. Since [K] = [I] ∪ {z} for some z ∈ [R], it follows that [R] r [I] is finite as well. Hence, we have rad (I) ⊇ rad ((X)). On the other hand, we have I ( K ⊆ R and so I ⊆ (X), which implies that rad (I) ⊆ rad ((X)). We conclude that rad (I) = rad ((X)). (b) Argue as in part (a); see Exercise 8.1.20. 

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Exercises 8.1.22 and 8.1.23 show how the following result can fail if A is not a field. Proposition 8.1.11. Let A be a field and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R, and let K be a monomial cover of I. If J is an ideal of R, not necessarily a monomial ideal, such that I ⊆ J ⊆ K, then either I = J or J = K. Proof. Proposition 8.1.5 implies that there is a corner element z ∈ CR (I) such that K = I + (z)R. Assume that J is an ideal of R, not necessarily a monomial ideal, such that I ( J ⊆ K. We will show that J = K. To this end, it suffices to show that z ∈ J. Let f ∈ J r I. Then f ∈ K = I + (z)R, so there are elements g ∈ I and h ∈ R such that f = g + zh. Let a ∈ A denote the constant term of h, and set h1 = h − a. In other words, h1 is the sum of all the non-constant terms of h. It follows that h1 ∈ (X). The equations f = g + zh and h = a + h1 combine as f = g + az + h1 z. By assumption, we have z ∈ CR (I) ⊆ (I :R (X)), so the condition h1 ∈ (X) implies that h1 z ∈ I ⊆ J. Since f ∈ J and g ∈ I ⊆ J, this implies that az = f − g − h1 z ∈ J. On the other hand, we have a 6= 0. Indeed, if a = 0, then h = h1 ∈ (X) and so hz ∈ I; the condition g ∈ I then implies that f = g + zh ∈ I, a contradiction. Since A is a field, we conclude that z = a−1 az ∈ J as desired.



The next result generalizes part of Corollary 5.6.8. Proposition 8.1.12. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R. (a) If K1 and K2 are distinct monomial covers of I, then I = K1 ∩ K2 . (b) If cR (I) > 2, then I is reducible. Proof. (a) The ideal K1 ∩ K2 is a monomial ideal. Proposition 8.1.5 implies that there are I-corner elements z1 , z2 ∈ CR (I) such that [Ki ] = [I] ∪ {zi } for i = 1, 2. Since K1 and K2 are distinct monomial ideals, we have [I] ∪ {z1 } = [K1 ] 6= [K2 ] = [I] ∪ {z2 }. It follows that z1 ∈ / K2 and z2 ∈ / K1 . We conclude that [K1 ∩ K2 ] = [K1 ] ∩ [K2 ] = [I] and so K1 ∩ K2 = I. (b) If cR (I) > 2, then I has two distinct monomial covers by Corollary 8.1.6, so the conclusion follows from part (a).  Exercise 8.1.24 shows that the following result is false when rad (I) 6= rad ((X)), even when I has a monomial cover.

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Proposition 8.1.13. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Let I be a monomial ideal in R such that rad (I) = rad ((X)). For each monomial ideal Q such that I ( Q, there is a monomial cover J of I such that I ( J ⊆ Q. Proof. Proposition 5.5.4 implies that there is an I-corner element z ∈ [Q] ∩ CR (I). Proposition 8.1.5 implies that J = I + (z)R is a monomial cover of I; in particular, we have I ( J. Since I ⊆ Q and z ∈ Q, we have J ⊆ Q, as desired.  Definition 8.1.14. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Let I be a monomial ideal in R. Let S(I) = [R] r [I] that is S(I) is the set of monomials of R that are not in I. If rad (I) = rad ((X)), then we set s(I) = | S(I), that is s(I) is the number of monomials of R that are not in I. Remark 8.1.15. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Let I be a monomial ideal in R such that rad (I) = rad ((X)). (a) The set S(I) is finite by Exercise 2.1.30, so s(I) is a non-negative integer. Moreover, we have 1 ∈ / I, and so s(I) > 1. (b) If K is a monomial cover for I, then Proposition 8.1.5 implies that there is a monomial z ∈ CR (I) such that [K] = [I]∪{z}. It follows that S(I) = S(K)∪{z} and s(I) = s(K) + 1. (c) Conversely, if K is a monomial ideal of R such that s(I) = s(K) + 1, then Proposition 8.1.5 implies that K is a monomial cover of I. See Exercise 8.1.29. Exercises. Exercise 8.1.16. Prove that the conclusion of Proposition 5.5.4 does not hold if J is not a monomial ideal by finding a monomial ideal I ⊆ R = C[X, Y ] such that rad (I) = (X, Y )R and a (non-monomial) ideal J such that I ( J and yet J contains no corner element of I. Compare this with Proposition 8.1.1(c), and find an element that is in J ∩ (CR (I))R but not in I. Exercise 8.1.17. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Let I be a monomial ideal in R. Assume that rad (I) = rad ((X)) and let J be a (possibly non-monomial) ideal such that I ( J. Let CR (I) = {z1 , . . . , zm }. Show that there exists an element f ∈ J r I suchPthat f is an A-linear combination of the corner m elements; that is, such that f = i=1 ai zi for some elements ai ∈ A. Exercise 8.1.18. Prove Corollary 8.1.6. Exercise 8.1.19. Verify the facts from Remark 8.1.8. Exercise 8.1.20. Prove Proposition 8.1.10(b). Exercise 8.1.21. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Set (X) = (X1 , . . . , Xd )R. Let I be a monomial ideal in R.

