An Introduction to Gröbner Bases

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PURE AND APPLIED MATHEMAIICS

,

HIiIey-Interseienee Series at Texts, Monographs, and Travel:

An

_

0

Introductlon

t0 _ Grobner Bases

Ralf Frébérg‘

An Introduction to

Grébner Bases

An Introduction to Grobner Bases

Ralf Friiberg Stockholm University, Sweden

JOHN WILEY & SONS Chichester - New York - Weinheim - Brisbane - Singapore - Toronto

Copyright © 1997 by John Wiley & Sons Ltd, Baffins Lane, Chichester, West Sussex P019 IUD, England

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John Wiley & Sons (Asia) Pte Ltd, 2 Clementi Loop #02—01, Jin Xing Distripark, Singapore 0512 John Wiley & Sons (Canada) Ltd, 22 Worcester Road, Rexdale, Ontario M9W 1L1, Canada Library of Congress Cataloging in Publication Data Fréberg, Ralf. An introduction to Grobner bases / Ralf Fréberg. p. cm. — (Pure and applied mathematics) Includes bibliographical references and index.

ISBN 0 471 97442 0 (alk. paper) 1. Grébner bases. 1. Title. II. Series: Pure and applied mathematics (John Wiley & Sons : Unnumbered)

QA251.3.F76 1997 512'.24—dc21

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Contents Preface

1

ix

Rings, Fields, and Ideals

1.1

Definition of a Ring

1.3

1.4

1.5

4

Rings and Fields ........................ 1.2.1 Integral Domains and Fields ..............

5 5

1.2.2

A Convention ......................

6

1.2.3 1.2.4

A Finite Ring, Zn ................... Polynomial Rings ....................

6 8

1.2.5 Zerodivisors ....................... Ideals and Rings ........................ 1.3.1 Definition of an Ideal ..................

9 10 10

1.3.2 1.3.3

The Ideal Generated by a Set ............. Principal Ideals and Euclid’s Algorithm .......

11 11

1.3.4 Euclidean Rings .................... 1.3.5 Ideals and their Calculus Equivalence Relations .....................

16 17 20

Field of Fractions of an Integral Domain ...........

21

Exercises ............................ Unique Factorization Domains Exercises ............................ Factor Rings and Homomorphisms .............. Prime Ideals and Maximal Ideals ............... Vector Spaces .......................... Exercises

22 23 25 25 29 31 34

Monomial Ideals 2.1 Sums and Products of Monomial Ideals 2.2 Intersections of Monomial Ideals ...............

37 38 38

1.6 1.7 1.8 1.9

2

1

............................

Exercises

1.2

1

CONTENTS

vi 2.3 2.4 2.5

Quotients of Monomial Ideals ................. Prime Ideals ........................... The Radical of a Monomial Ideal ...............

Grfibner Bases

3.1 3.2

Monomial Orderings ...................... 3.1.1 A Classification of Orderings ............. Dickson’s Lemma and Some Applications ..........

3.3

3.2.1 Dickson’s Lemma .................... 3.2.2 Applications of Dickson’s Lemma ........... The Reduction Process .....................

3.4

Definition of Gr6bner Bases ..................

3.5

Hilbert’s Basis Theorem and Noetherian Rings

3.6

Grobner Bases and Normal Forms

3.7 3.8

Reduced Grobner Bases .................... Construction of Grobner Bases ................

3.9

Free Modules and Syzygies

.......

39 40 41 43 45 46 47 47 48 49 5O 51

..............

..................

3.10 Syzygies of Sequences of Monomials 3.11 S—polynomials .......................... 3.12 A Criterion for Grobner Basis .................

3.13 The Buchberger Algorithm

..................

Algebraic Sets

4.1

Algebraic Sets and Ideals

4.2

Hilbert’s Nullstellensatz ....................

4.3

A Dictionary: Algebraic Sets H Radical Ideals .......

61 61 66 68

Primary Decomposition of Ideals

71

Solving Systems of Polynomial Equations 6.1 Systems with Only One Solution ............... Exercises ............................

79 79 80 80 80 81 84 86 87 88

6.2

Systems with Finitely Many Solutions ............ 6.2.1

6.3

Decomposition of the Ideal

6.2.2 Decomposition of the Ring Solving Zero-dimensional Systems ............... Exercises

6.4

..............

............................

Systems of Higher Dimension ................. Exercises

............................

CONTENTS 7

Vii

Applications of Griibner Bases 7.1 Membership Problems ..................... 7.1.1 Ideal Membership ................... 7.1.2 Radical Membership ..................

7.2

7.1.3 Subalgebra Membership ................ Calculation in Factor Rings of Polynomial Rings ......

7.3

Elimination ...........................

7.4

Ideal Operations ........................

7.5

7.4.1 Intersection of Ideals .................. 7.4.2 Ideal Quotient ..................... Supplementary Exercises ....................

Homogeneous Algebras 8.1 Homogeneous Ideals and Algebras Exercises

8.2

............................

Homogenizing and Dehomogenizing .............. 8.2.1 Homogenizing Polynomials .............. Exercises ............................

8.2.2

Homogenizing Ideals ..................

Exercises ............................ 8.2.3 Dehomogenizing Polynomials ............. 8.2.4 Dehomogenizing Ideals ................. 8.2.5 Homogenization versus Dehomogenization ...... Exercises ............................

8.3

Grobner Bases for Homogeneous Ideals ............ Exercises ............................

Projective Varieties

9.1

Projective Closure of an Algebraic Set ............

10 The Associated Graded Ring

107 108 111 111 115 117 121 123

127 127 129 11.3 Geometric Meaning of Hilbert Series ............. 132

11 Hilbert Series

11.1 Formal Power Series ...................... 11.2 Hilbert Series ..........................

139 139 Exercises ............................ 141 12.2 More General Orderings within More General Rings . . . . 141 12.3 Grobner Bases for Noncommutative Rings .......... 142

12 Variations of Grobner Bases

12.1 Grobner Bases for Modules ..................

CONTENTS

viii

Exercises ..................... 12.4 Differential Gr6bner Bases ............ ....... 12.5 SAGBI Bases ................... ....... 13 Improvements to Buchberger’s Algorithm

13.1 Choice of Ordering ................ ....... 13.2 Strategies ..................... ....... 13.3 Unnecessary Reductions ............. ....... 13.4 Homogenizing ................... ....... 13.5 Factorizing .................... ....... 13.6 Coefficients .................... ....... 13.7 Using the Hilbert Series ............. ....... 13.7.1 Calculation of Hilbert Series ....... ....... 13.8 Change of Ordering 13.9 Tracing ...................... .......

147 147 149

149 150 150 151 151 151 152 152 153 154

14 Software

155

15 Hints to Some Exercises

157

16 Answers to Exercises

163

17 Bibliography

171 171 171

....................... ....... 17.2 Articles ...................... ....... 17.1 Books

Index

175

Preface When talking to students in mathematics at the beginner’s level, one often gets the questions: How is it possible to do research in mathematics? Isn’t all mathematics already explored? It is rather hard to be able to explain

that the converse is true, that the research and knowledge increases at greater speed every decade. One reason for the difficulty to discuss this is of course that the level of abstraction of mathematics has constantly increased. This is true for most parts of mathematics, in particular for commutative

algebra and (especially for) algebraic geometry. Since the 1960s there is however also a trend in the opposite direction in these fields. Interest in constructiveness and concrete calculations is also now increasing. One important reason for this is the growth in the capability of computers. It has become possible to carry out calculations that one could only dream of earlier. It is also possible to perform experiments on a large scale. But at least as important is the development of algorithms and of software. The

work of Bruno Buchberger on Grobner bases was a major breakthrough on the algorithmic side. The basic theory of Grobner bases (but not all of its applications) is as unusual as any piece of elementary modern mathematics. The Grobner basis program Macaulay by Dave Bayer and Michael Stillman has had a large impact. It has shown that calculations may be made efficiently, and the many scripts written by the authors and by David Eisenbud have shown the way to attack a wide range of problems. Following these pioneers has come a strong development on both the algorithmic side and - the software side. All major computer algebra systems are now able to

calculate Grobner bases, and there are many good specialized programs. This book is intended to be a concrete introduction to commutative algebra through Grobner bases. The only prior knowledge we assume is a first course in algebra containing complex numbers and some elementary linear algebra. We have emphasized constructiveness above abstraction, and tried to give exemples for all new concepts and results. To make the theory

accessible to a wider audience, we have intentionally confined ourselves to the more elementary parts. At certain points we suggest some further reix

PREFACE

x

ading. And in the bibliography (which easily could be ten times as long), we have tried to limit ourselves to more accessible articles, although some

do require further mathematical knowledge. The audience we have in mind is, besides math students, students in technology and computer science, as well as people from applications inte-

rested in learning about this technique. Much of the basic material is classical, and we have not found it easy

to draw a time line for attributing results to their inventors. Thus we have taken the easy way out; the only times we have mentioned literature in the

text are when we have suggested further reading.

Chapter 1

Rings, Fields, and Ideals In this chapter we will introduce some basic concepts from commutative algebra that are important for the understanding of the following chapters. Our main object of study, will in the forthcoming chapters be polynomials in several variables with coefficients in an arbitrary field. Fields are patter— ned after, and are generalizations of, the complex numbers (3. The set of

polynomials in n variables with coefficients from (C is an example of a ring that plays an important role in both algebraic geometry and analysis. A ring is an algebraic system which is a generalization of the integers Z.

1.1

Definition of a Ring

The set of integers Z is an algebraic system with two operations, addition and multiplication. These operations satisfy rules such as a+ b = b+ a, and ab = ba for all integers a and b. Moreover, for every integer a there is an

integer —a such that a + (—a) = 0. It is worth noting that the quotient a/ b is usually not an integer. Roughly speaking, a ring is a set of objects that can be added, subtracted, and multiplied, more or less like integers. We will now define precisely what a ring is and check how well this definition agrees with our prototype Z.

DEFINITION 1 (RING). A ring is a set R with two binary operations, denoted “+” and “-”, such that for all elements a, b, c in R the following conditions are satisfied.

(a) (closed) a+bER,anda-b€R (b) (commutative) a + b = b + a

2

CHAPTER 1. RINGS, FIELDS, AND IDEALS (c) (associative) (a + b) + c = a + (b + c), and (a - b) -c = a - (b ' c)

(d) (additive identity) There is an element in R called zero, and written 0, such that 0 + a = a.

(e) (additive inverse) Every element a has an additive inverse —a such that a + (—a) = 0.

(f) (distributive) a-(b+c)=a-b+a-c, and (a+b).c=a-c+b-c It is easy to see that the set of integers forms a ring with the usual interpretation of the operations “+” and “~”. What the definition does not guarantee is the existence of a multiplicative identity, i.e., an element 1 in R such that a - 1 = 1 - a = a. Neither can we be sure that a - b = b - a for arbitrary rings. These observations lead us to the following two definitions.

DEFINITION2 (RING WITH UNIT ELEMENT). If there is an element 1 in the ring R such that a - 1 = 1 -a = a for every a in R, then R is called a ring with unit element.

DEFINITION 3 (COMMUTATIVE RING). The ring R is said to be commutative if the multiplication of R is such that a - b = b - a, for every a, b E R. We will now derive some immediate consequences of Definition 1 and check how well they agree with our prototype Z. Our goal is to establish rules so that we can compute with elements in rings in much the same

manner as we compute with integers, keeping in mind that in general a-b 5£ b ~ a. First we will show that there cannot be more than one zero element. Suppose that 0’ + a = a for every a in R. Then, in particular, 0’ + 0 = 0,

but 0’ + 0 = 0+0’ by (b), and 0+0’ = 0’. As 0’ +0 equals both 0 and 0’, we conclude that 0’ = 0. Similarly, assume that 1 and 1’ are both unit elements, then 1 = 1-1’ = 1’. Hence, a ring has at most one unit element. We will establish another uniqueness result, namely if a and b are given, the equation a + :1: = b has a unique solution. For by (e) every ring element

a has an additive inverse, i.e., an element —a such that a + (—a) = 0. Therefore, since a + (—a + b) = (a + (—a)) + b = 0 + b = b, it follows that —a + b is one solution. Again, supposing b = a + m, we have —a + b =

—a+(a+:v) = (—a+a)+a: = (a+(—a))+:1: = 0+$= (1:, which shows that —a + b is the only solution. A special case is the equation a + x = 0, with unique solution :1: = —a. If we apply this for the equation 0 + 0 = 0 we see that 0 = —0 (think of the equation as 0a + 03 = 0, with solution 0,, = —0a). Similarly, from

1.1. DEFINITION OF A RING

3

—a + a = 0 we see that a = —(—a). Again, by (b), (c), and (e) we have (a + b) + (-a) = b, and if we add —b to both sides of this equation we find that (—a) + (—b) is the solution to (a + b) + a: = 0, so we find that (-—a) + (—b) = -(a + b). It is also true that a - 0 = 0 - a = 0 for arbitrary rings, i.e., the additive zero is also a multiplicative zero, or annihilator. To prove this, we write 0 - a in two different ways and use the fact that the equation a + x = b has

a unique solution. To begin with 0 - a = (0 + 0) - a = 0 - a + 0 - a, but it is also true that 0 - a = 0 - a + 0. It follows that 0 ~ (1 = 0. We can prove that a - 0 = 0 in a similar way.

There is a ring with only one element, called the trivial ring. By ((1) a ring contains a zero element, and therefore the single element must be 0. In this ring 0 + 0 = 0 ' 0 = 0. Hence, 0 is also the unit element. In any other ring with unit element, R, we have 0 aé 1, since if 0 = 1 we get r=r-1=r-0=0f0ranyr€R,soOistheonlyelementinR.

From the distributive laws (0, the equalities a - b+ (—a) - b = (a + (—a))b: 0-b=0, anda-b+a-(—b) =a~(b+(—b)) =a-0=0, weseethat —(a - b) = (—a) ~b = a - (—b). If we now take the negatives of both sides of the equality —(a - b) = (—a) -b we obtain a - b = —(—a) - b = (—a)(—b). Suppose that the ring has a unit element, then a + (—1) - a = 1 - a + (—1)~a = (1 +(—1))-a = 0 - a = 0. Thus, (—1) «a = —a. In particular if

a = (-1), (-1) - (-1) = -(-1) = 1 The above observations permit us to compute with negatives, zero, and one as we have always done when working with the integers. For conve—

nience a + (—b) will henceforth be written a — b. Some immediate consequences of this notation are —(a — b) = b — a, c- (a — b) = c - a — c - b, and (a — b) - c = a - c — b - c. For neater notation, let us define ai for any nonzero ring element a and any nonnegative integer i by a0 = 1 (if the

ring has a unit element), a1 = a, a2 = a - a, a3 = a - a2, and generally a" = a - a”_1. We leave it to the reader to verify that the usual rules of exponents apply, i.e., for any positive integers in and n the product am -a” = am+", and (am)" = am". Similarly, in any ring we shall abbreviate expressions like a + a by 2a and more generally for any positive integer

n put na=a+a+---+a (n terms). Whenn< 0, sayn: -m,m > 0, we define na to mean —ma. It is easily verified that ma + mb = m(a + b), and m(na) = (mn)a for arbitrary integers m, n. For ease of notation we will often drop the dot in a « b and simply write

this product as ab. We have also seen that calculations with 0 and 1 do not give any surprises. We will in the sequel use the notation 0 and 1, respectively.

It is high time to pause and take a look at some concrete examples of rmgs.

4

CHAPTER 1. RINGS, FIELDS, AND IDEALS

EXAMPLE 4. The familiar number systems Z, Q (the rational numbers), ]R (the real numbers), and (C (the complex numbers) are all rings under the usual operations “+”, “— , , and zero element 0. Each one of these rings is in fact a commutative ring with unit element 1. El When we use 0 as we did in this example, it is understood that 0 is to be interpreted as one of Oz, 0Q, 0R, 0c, the zero element of Z, Q, IR, and C respectively, depending on the context. If there is no danger of confusion, we will feel free to use this style. In situations where it is not clear which element or operation we are talking about, subscripts will be used, e.g., 1 3, +3.

EXAMPLE 5. Let R be the set of even integers under the usual operations of addition and multiplication. Then R is a commutative ring but has no unit element.

I]

EXAMPLE 6. Let R be any ring and 71 any positive integer. The set of n X n matrices with entries Tij E R forms a ring, Rn, under the usual operations,

i.e., the sum of two matrices C = A +3" B is a matrix with elements Cij = aij +R bij, and the product 0 = A 13,. B is a matrix with elements cij = 22:1 (1“, -R bkj. The zero element in Rn is the n X n matrix whith every entry being equal to On. We note that if n > 1, and if R has more than one element, then R4,I is noncommutative, even though the ring R itself may be commutative. If the R has a unit element, so has Rn. The unit element 1 R. is the matrix (Tij) With Tij = 0R for 7: 55 j, and Tij = 1R for ’1: = j.

E]

EXAMPLE 7. We have seen that Q, R, and (C are commutative rings with unit elements. These number systems have more structure than most rings since we can divide with nonzero elements. Such systems are called fields,

see below.

[1

Exercises

1. Show that Q[\/§] = { a + b\/§ I a, b E Q} is a commutative ring with unit element. 2. Show that the set of matrices

{(5:12)

a,b€lR}

is a commutative ring with unit element with the usual addition and multiplication of matrices.

5

1.2. RINGS AND FIELDS

3. Show that the set of pairs of real numbers {(a, b) | a,b 6 IR} with addition (a, b) + (c, d) = (a+c, b+d) and multiplication (a, b) ~ (0, d) = (ac, ad + bc) is a commutative ring with unit element.

1.2

Rings and Fields

In this section we will continue to study different examples of rings and introduce some special classes of rings.

1.2.1

Integral Domains and Fields

DEFINITION8 (INTEGRAL DOMAIN). A commutative ring with unit element in which ab = 0 implies that a = 0 or b = 0 is called an integral domain. EXAMPLE 9. The rings Z, Q, R, and C are integral domains.

El

DEFINITION 10 (FIELD). A field is a commutative ring with unit element in which every element a aé 0 has a multiplicative inverse a‘l, so that

aa“1 = 1. For practical reasons we exclude the trivial ring, {0}, from the set of fields. The multiplicative inverse of a. is unique, since if ab = 1, we get

a‘1(ab)

=

a—1 - 1 = a”1, and

a”1(ab)

=

(a‘la)b = (aa'1)b = 1 - b = b,

hence b = a‘l. From aa— 1 = a‘la = 1 we also get (a‘1)‘1 = a. PROPOSITION 11. A field is an integral domain.

Proof. Ifab = 0 and b 76 0 we get a = a-1 = a(bb’1)=(ab)b‘1 = 0-b‘1 = 0. C]

The most familiar examples of fields are the rational numbers Q, the real numbers R, and the complex numbers C, but there are also fields with a finite number of elements that are of great importance. And now we prepare for the study of such fields.