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(a) Assume that rad (I) = rad ((X)). Prove that I has a unique monomial cover if and only if I = PR (z) for some monomial z ∈ [R]. (b) Prove that one implication in part (a) is still true when we drop the assumption that rad (I) = rad ((X)). (c) Give an example to show that one implication in part (a) is false when we drop the assumption that rad (I) = rad ((X)). Exercise 8.1.22. Let R be the polynomial ring R = Z[X, Y ] in two variables. Construct a monomial ideal I with a monomial cover K such that there exists a (non-monomial) ideal J such that I ( J ( K. Exercise 8.1.23. Let A be a commutative ring with identity. Prove that the following conditions are equivalent: (i) the ring A is not a field; (ii) for every polynomial ring R = A[X1 , . . . , Xd ] and every monominal ideal I ⊆ R and every monomial cover K of I, there exists a (non-monomial) ideal J such that I ( J ( K; and (iii) there exists an integer d > 1 and a monomial ideal I in the polynomial ring R = A[X1 , . . . , Xd ] and a monomial cover K of I such that there exists a (non-monomial) ideal J such that I ( J ( K. Exercise 8.1.24. Construct an example to show that Proposition 8.1.13 is false without the assumption that rad (I) = rad ((X)), even if we assume that I has a monomial cover: Find monomial ideals I and Q such that I ( Q and I has a monomial cover, but Q contains no monomial cover of I. Exercise 8.1.25. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R. Prove that if K is a monomial cover of I, then rad (I) = rad (K). (Hint: Exercise 5.3.27 may be helpful.) Challenge Exercise 8.1.26. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R such that rad (I) = rad (X). Assume that K is a monomial cover of I. What can you say about the difference between cR (I) and cR (J)? Is one always larger than the other? Is the difference always bounded? Exercise 8.1.27. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R. Prove or disprove: If I is reducible, then there exist monomial ideals J, K such that I ( J and I ( K and I = J ∩ K. Exercise 8.1.28. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let z be a monomial in [R]. Give necessary and sufficient conditions for the principal ideal (z)R to be irreducible. Exercise 8.1.29. Verify parts (b) and (c) of Remark 8.1.15. Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output. 8.2. Saturated Chains of Monomial Ideals Definition 8.2.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in

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R. A finite saturated chain of monomial ideals (from I to R) is a finite chain of monomial ideals I = I0 ( I1 ( I2 ( · · · ( In = R such that Ij is a monomial cover for Ij−1 for j = 1, . . . , n; the length of this chain is n, the number of proper containments in the chain. Remark 8.2.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R. (a) It is straightforward to show that there is a finite saturated chain of monomial ideals of length n = 0 if and only if I = R. (b) Assuming that I 6= R, that is, that I ⊆ (X) = (X1 , . . . , Xd )R, it is straightforward to show that if there is a finite saturated chain of monomial ideals

I = I0 ( I1 ( I2 ( · · · ( In = R then In−1 = (X). (c) Example 8.2.3 shows that some monomial ideals I have a finite saturated chain of monomial ideals from I to R and some do not. See Proposition 8.2.4 for a general criterion. Example 8.2.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in d variables. Set I = (X 2 , Y 3 )R and J = (X 2 Y, Y 2 )R. First, we construct a finite saturated chain of monomial ideals from I to R. 1 Here is the graph of I, with the only corner element indicated by the symbol .