PROPOSITION 12. An integral domain with only a finite number of elements is a field.

6

CHAPTER 1. RINGS, FIELDS, AND IDEALS

Proof. Let R = {2:1,a:2, . . . ,xn} be a finite integral domain and suppose

w,- 96 0. We claim that {mm | k = 1,. . . ,n} contains all elements of R. To show this, it is sufficient to show that smack aé mix, if k 7E I. This is true, since if mix], = man then $13,155,, — an) = 0 which gives an, — x; = 0 since a: gé 0, so :rk = :31. Since every element in R can be written as mix], for some k, we get that mix], = 1 for some k so at, has a multiplicative inverse. El

DEFINITION 13 (SUBFIELD). Suppose k g K are fields, and the operations in k and K agree on the elements of k, i.e., that a +1c b = a +K b and a-k b = a~K b if a, b E k. T Then k is called a subfield of K, and K is called an extension field of k. EXAMPLE 14. Q is a subfield of IR, and R is an extension field of Q.

D

We will use the word subring in a similar way as we do subfield. Thus, that R is a subring of S, does not only mean that R Q S as sets, but that we add and multiply elements in R in the same way regardless whether we think of them as elements in R or in S. Exercises

1. Show that if a and b are nonzero elements in a field, then (ab)‘1 =

a‘lb‘l. 2. Show that the rings in Exercise 1 and Exercise 2 in Section 1.1 are fields. 1 .2.2

A Convention

From now on, almost all the rings that we will consider will be commutative,

and will all have a unit element. Therefore, we lay down the following convention: Unless explicitly stated otherwise, the word ring will from now on mean a commutative ring with

unit element.

1.2.3

A Finite Ring, Zn

We will now give an important example of a finite ring. Let n be a positive

integer. We will denote the ring by Zn and its elements by [0], [1], . . . , [n—l], so there are n elements in Zn.

The addition of elements is defined by

[2'] + [j] = [k], where k is the remainder when i + j is divided by n, and 0 S k g n — 1. The multiplication is defined similarly, [i] - [j] = [m], where

1.2. RINGS AND FIELDS

7

Table 1.1: Addition and multiplication in Z4.

m is the remainder when ij is divided by n, O S m g n — 1. It is clear that

addition and multiplication are commutative, that [0] is the additive zero element, and that [1] is the multiplicative unit. It also follows that [n — i] is the additive inverse to [i] if i > 0. It remains to show that addition and multiplication are associative, and that the distributive law holds. Consider

(m - [m - [Ir]. He [r1 . [j] = in], where er = elem. Hence ([i][j])[k] = [r2],

where rlk: = qgn + T2. Thus ijk = kqln + qzn + T2 = (kql + q2)n + r2,

so r2 is the remainder when ijk is divided by n. In the same manner we

see that [i]([j][k]) = [r2]. Similarly one shows that both ([i] + [j]) + [k] and [i] + ([j] + [19]) equal [1"], where r is the remainder when i + j + k is divided by n. Next consider [i]([j] + [19]). Here [j] + [k] = [31], where

j + k = qln + .91. Hence [i]([j] + [16]) = [32], where 2‘31 = qzn + 32. Thus {(j + k) = iqln + ([271 + .92 = (iql + q2)n + 32, so 32 is the remainder when

i(j + k) is divided by n. In the same manner we see that [i][j] + [i][k] = [32]. The proof that Zn is a ring is complete. As a concrete example we give the tables for addition and multiplication in Z4, Table 1.1 We now show that the ring Zn is a field if and only if n is a prime number.

Suppose n = nlnz is composite, i.e., not a prime. Then [n1][n2] = [0] in Z”, so Zn is not an integral domain and can therefore not be a field according

to Proposition 11. Now suppose p is a prime. If [a][b] = [0] in Zp, then by definition ab is divisible by p, which gives that either a or b is divisible

by p. Hence either [a] or [b] equals [0] and hence Zp is an integral domain. That Zp is a field follows from Proposition 12. For the record we formulate this in a proposition.

THEOREM 15. The ring Zn is a field if and only if n is a prime number. Exercise 1. Construct tables for addition and multiplication for Z6 and Z7. Determine the multiplicative inverses to each element 95 0 in Z7.

8

CHAPTER 1. RINGS, FIELDS, AND IDEALS

1 .2.4

Polynomial Rings

The set of polynomials a0 + ala: + (122:2 +

+ anwn,ai E Q, n finite, is

denoted Q[:1:]. With natural addition and multiplication 2 am" + Z biilti = 2014‘ + In)“ 1:

i

1;

and Zaixi-Zni = 2010 -b,-+a1 -bi_1 +~~+ai-bo):ri

i

i

i

it is easily verified that Q[z] is a ring. In fact, if R is any ring, then the set R[x] of polynomials 7‘0 + no: + ~ - ~ + mm" with coefficients 1"; e R is a ring with natural addition and multiplication. We can identify the ring R

with the polynomials of degree zero in R[:v]. Thus R is a subring of R[:1:], in particular they have the same unit element.

Since R[a:] is a ring, it makes sense to consider the set (R[:I:])[y], i.e., polynomials in y with coefficients in R[a:]. We see that

(R[w])[y] = He + fly + ---+ fny"| fi 6 ll} =

{TO + 1'11: + my + r3x2 + r4$y + r5112 + - - }

{go+glx+gzx2+---+ymm’" I 9i 6 RM}

(R[y])[$]So (R[x])[y] = (R[y])[:v], and can therefore be denoted by R[:L',y] without any risk for confusion. Likewise, we can define the polynomial ring over R in 77, variables $1,232, . . . ,xn. This polynomial ring is denoted by

R[a:1,x2,...,xn]. DEFINITION l6 (DEGREE OF A POLYNOMIAL). The degree of an element m = 2:11 - - 4:1: in a polynomial ring R[:1:1, . . .,:1:,.] is deg(m) = i1 +.- - - +ip. The degree of a nonzero polynomial f(1:1,...,:1:n) = Zri1,,,,,inw§1--m§:

equals deg(f) = max { deg(a:;‘ ”xi: l 131W,“ 76 0 }. A polynomial of degree 0 is called a constant. EXAMPLE 17. The degree of f(x,y,z) = 2 + $2 + :L‘yz E Z[$,y, z] is 3.

D

We can consider the ring R as a subring of R[m1, . . . , 13”] consisting of

constant polynomials. REMARK. By far the most important classes of rings for us, in this book,

will be polynomial rings over a field, i.e., rings of type k[:1:1, . . . , con], where k is a field.

1.2. RINGS AND FIELDS

9

Exercises

1. Show that R[:1:] is an integral domain if R is an integral domain.

2. Show that if f,g e R[a:1, . . . ,wn]\{0} and f+g 76 0, then deg(f+g) S max(deg(f), deg(g)) and deg(fg) g deg(f)+deg(g). If R is an integral domain, show that deg(fg) = deg(f) + deg(g) but that there might be inequality if R is not an integral domain.

3. Show that the only invertible elements in R[x], where R is an integral domain, are the constants which are invertible as elements in R.

Show that there might be other invertible elements in R[.7:] if R has zerodivisors. 4. Let k be a field and let R be the subset

{ao+a1x+~--+akx"|a1 :0} of k:[z]. Show that R is a subring of k[:r]. 1 .2.5

Zerodivisors

Although a ring is a straightforward generalization of the integers, not every fact that we have become accustomed to in the ring of integers holds in general rings. As we have seen and as shown by the examples below, it is possible that a - b = 0 with neither a nor b being zero. DEFINITION 18 (ZERODIVISOR). An element r 96 0 in a ring R is called a zerodivisor in R if rs = 0 for some element 3 E R, s 96 0. An element 1" 7E 0 which is not a zerodivisor is called a nonzerodivisor. EXAMPLE 19. An integral domain is a ring in which every nonzero element is a nonzerodivisor. [3

EXAMPLE 20. The element [2] in Z4 is a zerodivisor since [2][2] = [0].

D

EXAMPLE 21. Let R = { (a, b) | a, b 6 Z } with addition, and multiplication defined by (a, b)+(c, d) = (a+c, b+d) and (a, b)(c, d) = (ac, bd), respectively. It is not difficult to check that R becomes a ring (often denoted Z x Z) with zero element (0,0) and unit element (1,1). Here we have (1, 0)(0, 1) = (0,0), E! so (1,0) is a zerodivisor. DEFINITION 22 (CHARACTERISTIC). The characteristic of a ring R, here denoted char(R), is the smallest positive integer n such that n . 1 = 0. If no such integer exists, the characteristic of the ring is 0. EXAMPLE 23. The characteristic of Z,Qx] and R[:1:1,...,2:m] is 0. The characteristic of Zn and Zn[x1, . . . ,wm] is n.

El

10

CHAPTER 1. RINGS, FIELDS, AND IDEALS

Exercises

1. If char(R) = p 7E 0 for an integral domain R, show that p is a prime number. 2. Show that the ring in Exercise 3 in Section 1.1 has zerodivisors.

1.3

Ideals and Rings

In this section we will introduce the important concept of ideal. An ideal is a special kind of subset of a ring, which behaves well with respect to the ring operations. Furthermore we will define and study certain important classes of rings. Finally we will introduce some operations, such as sum

and product, on the set of ideals of a ring, and study the rules for these operations.

1.3.1

Definition of an Ideal

A subset a of a ring R is called closed under addition if a,a’ E a implies that a+ a’ 6 a. It is closed under multiplication with ring elements if a E a and r E R implies that a - r E a. A nonempty subset of a ring which is closed under these two operations is called an ideal.

DEFINITION 24 (IDEAL). A nonempty subset a of a ring R is called an ideal of R if

1. a,a’ E u=>a+a’ E a, 2. aEa,r€R=>a-r€a. We note that every ideal is closed under the operation of taking additive

inverses, i.e. if a E a, then —a E a, since —a = a- (—1). It also follows that every ideal contains the element 0, since 0 = a - 0. The set consisting

of only 0 is an ideal, as is the whole ring R. An ideal a coincides with R if and only if 1 E (1, since if 1 E a, then r = 1 -r E a for every 7' E R. Exercises

1. Show that {3n| n E Z} is an ideal in Z. 2. Let k; be a field. Show that {f E k[a:]| f(1) = 0} is an ideal in k[z]. 3. Let (01,.. .,cn) 6 k", where k is a field. Show that {f6k[:1:1,...,a:n]|f(cl,...,cn)=0} is an ideal in k[xl,...xn]. 4. Determine all ideals in Z6.

1.3. IDEALS AND RINGS

1.3.2

11

The Ideal Generated by a Set

If S is an arbitrary nonempty subset of a ring R, then the set of finite linear

combinations of elements in S, {r131 + - - - + rnsn| n- E R, 3) E S}, is an ideal denoted by (S). If for an ideal a there is a finite set .S' = {51, . . . , 3”} such that a = (S), then a is said to be finitely generated and denoted ($1,...,Sn).

Exercises

1. Show that the ideal in Exercise 1 in Section 1.3.1 is (3). 2. Show that the ideal in Exercise 2 in Section 1.3.1 is (a: — 1). 3. Show that the ideal in Exercise 3 in Section 1.3.1 is (ml—01,...,xn—cn).

4. Show that a ring R is a field if and only if the only ideals in R are {0} and R.

1.3.3

Principal Ideals and Euclid’s Algorithm

Let 1' be an element in a ring R. The set of all multiples of r, {r - r’ I r’ E R}, constitutes an ideal. Such ideals are called principal ideals and 7‘ is called a generator for the ideal. The principal ideal generated by 1' is denoted ’I‘R or

(r). The two ideals R and {0} are both principal, R = (1) and {0} = (0). We will now show that all ideals in Z are principal ideals. Suppose that

a is an ideal in Z different from (0). Then a contains both positive and negative elements since a E (1 implies that —a E a. Let n be the smallest positive number in a. We claim that a = (n), Le, a consists of all multiples of n. Let b be an element in a. Then, r, 0 g r < n, or stated otherwise 7‘ = both b and —qn belong to a, so 1‘ e a. positive number in a which forces 7‘ to . of 77..

b = qn + r for some integers q and b — qn. Since n E a it follows that But we picked n to be the smallest be zero, so b = qn + 0 is a multiple

Similar reasoning shows that every ideal in Q[:c] is principal. Let a be an ideal in QM different from (0), and let f aé 0 be a polynomial in a of lowest degree. If g is another polynomial in a, we can find polynomials q

and r such that g = qf + r with deg(r) < deg(f) or r = 0. As above, we see that r E a and conclude that 1" must be the zero polynomial since f was a polynomial in a of lowest degree.

We will now show that the the same conclusion can be drawn for k[:1:], where k is an arbitrary field.

12

CHAPTER 1. RINGS, FIELDS, AND IDEALS

THEOREM 25 (THE DIVISION ALGORITHM). Let k be any field and suppose

f, g E k[:1:], f :fi 0. Then there are uniquely defined polynomials q,1‘ E k[a:] such that g = qf + r with deg(r) < deg(f) or r = 0. Proof. We first show the existence of q and r. If g = 0 we can choose q = r = 0. Otherwise, let f = aux” + +ao,an 76 0 and let 9 = bmzcm + ---+ bo,bm 96 0.

If m < n we can choose q = 0 and 1' = g.

If m _>_ n we see that g = bma;1z7'_"f + n, where deg(r1) < deg(g) or T1 = 0. If n 95 0 and deg(r1) > deg(f), say r1 = ckx’“ + +co, we can continue and write r1 = cka;1xk‘"f + 7‘2 with deg(r2) < deg(r1)

or T2 = 0, so 9 = (bmaglxm_" + cka;1z’°‘")f + T2. It is clear that in a finite number of steps we get a remainder which either is zero or has a smaller degree than deg(f). It remains to be shown that q and r are unique. Suppose g = qf + 7' = qlf + n. Then (q —— q1)f = r1 -— r. We have deg((q — q1)f) 2 deg(f) if (q -— ql) 75 0, which is a contradiction since deg(r1 — r) < deg(f). Thus q = q1. This gives 0 = 0 - f = 7'1 — r, so 7'1 = 7'.

D

In the proof above it is essential that k is a field. There is no division

algorithm in Z[a:] as the example 9 = x2 and f = 21 shows. It is not easy to see a generalization of the division algorithm to k[z1, . . . , 2:7,]. One can say that the purpose of Gréibner bases is precisely to get a way to extend the division algorithm to polynomial rings in several variables.

We will now describe the Euclidean algorithm in k[av_]. Let f and g be nonzero elements in k[$], k a field. The division algorithm gives ql and n such that g = qlf + 1'1 with deg(1'1) < deg(f) or n = 0. If n 76 0 the division algorithm gives (12 and r2 so that f = qzrl + T2 with deg(r2) < deg(r1) or 1‘2 = 0. If r2 gé O we get r2 = q3r2 + T3 a.s.o. If 1'3 76 0 we have

deg(r3) < deg(r2) < deg(r1). Since the degrees decrease, this process is finite, so we must eventually get a remainder which is zero.

9

=

41f + 7‘1

f

=

t127‘1 + T2

7‘1

=

(137'2 + T3

TN—2

=

qNTN—I + TN

TN—l

=

(1N+17‘N

We claim that the last nonvanishing remainder rN divides both f and g. The last equation gives that m; divides rN_1. The second equation from

1.3. IDEALS AND RINGS

13

the bottom shows that my divides rN_2, since rN divides both terms on the

right-hand side. Continuing like this upwards we eventually get the claim that rN divides both f and g. Thus we have shown that rN is a common divisor to f and g. Now suppose that h is another common divisor to f and g. The first equation shows that h divides r1 = g—q1 f, since h divides both 9 and f. The second equation then shows that h divides r2 = f — qgrl, since

h divides both f and r1. Continuing downwards like this we eventually get that h divides rN. Thus we have shown that any common divisor to f and g must divide rN. Since rN is a common divisor to f and g, and since any other common divisor to f and g divides rN, it is natural to say that my is

the greatest common divisor to f and g, rN = gcd(f, g). DEFINITION 26 (GCD). Let f and g be nonzero polynomials in k[:z:]. Then h is a greatest common divisor of f and g, gcd(f, y), if h divides both f and g, and any other polynomial which divides both f and g also divides h.

THEOREM 27. The last nonvanishing remainder in the Euclidean algorithm performed on f and g is a greatest common divisor to f and g. If hl and hg both are greatest common divisors to f and 9, then h1 = ch; for some c E k. Proof. We only have to prove the last statement. Since hl is a common

divisor to f and g and fig is a greatest common divisor to f and g we get that h1 divides h2, h2 = qlhl. Changing the roles of h1 and h2 we get h = q2h2. Thus h2 = q1q2h2, so q1q2 = 1, hence q1 = q? 6 k. E]

The Euclidean algorithm can be used to show that the greatest common divisor h to f and g is a linear combination of f and g, i.e. that there are polynomials h1 and h; such that h = hl f + hzg. The second equation from the bottom in the Euclidean algorithm shows that 1w 2 1 -rN_2 —qN -rN_1, hence rN is a linear combination of rN_2 and rN_1. The next equation from the bottom shows that rN_1 = 1 ‘rN_3 —— qN_1 - rN_2. Thus rN = —qN -rN_3 + (1 + qN_1)rN_2, hence rN is a linear combination of rN_3 and rN-2. Continuing upwards like this we eventually get the claim that m; is a linear combination of f and g. This result is most easily described in. terms of ideals.

THEOREM 28. Let f and g be two nonzero polynomials in a polynomial ring k[:I:] over a field It. Then (f, g) = (gcd(f,g)) Proof. Let h = gcd(f, 9). We have shown that h is a linear combination of f and g which gives h e (f, g). This gives that (h) g (f, g), since (h) = {rh| r E k[w] }, and if h 6 (fig), then rh E (f,g). On the other hand, both f and g are multiples of h, thus f, g e (h), which gives (f, g) = {r1f+r2g|r1,r2Ek[a:]}§(h). El

14

CHAPTER 1. RINGS, FIELDS, AND IDEALS

Let f be a polynomial of positive degree in k[a:] and let 0 e k\{0}. Since f = c((1/c)f) = (1/c)(cf), we see that c and cf divide f. These divisors are called trivial divisors of f. If f is a polynomial of positive degree and the only divisors of f are the trivial ones, then f is called irreducible. All

polynomials of degree one are irreducible, but there might be others, such

as x2 + 1 in RM. Let f E k[a:],deg(f) > 0. If f is not irreducible then f = f1f2 with deg(fi) < deg(f),i = 1,2. If some fi is not irreducible then this f,- can be factored further in nontrivial factors. Since the degrees of the factors decrease, it is clear that we eventually get f = fl - - - fN with f,irreducible. We will now show that this factorization is essentially unique.

LEMMA 29. If f is irreducible and f divides f1 f2, then f divides f1 or f divides f2. Proof. Suppose that f does not divide f1. Then gcd(f, f1) = 1 so 1 = h1 f + h2f1 for some polynomials h1,h2, which gives f2 = In ff2 + h2f1f2. Since f divides both terms on the right-hand side, we get that f divides

E!

f2. A polynomial f = chm,“ + ck_1a:k_1 + - - - + co is called manic if ck = 1.

THEOREM 30. Suppose deg(f) > 0. Then f = cf; --~fk, where fi are monic irreducible polynomials. This factorization is unique up to the ordering of the f.- ’s. Proof. We only need to show unicity. Let f = c1f1---f,c = C2g1---gm. Being the coefficient of the largest power of a: in f, it is clear that c1 = C2. Since f1 divides 91 - - -gm, we get that f1 divides one of the gi’s. By reordering the gi’s we can suppose that f1 divides 91. Since both f1 and 91

are monic and irreducible, we get that f1 = 91. The equality f1f2 - - - fk = f192"‘9m gives f1(f2"'fk—92"‘gm) = 0, $0 f2"‘fk = g2~~gm and we

can continue in the same manner to get the desired conclusion.

El

Let f and g be monic polynomials of positive degree and let f1, . . . , fN be

all the monic irreducible factors that occur in the factorizations of f or 9. Then we can write f = ff“ mfg,” and g = 151...s with ahbi 2 0. It follows from the theorem that f divides 9 if and only if a,- g b,- for all i.

Thus h = ffmnml’b‘)

ffimumb”) divides both f and g, and any common

divisor of f and 9 must divide h. Hence h = gcd(f, g).

EXAMPLE 31. If f = (a; — 1)3(x — 2)(ar:2 + 1) and g = (x — 1)2($ — 3)(:::2 + 1)(:r2 + 2), then gcd(f,g) = (a: — 1)2(:I:2 + 1). El

1.3. IDEALS AND RINGS

15

DEFINITION 32 (LCM). Let f and g be nonzero polynomials in k[$]. Then h is a least common multiple of f and g, lcm(f, 9), if both f and g divide h and if any other polynomial which is a multiple of both f and g is a multiple of h. We conclude this section with an application of the division algorithm. The following theorem is usually called the Factor Theorem. If f = akzrk +

+ ac and c E k then the evaluation of f at c is the element f (c) = akck +---+ao E k.

THEOREM 33 (THE FACTOR THEOREM). Let k be a field and suppose that f (x) E k[:v]. Let f(c) E k denote the evaluation of f at c. Then f (c) = 0 if and only ifa: —— c divides f(m) Proof. The division algorithm gives q,r E k:[:c] such that f = q(:1: — c) + r and deg(r) < deg(x — c) = 1 or r = 0. Hence r e k. Evaluation at c gives f(c) = 0 + r, so f(c) = 0 if and only if r = 0, hence if and only if a: — c divides f.

D

COROLLARY 34. A polynomial of degree n in k[.7:] cannot have more than n zeros.

Proof. If c1, . . . ,cr are zeros to f then (a: — cl) - - - (a: - cr) is a factor of f which is impossible if r > deg(f). D We also remind the reader of what is usually called the Fundamental

Theorem of Algebra. THEOREM 35 (THE FUNDAMENTAL THEOREM OF ALGEBRA). Each poly-

nomial f E (C[:I:] of degree at least one has a zero in C. COROLLARY 36. Any polynomial f e C[x] with deg(f) = n > 0 can be factored in linear factors, f = 0(2: — cl) - - - (:1: — cn). COROLLARY 37. The polynomials :1: — c are the only monic irreducible po-

lynomials in C[:1:]. _ If (a: — c)“ divides f (1:) but (a: — c)m+1 does not, we say that c is a root of multiplicity m to the equation f (x) = 0. COROLLARY 38. An equation f(m) = 0,f E C[a:],deg(f) = n, has exactly n roots, provided the number of roots are calculated with multiplicity. Fields with the property of the theorem above, and thus of the corollaries, are called algebraically closed, hence (C is an algebraically closed field.

It can be shown that any field has an extension field which is algebraically closed.

CHAPTER 1. RINGS, FIELDS, AND IDEALS

16 Exercises

1. Show the following division algorithm in Z. Let m, n E Z, n 7’: 0. Then there are q,r E Z with 1' < |n| such that m = qn + r. 2. Define the gcd of two elements in Z analogously to Definition 26. 3. Prove the analogue of the Euclidean algorithm for Z. 4. Prove the following analogue of Theorem 27. The last nonvanishing remainder in the Euclidean algorithm performed on m and n is a greatest common divisor of m and n. If m and 712 are both greatest common divisors to m and n, then 1n = :tn2.

Show also that if

d = gcd(m, n) there are a, b E Z such that d = am + bn. Determine gcd(a/:4 + 3:1:2 + 2, $3 — 2:1:2 + :1: — 2) in QM. Write gcd(:r4 + 3.22 + 2,1:3 — 22:2 + a: — 2) as a linear combination of 2:4 + 3:1:2 + 2 and

1:3 — 2:52 + a; — 2. . Let f and g be monic polynomials of positive degree and let f1, . . . , fN

be all the monic irreducible factors that occur in the factorizations of f or 9. Let f = ff“ ---f,‘f,” and g = ff‘ ---fg," with abbi 2 0. Show

that lcm(f, g) = ffmahm . . . fg‘aXWM'm). . If f = (:1: — 1)3(:1: — 2)(:l:2 + 1) and g = (:1: — 1)2(:c — 3)(:r:2 + 1)(al:2 +2), determine lcm(f, g). . If 7' > 2 we make the following recursive definition of gcd (f1, . . . , fr). We define gcd (f1, . . . ,f,.) to be gcd (gcd (f1, . . . , fr_1), fr). Show that gcd(f1...,fr) divides fi for i = 1,...,r. Show that if h divides fz- for i = 1,. . . ,r then h divides gcd (f1, . . . ,fr). Conclude that gcd (f1, . . . , fr) does not depend on the order of the fi’s. Show that (flv"'af1‘)=(n(f1:'-"f7')>'

1.3.4

Euclidean Rings

We can generalize the reasoning for Z and k[a;] in the following manner.

DEFINITION 39 (EUCLIDEAN RING). Suppose that R is an integral domain in which to each nonzero element a there is a nonnegative integer d(a) so that (1) d(ab) g d(a) if a,b 76 0 (2) For each pair a, b of elements, a 76 0 there are elements q, r so that b=qa+r

where either 1" = 0 or d(r) < d(a). Then R is called a Euclidean ring.

1.3. IDEALS AND RINGS

17

EXAMPLE 40. HR = Z we use d(a) = |a|. HR = k[:r] we use d(f) = deg(f). Thus Z and k[a:] are Euclidean rings. I] In the same way as for Z and k[:r] we can prove the following theorem. THEOREM 41. In a Euclidean ring every ideal is principal.

Proof. Let a 76 (0) be an ideal and let a e a be an element with d(a) minimal for elements in a. If b E a we have b = qa + r, and 7' = b — qa E a.

Since d(a) is minimal we must have 1‘ = 0. Hence every element in a is a multiple of a. El Exercises

1. Consider the Gaussian integers Z[i] = {a + ib| a,b E Z }. Put d(a + 2b) = a2 + b2. Show that for a,,6 E Z[z'] we have d(afi) = d(a)d(,6). Show that Z[z'] is a Euclidean ring. Hint: Suppose a = a + ib and fl = c+id, fl 76 0, are Gaussian integers. Show that a/fl = 7+q1 +iq2 for some 7 E Z[i] and some q1,q2 E Q with |q1] S [til/2, |q2| g |b|/2. 2. Show that in a Euclidean domain d(a) is minimal if and only if a is invertible.

3. Show that in a Euclidean domain one has d(a) = d(b) if a = bu for some invertible element u.

1.3.5

Ideals and their Calculus

We will now introduce some operations on the set of ideals in a ring. We will later show that, for ideals in a polynomial ring, these operations will

have geometric meanings. If a and b are ideals in R then

1. a+b={a+b|a€a,b€b} 2q 3. a-b={2f=1a,~bi aiEa,b,-Eb}

4. a:b={rER|rbEa

Vbeb}

are all ideals. We prove this for a : b and leave the other as exercises. If r1,r2 E a : b then (r1 +7'2)b = r1b+r2b E a for every b E [3 since rlb,r2b E b and a is closed under addition. Similarly, if 1'1 6 a : b and r2 6 R, then (r2r1)b = r2(r1b) E a for every b E b since rlb E a and a is closed under

multiplication by elements in R.

18

CHAPTER 1. RINGS, FIELDS, AND IDEALS

DEFINITION 42 (THE RADICAL OF AN IDEAL). The radical of an ideal a in a ring R is the set \/E = {re RI 7-" e afor some n}, where n is a positive integer.

DEFINITION 43 (RADICAL IDEAL). An ideal b is a radical ideal if f = b. Note that a g fl since 7’ E a implies that r" E a for n = 1, and so by

definition, 7‘ 6 ME. THEOREM 44. If a is an ideal, then fl is an idea] which contains a. More-

over, \/5 is a radical ideal. Proof. We have already shown a g fl, so all that remains is to show J5 is an ideal which is radical. First, we have to prove that J5 is an ideal. Suppose that 11,13 6 J5. Then, by definition, there are two positive integers n1 and 712 such that

1111,1312 6 J5. If we expand the sum (71 + r2)"1+"2’1 with the binomial theorem, we see that every term is a multiple of some rinlrg” with m1 + m2 = m + 712 — 1. Since either m1 2 m or m2 2 in, either if” or r3" is

in a. Therefore, since all its terms are in a, (1'1 + r2)"1+”2‘1 E a. We have proved that n + m 6 fl, but we still have to show that if n E \/E and r E R, then rm 6 J5. We have 1'? E a for some n, hence rnr? = (M1)" 6 a,

so rm 6 J5, hence \/E is an ideal. We have J5 Q \/\/3. Let r E \A/E. Then T" E \/E for some n, hence rm” = (rn)m e a for some m, which gives r 6 J5, so a is radical. I] There are lots of formulae connecting the different operations on ideals. We now prove some which will be needed later, and leave others as exercises. PROPOSITION 45. Leta, b, andc be ideals in aring R. Then H

(anc)+(bnc)g(a+b)nc 2. (a:c)+(b:c)§(a+b):c 3. a:(b+c)=(a:b)n(a:c)

4. a§b=>\/E§\/b_ 5

vii—kw:

6 .fim/E2m/Em/F Proof.

1. Let n Eanc and r2 6 bnc.Thenr1 E aand 7'2 E 650

r1+r2€a+b.SincerlEcandrgecwegetr1+r2€c

2. LetrEa:c+b:c. Thenr=r1+r2whererlc€aandrgceb for evercc. Thus rc=r1c+r2ce (a+b) for evercc, so

rE(a+b):c.

1.3. IDEALS AND RINGS

19

Let r e a : (b + c). Then r(b+ c) E a for every b E b and every c E c. In particular rb e a for every b E [3 since 0 E c. In the same way we seethatrce aforevercc. HencerEazbandrEazcso

r6 (a:b)r‘l(a:c) Nowsupposere (a:b)fl(a:c). ThenrbE afor every b E b and TC 6 a for every 0 E c. Hence r(b + c) = rb + TC 6 a for everybE b andeverc c, sore a: (b+c). .IfrEfi,thenr"€a§bforsomen,sor€x/b_. 5. Since a" g a we have Vak Q fl. Ifr E x/E, then r" e a for some n.

Then 1"“ = (7'")" 6 CU“, so 1‘ 6 Va’“. . Let r E m/Enx/E. Then 7‘" E fiflx/Efor some n. Then 1‘7”“ 6 a and T” E b for some k, hencer E fiandr E x/E, sor E firm/E. The other inclusion follows from (4). El

Exercises Let a, b, and c be ideals in a ring R. Show that 1. a+ b,afl b, and a- b are ideals. Ifa = (S) and b = (T), show that

u—nn—tv—I Nan—‘9

PWNQP‘PPN

a+b=(SUT) and a-b=(st|sES,t€T).

u+b=b+a, a+(b+c)=(a+b)+c ub=ba, a(bc)= (ab)c, a(b+c)=ab+ac a§a+b, abgaflb

(u:b)b§u, agazb (a:b):c=a:(bc)

W=\/\/E+x/E

x/Ex/F=x/E Nah/WE Vazb=\/E:\/E

. Determine «(72) in Z. . Prove that if (m) and (n) are ideals in Z, then

(a) 0)) (C) (d)

(m)+(n) = (gcd(n,m)) WWW) = (lcm(n,m)) (m)-(n> = (mn) (W) = (fl) = (m/ gcd(m,n))

20

CHAPTER 1. RINGS, FIELDS, AND IDEALS

13. Prove that if (f) and (g) are ideals in k[a:], then

(a) (b) (C) (d) 1.4

(f)+ (g)= (gcd(fyg)) (f)0 (9)=(1Cm(f,g)) (f) (g)= (f9) (f)- (9>= (f/gcd(f,9)) Equivalence Relations

Let M be a set and M x M the set of ordered pairs of elements in M. A subset S of M x M is called a relation on M and we write a ~5 b if the

pair (a, b) E 5'. DEFINITION 46 (EQUIVALENCE RELATION). A relation 5' is called an equivalence relation if

1. a ~S a for all a in M (reflexivity) 2. a ~5 b => b ~s a (symmetry) 3. a ~s b and b ~3 0 => a ~s c (transitivity) EXAMPLE 47. The relation a ~ b if a and b have the same parity, i.e., if a and b are both odd or both even, is an equivalence relation on Z.

El

PROPOSITION 48. Let S be an equivalence relation and let [a] be the subset ofM consisting ofall elements equivalent to a, i.e., [a] = {bl b ~5 a}. Then

b ~s a implies [a] = [b]. Proof. If c E [a] then c ~5 a and since a ~s b we get c ~5 b, hence c E [b]. Similarly we get the other inclusion.

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Thus M is partitioned in subsets called equivalence classes. Since a E [a], every element belongs to one and only one equivalence class. An element

in a class [a] is called a representative for [a]. EXAMPLE 49. The equivalence relation on Z, (1 ~ b if a and b have the

same parity yields the two classes, the odd numbers and the even numbers. Every even element is a representative for [0] and every odd element is a

representative for [1].

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We now give a new description of Zn in terms of an equivalence relation on Z. Let n be a positive integer. Every element m can be written m = q'n. + r, for some quotient q and some remainder r, 0 S r < n. We get that the relation m1 ~ m2 if m1 and m2 have the same remainder when

1.5. FIELD OF FRACTIONS OF AN INTEGRAL DOMAIN

21

divided by n is an equivalence relation. Another way to define this equivalence relation is to say that m1 ~ m2 if m1 — m2 is divisible by n. The finite set of equivalence classes constitute Zn. The equivalence classes are

[0], [1], . . . , [n -— 1], where the class [1'] consists of all integers with remainder i when divided by n. The addition in Zn is then defined by [a] + [b] = [a+ b] and the multiplication by [a] [b] = [ab]. These definitions seem to depend on the choice of representative for the classes, and we must show that they do not. If we choose other representatives a1 and b1, where a1 ~ a and

b1 ~ b, we must show that [al +b1] = [11+ b], i.e., that a1 + b1 ~ a+ b. But a; ~ (1 if and only if n divides a1 — a and b1 ~ b if and only if n divides

b1 - b. Then ((11 + b1) — (a + b) = (a1 — a) + (b1 — b) is divisible by n, so (11 + b1 ~ (1 + b. Next we define a multiplication on Zn by [a] ' [b] = [ab]. Again, we must show that 0.1 ~ a and b1 ~ b implies albl ~ ab. Now albl —ab = a1b1 —a1b+a1b—ab = a1(b1 ——b)+b(a1 —a). Hence, if both b1 —b and a1 -a are divisible by n we get that al b1 —ab is divisible by n. It is now easy to check that all axioms for a ring are satisfied for Zn. For example, we

see that ([a] - [bD ' [C] = [ab] ' [C] = [(61106] = [000)] = [a] - [be] = [a] - ([bl ' [0])-

1.5

Field of Fractions of an Integral Domain

The field Q is the smallest field containing Z. If we want to construct Q starting from Z, we could introduce inverses n—1 to all nonzero elements

in Z and let Q = {mn‘1 I m E Z,n E Z\ {0}}. A problem is that this representation is not unique, e.g., 2 ~ 3‘1 = 4 ~ 6—1. A solution to this problem is to consider pairs of integers and introduce an equivalence relation

on the pairs so that (2,3) ~ (4,6) say. Then Q is defined to be the set of equivalence classes of such pairs. We will perform such a construction for any integral domain. Let R be an integral domain. We will construct a field, called the fraction field of R, that contains R and that does not have any subfield containing R. Consider the set of pairs {(r1,r2) | 11,17 6 R, 7'2 76 0 }, and

introduce a relation on this set of pairs by “(T1,1‘2) ~ (r’1,r’2) if r1 r5 = r’lrz”. We leave it to the reader to verify that this is an equivalence relation. The set of equivalence classes associated with this equivalence relation

can be made into a ring if we define the sum and product as follows. Let (r1,r2), and (r’l, 7J2) be representatives for two equivalence classes, and define the sum of these classes to be the class containing (rlré + r’1r2,r2r§), and define the product as the class containing (r1 r’1,r2r’2). To show that these operations are well defined one must check that the sum and product does not depend on the choice of representatives. Thereafter it must be

22

CHAPTER 1. RINGS, FIELDS, AND IDEALS

verified that the axioms for a ring hold. All this is left to the reader. The zero element in this ring is the class containing (0, 1) and the multiplicative unit element the class containing (1, 1). The class [(r1,T2)l is different from the zero element if, and only if, n 76 0, and if so [(r1,r2)] - [(r2,r1)] = [(r1r2,r1r2)] = [(1,1)]. So every element different from the zero element has a multiplicative inverse, hence

the ring is a field. We denote this field by Q(R). The map f : R —) Q(R), f : r I—> [(131)] is bijective, and shows that there is a copy of R inside Q(R). It is easy to check that, if we identify R with its image in Q(R), R is a subring of Q(R). This means that addition and multiplication of elements are the same, regardless whether we do the operations in R or in

Q(R). For example is [(r,1)] + [(3,1)] = [(r + s, 1)] in Q(R). Since we have

identified 7‘ E R with [(731)] E Q(R), the addition is the same in R as in Q(R). Similarly for multiplication. Any subfield of Q(R) that contains R (with the identification we have made) will contain all equivalence classes [(1,r2)],r2 76 0, since these are inverses to elements in the image of the elements [(r2,1)] in R. Hence, the subfield will contain [(11, 1)] - [(1,1'2)] = [(11, 7'2)], and so is all of Q(R), EXAMPLE 50. If R = k[m] then Q(R) is denoted k(:1:) and consists of all elements f/y, where f, g E k[a:], and g aé 0. Here we have identified [(f, g)] with f/g. E]

Exercises

1. Let R be an integral domain and consider the set of ordered pairs { (11,13) [ 1'1,r2 E R,r2 76 0 } Show that the relation (11,13) ~ (r[,r’2) if 117'; = fin is an equivalence relation.