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2 elements designated by the symbol .

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This ideal has two monomial covers, the ideals I4 = (X, Y 2 )R and I40 = (X 2 , Y ), which we graph next. .. .O

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0 1 2 3 The chain then terminates with the ideal I6 = R. Using this algorithm, one can show that there are only finitely many possible finite saturated chain of monomial ideals from I to R. Furthermore, once they are calculated explicitly, one can check that they all have length 6; see Exercise 8.2.6. Note that [I] = 6. This is not a coincidence; see Proposition 8.2.5. Next we show that there is no finite saturated chain of monomial ideals from J to R. Here is the graph of J, with the only corner element indicated by the symbol 1

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chain of monomial ideals from J to R. We graph J1 next with the J1 -corner elements 2 designated by the symbol . .. .O

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This ideal has no corner element and hence no monomial cover. Therefore, the process cannot continue. Since each choice up to this point was unique, this shows that there is no finite saturated chain of monomial ideals from J to R. Proposition 8.2.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let (X) = (X1 , . . . , Xd )R, and let I be a monomial ideal in R. The following conditions are equivalent: (i) there is a finite saturated chain of monomial ideals from I to R of length s(I); (ii) there is a finite saturated chain of monomial ideals from I to R; and (iii) we have rad (I) ⊇ rad ((X)). Proof. (i) =⇒ (ii): Trivial. (ii) =⇒ (iii): Assume that there is a finite saturated chain of monomial ideals I = I0 ( I1 ( I2 ( · · · ( In = R from I to R. If n = 0, then I = R by Remark 8.2.2(a), and we are done. Thus, we assume that n > 1. Remark 8.2.2(b) explains the final equality in the next sequence rad (I) = rad (I0 ) = rad (I1 ) = rad (I2 ) = · · · = rad (In−1 ) = rad ((X)) and Exercise 8.1.25 explains the rest. This gives the desired conclusion.

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(iii) =⇒ (i) Assume that rad (I) ⊇ rad ((X)). We prove by induction on s = s(I) that there is a finite saturated chain of monomial ideals of length s from I to R. If s = 0, then I = R, so there is a finite saturated chain of monomial ideals of length 0 from I to R. Inductively, assume that s > 1 and that for every monomial ideal J, if rad (J) ⊇ rad ((X)) and s(J) = s − 1, then there is a finite saturated chain of monomial ideals of length s − 1 from J to R. Since s(=)s > 1, we have I 6= R, that is I ⊆ (X). The condition rad (I) ⊇ rad ((X)) then implies that rad (I) = rad ((X)). It follows that I has a corner element z ∈ CR (I). Proposition 8.1.5 implies that the ideal I1 = I + (z)R is a monomial cover of I. Remark 8.1.15(b) says that s(I1 ) = s(I) − 1 = s − 1, and furthermore, we have rad (I1 ) ⊇ rad (I) = rad ((X)). Our induction hypothesis implies that there is a finite saturated chain of monomial ideals of length s − 1 I1 ( I2 ( · · · ( Is = R from I1 to R. Adding the containment I ( I1 to the beginning of this chain yields a finite saturated chain of monomial ideals from I to R of length s, as desired.  Proposition 8.2.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let (X) = (X1 , . . . , Xd )R, and let I be a monomial ideal in R such that rad (I) ⊇ rad ((X)). Then every finite saturated chain of monomial ideals from I to R has length s(I). In particular, every finite saturated chain of monomial ideals from I to R has the same length. Proof. We proceed by induction on s = s(I). Base case: s = 0. In this case, we have I = R, so there is only the trivial chain I = I0 = R. Induction step. Assume that s > 1 and that for every monomial ideal J in R such that rad (J) ⊇ rad ((X)) and s(J) = s − 1, every finite saturated chain of monomial ideals from J to R has length s−1. Consider an arbitrary finite saturated chain of monomial ideals from I to R I = I0 ( I1 ( I2 ( · · · ( In = R. Since s > 1, we have I 6= R, and so n > 1. The ideal I1 is a monomial cover of I, so Remark 8.1.15(b) says that s(I)1 = s(I) − 1 = s − 1. Since rad (I1 ) ⊇ rad (I) ⊇ rad ((X)), our induction hypothesis says that every finite saturated chain of monomial ideals from I1 to R has length s − 1. In particular, the chain I1 ( I2 ( · · · ( In = R. has length s − 1. By construction, it has length n − 1, so we have n − 1 = s − 1. It follows that n = s, as desired.  Exercises. Exercise 8.2.6. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set I = (X 2 , Y 3 )R. Find all the finite saturated chains of monomial ideals from I to R. Verify that each one has length 6. Exercise 8.2.7. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables.