2. On the set of equivalence classes, define [(a, b)]+[(c, d)] = [(ad+bc, bd)], and [(a, b)] - [(c, d)] = [(ab, cd)]. Show that these operations are well defined, i.e., check that the sum and product do not depend on the choice of representatives. 3. Show that the set of equivalence classes associated with the equivalence relation in Exercise 1 with sum and product defined as in Exercise 2

form a ring with [(0, 1)] as the zero element, and [(1, 1)] as the multiplicative unit element.

4. Describe the fraction field of the ring Z[i] of Gaussian integers; see Exercise 1 in Section 1.3.4.

1.6. UNIQUE FACTORIZATION DOMAINS

1.6

23

Unique Factorization Domains

We have seen that in a polynomial ring k[:r] over a field k, every polynomial of positive degree is a product of irreducible polynomials, and that this factorization is essentially unique. We will show that this is true also for

polynomial rings in several variables over a field. It is not harder to be a bit more general; we will prove that if a ring R has unique factorization,

then R[a:] has unique factorization. Note that if u is an invertible element in a ring R then fg = (uf)(u‘lg), so we cannot expect perfect uniqueness if the ring has invertible elements other than 1. DEFINITION 51. A nonzero element r in an integral domain R is said to be irreducible in R if r = 7'17'2 implies that n or T2 is invertible.

DEFINITION 52 (UNIQUE FACTORIZATION DOMAIN). An integral domain I? is called a unique factorization domain if every element in R which is neither invertible nor 0 is a product of irreducible elements, and this factorization is unique in the following sense. If r = r1 . - -rk = $1 - - . sm are two factorizations in irreducible elements then k = m, and after renumbering, for each 1', Ti 2 uisi for some invertible elements u,- E R. To prove the main result we need some lemmas. LEMMA 53. Let p be an irreducible element in a unique factorization domain R. If ab is a multiple of p, then a or b is a multiple of p. Proof. We have ab = pc for some 0. Let a = p1---pk,b = q1---q,,c = 7'1 ~ - - I'm be factorizations in irreducible elements. Then p1 - - ~pkq1 - - ~q, 2 pm ---r.,,,, and the uniqueness gives that up = p,- or up = qi for some 1'.

Hence either a or b is a multiple of p.

E]

LEMMA 54. If R is a unique factorization domain, then two elements a and

b in R have a greatest common divisor gcd(a, b) Which is unique up to a multiplication with an invertible element. Proof. If c divides both a and b, then each irreducible factor of c divides both a and b. Hence the product of all irreducible factors which occur in

the factorizations of both a and b is gcd(a, b).

CI

DEFINITION 55 (CONTENT OF A POLYNOMIAL). Let R be a unique facto-

rization domain, and let f (m) = a0 + - - - + am" 6 R[x] Then the content of f is C(f) = gcd(ao, . . . ,ak).

LEMMA 56. If R is a unique factorization domain and f,g E R[:r], then

C(fg) = C(f)0(9)-

24

CHAPTER 1. RINGS, FIELDS, AND IDEALS

Proof. Let f = a0 + - - - +akx" and g = be + - - - +bma:’". Then c(f) divides all a,- and c(g) divides all bj, hence c(f)c(g) divides all ajbj and thus also all coefficients of f9. We have proved that c(f)c(g) divides c(f9). Suppose that there is an irreducible element p which divides c( f9) but does not divide c(f)c(g), i.e., does not divide either c(f) or c(g). Suppose that p divides a0, . . . ,ai_1 but not ai, and that p divides b0, . . . , bj_1 but not bj.

Consider the coefficient a0b¢+j + ~ ' - + aibj + - - - + ai+jb0 of mi“ in fg. We see that p divides all terms in this sum except aibj, which gives that p does not divide the sum, a contradiction.

U

THEOREM 57. If R is a unique factorization domain, then R[a:] is a unique factorization domain.

Proof. Let f (m) E R[a:] be a polynomial which is neither invertible nor 0. If deg( f) = 0 then f E R, so f is a unique product of irreducibles, since R has unique factorization. It is clear that this is a unique factorization

into irreducibles also in R[:c]. If deg(f) > O we write f = c(f) f1. Then c( f1) = 1. Let k = Q(R) be the fraction field of R. Every polynomial

in R[:c] can be considered as a polynomial in k[z]. We know that k[a:] is a. unique factorization domain by Theorem 30 in Section 1.3.3. Thus f1(:r) has a unique factorization into irreducible factors in k[a:]. It is clear that all we have to prove is the following. If f (z) e R[:1:] is a polynomial with content 1, and f (x) is irreducible in R[:c], then f (x) is irreducible in k[a:]. So suppose that f(a:) = g(m)h(m) in k[a:] and deg(g) < deg(f),deg(h) < deg(f). Then g($) = (cl/02)G(:z:) and h(:1:) = (63/04)H(w) for some 01- E R and some

G, H E R[z] with contents 1 (factor out the least common denominators of the coefiicients of g and h, respectively, to get polynomials in R[x], and then factor out the content). Then (C264)f(:13) = (c1C3)G(a:)H(2:) and c(G'H) = 1 by Lemma 56. Thus f (3:) = G(:I:)H(:1:) is a factorization in R[$]. I] COROLLARY 58. If k is a field then k[:c1, . . . ,mn] is a unique factorization

domain. Proof. This is immediate by induction on 77..

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REMARK. From the proof of the theorem it follows that if f (m) is irreducible in R[x] of positive degree, then f (2:) is irreducible in k[:1:]. So for example if f (:c) e Z[:c] can be factored with factors of lower degree in QM, then f (m) can be factored with factors of lower degree in Z[:r].

1.7. FACTOR RINGS AND HOMOMORPHISMS

25

Exercises

1. Show that x3 + a: + 1 is irreducible in Q[z]. 2. Prove Eisenstein’s irreducibility criterion: Let f (x) = a0 + an: + - - ~ + aux" e Z[a:]. If for some prime p we have that p divides a,- for i < n, 1) does not divide an, and 122 does not

divide a0, then f (2:) is irreducible in Z[w]. 3. Show that $7" — p is irreducible in Q[:c] for any m if p is a prime. 4. Show that q(a:) = sup—1 + sup—2 + - - - + z + 1 is irreducible in Q[a:] if p is a prime. 5. Determine all monic irreducible polynomials of degree at most 3 in

Z2[:r]. Determine all monic irreducible polynomials of degree at most

2 in Z3[a:]. Show that .165 + x + 1 is irreducible in Z2[£L‘]. Show that 1:5 + 21: + 2 is irreducible in Z3[x].

1.7

Factor Rings and Homomorphisms

We will now imitate the construction of Zn from Z to get, from any ideal a in any ring R, a new ring, the factor ring of R with respect to a. Let a be an ideal in a ring R. Then the relation a1 ~ (12 if (11 — a2 6 a is an equivalence

relation on R. An equivalence class [(1] consists of the set {a + a’ | a’ E a} and is denoted a + at. These equivalence classes are often called cosets of a. If a + a = b + a, i.e., if a — b E a we say that a is equivalent to b modulo (1.

The set of equivalence classes is denoted R/ a. We make R/a into a ring by defining (a1 +a)+(a2+a) = (a1 +a2)+a, and (a1 +a).(a2+u) = a1 -a2+a. As for Zn = Z / (n), one easily checks that the addition and multiplication are well defined. With these operations R/a becomes a ring, the factor ring of R modulo a.

EXAMPLE 59. Let f(:1:) E Q[a:] be a polynomial of degree d. Any element g(:I:) E Q[:r] can be written g(:1:) = f(1:)q(a:) + r(a:) with deg(r(:z:)) < d or r(:1:) = 0. Hence, any element in Q[:z:] is equivalent to either a polynomial of

degree less than d or to 0 modulo (f (23)) On the other hand, two different polynomials f1(:c) and f2(x) of degree less than d cannot be equivalent modulo f(a:) since f1(a:) — f2($) cannot be divisible by f(9:) for degree reasons. Hence, the equivalence classes different from (f (x)) are in 1—1 correspondence to the set of polynomials of degree less than d.

I]

26

CHAPTER 1. RINGS, FIELDS, AND IDEALS

EXAMPLE 60. Let f (x) = x2 + 1 E Q[a:].

The equivalence classes are

{ aw + b + (x2 + 1) | a, b 6 Q}. Addition and multiplication are given by (a1at+b1+ S be a homomorphism. Then ker(f) is an ideal in R. If f is surjective, then S is isomorphic to R/ ker(f). Proof. Let a = ker(f). That a is an ideal follows from the fact that if r1,r2 E a and r E R, then f(r1 +r2) = f(r1)+ f(rz) = 0 + 0 = 0 and firm): f(r)f(r1) = f(r) -0 = 0. We now define a map 9 : R/a ——) S by g(r+a) = f (r). The first thing we have to check is whether 9 is well defined. This means, if we happen to choose another representative for r + a, say

3 + a = r + a, is it then true that f (r) = f (5)? That this is true follows

1.7. FACTOR RINGS AND HOMOMORPHISMS

27

from the fact that s+a = r+a implies r —s E a so f (r—s) = 0, which gives

f (r) — f (s) = 0, hence f(r) = f (s). We have shown that g is well defined. That 9 is a homomorphism follows from the following calculations:

g((?‘1+a)+(r2+a)) = g (7‘1 + 1‘2 + a) = f (7‘1 + 7‘2) f (7‘1) + f(T2) 901 + a) + 9(T2 + 0) 9((7‘1 + a)(7‘2 + 0)) = 9(T17‘2 + a) = f(Tirz)

= f(r1)f(1‘2) =

g(r1 + a)g(r2 + a)

It remains to show that g is bijective. That 9 is surjective follows from

the asumption that f is surjective. Suppose g(r1 + a) = g(1'2 + a). Then f(rl) = f(r2) so f(rl — r2) = 0, hence r1 — r2 6 ker(f). But then n + a = 1'2 + a, so 9 is also injective.

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We have defined polynomials in k[x] as formal expressions aux" + - - - + a0, a; E k. The addition and multiplication of these formal expressions have been defined so they satisfy the ring axioms. Some readers are perhaps more used to considering polynomials as functions, e.g., from R to R. If the field has an infinite number of elements, there is no great risk of confusion, since

f = g in k[m] if and only if f = 9 considered as functions on k. For finite fields one has to be a bit more cautious. A function (1),: : k ——) k is called a polynomial function if there is a polynomial f = aux" + - ~ - + no so that

¢f(c) = anc" +

+ a0 for c E k. The sum and product of polynomial

functions are polynomial functions, and it is easy to see that the set of

polynomial functions constitute a ring Pol(k). PROPOSITiON 62. The map (f) : k[:c] —> Pol(k),f 5—) the is a homomorphism. If the number of elements in k is infinite then ¢ is an isomorphism; if k is finite then (i) is not an isomorphism. Proof. That 45 is a homomorphism is immediate from the definition. We

have ker(¢) = {f E k[z] | f(c) = 0 for all c E k}. If k is infinite it follows from Theorem 33 (the Factor Theorem) that only the zero polynomial has all elements in k: as zeros. But if k is finite, k = {C1, . . . ,cn}, then ker(¢) =

((m-ci)---(x-cn))-

n

CHAPTER 1. RINGS, FIELDS, AND IDEALS

28 Exercises

1. Let k be a field and let 0 E k. Show that ¢ : k[a:] —-+ k,¢(f) = f(c) is a homomorphism. Determine ker(qS). . Let kbeafield and let 01,...,Cn e k. Show that ¢:k[m1,...,:rn] —~) k, ¢(f) = f(c1,. . .,cn) is a homomorphism. Determine ker(qb). . Let f : Q(Z) —> Q, [(m,n)] H m/n. Show that f is an isomorphism. . Let R be an integral domain and k a field, and suppose that R is a subring of k. Show that It has a subfield isomorphic to Q(R). . Show that a field of characteristic 0 has a smallest subfield which is isomorphic to ()3. Let p be a prime number. Show that a field of characteristic p has a smallest subfield which is isomorphic to Zp.

6. Show that 1R[:1:]/ (m2 + 1) 2 (C. . Show that R = Z2[m,y]/ ($2,134,112) is a ring with 8 elements. Determine a multiplication table for R.

. Show that Q[a;]/ (m2 — 2) 2 Q[\/§] . Show that R = Z3[:c]/ (x2 + 1) is a field with 9 elements. Determine a multiplication table for R. 10. Show that the homomorphism f is injective if and only if ker(f) = (0). 11. Show that if f is a surjective homomorphism, then f (1) is the multi-

plicative unit element in im(f). 12. Let R be the ring in Exercise 2 in Section 1.1. Let f : R —) C be defined by f(( _2

2 )) = a + ib. Show that f is an isomorphism.

13. Let R be the ring in Exercise 3 in Section 1.1. Show that the map

f : R —> IR, where f is defined by f ((a, b)) = a, is a homomorphism.

Determine ker(f). Show that R 2 R[a:]/ (2:2). 14. Let (i) : k[a:1, 9:2] —) k[$] be defined by ¢(f(a:1,w2)) = f(x2,:1:3). Show that d) is a homomorphism. Show that the image of 45 is the ring R in

Exercise 4 in Section 1.2.4. Show that R 2 k[:c1,:1:2]/ (a)? — x2). 15. Suppose that (b : R -—) S is a surjective ring homomorphism. Show that there is a 1-1 correspondence between ideals in R which contain

ker¢ and all ideals in S. Show that if a is an ideal in R containing ker ¢, then R/a 2 S/¢(a). 16. Let ()5 : Z[a:] —) Zp[.'1:], where p is a prime, be defined by ¢(ao + 0.12: +

'--+ aux") = [a0] + [a1]z + ' - - + [an]:z:”. (a) Show that d) is a surjective homomorphism.

1.8. PRIME IDEALS AND MAXIMAL IDEALS

29

(b) Show that if ¢(f) is irreducible, then f is irreducible.

(c) Show that .775 — m2 + 1 is irreducible in Q[:c]. (d) Show that 11:4 — 32:3 + 3:1:2 + 7 is irreducible in QM. 17. This exercise shows that factorization in QM is algorithmic. Let f (m) E Z[:1:] be a polynomial of degree n. If f is reducible it has a factor g of degree at most [11/2] (the largest integer S n/2). Take any integers (11,. . . , rim/214.1. We have that g(ai) divides f (at), so there are only a finite number of possibilities for 9(a)). A polynomial of degree k is determined by its values in k +1 points. Thus there are only finitely many possible factors 9 to test. Use this to factor x4 — 9:3 + 2102 — :1: + 1

in QM.

1.8

Prime Ideals and Maximal Ideals

Two particular types of ideals will be very important in the sequel, prime ideals and maximal ideals. In polynomial rings they will correspond to

building blocks for geometric sets. DEFINITION 63 (PRIME IDEAL). An ideal p 96 R in a ring R is a prime ideal if N E )3 implies that r E p or s E p. LEMMA 64. We have that p is a prime ideal if and only if a1---ak g )3 implies that some ai g p. Proof. Suppose p is a prime ideal. By induction on k it is clear that it

suffices to consider the case when k = 2. Let (11:12 2 and

that a 0 pi Z Uj¢ipj for all i (so all pi’s are really needed). Take for each

30

CHAPTER 1. RINGS, FIELDS, AND IDEALS

i an ca 6 (an pi) \ Uj¢ipj. Then we get that a = a1 + a2---as E a. We

have a2 - - 'as (7! p1 (here we use that in is a prime ideal) so (1 ¢ 131 since a1 6 p1.1f2'> 1 then a1 ¢ )3, and (Lg-”as E p,- so a ¢ pi, which gives usa contradiction. El

DEFINITION 66 (MAXIMAL IDEAL). Let R be a ring. An ideal m 96 R in R is a maximal ideal of R if the only ideal of R which strictly contains in is

the whole ring R. Prime ideals and maximal ideals can be characterized by means of their

factor rings. THEOREM 67. An ideal p aé R in a ring R is a prime ideal if and only if R/p is an integral domain. An ideal m yé R is a maximal ideal if and only if R/m is a field. Maximal ideals are prime. Proof. If a is an ideal in R, we denote the element 1' + a in the factor ring

R/a by F. In particular the coset a in R/a, which is the zero element in R/a, is denoted 6. Suppose p is an ideal and that rs E p. In R/p this means

that is = O. The equation rs = 0 is equivalent to 1“ = G or s = 0 if and only if R/p is an integral domain, which means that rs E p is equivalent to r E p or s E 33 if and only if R/p is an integral domain. Suppose m is a

maximal ideal. If r 9! m we get m + (r) = R since m + (r) strictly contains m. Thus 1 = m + rs for some m 6 m and s E R. In R/m this gives l = is so any nonzero element F in R/m has a multiplicative inverse, so R/m is a

field. Conversely, if R/m is a. field and F 7E 0 in R/m, then 715 = l for some

3‘ 6 R/m, which gives rs+m = 1 for some m E m, hence m+ (r) = (1) = R for any 7' ¢ 111. Thus an ideal which strictly contains m equals R and m is a maximal ideal. Since fields are integral domains we get that maximal ideals are prime.

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EXAMPLE 68. The prime ideals in Z are (p), where p is a prime number and (0). All these but (0) are maximal. The prime ideals in k[w], where k is a field, are (f (23)) Where f is irreducible and (0). All these but (0) are maximal.

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Exercises

1. Let (c1, . . . ,cn) E k" where k is a field and k" is the set of n-tuples of elements in k. Show that the ideal (:31 — c1, . . . ,xn — Cu) is a maximal

ideal in Man, . . . ,xn].

2. Show that (ref + 1,202) is a maximal ideal in the ring R[:rl ,m2].

31

1.9. VECTOR SPACES 3. Show that (x? — 2:3) is a prime ideal in k[zl,:1:2].

Hint: Exercise

Exercise 14 in Section 1.7.

4. Show the following improvement of Lemma 65. Let a Q U?l: lui. If all but at most two of the ai’s are prime ideals, then a Q a, for some i.

5. If p is a prime ideal, then fl = p.

6. Let R be the ring in Exercise 3 in Section 1.1. Show that {(0, b)} is a maximal ideal. 7. Let (1) : R —> S be a surjective ring homomorphism. Show that, in the

correspondence of ideals in R and 3 (see Exercise 15 in Section 1.7), we have that prime ideals correpond to prime ideals and maximal ideals to maximal ideals.

1 .9

Vector Spaces

We suppose that the reader is familiar with the concept of real vector spaces. It is possible to define vector spaces over any field, and the elementary

theory gives no surprises. We therefore refrain from giving proofs of the most elementary facts.

DEFINITION 69 (VECTOR SPACE). A vector space over a field k is a set V with an operation + and admitting a multiplication with elements in k. The elements in V are called vectors. The following rules are satisfied:

9°?‘P’P‘F‘9’5‘3

1. v1+v2€Vifvi€V cVic/candvEV

111 +112 =v2 +v1 and v1 +(v2 +123) = (111 +02)+v3 ifvi E V

There exists a zero vector 0 such that v + 0 = v if v E V Each vector v has an additive inverse —v such that v + (—v) = 0 1 - ’U = v if v E V and 1 is the multiplicative unit element in k

(c102)v = c1(czv) if c,- E k and v E V C1(’l)1 + v2) = c1111 + cum and (cl + cz)vl = 01121 + 02121 if c,- E k and W E V

A set of vectors {11,-},eI is said to be linearly independent if any linear combination c1v1 + + cnvn = 0 of elements in {11,} implies that all coefficients cz- = 0. A set of vectors {vihel is called a generating set for V if every element in V is a finite linear combination of elements in

{1),}. A set which is both linearly independent and a generating set is

32

CHAPTER 1. RINGS, FIELDS, AND IDEALS

called a basis for V. If V has a finite basis then V is called finite dimensional, and in that case every basis has the same number of elements. This number is called the dimension of V. Any linearly independent set can be extended to a basis and any generating set has a subset which is a basis. The set k" of n—tuples (c1,...,cn) of elements ci 6 k is a

vector space with addition (01,...,cn)+(c’1,...,c’n) = (cl +c’1, . . . , c1. +cg) and multiplication c(c1, . . . ,0") = (ccl, . . . ,ccn). It" has dimension it since {(1,0,...,0),(0,1,...,0),...,(0,0,...,1)} is abasis. EXAMPLE 70. If k g K are fields then K is a vector space over k. As an example, C is a vector space over R of dimension 2 with basis {1, i}. D EXAMPLE 71. The ring k[a:] is a vector space over It with basis {1,m,:1:2, . . .}. More generally, k[:1:1_, . . . , 91“] is a vector space over It with a basis consisting

of all monomials {161‘ mp3;- }.

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EXAMPLE 72. Let f E k[$] be of degree d. Then k[a:]/ (f) is a vector space

over k of dimension d with basis B = {1 + (f) ,2: + (f) ,...,md‘1 + (f)}. This follows from the fact that every element in k[:v]/ (f) can be written as g(:1:) + (f) with deg(g) < d, hence B is a generating set. Furthermore the elements in B are linearly independent, since a nontrivial linear combination

which is the zero element (f) in k:[$]/ (f) would yield an element of degree < d which is a multiple of f.

[3

DEFINITION 73 (SUBSPACE). A nonempty subset of a vector space V, that is closed under addition and multiplication with field elements is called a subspace of V. If V is a subspace of W, we define an equivalence relation on W by wl ~ 1112 if wl — 102 E V. The set of equivalence classes { w + VI in E W}

constitute the factor space W/V with addition (wl + V) + (1.02 + V) = (wl +w2) + V and multiplication c('w + V) = cw + V. A map f : V —) W between two vector spaces is called linear or a homomorphism if f (121 +122) = f(v1) + f(v2) and f(cv) = cf(v) for all v1,v2,v E V,c E k. Iff : V —> W is a homomorphism then ker(f) = {v E V| f(v) = 0}, and ker(f) is a subspace of V. A surjective homomorphism is called an isomorphism if it is bijective. Two vector spaces V1 and V2 are called isomorphic, V1 2 Vg if there exists an isomorphism f : V1 —) V2. If f : V ——> W is a surjective

homomorphism then W 2 V/ ker(f), and f is an isomorphism if and only if ker(f) = {0}. THEOREM 74. Suppose V is finite dimensional and that f : V —) W is

surjective. Then dim(ker(f)) + dim(W) = dim(V).

1.9. VECTOR SPACES

33

Proof. Take a basis {e1,...,ek} for ker(f) and extend it to a basis for V, {e1,...,ek,...,en} for V. We claim that B = {f(ek+1),...,f(en)} is a basis for W.

Since f is surjective any element in W has the form

f(clel +- - -+cnen) = c1f(el)+- - -+cnf(en) = ck+1f(ek+1)+- - -+cnf(en), thus B is a generating set for W. Suppose ck+1f(ek+1) +- - - +cnf(en) = 0. Then f(ck+1ek+1 + - - - +cnen) = 0, so ck+lek+1 + - ~+c,,e,, e ker(f), thus ck+1ek+1 + - - - +cnen = 0161 +~-+ckek, so 0161 + - ' - +cke;c —ck+1ek+1 —

-~—— onen = 0. Since {31, . . . ,en} is a basis for V we get that c,- = 0 for all

i, hence { f (eh-+1), . . . , f (en)} are linearly independent.

El

DEFINITION 75 (DIRECT PRODUCT). Let V1,V2,... be a sequence (finite or infinite) of vector spaces over k. Let H V, = {(121,112, . . .) | v,- E V,- }. If we define addition in H V,- and multiplication of an element in H V,- with an element in k componentwise, (121,122, . . .) + (1111,1122, . . ) = (1)1 + w1,v2 +

102,...) and c(o1,v2,...) = (co1,cv2,...), then HV, is a vector space, the direct product of the Vi’s. If we have a finite sequence V1, . . . , V", the direct product is often denoted I] x x V“. If furthermore V,- = V for all i then UV,- is denoted ‘1']!

DEFINITION 76 (DIRECT SUM). Let V1,V2, . .. be a sequence (finite or infinite) of vector spaces over k. Let EBV,‘ = {(v1,v2, . . .) | v,- E V,- }, where v,- = 0 for all but finitely tiplication of an element (mm-2,...) + (w1,w2,...) (cv1,cog, . . .), then 69V,- is

many 1'. If we define addition in EBV, and mulin 63V, with an element in k: componentwise, = (121 + 101,122 + 1122,...) and c(v1,v2,...) = a vector space, the direct sum of the Vi’s.

In case the sequence is finite, the condition that v,- = 0 for all but finitely many 1' is an empty condition. Hence, in this case the direct sum coincides with the direct product.

THEOREM 77. If V and W are finite dimensional, then dim(V x W) = dim(V) + dim(W). Proof. If {e,} is a basis for V and {6;} is a basis for W, we claim that {(e,, 0)}U {(0, (32)} is a basis for V X W. Suppose Z ci(e,-, 0) +2 c§(0, 6;) = (0,0). The left-hand side equals (2656130262) so Zcie, = 0 in V and 2 age; = 0 in W. Since {e,} are linearly independent in V, we get c,- = O for all 2'. Similarly c; = 0 for all i, so {(ei, 0)}U{(0, 82)} is linearly independent. Finally {(ei, 0)} U {(0, 62)} is a generating set for V x W since an element in V x W is of the form (v,w) with v E V and w 6 W. Since {(2,} generates V, we have I) = zcie, for some c, and similarly we have m = 2: age; for

some c,’,-. Hence (v,w) = (Z ciei,c§e’,) = Zci(e,~,0) + ZCKO, e2).

El

34

CHAPTER 1. RINGS, FIELDS, AND IDEALS

The most important property of 69% is that the coordinates are unique, if (121,112, . . .) = (1111,1122, . . .) then vi = w,- for all 2'. Our construction, from

a (finite or infinite) set of vector spaces Vi define a new vector space 63V“ is sometimes called the external direct sum of the Vi’s. Now suppose that we have subspaces V1, Vz, . .. of a vector space V. The sum 2 V,- of the vector spaces V; consists of all sums Eel-vi, where vi 6 V,-,c,- E k; and ci = 0 for all but finitely many i. It is easy to see that 2 V5 is the smallest subspace of V that contains all Vi. We now suppose that V = Z V,- and that for all

i, V.- n 2].?” V]- = {0} Then we claim that V ': 63%, and V is called the internal direct sum of the Vi’s. To prove this claim, we consider the map

¢ : 63%- —> V defined by ¢((v1,v2,...)) = '01 + 122 + - - -. (Even if there is an infinite number of W’s, this sum is a finite sum.) It is clear that ¢ is a

homomorphism and that 915 is surjective. Suppose ¢((v1,v2, . . .)) = 0. Then Ui+vz+- -- = 0 80111 = —01—02—' ' "Ui—1_vi+1_"‘, SO ”i E KNEW-V,which gives vi = 0. This means that ([5 is also injective. Note that if V is

the internal direct sum of {V2}, then every element 12 E V can be written 12:01 +v2+---,v,~ E V,» inauniqueway.

EXAMPLE 78. Let V = k[a:]. Then V} = { cXi I c E k } is a one-dimensional vector space with basis Xi and V is the (internal) direct sum of the Vi’s.

[I

For future reference we formulate a lemma. LEMMA 79. Let Vi be subspaces of a vector space V. Then V is the internal direct sum of the V.- ’s if and only if each element in v E V can in a unique way be written as a finite sum 1) = 21);, Where v,- E W. Proof. We have already shown “only if”, so suppose each element in V has a unique expression 2 vi, 1).- 6 Vi. Then Vinzfii V]- = {0} since otherwise,

if vi 6 Vi F‘I 21¢,- V131).- 76 0, we would have a contradiction to unicity.

El

Exercises

1. Show that the zero vector and the inverse are unique and that 01) = 0 if v E V.

2. Prove that for a vector space V over k, we have dim(V) = n if and only if V 2 k".

3. Determine a basis for, and the dimension of, Qh/i] (cf. Exercise 1 in Section 1.1) as a vector space over Q. 4. Determine a basis for, and the dimension of, the ring in Exercise 7 in

Section 1.7 as a vector space over the field k.

1.9. VECTOR SPACES

35

5. Determine a basis for, and the dimension of, the ring k[:1:,y]/ (223,314) as a vector space over the field k. 6. Show that if V1,V2, . .. are vector spaces over k, then HVi and 69%are vector spaces over k. 7. Show that if V1,V2, . .. are subspaces of a vector space V, then 2 Vi is the smallest subspace of V that contains all Vi.

Chapter 2

Monomial Ideals In this chapter we will consider a particularly simple type of ideal in a

polynomial ring, called monomial ideals. A monomial ideal is an ideal generated by a set of monomials, i.e., elements in k[:r1, . . . ,m,,] of the form xi" ”3:5,". In all examples we give, the monomial ideals will be finitely generated. This is no coincidence. We will show in the next chapter that all monomial ideals (in fact all ideals in k[:r1, . . . ,wnD are finitely generated. Monomial ideals will be important in the sequel, and we will now study them in order to be familiar with them. Another good reason to study monomial ideals is that they are much easier to understand and to compute with than arbitrary polynomial ideals, so they constitute a good class of ideals to start with. For example, if (f1, . . . , fr) is an ideal and f,- divides fj for some 2' 725 j, then fj is not needed as a generator. For monomial ideals this also gives a sufficient condition to get a minimal generating set. Thus,

since it is trivial to see if a monomial divides another, it is easy to reduce a generating set to a minimal generating set, minimal in the sense that no generator can be cancelled without changing the ideal. The essential feature of monomial ideals, which explains why they are easier to handle,

is formulated in the following theorem. THEOREM 1. Let a be a monomial ideal in k[:r1, . . . ,mn]. Let f = Z cimi,

where ci 6 k \ {0} and m,- are diflerent monomials. If f E a then m,- E a for each 1'. Proof. We introduce the concept of fine grading or multigradmg of the

polynomial ring k[:1:1,...,a:n]. If cm = 6:1}? ~--xf;',c E k \ {0}, we set mdeg(cm) = (i1,...,in). Let a = (n1,...ns) be a monomial ideal, and suppose f = Eel-mi E a. Then f = glnl + +gsns for some 9,- = ECU-mi}- E k[a:1,...,$n]. Let cim,» be a nonzero term in f. Then am,-

37

38

CHAPTER 2. MONOMIAL IDEALS

equals the sum of all elements cijmijni which are of the same multidegree as cimi. Hence cimi is a linear combination of the ni’s, so cimi E a. D

2.1

Sums and Products of Monomial Ideals

Remember that for any two ideals, a = (ah-Hid?) and b = (b1""’b')’ the sum is a+b=(a1,...,ar,b1,--', b8>’

and the product is ab = (01b1,. ..,a1bs,...,arb1,...,arbs) ,

see Exercise 1 in Section 1.3.5. Throughout the chapter we will look at some concrete computations

involving the two monomial ideals a = ($3, $31,314) and b = (2:2, my?) Thus, for our particular monomial ideals, we get

a + b = (x3,xy,y4) + (11:2,xy2) = (x3,xy,y4,x2,a:y2). But :1:2 divides x3 and any divides zy2, so we can remove the generators 2:3, and may2 without changing the ideal. The final answer thus becomes

a + b = (x3, arm/4) + (w2,zy2) = (w3,wy,y4,z2,zy2) = ($21,222,114)The computation of the product ab is also straightforward.

ab = (m3,my,y4) ($2,1Iy2) = (x5,w4y2,$3y,m2y3,$2y4,my6) = (x5,x3y,w2y3,wy6). 2.2

Intersections of Monomial Ideals

It is a bit harder to determine the intersection and quotient of monomial ide-

als. We note that if the ideals a = (m) and b = (n) are both principal, i.e., generated by a single element, then their intersection a n b = (lcm(m,n)).

Thus, (2123/) 0 ($113) = (lcm(w2y,zy3)> = ($2y3). For any three ideals, a, b, and c, we have seen that

(a+b)flc2(aflc)+(bnc), see Proposition 45 in Section 1.3.5. But if a, b, and c are monomial ideals, the relation becomes an equality,

(a+b)nc=(anc)+(bnc).

2.3. QUOTIENTS OF MONOMIAL IDEALS

39

To prove this, let f = fl + f2 6 c with f1 E a and f2 6 b. Since a and b are monomial ideals, each monomial occuring in f1 belongs to a and each monomial occuring in f2 belongs to b. If the same monomial occurs in both f1 and f2, say clm is a term of f1 and 02111 is a term of f2, we can write

f1+ f2 = (fl — mm) + (f2 + 01m) where f1 — 01m 6 a and f2 + 01m 6 b by Theorem 1. Hence we may assume that the monomials occuring in f1

are different from those occuring in f2. Since c is a monomial ideal, we get that each monomial occuring in f1 or f2 belongs to c. Hence f1 6 c and f2 6 C. Thus f1+f2 E (ant) +(bnc). Consequently, for monomial ideals we get

(1111,. ..,mr) fl (n1,...,ns) = 2:: (mg) m (11,-) = Z: (lcm(m,-,nj)). i=1 j:1 i=1 j=1

In particular, for our monomial ideals a = ($3,331,314), and b = (:52, 2:312), we find that

(r3,zy,y4)n($2,my2)

=

(lcm(m3,z2),lcm(x3,xy2),...,lcm(y4,my2))

= (x3,z3y2,z2y,wy2,x2y4,zy“) =

2.3

(x3,m2y,xy2>.

Quotients of Monomial Ideals

We now turn to the quotient a : b of monomial ideals. For arbitrary ideals a, b, and c, we have seen that the following relation holds:

(a+b):c2(a:c)+(b:c), see Proposition 45 in Section 1.3.5. (m1, . . . ,mr) and a monomial n

But for the monomial idea] a =

(m1,...,mr):(n)=(m1):(n)+---+(m,):(n). This can be seen as follows. It suffices to check f 6 (m1, . . . ,mr) : (n) for monomials, since fn 6 (m1, . . . ,m,) if and only if mn 6 (m1, . . . ,m,) for

every monomial m that occurs in f. If mm 6 (m1, . . . ,m,) then mn is a multiple of some mi, i.e., mn 6 (mi), hence m e (mi) : (n). For arbitrary ideals a, bl, b2, we have seen that a:(b1+b2)= (02130061252),

see Proposition 45 in Section 1.3.5, so by induction we get a 2 (b1 +"'+bs) =nf:1(a:b,-).

40

CHAPTER 2. MONOMIAL IDEALS

Let a: (m1,...,mr), b = (n1,...,ns), and b,- = (11,-). Then [1 = b1+m+ b3. It is easily seen that for any two monomials, m and n, the quotient

ideal (m) : (n) = (m/ gcd(m, 71)). We conclude that =

(1:!)

a:(b1+---+bs)

a3((Tli)+"'+(ns)) (t1=(n1))fl“'rl(az (725)) ((m1,...,mr) : (n1)) Fin-D ((m1,...,mr) : (115))

= ((m1)=(n1)+"-+(mr)=(”1W7"n ((m1)=(ns)+'“+ jk_1 for some k. In

other words in -< n if deg(m) < deg(n) or if deg(m) = deg(n) and the last variable with difierent exponents in m and n has higher exponent in m. There are different ways to vary these orderings to get new ones. All examples we have given imply the ordering :13" < 231-1 ~< -< :31 of the

variables. The ordering of the variables can be varied in 17.! ways which all induce new orderings. Another way to get a new ordering is to divide the variables into two or more blocks and compare each block in order. These orderings are called block orderings or product orderings. As an example, let 31 = {9:1, . . .,zk} and B2 = {n+1 m.,1:n} Then every monomial can be written in = mlmg with all variables in mi in block Bi,i = 1,2. To compare two monomials m and n we make the partition of them in blocks, m = mlmg and n = nlng. We let in -< n if m1 ~ 0. We leave it as an exercise to show that such an ideal is (wt-1,. . . Jig-primary. Collecting all ideals belonging to the same prime ideal and performing their intersection, we obtain an intersection of primary ideals with difierent primes. Making the intersection irredundant gives us a primary decomposition.

EXAMPLE 18.

a = (m3,x2y,:1:y2z) =(m3,m2,m)fl

and (2:,y) We have ($2,312) 0 (m3,y) = ($3,232y,:1:3312,112): (x3, 2:2y, y ) Thus a = (ac) 0 (:53, $231,312) n (:32, z) is a primary decomposition of a. The associated prime ideals are (2:) , (as, y), and (3,2).

D

77 Exercises

1. Show that a monomial ideal in k[:1:1, . . . ,xn] is irreducible if and only if

it is of the form (932111,. .. ,xze"). . Show that an ideal q in a Noetherian ring is primary if and only if ab 9 q, a s; q implies that b" Q q for some n.

. Show that (m3,my,y4) is primary and that ($3,114) is not primary in k[.7:, y].

. Show that (2:2,xy) = (:13) n (9:2, 3/ + arc) in k[:r,y] are different primary decompositions for every (1 E h.