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141

(a) Set I = (X 2 , Y 3 , Z 4 )R. Prove that I has a finite saturated chains of monomial ideals from I to R and that every such chain has the same length. What is the length? (b) Repeat part (a) for the ideal J = (X 2 , Y 3 , Z 4 , XY 2 Z 3 )R. Definition 8.2.8. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I and J be monomial ideals in R such that I ⊆ J. A finite saturated chain of monomial ideals from I to J is a finite chain of monomial ideals I = I0 ( I1 ( I2 ( · · · ( In−1 ( In = J such that Ij is a monomial cover of Ij−1 for 1 6 j 6 n. The length of such a chain is the number of containments, in this case, n. Exercise 8.2.9. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let (X) = (X1 , . . . , Xd )R, and let I and J be monomial ideals in R such that I ⊆ J. (a) Assume that J 6= R. Prove that if there exists a finite saturated chain of monomial ideals from I to J, then rad (I) = rad (J). (b) Prove that if rad (I) = rad ((X)), then there exists a finite saturated chain of monomial ideals from I to J. (c) Does the converse of part (a) hold? Justify your answer. Exercise 8.2.10. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let (X) = (X1 , . . . , Xd )R, and let I and J be monomial ideals in R such that I ⊆ J. Assume that there exists a finite saturated chain of monomial ideals from I to J. (a) Prove that the length of each finite saturated chain of monomial ideals from I to J is exactly the number of monomials in J − I, that is, it is the cardinality of the set [J] r [I]. (b) Prove that if rad (I) = rad ((X)), then the length of each finite saturated chain of monomial ideals from I to J is exactly s(I) − s(J). Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output.

8.3. Parameter Ideals and Containments In this section, we consider the question of which parameter ideals contain a given monomial ideal. Definition 8.3.1. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let I be a monomial ideal in R and set P(I) = {parameter ideals Q ⊂ R | I ⊆ Q}.

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Example 8.3.2. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set I = (X 2 , XY 2 , Y 3 )R. .. .O

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Certainly, the parameter ideals making up the irredundant parametric decomposition of I are in P(I). PR (XY ), PR (Y 2 ) ∈ P(I) Also P(I) contains any parameter ideal that would make the parametric decomposition redundant, that is, the parameter ideals containing PR (XY ) or PR (Y 2 ). Using Exercise 5.1.13 this implies that PR (XY ), PR (Y 2 ), PR (Y ), PR (X), PR (1) ∈ P(I). In other words, for every monomial z ∈ S(I), we have PR (z) ∈ P(I). The next result shows that this is true in general when rad (I) = rad ((X)), and, moreover, that this describes P(I) completely. Proposition 8.3.3. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let (X) = (X1 , . . . , Xd )R, and let I be a monomial ideal in R. (a) We have P(I) = {PR (z) | z ∈ S(I)}. (b) If rad (I) = rad ((X)) and Q ∈ P(I), then there is a corner element z ∈ CR (I) such that I ⊆ PR (z) ⊆ Q. (c) If rad (I) = rad ((X)), then the minimal elements of P(I) are exactly the ideals PR (z) such that z ∈ CR (I). Proof. (a) To prove the containment P(I) ⊆ {PR (z) | z ∈ S(I)} let Q ∈ P(I). Then Q is a parameter ideal, so there is a monomial z ∈ [R] such that Q = PR (z). The condition I ⊆ Q = PR (z) implies that z ∈ / I by Exercise 5.1.12, that is z ∈ S(I), as desired. To prove the containment P(I) ⊆ {PR (z) | z ∈ s(I)} consider a parameter ideal PR (w) such that w ∈ S(I). Then w ∈ / I, so Exercise 5.1.12 implies that I ⊆ PR (w). This means that PR (w) ∈ P(I), as desired. (b) Assume that rad (I) = rad ((X)), and let Q ∈ P(I). Part (a) implies that there is a monomial w ∈ S(I) such that Q = PR (w). Since w ∈ S(I), Proposition 5.5.3 yields a monomial g ∈ [R] such that gw ∈ CR (I). In particular, we have gw ∈ S(I), so part (a) gives the first containment in the next sequence I ⊆ PR (gw) ⊆ PR (w) = Q.

8.3. PARAMETER IDEALS AND CONTAINMENTS

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The second containment is from Exercise 5.1.13. (c) Assume that rad (I) = rad ((X)), and let Q be a minimal element of P(I). (Note that the condition rad (I) = rad ((X)) implies that S(I) is a finite set, so the set P(I) is finite. Hence P(I) has minimal elements.) Part (b) implies that there is a corner element w ∈ CR (I) such that I ⊆ PR (w) ⊆ Q. The minimality of Q implies that Q = PR (w), as desired.  We have seen in Exercise 5.6.19 that a containment of monomial ideals I ⊆ J does not necessarily imply a containment of sets of corner elements. Next, we describe what does happen, beginning with a motivating example. Example 8.3.4. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set I = (X 6 , X 5 Y 2 , Y 5 )R and I = (X 3 , X 2 Y 3 , Y 4 )R. The graphs of these ideals appear together in the next diagram, along with their corner elements. .. .O

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Note that every corner element of the larger ideal J “catches” at least one corner element of the smaller ideal I. Also note that not every I-corner element is “caught” by a J-corner element. Here is [5, (6.4)] for our situation. Proposition 8.3.5. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X1 , . . . , Xd ] in d variables. Let (X) = (X1 , . . . , Xd )R, and let I and J be monomial ideals in R. Assume that I ⊆ J ⊆ (X) and rad (I) = rad ((X)). Let CR (I) = {z1 , . . . , zm } and CR (J) = {w1 , . . . , wn }. For i = 1, . . . , n there is an index j such that zj ∈ (wi )R and PR (zj ) ⊆ PR (wi ). Proof. The containment I ⊆ J implies that S(J) ⊆ S(I), and so P(J) ⊆ P(I) by Proposition 8.3.3(a). The condition wi ∈ CR (J) implies that PR (wi ) ∈ P(J) ⊆ P(I). Hence, Proposition 8.3.3(b) yields an I-corner element zj ∈ CR (I) such that PR (zj ) ⊆ PR (wi ). Exercise 5.1.13 implies that zj ∈ (wi )R.  Exercises. Exercise 8.3.6. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y ] in two variables. Set I = (X 4 , X 2 Y, Y 5 )R and J = ((X, Y )R)3 .

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(a) (b) (c) (d)

8. CHAINS OF MONOMIAL IDEALS

List the elements of P(I). List the minimal elements of P(I). List the corner elements for I and J. For each w ∈ CR (J), find an element z ∈ CR (I) such that z ∈ (w)R.

Exercise 8.3.7. Let A be a commutative ring with identity and let R be the polynomial ring R = A[X, Y, Z] in three variables. Repeat Exercise 8.3.6 with the ideals I = (Z 4 , Y 2 Z 3 , Y 3 , XY Z, XY 2 , X 2 )R and J = ((X, Y, Z)R)2 . Computer Exercises. Outline how to. . . . Make sure to explain how to interpret the output.

Bibliography 1. CoCoATeam, CoCoA: a system for doing Computations in Commutative Algebra, Available at http://cocoa.dima.unige.it. 2. D. Eisenbud, Commutative algebra, Graduate Texts in Mathematics, vol. 150, Springer-Verlag, New York, 1995, With a view toward algebraic geometry. MR 1322960 (97a:13001) 3. D. R. Grayson and M. E. Stillman, Macaulay 2, a software system for research in algebraic geometry, Available at http://www.math.uiuc.edu/Macaulay2/. 4. G.-M. Greuel, G. Pfister, and H. Sch¨ onemann, Symbolic computation and automated reasoning, the calculemus-2000 symposium, ch. Singular 3.0 — A computer algebra system for polynomial computations, pp. 227–233, A. K. Peters, Ltd., Natick, MA, USA, 2001. 5. W. Heinzer, L. J. Ratliff, Jr., and K. Shah, Parametric decomposition of monomial ideals. I, Houston J. Math. 21 (1995), no. 1, 29–52. MR 1331242 (96c:13002) 6. T. W. Hungerford, Algebra, Graduate Texts in Mathematics, vol. 73, Springer-Verlag, New York, 1980, Reprint of the 1974 original. MR 600654 (82a:00006)

145

Index

Qn

i=1 ri , 3 rad (I), 15 Pn ri , 3 Pi=1 n j=1 Ij , 10 nZ, 7 nr, 2 r − s, 2 rJ, 13 rS, 14 rn , 2 r−1 , 3 r1 + · · · + rn , 3 r1 · · · rn , 3