It can be shown that p = (f1,f2,f3) = (y2 —- $2,342 — $3,22 — 1:21;) is a prime ideal in k[a:,y, z]. Show that p2 is not a primary ideal and that \/p—2 = p. Thus powers of prime ideals are not necessarily primary, in particular it is not sufficient that J5 is prime for a to be primary. Hint:

Consider f3 — f1f3. Show that the ideal (wf1,...,x2") ,m > 0 in k[:1:1,...,:1:n] is primary t0($1,...,:1:k).

Make a primary decomposition of a = (wzy,y2z, 9:22). Show that a monomial ideal in k[a:1, . . . ,xn] is primary if and only if it

is of the form (2:31, . . . , my: , m1, . . . , mt), where the mi’s are monomials in {mi1,...,xik}.

Chapter 6

Solving Systems of Polynomial Equations We start in the simplest way with a system of polynomial equations in one variable. We allow complex coefficients and look for complex solutions. Thus we are interested in the solutions of

13(91): f2(93) =

= fk($) = 07130106 Clwl:

i.e., in n((f1, . . . ,fk)). Now (f1,...,fk) = (gcd(f1,...,fk)), see Theorem 28 in Section 1.3.3, which is a principal ideal. Thus the system f1(x) = f2(:c) = - -- = fk(:1;) = 0 is equivalent to one single equation, gcd(f1, f2, . . . , fk) = O.

The equation f (2:) = O has n solutions, where n = deg(f), see Corollary 38 in Section 1.3.3, provided that the solutions are counted with multiplicities.

Thus 1:4 — 2:2 = 2:2(23 — 1)(a: + 1) = 0 has four solutions: :1; = 0 with multiplicity 2 and a: = :l:1, each of multiplicity 1. An algebraic way to describe the degree of a polynomial f (9:) is deg(f) = dimc (C[a:]/ (f(:1:)), since

{1 + (f($)) #6 + (f($)) ,--.,$"‘1 + (f($))} is a (C—basis for (C[a:]/(f(:1:)); see Example 72 in Section 1.9. Thus the number of solutions to f (x) = 0 equals dimc (C[m]/ ( f (2:)).

6.1

Systems With Only One Solution

We now turn to the general (and more interesting) situation of several variables. We have seen that points in C“ correspond to maximal ideals

in (C[a:1,...,:1:n], ifP = (01,...,cn) then i({P}) = ($1 —c1,...,2:n —cn); 79

80 CHAPTER 6. SOLVING SYSTEMS OF POLYNOMIAL EQUATIONS see Corollary 14 in Section 4.2. Hence if the system f1 (1.1,” .,mn))=

fk(a:1, . . . ,xn) = 0 has P as its only solution, i.e., 0((f1,.. .,fk)) — — {P}=, then (f1, . . . ,fk) = (x1 — c1, . . .,xn -— on); see Example 3 in Section 4.1 Let us denote ($1 — c1, . . . ,mn — C") by mp. We have seen that the ideals a, for which J5 = mp, are exactly those which satisfy m'}, _C_ a g mp for some 3; see Lemma 16 in Chapter 5. We will next discuss the concept of multiplicity of a solution. We start with some examples. The system a: = y2 = 0 corresponds to the intersection

of the y-axis a: = 0 and the x-axis y = 0 “counted twice”. It is thus natural to say that (0,0) should have multiplicity 2 as a solution to this system. Similarly (0,0) should have multiplicity 4 as a solution to :1:2 = y2 = 0, since both the rc-axis and the y-axis are double. We note that

dimc (C[1:,y]/ (Ly?) = 2 and dimc (C[z,y]/ ($2,342) = 4. DEFINITION 1 (MULTIPLICITY OF A SOLUTION). Let a = (fl, . . . , fk) be an ideal such that the radical of a is fl = (1:1 — c1, . . . ,xn — on). Then the

multiplicity of the solution (c1, . . . ,cn) to the system f1 =

= fk = 0 is

dimcC[:v1,. . . ,mn]/u.

We should check that this dimension is finite. We know that mi, g a for some 3, and hence that dimc (C[a:1 , . . . , :L'n]/a S dimc (C[:1:1,. . . ,mn]/m§3. The latter is easily seen to be finite, since after a coordinate transformation

we can assume that P = (0,0, . . . ,0), so mp = ($1, . . . ,xn). The monomials of degree < s constitute a (C—basis for dimc (C[:c1 , . . . ,xn]/ (x1, . . . ,mn)s, and there are only finitely many such monomials. Exercises

1. Show the statements above: that dimc (C[.7:,y]/ ($12,112) = 2 and that dimc C[:L',y]/ ($2,312) = 4.

). 2. Show that in fact dimc (C[.'I:1 , . . . KIM/mi, = (n+s—1 Tl

6.2

Systems with Finitely Many Solutions

We will now study systems of equations with finitely many solutions. Since points are considered to be zero dimensional, such systems are often called zero-dimensional systems.

6.2.1

Decomposition of the Ideal

Now suppose that f1 = -~ - = fk = 0 has finitely many solutions P1, . . . , P3.

Leta=(f1,...,fk). Then b(a)={P1,.. .,Ps}so\/_= mplfl ---nmp,;

6.2. SYSTEMS WITH FINITELY MANY SOLUTIONS

81

see Proposition 6 in Section 4.1. We have proved that any ideal a has a primary decomposition, a = q1 n- - -nqt with W = pi, a prime ideal. Then

x/E=\/q1_n-‘--n—qt=¢q—m~-n¢q7=mn~~npt. Thus p1 fl-nflpt =mpl fln-flmp“ which gives

v(pln---nm)=v(p1)U---Uv(pt)={P1,...,Ps}= 0011131) U ' - - U0(mp‘) = t)(mp1 n .. ' 0 mp1).

We have that 003,-) is an irreducible set since :3,- is a prime ideal. The only

irreducible subsets of {P1, . . . ,P3} are the one-point subsets {Pj}, so we get that 0(1),) = {P3} for some j, i.e.

Pi = i(v(m)) = i({Pj}) = mp.We have proved the first part of the following theorem. THEOREM 2. Suppose that f1 = = fk = 0 has a finite set of solutions {P1,...,Ps}. Then (f1,...,fk) = q1 n---nqs, with fl = mp1. fori = 1,...,s. The ideals q,- are unique.

Proof. A11 primary components are isolated, so they are unique; see Chap[:1 ter 5, Proposition 14.

6.2.2

Decomposition of the Ring

Next we will see that writing the ideal a as an intersection of simpler ideals can be used to decompose the ring (C[:1;1, . . . ,xn]/a as a product of simpler rings; we claim that C[m1,...,:1:,,]/a:(C[:I:1,...,a:n]/q1 x

> 51:2 >

> (En. Suppose dimcC[:1:1,...,:rn]/a < co and

that b = l(a) in Lex. Then there is an mn E N such that :12?" E b, hence there is an f in the Grobner basis with lm(f) = will". But the only monomials that are smaller than a)?" in this ordering are avid < m". Hence f E (C[:I:n],deg(f) = mm so there are only at most mn different possible values of 3,, in a solution to the system. By choosing Lex with say > wn_1 we see in the same way that there are only xn >- 9:1 > {1:2 >

finitely many possibilities for xn_1 in a solution to the system a.s.o.. We have proven the following theorem. THEOREM 7. Let a = (fl, . . . , fk). Then there are finitely many solutions to f1 = = fk = 0 ifand only ifdimc (C[:1:1,...,:cn]/a < 00. The number of solutions, counted with multiplicity, is dimc C[:z:1 , . . . , :rn]/a.

6.3

Solving Zero-dimensional Systems

We now turn to the question of how to actually get the solutions to a system of polynomial equations. We can reduce the problem to solving polynomial equations in one variable. In one variable we have no help from Grobner bases. One method comes from the reasoning above. For each variable 9:,- we can get an equation, gi($i) = 0, that the sci-coordinate ci must satisfy for any solution (c1,cz,...,cn) to the system. If deg(g,-) = m,- we have m1 m2 - - -m,, candidate solutions and we can plug them into the system to see if they are genuine solutions. This is theoretically possible, but practically impossible in many cases as it could mean calculating n diffe-

rent Grobner bases in Lex. Theoretical results and overwhelming practical evidence show how it is a very time-consuming task to calculate Grobner bases in Lex. It is almost always many times slower than a correspon—

ding calculation in Degrevlex. We will now describe a method where one calculation in Lex suffices, but we will later indicate that there are even faster methods. Hence consider the system f1 = f2 = -- - = fk = 0, where fi 6 C[:r1,...,wn], and suppose there are finitely many solutions

(this will be checked automatically during the calculations). Choose Lex with x1 >

> 11:“.

As above, the Grobner basis contains an element

gn(:t:n) in one variable. If we consider the reduced Grobner basis there is only one such polynomial, otherwise the polynomials of higher degree could

be reduced by those of lower degree. Denote the solutions to gn(mn) = 0 by 01,. . -,C-m,,- There is also an element gn_1 with lm(gn_1) = 51:211— 1 in

the Grobner basis. The only monomials smaller than 10:11” that can oc— cur in l(a) are £14935, with j < mn_1,k < m". Hence there can be only finitely many elements G1(a:n_1,:z:n),. . .,Gl(zn_1,.1:n) in the Grobner ba-

6.3. SOLVING ZERO-DIMENSIONAL SYSTEMS

85

sis corresponding to such monomials. Plug in :5” = cl in each Gi. Then we get a finite set of polynomials Gi(a:n_1,cl) in one variable. The system G1(a:n_1,cl) = = G[(xn_1,cl) is equivalent to one equation gcd(G’1(xn_1,c1),...,G;($n_1,cl)) = 0. This equation has only finitely many solutions 01,1, . . . , 01,m,._1~ Repeat this process for C2, . . . , cm“. There

is also an element gn_2 with lm(gn_2) = $212? in the Grobner basis. The . mn_2 j k 1 only monom1als smaller than xn_2 that can occur in l(a) are xn_2mn_1xn with j < mn_2, k < mn_1,l < mn. Hence there can be only finitely many elements H1(zn_2,a:n_1,mn), . . . ,Hm(mn_2,zn_1,:l:n) in the Grobner basis corresponding to such monomials. Plug in 3% = cl,mn_1 = 01,1 in each Hi. Then we get a finite set of polynomials Hi(xn_2, c1,1,c1) in one variable. The system

H1($n—2,01,1,01) =

= Hm($n—2,Ci,1,01)

is equivalent to one equation

gcd(H1 (In—2,01,1,01), - - - ,Hm($n—2,ci,1,ci)) = 0This equation has only finitely many solutions 02,1, . . ”0211,12. Repeat this

process for each pair (cm, Cj). It is clear that induction gives us all solutions to the system. EXAMPLE 8. Consider the system

11:? + $3 + 1:3 — 3

=

0

$1132.73 - 1

=

0

(611:3 +332 —2

=

0.

The ideal

(a3? + $3 + $3, — 3, m1m2x3 — Lawn + $2 —- 2) has in Lex with 2:1 > 932 > $3 the reduced Grobner basis

{$1 + 32953 — 1/2z3 + 2x; — 7/2w3,x§ — 22:2 + 1, 11:21:3— 2:2 + 1/2x§ — 2:133 + 3/2,:rg — 393% + 32:3— 1}. From the last equation we get $3 = :|:1. If we substitute x3 = 1 in the first

three equations, we get :01 + $2 — 2 = mg — 2.7:; + 1 = 0, so 932 = m1 = 1. If we substitute .733 = —1, we get 9:1 — 222 + 2 = mg - 21:2 + 1 = 0 = 0, so 2722 = 1,1101 = —1. Hence the system has the two solutions (1,1,1) and

(—1, 1, —1). (One can show that they both have multiplicity 4.)

D

86 CHAPTER 6. SOLVING SYSTEMS OF POLYNOMIAL EQUATIONS Exercises

1. Solve the system of equations x3+x2y—wy—y2

=

0

xy—x—y+1 x2—y2

= =

0 0

by calculating the Grobner basis of

($3 +x2y—xy —y2,a:y —a:—y+ 1,91;2 —y2) in Lex. 2. Solve the system wy—z

=

0

zz—y

=

0

312—2:

=

0

(133+:z:y+y2—z2

m2—xz+y+22-—2

=

x4+xyz+yz+z2—2 xz—y—2z+1

pooo

3. Solve the system

4. Theideal

a: (a+b+c+d+1,ab+bc+cd+d+a,abc+bcd+cd+ad+ab, abcd + bcd + cda + dab + abc, abcd — 1) has ideal of leading monomials l(a) = ((1, b2, bc2, bcd, M2, 03, cdz, d3) in

a+b+c+d+1

ab+bc+cd+d+a

=

abc+bcd+cd+ad+ab

=

abcd+bcd+cda+dab+abc

=

abcd—l

OOOOO

Degrevlex. Determine the number of solutions to

6.4. SYSTEMS OF HIGHER DIMENSION

87

5. Determine the number of solutions to

(131:1:2 — 1

=

0

11:2:E3 - 1171

=

0

$3 — $2 — 1

=

0

6. A point ((11, . . . ,an) is called a critical point for f(rc1, . . . ,xn) if

8

6—;(a1,...,an) = 0

for all 1'. Determine the critical points for f (2;, y, z) = x3 + 22:312 + y3 + yz + 22.

6.4

Systems of Higher Dimension

For zero—dimensional systems it is clear what one should mean by solving the system, at least if all solutions have multiplicity 1. A solution is then

simply a complete list of points satisfying the equations. For systems with infinitely many solutions the situation is more difficult. It is not even clear what one should mean by “solving” such a system. There is no natural easiest way to present the solutions. Despite that, one can sometimes use Grobner bases to get an easier presentation of the solutions than given by the system itself. As for zero-dimensional systems, the Lex ordering is best suited to present the solutions. EXAMPLE 9. Consider the system x2—y2+x+y—z

=

0

a32+2y2—2$+y—z

=

0

x3 —x2z—zy2 —2y22+:c2 +11:y+:z:2z—yz+z2

= 0

A Grobner basis for

(x2 —y2+m+y—z,zg+2y2—2m+y—z, x3 —:z:22——:ry2 —2y2z+a:2 +zy+mz—yz+z2) is {ac — y2,y4 + y — 2}. Thus the solutions are given by =t4+t

y=t x=t2.

88 CHAPTER 6. SOLVING SYSTEMS OF POLYNOMIAL EQUATIONS Exercise

1. The method of Lagrange multipliers tells that if a maximum (or mini-

mum) of f (.731, . . . ,2") subject to the condition g($1, . . . , :6”) = 0 exists, it is attained at a point where aialf + Ag) = 0 for some A.

(a) Determine max(a:y2z3) when a: + y + z — 1 and 2:,y, z > 0. (b) Determine the point on the surface my + 9:2 + yz = 1 that is closest to the origin.

Chapter 7

Applications of Grobner Bases

Solving systems of polynomial equations, which we treated in the previous chapter, is one of the most important applications of Grobner bases. In this chapter we will consider some other applications within mathematics. We stress that it is constructiveness we are interested in; we want to show that certain operations may actually be performed constructively (by hand

or with a computer). We recall some important facts that we will use.

Let a be an ideal

in k[m1,...,.1:n] and let G be a Grobner basis for a. Any element f has a unique normal form with respect to G; NF(f) is a linear combination, with coefficients in k, of monomials outside the ideal of leading monomials

1(a). We have for all f that f — NF(f) E a. In many problems we need a specific ordering of the monomials. Usually we divide the variables into two groups, say {m1,...,$m} and {y1,...,yn} in k[m1,...,xm,y1,...,yn], and we want all monomials (except the monomial 1) in the first group to be larger than any monomial in the last group, or similarly, that all x,- are larger than any monomial in the yj’s. This is achieved by the Lex ordering with x1 >> (cm, > y1--- > 1111 but there are several other orderings satisfying our need; see Section 3.1.

89

90

CHAPTER 7. APPLICATIONS OF GROBNER BASES

7.1

Membership Problems

7. 1. 1

Ideal Membership

We have already touched on the problem of how to judge whether an ele-

ment belongs to a given ideal. We recall the method. Let a = (fl, . . . , f,) be an ideal in Man, . . . ,xn] and let 9 E k[w1, . . .,xn]. Determine a Grobner basis G of a. We recall that each element in Man, . . . ,wn] has a unique

normal form with respect to G, and that NF(f) = 0 if and only if f E a. EXAMPLE 1. Suppose we want to show that

y3 + 23 e (x2,xy + y2,zz + 22) = a.

A Grobner basis for a in Degrevlex is {332, my +1112, 232 +22, y3, 23}. Since 113 and 23 belong to the Grobner basis we have NF(y3 + 23) = 0. Thus 3/3 + 23 belongs to a. We note that, if we remember the calculation of the Grobner basis, we can actually get an expression

y3+z3 =f1 '1172 +f2 - ($y+y2)+f3-(:rz+z2). The S—polynomial

5(z2,my+y2)=y-x2-z-(xy+y2)=—wy2=-y-($y+y2)+1-y3, so

343 = y - x2 + (—2: HOW +92)Similarly 23 = 2-922 + (—1'+z)(a:z+z2) so 3 3 _ 2 2 2 y +2 — (ZN-Zn +(y—z)(my+y )+(z—:c)(a:z+z ). El

Exercise

1. Determine the smallest power of 2 which belongs to the ideal a =

(2:2 + 11:2, my + 312, yz + z2>. Express this smallest power of 2 as a linear combination of the generators of a.

7.1. MEMBERSHIP PROBLEMS

7.1.2

91

Radical Membership

If we assume that we are in a polynomial ring over (C, we can answer

the following question: Does g E x/(f1, . . ., fr)? Note that, according to Corollary 13 in Section 4.2, we can also formulate the question like this: Is every solution to f1 = = fr == 0 also a solution to g = 0? By using a trick we will reduce this problem to an ideal membership problem in a ring with an extra variable.

PROPOSITION 2. Let (f1, . . .,f,.) be an ideal in Cx1,...,xn]. Then an element 9 E (C[$1,...,$n] belongs to x/(f1,...,fr) if and only if the ideal b = (f1,...,fr,1—yg) in (C[a:1,...,2:n,y] equals C[:L‘1,...,:L‘n,y], i.e., if and only if 1 E b. This is true if and only if the reduced Gréibner basis of

b is {1}. Proof. This is exactly what we proved in Hilbert’s Nullstellensatz; see Cor-

ollary 13 in Section 4.2. [1

EXAMPLE 3. Let us show, using this method, that y 6 «(372,231; +y2). Consider the ideal b = (222,933; + 312,1 — yz) in (C[a;,y,z]. A Grobner basis for b is {1}. El Exercise

1. Determine \/E if a = (x2 + waxy + y2,yz + 22) in (C[a:,y,z]. 7. 1 .3

Subalgebra Membership

Let f1,...,f,. E k[a:1,...,a:n]. The subalgebra k[f1,...,fr] of k[:1:1,...,:1:n] generated by f1, . . . , fr is the set of polynomials f (f1, . . . , fr), where f is a polynomial in r variables. Hence k[f1, . . . , fr] consists of all polynomials which are the result of a substitution y1 = fl, . . . ,y, = fr in some poly—

nomial f(y1, . . . ,yr) 6 k[y1, . . . ,yr]. Another way to phrase this is to say that k[f1, . . . , fr] is the image of the homomorphism g5 : k[y1, . . . ,yr] ——) k[m1, . ..,z.n] defined by ¢(y,-) = fi(2:1,...,1:n). Thus

H132]:{“°+a2$2+a4$4+“-+a2k$2kIaiek}, and

“9324/31 = {Z aiflz‘y3’ Ham 6 kl It is clear that k[f1, . . . ,fr] is a ring.

92

CHAPTER 7. APPLICATIONS OF GRCBNER BASES

Let f1, . . ., fr, 9 E k[m1,...,mn]. The subalgebra membership problem is to decide whether or not 9 E k[f1, . . . , fr]. We will solve this and further— more, if g E k[f1, . . . , fr], determine a polynomial h(y1, . . . ,yr) such that g =h(f1,...,fr).

PROPOSITION 4. Let f1, . . . , fmg e k[x1, . . . ,mn] and consider the ideal b = (y1 — f1, . . . ,yr — fr) in Man, ...,:cn,y1,...,yr]. We use an ordering ofthe

monomials in k[zl, . . . ,xn,y1,... ,yr] such that each am is larger than every monomial in {y1, . . . ,yr}. Then 9 E k[f1, . . . ,f,] if and only if the normal form of g with respect to a Grb'bner basis of b belongs to k[y1, . . . ,yr]. If so, and the normal form ofg is h(y1, . . . ,yr), then 9 = h(f1, . . . ,fr). Proof. Consider the homomorphism ¢ 2 k[$l)"'1$nay19- ' '1y1‘] _-) k[$1,.. -,$n]

defined by ¢(f($1,---,$n,y1,...,yr)) =f($17"'1$n:.f11"'1f1')'

It is clear that ker(¢) = b. Suppose g E k[f1,...,fr], say 9 = h(f1, . . .,fr). Then 9 — h(y1, . . . ,yr) 6 b, which gives that NF(g) = NF(h(y1,...,y,)). But every polynomial in k[y1, . . .,y,] will, in the ordering we have cho-

sen, reduce to polynomials in k[y1, . . . ,yr], so NF(g) E k[y1, . . . ,yr]. Since .9 _h(yla"'7y7') E b we get 9 = (15(9) : ¢(h(y1:mayr)) = h(f11'--vf1‘)'

Conversely, if NF(g) = h(y1,...,yr) we have 9 — h(y1,...,yr) E b, so

¢(g-NF(9))=g—h(f1,---,fr)=0-

El

EXAMPLE 5. We will show that z8 — m2 E k[a:4 + 9:2, 2:3]. Consider the ideal b = (y1 —— x4 — $2,312 — $3) in k[x,y1,y2]. The Grébner basis for b in the Lex ordering with :1: > yl > 312 is

G = {rt2 + 931/2 - y1,xy1 + 9:113 — yiyz — 312, $213 - xyz +21? - y1y§ - 2313,11? - 3111113 - 11%— .113}The normal form of x8 — m2 with respect to G is y? — 231% — yl. Thus as — 2:2 = (51:4 + :22)2 — 2(:1:3)2 — (2:4 + x2) 6 k[a:4 +m2,a:3]. El

A polynomial f (931,232, . . . ,xn) is called symmetric if any permutation of the variables leaves f fixed. Examples of symmetric polynomials are the

7.2. CALCULATION IN FACTOR RINGS OF POLYNOMIAL RINGSQ3 following, called the elementary symmetricfunctions: 0’1

=

0’2

=

$1+$2+"'+$n E

(Bittj

15i k:[2:1, . . .,a:n] where ¢(yi) = fi. Hence k[f1,...,fr] 2 k[y1,...,y.,]/ ker(¢) and ker(¢) is the intersection of the ideal (yl — f1, . . . ,yr — fr) in k[a:1,...,xn,y1,...,yr] with k[y1,...,yr], i.e., ker(¢) = (y1 — f1, . . . ,yr — fr) 0 k[y1, . . . ,yr]. Hence we can determine the presentation of k[f1, . . . , fr] as a factor ring of k[y1, . ,yr]. Consider for

example k[x3,x5,z7]. We determine b =(y1 — $37192 - $5,313 _ 1:7)0 klyliy27y3]

by calculating a Gr6bner basis for b in Lex with a: > 111 >- 112 >- 7,13. The result is

{m3 — 2J1, $2.141 — y2, xys — y1y2,xyi- 213,113 - yiy2,y2y3 — yi‘, yiys — 113,113 - 31?} Hence

k[w3, $5,217] 1' k[y1,y2,y3]/ (y? - yiy2,y2y3 — yiyiya — 213,113 - y?)-

96

CHAPTER 7. APPLICATIONS OF GRGBNER BASES

Note that we do not get a minimal generating set for the ideal since

343 - 11?: -y2(y1y3 — yi) + y1(y2y3 — 111‘). so ya — y? is not needed as a generator.

El

Exercise

1. Write k[:1:3, :37, 3:8] as a factor ring k[y1, y2,y3] /a. Determine a minimal set of generators for a.

7.4

Ideal Operations

Given two ideals a = (f1,...,fr) and b = (91,...,gs) we want to get a generating set for a + b, a - b, a n b, and a : b.

For the sum and the

product there are no problems: a + b = (f1,...,fr,g1,...,gs) and a- b = (f1g1, . . . , frgs). The intersection and quotient of ideals are, however, not equally easy.

7.4.1

Intersection of Ideals

We will reduce the problem of calculating the intersection of ideals to an elimination problem by adding a new variable. Hence we can use Proposition 8 to solve the problem of getting a generating set for the intersection

of two ideals.

PROPOSITION 10. Let a = (f1,...,f,) and b = (91,...,gs) be ideals in k[w1, . . . ,zn]. Consider the ideal

c=(yf1,--.,yfr,(1-y)91,---,(1-y)gs) in k[:1:1,...,:rn,y]. Then aflb=cflk[:z:1,...,:1:n]. Proof. Suppose f 6 an h. Then yf 6 ya and (1 —y)f e (1 — y)b so

f=yf+(1-y)f€ya+(1-y)b=c Suppose f E cflk[w1,...,wn]. Then f: h1($la"',$n7y)yf1($1>"-7xn) +"'+hr($1y"'7$nyy)yfr($la---axn)+

hr+1($l;---,$my)(1 _y)gl($11”'!xn) + ”'+

7.4. IDEAL OPERATIONS

97

hr+s($1; - - - )wna y)(1 _ y)gs($1) ' ‘ ' ) $n)‘

Since f 6 Han, . . . ,xn] we get f = hr+1(1'1,...,$n,0)gl(.’l§1,...,$n)+"‘+

hr+s(z1,...,xn,0)gs(x1,...,a:n) E b

and f = h1($1,...,$n,1)f1($1,...,$n) + "'+

hr(:c1,...,:rn,1)fr(:c1,...,a:n) 6 a. El

EXAMPLE 11. Let a =(:1:2 +y2,:cy) and b =(.17:2 — yz). Let c = (xzt + y2t,:1:yt,ta:2 — ty2 — 2:2 + (112). A GrObner basis for c in Lex with t >- as >- y is

{2z2t — 2:2 — y2,:z:yt,2y2t + $2 + 312,223 — $312,193; — 3/3}. Hence a n b = ($3 — $112,323; — ya).

El

Exercise

1. Determine (11:3 + :c,:vy + yz) n (y3 + 11,221; + 2:2) in k[a:,y]. 7.4.2

Ideal Quotient

Let a be an ideal and b = (91,. . . ,gs); see Proposition 45.3 in Section 1.3.5. Then a : b = 0: 1a : (91»). Hence it is sufficient to consider the case when b is principal, since we know how to perform intersections.

PROPOSITION 12. Ifan (g) = (h1,...,ht) then a: (g) = (hl/g,...,ht/g). Proof. Let f 6 a : (9). Then fg 6 (10(9). Let h’ 6 an (9). Then h’ = gh

for some h and h 6 a : (9). Thus h’ 6 (10(9) ifand only ifh’/g E a : (g).

E!

EXAMPLE 13. Let a = (2:2 +y2,a;y) and b = ($2 —y2). Since a n b = (2:3 -— $312,293; — y3) we get a: b = (2:,y). Exercise

1. Determine (x3 + x,my + 3/2) : (313 + y,my + :52) in k[:c,y].

El

CHAPTER 7. APPLICATIONS OF GROBNER BASES

98

7.5

Supplementary Exercises

1.

Show that 11:53; — 2234112 e (2:313 — 22:31; + 3102312, $3312 + 5133/3).

2.

Show that 2:10 + 1:16 E Z2[:c4,a:6 + 977,1:10].

3. Show that

(/(32 —x+xz—z,:ry—y+y2,y+yz+z2 +22+1) = (m— 1,y,z+ 1).

Let b = (my — y,:rz —:L',yz -— 2). Determine bfl k[:v,y] and b n k[:1:]. Let R = k[a:1,...,xn]/ (91,...,gs) = k[2:1,...,xn]/a. Define a homomorphism ¢ : k[y1, . . .,yr] —) R by ¢(yi) = fi+a. Show that ker(¢) =

((1/1 -f1,---,yr-fr)+a)flk[y1,...,yr].

Show that R = Q[a:]/ (a:3 + a: + 1) is a field and a vector space of dimension 3 over Q. Show that any element a E R is a root of a polyno-

mial equation of degree at most 3 with coefficients in Q. (Hint: Since dimQ R = 3, we get that 1, a, a2, a3 are linearly dependent.) Determine an equation for (the image of) :1: + 1.

. Write k[:1:4 + $2,053] as k[yl,y2]/a-

. Write k[:c4, x3y,a:y3,y4] as a factor ring of k[y1, y2,y3,y4]. . Show that (y2 — $2,312 - a: 3,22 —m2y) is a prime ideal in k[x,y,z]. Hint: Write k[t3, t4, t5] as a factor ring of k[:1;,y, z].

Chapter 8

Homogeneous Algebras We will in this chapter study the class of ideals in a polynomial ring which

are generated by homogeneous elements (and their factor rings). We will show that this class is closed under the usual operations on ideals and that

the primary components of such an ideal can be chosen to be generated by homogeneous elements. In the second part of the chapter we will study the

connection between an arbitrary ideal and its “homogenization”, which is an ideal generated by homogeneous elements in a ring with one more variable. Finally we will study Grobner bases of ideal, generated by homogeneous elements.

8.1

Homogeneous Ideals and Algebras

Let A = k[:1;1, . . .,:Un]. An element f E A is called homogeneous if f is a linear combination (with coefficients in k) of monomials of the same degree.

Thus .733 + 2:123;2 is homogeneous but 3:3 +2arsy2 —:::y is not. The homogeneous elements of degree 2' in A constitute together with {0} a vector space A,over k: with the set of monomials of degree 1' as a basis. Furthermore AiAj g AiH, i.e., a product of two homogeneous elements of degree 2' and j, respectively, is homogeneous of degree 1' + j. Every element f E A \ {0} can be uniquely written as f = f0 + f1 + ' - - + fk, where fi 6 Ai. The fi’s are called the homogeneous components of f. Another way to express this is to say that A is the direct sum of the Ai’s, A = GBiZOAi; see Lemma 79

in Section 1.9. An ideal a g A is called homogeneous if a has a generating set consisting of homogeneous elements. EXAMPLE 1. If a is generated by monomials, then a is homogeneous.

99

D

100

CHAPTER 8. HOMOGENEOUS ALGEBRAS

LEMMA 2. An ideal a is homogeneous if and only if the following condition holds. If F e a then every homogeneous component of F belongs to a.

Proof. Suppose that a = (f1,..., fk) with f; homogeneous for all i and let F E 0. Then F = g1 f1 + + gkfk. Let Fj be the homogeneous

component of degree j of F. Then F]- = gifdeg f1)f1 + + gig—deg mfk where 9” deg f‘) is the homogeneous component of degree (j — deg fi) of 91-. Hence F,- e a. For the converse let a be generated by h1,. . . ,hk. Each homogeneous component hgj) of each h,- belongs to a. Then a is generated

by the hgjl’s.

u

REMARK. Another way to express the statement of the lemma is the following. Let a be a homogeneous ideal and let ai be the k-vector space of homogeneous polynomials of degree i in a. Then a = 691-3004.

DEFINITION 3 (HOMOGENEOUS ALGEBRA). A ring R = k[a:1, . . . ,xn]/a is called a homogeneous algebra if a is homogeneous.

We will now see that there exists in a homogeneous algebra a direct sum decomposition in homogeneous components, much like in the polynomial ring.

LEMMA 4. If R = k[a:1,.. .,_:::,.]/a is a homogeneous algebra then any element in R can be written f: f0 + -+ fk, Where fi is homogeneous of degreez' or zero, and where 1‘: is unique Equivalently, R: EBRi where R,-

is the image of Ai in the natural map A —) A/a = R. On the other hand, ifR = k[z1, . . . ,mn]/a and R = @Ri Where Ri is the image ofAi, then a is homogeneous.

Proof. Since any f_ E A can be written f= f0 + -+ fk with fi 6 A.-, then any element_f E R can be written f: f0 + + fk, fi 6 A-. Suppose

f— —- fo+

+fk =yo+

+gk— — 9, SO fo-go+~ +fk-gk =0

This means that f —g = (f0 —go) +

+ (fk —gk) 6 (I. But 12- —g,- is

the homogeneous component of degree 1' of f — 9, so fi — g E 0. since a is homogeneous. Hence fi — g = 0, so f”; = g]. Now suppose that R = 69R.where Ri =imAi. This means that f0 +---+fk = fo+---+fk = Oifand

only if f1- = f) for every 1', i.e., f0 + - ~ + fk E a if and only if fi 6 a for all i, i.e., if and only if a is homogeneous.

El

COROLLARY 5. Let a g b be homogeneous ideals in Ha, . . .,a:n] and let R = k[:L'1,...,:vn]/a = EBQORt-“S' = k[1:1,...,a:n]/b = 691-205}. Then St- 2

¢(Rl~), where ¢ : R —> .S' is the homomorphism defined by ¢(f + a) = f + b.

8.1. HOMOGENEOUS IDEALS AND ALGEBRAS

101

Proof. That ¢(R,-‘) = s.- is clear. If f1 + a = f2 + a then f; — f2 6 a g b, so f; + b = f2 + b. Hence ¢ is well defined, and it is easy to check that d) is a homomorphism.

I]

The set of homogeneous ideals behave well under ideal operations.

LEMMA 6. If a and b are homogeneous ideals in k[:1:1, . . . ,zn] then a + b, ab, (1 n b, a : b, and J5 are homogeneous.

Proof. If a and b are generated by homogeneous elements then a + b and ab are generated by homogeneous elements. Suppose that any element in

a has all its homogeneous components in a and that the same is true for h. Then any element in an 6 has all its homogeneous components in a n b, so

an b is homogeneous by Lemma 2. If h = (fl, . . . , fk) with f,- homogeneous then a : b = (a : (f1)) n n (a : (fk)), hence it is sufficient to assume that b is principal. If a n (f) = (91,...,gm) with g,- homogeneous then a : (f) = (gl/f, . . . ,gm/f), cf. Proposition 12 in Section 7.4.2, and gi/f is homogeneous. Finally, let F = Fi, + + F5... 6 «El-,1?“ 6 1413,11 < - - - < im. Then F” e a for some n. The homogeneous component of lowest

degree in F" is F,’:. Since a is homogeneous we have F]: E a, so E, 6 J3.

Hence F — Fi, = F}, + - - - + Fin, 6 J5 and we are ready to use induction on m.

I]

The following criteria are useful when checking whether a homogeneous ideal is prime or primary. LEMMA 7. Suppose that a is homogeneous. 1. If ab 6 a, a ¢ a implies that b E a for homogeneous elements a, b, then a is a prime ideal.

2. If ab E a,a $4 (1 implies that b" E a for some n for homogeneous elements a, b, then a is a primary ideal.

Proof. 1. Suppose that fg E awhere f = fo+- ~ -+f;c andg = go+- - -+gm with f,, g,- 6 Ai. Suppose that f y! 0. Then f,- ¢ a for some i, let 2'0 be the smallest such 2'. The homogeneous component of degree i0 of fg is

fogio + - - - + fro go. Since fg E a and a is homogeneous we get that this component belongs to a. Since all terms except possibly fie go belong

to a, we get that fiogo, and thus g0, belongs to a. Then f (g — go) 6 a and we can repeat the same argument to get 91 6 a. We are ready to use induction on m.

102

CHAPTER 8. HOMOGENEOUS ALGEBRAS

2. Suppose fg E a where f = f5. +--'+fi;e and g = gj1 + ---+g,-, with fini homogeneous components. Suppose f ¢ (1. If fil E a then

(f -— fi1)g E u,f — fil ¢ a, so we can just as well assume that fil ¢ (1. If we can show that for each i there is an mj such that 9;“ e a if m = max{m,-}. The homogeneous component of lowest degree in fg is lgjl. Since fg E a and a is homogeneous we have filgj1 E a. Since fl'nl ¢ 0 we have gjm1 6 a for some m1. Now suppose we have proved

g].1 ,. . .,gj’:“ Ea. LetM: max{mj1,...,mj_,}. Consider

F = f(gj.+1+---+gj.)” = f(y—(gj1+---+gj.))” =

-L>(1))" '9‘(gj.+ ---+gj,)"-i= fZC i=0

(—1>"f(g.-.+ +gj..)“+f§::(:.L(—1))"““(ggl ~+g,-,)n-i. If we choose n = s(M—1)+1, the first term belongs to (gJJ-‘l", . . . ,gff) g a. The remaining terms belong to a since fg E (1, thus F E a. The homogeneous component of lowest degree in F is fil 91..+1 and this component belongs to a since a is homogeneous. Thus gj+ "1““ E a for some

mj5+1 and we are through by induction. El

REMARK. There is only one homogeneous maximal ideal in k[a:1, . . . ,zrn], namely 9)? = ($1 , . . . , :12”) consisting of all homogeneous elements of positive degree. All homogeneous ideals, except (1), lie in am. NOTATION 8. For any ideal a in k[$1, . . . ,xn] we denote by a* the largest homogeneous ideal in a, or similarly, the ideal generated by all homogeneous elements in a. We will now investigate to what extent properties of an ideal a are preserved in a*. LEMMA 9.