(S)R, 8 (s)R, 8 (s1 , . . . , sn )R, 8 −r, 2 0R , 2 1R , 2 2Z, 2 A[X, Y, Z], 6 A[X, Y ], 6 A[X], 5 A[X1 , X2 , X3 , . . .], 6 A[X1 , . . . , Xd ], 6 A[X], 6 I + J, 10 IJ, 12 I1 + I2 + · · · + In , 10 I1 · · · In , 12 CR (J), 94 S(I), 134 C, 1 N, 2 Q, 1 R, 1 Z, 1 Zn , 2 J [k] , 56 (I :R S), 14 cR (J), 95 Pfinite s∈S srs , 8 gcd, 44 Γ(I), 23 lcm, 39, 44 Xn , 3 C(R), 2, 7 D(R), 2 νR (J), 32 s(I), 134 [I], 21 γ(f ), 71 <, 18 P(I), 141 PR (f ), 87 z (k) , 124

ACC, 35 additive identity, 1 additive inverse, 1 antisymmetry, 18 ascending chain condition, 29, 35 associates, 45 associative law, 11, 12 associativity of addition, 1 associativity of multiplication, 1 binomial theorem, 4 cancellation, 2 chain of ideals, 141 chain of ideals, 7, 11, 132 closure under addition, 1, 7 closure under additive inverses, 7 closure under external multiplication, 7 closure under multiplication, 1 closure under subtraction, 2, 7 coefficient, 5 colon ideal, 14 commutative law, 11, 12 commutative ring with identity, 1 commutativity of addition, 1 commutativity of multiplication, 1 comparable, 18 constant polynomial, 5 constant term, 5 corner element, 94 147

148

Dedekind, 7 degenerate monomial ideal, 53 degree, 5, 6 distributive law, 13 distributivity, 1 divides, 4 divisibility order, 23 edge set, 115 empty product, 8 empty sum, 8 endpoints, 115 exponent vector, 22 field, 4 finitely generated, 8 Fundamental Theorem of Algebra, 79 Fundamental Theorem of Arithmetic, 45, 79 generalized associative laws, 3 generalized closure laws, 3 generalized closure law, 7 generalized commutative law, 3 generalized distributive law, 3 generating set, 8 Gordan, 37 graph, 115 graph of I, 23 greatest common divisor, 44, 49 Heinzer, v Hilbert, 1, 37 Hilbert Basis Theorem, 36 Hilbert function, 96, 98 Hilbert polynomial, 96, 98 ideal, 7 ideal generated by S, 8 ideal generated by s, 8 ideal generated by s1 , . . . , sn , 8 ideal numbers, 7 incident vertices, 115 indeterminate, 5 integral domain, 26 irreducible, 45 irreducible decomposition, 79 irreducible ideal, 70 irredundant generating sequence, 30 irredundant irreducible decomposition, 79 irredundant m-irreducible decomposition, 76 irredundant parametric decomposition, 89 leading coefficient, 5 least common multiple, 39, 44, 49 length of a saturated chain of monomial ideals, 136 lexicographical order, 32

INDEX

m-irreducible decomposition, 76 m-irreducible monomial ideal, 67 m-prime, 70 m-reducible monomial ideal, 67 maximal element, 29 monic, 5 monomial, 3 monomial cover, 129 Monomial Hilbert Basis Theorem, 28 monomial ideal, 21 monomial multiple, 22 monomial radical, 59 multiplicative identity, 1 multiplicative inverse, 3 nilpotent, 45 nilradical, 45 Noether, 1 noetherian, 36 non-degenerate monomial ideal, 53 occur, 6 parameter ideal, 87 parametric decomposition, v, 89 partial order, 18 polynomial, 5 poset, 18 prime, 45 prime ideal, 75 principal, 8 product of ideals, 12 radical of an ideal, 15 Ratliff, v reducible ideal, 70 reduction, 62 redundant generating sequence, 30 redundant irreducible decomposition, 79 redundant m-irreducible decomposition, 76 redundant parametric decomposition, 89 reflexivity, 18 relation, 17 saturated chain of monomial ideals, 136 Shah, v square-free monomial, 65 square-free monomial ideal, 65 stabilize, 35 subring, 4 sum of ideals, 10 support, 62 support of a polynomial, 71 theology, 37 total degree, 6 total order, 18 transitivity, 18 unique factorization domain, 46

INDEX

unit, 3 unit ideal, 10 variable, 5 vertex set, 115

149