1. Ifu g b then a* g b".

2. «a: = fi*. 3. If p is prime then 33* is prime. 4. If q is primary then (1* is primary.

Proof.

1. This is obvious.

8.1. HOMOGENEOUS IDEALS AND ALGEBRAS

103

2. Since a* g a we have x/EIT g J5 but x/a_* is homogeneous, see Lemma 6,

and J? is the largest homogeneous ideal in VS, so x/a—* g Vial. Let

F E VT. Since J? is homogeneous, all homogeneous components of F belong to VT.

Hence all homogeneous components Fij of F

have a power F5?” in a. If F has It homogeneous components, F =

m, + - --+F,—,, and m = max{m,-J. }, then Fk 1wehaveq,-:(g)=(1)ifi> 1,soa: (g)=q1 : (g) whichisimprimary. We consider the set of ideals a : (g) with g homogeneous such that a : (g) is SLR—primary. We have just proved that this set is nonempty, so this set has a maximal element, say a : (f); see Proposition 13 in Section 3.5.

We claim that a : (f) is a prime ideal. To prove that a : (f) is a prime ideal, it suffices to prove that if :r and y are homogeneous elements such that my 6 a: (f) ,x gt a: (f), then 3; e a : (f); see Lemma 7.1. Suppose my 6 a : (f),:r ¢ a : (f) and $,y homogeneous. Then y E a : (fzt) and a : (fat) aé (1) since f2: ¢ a. Since a : (fat) is homogeneous (Lemma 6), we have a : (fat) g 931. Since DRQ\/a:(z)§\/a:(fx)g\/fi=im, so x/a : (fit) = m, so a : (fr) is SDI-primary by Proposition 17 in Chapter 5. But a:(f:1:)= a : (f) since a : (f) g a : (fit), fzc is homogeneous, a : (fz) is EDI—primary, and a : (f) is maximal. Thus 3; E a : (f) and a : (f) is a prime ideal. But then a : (f) = x/a: (f) = 9R, which gives f9)? g a. For the factor

ring R this means f-fi = 0. Now suppose that 9)? does not belong to a. By Theorem 15 in Chapter 5 it suffices to show that there is a homogeneous element in WI which does not belong to 13; U - - ~U p 3. Suppose the contrary,

that m g in U U 1),. By deleting some pi if necessary, we can assume that an g Ui¢jpi for every j. Take for each j a homogeneous element 17,- E ‘11i ¢ Uifipi. Let degml = d1 and degzzmxs = d2. Consider

the element 27‘:2 + (x2 - - - $5)d1. This is a homogeneous element (of degree d1d2). Since 2:1 ¢ Ui¢1pi and all pi’s are prime ideals, we have at?2 ¢ Ui¢1p,. Since 221 E Uf=1pi we must have x1 6 p1, and thus m‘f‘ 6 331. As above {122 E p2,...,:1:s 6 33;. Thus $2"'.’l}s E p,- lfi > 1, SO ($2-"$s)d1 E p,- if

t> 1. As above $2,...,xs ¢ p1, so (ar2~--:t,,)dl ¢ p1 and 531:2---.1c$)d1 E 33,-

if i > 1. We claim that the homogeneous element y = 1:12 + (1132 - - - 1:8)‘11

8.2. HOMOGENIZING AND DEHOMOGENIZING

105

does not belong to any pi, since ify 6 p1 then 31—:s2 = (2:2 ---a:s)d1 6 m, a contradiction, and if y E p.- for some i > 1 then y — (m2 - - - 233)“l1 = z‘fz 6 pi, a contradiction. Finally suppose that k is an infinite field. For each z' we have that the elements of pi of degree 1 constitute a k-subspace Vi of the

space V of all elements of degree 1, and that Vi 96 V. If all elements of degree 1 belong to some pi, we get V = Uf=ll/,~. But a vector space over an infinite field cannot be a union of a finite number of proper subspaces. El Exercises 1. Show that a vector space over an infinite field cannot be a union of a finite number of proper subspaces. Show that a vector space over a finite field can be a union of a finite number of proper subspaces.

2. Let R = Z2[m,y]/ (11:23; + 2:312). Show that there is no nonzerodivisor in R which is homogeneous of degree 1.

Determine a homogeneous

nonzerodivisor of degree 2.

8.2

Homogenizing and Dehomogenizing

For each polynomial (or ideal) in k[:1:1, . . .,:1:n] there is a natural way that to construct a homogeneous polynomial (or homogeneous ideal) in a poly—

nomial ring k[a:1, . . . , zn, y] with one more variable.

8 .2 . 1

Homogenizing Polynomials

Let f = f0 + ---+ fk e k[a:1,...,:cn] = A, f;- homogeneous of degree 2',

fk 9E 0. DEFINITION 13 (HOMOGENIZATION OF A POLYNOMIAL). The homogenization of f with respect to y is

h(f) = fol/k + fiyk—1 + ' ' ' + fie—1y + fk E k[$1,«--,$my]Thus for example h(1 + mm + $3) = y3 + 11:11:23; + :33. It is practical to have a formula for homogenizing.

LEMMA 14. We have h(f(:1:1,...,xn)) = ydegff(a:1/y,...,zn/y). Proof. Iff=fo+---+fk,fk760,wehave

W) = foy’c + - - - + fk = y"(fo + fl/y + - - - + fie/y") and it is easily checked that fi/yi = f,-(:1:1/y....,:rn/y).

El

106

CHAPTER 8. HOMOGENEOUS ALGEBRAS Homogenization works well with multiplication.

LEMMA 15. We have h(fg) = h(f)h(g). Proof. We have deg(fg) = deg f + deg 9, so

h(fg) = ydegf+deggfg($1/y, - - - ,In/y) = ydegf+deggf(w1/y, - - - ,xn/y)g($1/y, . . .,=cn/y) =

ydeg’f($1/y,~- -,xn/y)yd°“g(w1/y,-.-,xn/y) = h(f)h(g)El

Homogenization works less well with sums, unless deg f = deg g =

y2 + $1$2,h(9) = $131 — $1$2,h(f +9) = 3; +931 75 h(f) + My)-

Elll

EXAMPLE 16. If f = 9:1,g = 22% then h(f) = f, h(g) = g, and h(f +g) any + at? ¢ h(f) + h(g). If f = 1 + 33132,!) = 2:1 — 21:52, we have h(f)

||

deg(f + 9).

LEMMA 17. We have h(f + g) = ydeg(f+y)—deg fh(f) + ydeg(f+9)-deggh(g). In particular, ifdeg f = degg = deg(f+g), we have h(f+g) = h(f) +h(g). Proof. We have

h(f + g) = yd°g(f+9’(f(z1/y,- - -,wn/y) + 9(x1/y, - - - ,wn/y» = ydeg(f+y)-deg fydeg ff(:c1/y, . . _ ,IIJn/y)+

ydeg"+g)‘deggydeggg(x1/y, - - - , run/y) = ydeg(f+9)-deg fh(f) + ydeg(f+g)—deg ”h(g).

Exercise

1. Let f (9:) E k[x]. irreducible.

Show that f is irreducible if and only if h(f) is

8.2. HOMOGENIZING AND DEHOMOGENIZING

8.2.2

107

Homogenizing Ideals

DEFINITION 18 (HOMOGENIZATION OF AN IDEAL). Let a be an ideal in A = k[:1:1,. . .,a:n]. Then h(u) is the homogeneous ideal in A[y] which is generated by all h(f), f E a. For principal ideals a = (f) it is easy to determine h(a). PROPOSITION 19. Ifu = (f) then h(a) = (h(f)).

Proof. It is clear that (h(f)) g h((f)). Ifg E (f) then 9 = fgl for some

$513131 h(9) = h(fgl) = h(f)h(91) 6 WW» by Lemma 15, hence h((f)) Q

hf .

[3

EXAMPLE 20. If a = (f1,...,fk) then (h(fl),...,h(fk)) g h(a) but there

might be strict inequality. If a = ($1172 - 1,2:f — 9:2) then 53% — mly E h(a) but .733 — any ¢ (x1232 — yam? — 31:52).

El

Exercises

1. Show that if a = (11:32 -— 1,:1;¥— 2:2), then 1:3 — any 6 h(a) but 33% —

ziy ¢~ mn > y is

5. Determine the reduced Grobner basis in Deglex with :1: > y > z of (my — 2,232 — y,yz — x) by determining the reduced Grobner basis in

Deglex with a: > y > z > t of (my - zt,mz -— yt,ya: — (at). 6. Let a = (f1,...,fk) be an ideal in k[:c1,...,:1:n]. Show that if G = {91, . . . ,gs} is a Grobner basis of (h(f1), . . . ,h(fk)) in Degrevlex with 2:1 > > m" > y, then {h(a(gl)), . . . ,h(a(gs))} is a Grobner basis of h(a) (not necessarily reduced even if G’ is reduced).

Chapter 9

Projective Varieties We have seen that extending the field from R to (C greatly simplified the correspondence between geometric objects (affine varieties) and algebraic objects (ideals in or factor rings of polynomial rings). We are now going to introduce a new kind of geometric objects, projective varieties, which

will make many geometric statements more clear or “symmetric”. (We give some examples to show what we mean below.) This will be achieved by adding to the affine space k" points “at infinity”. We start by describing

the projective line P1 (R), which is obtained by adding one point 00 to IR. We give the elements in P1(lR{), which we call points, coordinates in the following way. Define an equivalence relation on the set R2 \ {(0,0)} by (9:,y) ~ ($1,311) if (:c,y) = (tx1,ty1) for some t 75 0. Then P1(]R) is the set of equivalence classes. We will write points in P1 (R) as (m : y), where (1:, y) is a representative for the class (a: : 3;). Thus (1 : 2) = (3 : 6). If one representative (3:, y) for a class has y gé 0, then every representative for this class has second coordinate ;£ 0, and there is a unique representative for

the class with second coordinate 1. We identify 1' 6 R with the class (1‘ : 1). The only remaining class is (1 : O), which we call the point at infinity. We are now ready for the definition of P”(k). DEFINITION 1 (THE PROJECTIVE SPACE). Let k be a field. Then P”(k) consists of the equivalence classes of elements in kn+1\{(0,0, . . . , 0)}, where the equivalence is defined by

($1,---,$n+1) ~ (y1:---1yn+1) if ($1:~--:-’13n+1) = (ty1,--«,tyn+1) for some t 96 0.

We write the equivalence classes (2:1 :

: $n+1) where (:31, . . . ,xn+1)

is a representative for the class. We identify (3:1, . . . ,wn) E k" with (2:1 :

117

118

CHAPTER 9. PROJECTIVE VARIETIES

: xn : 1) E P"(k), and the remaining points in P"(k), i.e., those with last coordinate 0, are called the points at infinity.

As an example P2 (R) is R2 extended with one point (a : b : 0) at infinity for each line through the origin with direction (a, b) 76 (0,0). Note that, since (a : b : 0) = (—a : —b : 0), each line contributes only one point at infinity, travelling along a line one approaches the same point at infinity in

both directions. EXAMPLE 2. Let f(a:,y) E 1R[x,y] and consider the affine curve h(f) =

{(0,17) 6 IR2;f(a,b) = 0}- Let h(f) = Zdegfflx/zw/Z) E R[$,y,z} be the

homogenization of f. Since h(f) is homogeneous, we have that

h(f)(ta. tb, to) = tdegfhuxa, b. a). so if h(f) is zero on a representative ((1, b, c) for a point (a : b : c) in P2(IR), then h(f) is zero on every representative for (a : b : 0). Thus the statement

that h(f) is zero on a point in P2(]R) is well defined. We call the set of points in P2 (R) for which h(f) is zero the projectivization of the curve

1](f)-

1:

EXAMPLE 3. A line I in R2 has an equation Ax+By+C = 0 with (A, B) 96

(0,0). The projectivization of l is the set of points in P2(]R) satisfying the equation A2: + By + C2 = 0. It is natural to make the following more

symmetric definition. A projective line in P2(]R) is the set of points in

P2 (R) satisfying an equation Ax + By + Cz = 0 with (A, B, C) 76 (0,0,0). Besides the projectivizations of afline lines there is only one more, namely 2 = 0. This line consists of all points at infinity, and is thus called “the

line at infinity”. If Ax + By + C = 0 and A13: + Bly + 01 = 0 are two nonparallel affine lines intersecting in the point (a, b), then the projective lines A1: + By + 02 = 0 and A1w+Bly +012 = 0 intersect in the projective

point (a : b : 1). If Az+By+C = O and Ax+By+Cl = 0 are two parallel lines, and thus do not intersect, then the projective lines A3: + By + 02 = 0

and A1: + By + 012 = 0 intersect in the point (B : —A : 0) at infinity. The intersection between A1: + By + 02 = 0 and z = 0 is also (B : —A : 0) (if (A, B) 76 (0,0)). Thus any pair of projective lines intersect in one projective point.

El

EXAMPLE 4. The affine hyperbola my = 1 in 1R2 corresponds to the pro-

jective curve my = 22. We have added the two points (0 z 1 : 0) and (1 : 0 : 0) corresponding to the two directions of the asymptotes to the affine points. The affine parabola y = $2 corresponds to the projective curve yz = $2, which has one point (0 : 1 : 0) at infinity. The affine

circle .732 + y2 = 1 corresponds to the projective curve .752 + y2 = 22

119 which has no point at infinity. Note that if we introduce new variables

by X = 2:,Y = z — y,Z = z + y, the equation becomes X2 = YZ. If we now forget that we have used one variable for homogenization and treat all variables equally, we see that there is no difference between the three

projective curves. The afline curves just describe different aspects of the I]

same projective curve.

The theory of projective sets runs in parallel to the theory of affine

algebraic sets. Let P = (c1 : : on“) E P“(k). If f is a homogeneous polynomial such that f(c1, . . . ,cn+1) = 0, then f(tc1, . . . ,tcn+1) = 0 for all t E k. We say that f vanishes on P or that P is a zero of f. DEFINITION5 (IDEAL OF A SET). If X g P"(k) then 3(X) denotes the homogeneous ideal, generated by all homogeneous polynomials, which vanishes for every c E X.

DEFINITION 6 (PROJECTIVE SET). If a is a homogeneous ideal of the po— lynomial ring k[a:1, . . . ,a], then the set of all projective points that are

zeros of all polynomials in a is denoted ‘B(a) and is called the projective set or variety defined by :1.

EXAMPLE 7. Ifa = ($1,...,a:n+1) then ‘13(a) = (2). Therefore (2:1, . . .,xn+1) is often called the irrelevant maximal ideal. More generally, if a is a homogeneous ideal with \fd = (1:1,. .. ,zn+1) then ‘B(a) = (2), since for each i we have 51:1.“ 6 a for some positive m, hence (0,0,. . . ,0) is the only common

zero, and (0,0, . . . ,0) does not give a point in P"(k).

El

PROPOSITION 8. (a) IfX1 g X2 _C_ P"(k) then 3(X2) g 3(X1). Ifa g b are homogeneous ideals in k[a:1,...,xn+1] then was) Q 11(a). (b) For any subset X g P"(k) we have X g m(3(X)), with equality if and only if X is a projective set.

(c) Ifa is a homogeneous ideal in k[a:1,. . .,$n+1] we have m(a) = ‘ZIfi/E). (d) If a and b are homogeneous ideals in Man, . . .,xn+1] then 2101 n b)

=

Q30!) U ‘Il(b),

‘13(ab)

=

Q3(a) U ”(6),

‘Il(a+b) = momma). (e) Ifa is a homogeneous ideal in M11, . . . ,zn+1] then a Q 3(Q3(a)) and the inclusion may be strict. Proof. Parallel to the proof of Proposition 6 in Section 4.1.

CI

120

CHAPTER 9. PROJECTIVE VARIETIES

DEFINITION9 (IRREDUCIBLE). A projective set X g P"(k), k a field, is said to be irreducible if X is not the union of two strictly smaller projective sets. A projective set which is not irreducible is called reducible.

PROPOSITION 10. Let X be a projective set in P"(k). Then X is irreducible if and only if 3(X) is a prime ideal. Proof. Parallel to the proof of Proposition 8 in Section 4.1. THEOREM 11. Let X1 3 X2 3

U

be a strictly descending sequence of

projective subsets of P"(k). Then the sequence is finite. Another way to state this is to say that any nonempty set {X0} of projective subsets of P"(k) has a minimal element, i.e., an element Xa0 not containing some other Xa.

Proof. Parallel to the proof of Theorem 9 in Section 4.1.

D

THEOREM 12. Every projective set in P"(k) is a finite union of irreducible projective sets, X = X1 U - - - U Xm, X.- irreducible. If we make this union irredundant, i.e., ifno X,- is contained in some Xj, j gé i, then the irreducible components are unique. Proof. Parallel to the proof of Theorem 10 in Section 4.1.

E]

THEOREM 13 (HILBERT’S PROJECTIVE NULLSTELLENSATZ, WEAK FORM). Let a be a homogeneous ideal in (C[:c1 , . . . ,wn+1], with W? strictly contained

in (11:1, . . . ,$n+1)- Then ‘13(a) 75 (0. Proof. Consider the affine algebraic set 0(a) in A"+1((C). As an affine algebraic set, 0(a) is called the afl‘ine cone of the projective set 23(a). The afiine Nullstellensatz, Theorem 11 in Section 4.2, shows that 0(a) is nonempty. Since \/E 76 (2:1,...,a;n+1), n(a) does not consist of only (0,0,...,0), so Q3(a) is nonempty. El COROLLARY 14 (STRONG PROJECTIVE NULLSTELLENSATZ). For any homogeneous ideal a Q C[m1,...,a:n+1] with J5 strictly contained in the

ideal ($1,...,$n+1) we have 3(m(u)) = J5. Proof. As in the proof of the affine case, Corollary 13 in Section 4.2, we

have fl g 3(‘13(\/E)) = ammo). Suppose that f = fo+-~+fs E 3(‘B(a)), fi homogeneous of degree 2'. Then f(:c1,...,xn+1) = 0 for every ($1 :

:

$714.1) 6 3(‘B(a)), so f(ta:1,...,tzn+1) = fo +tf1(a;1,...,:1:n+1) +

+

tsfs(x1, . . .,xn+1) = 0 for every t E (C. Since (C is infinite, this polynomial in it must equal the zero polynomial, hence fi(a:1,. . .,a:n+1) = 0 for every

2'. Thus 301101)) : i(n(a)) and the claim follows from the affine case.

El

9.1. PROJECTIVE CLOSURE OF AN ALGEBRAIC SET

121

COROLLARY 15. IfX is a projective set in P"((C) then m(3(X)) = X. If a = J5 and is strictly contained in ($1, . . .xn+1) then 3(m(a)) = a. Proof. Parallel to the proof of Corollary 14 in Section 4.2.

D

Exercises

1. Prove Proposition 8. Use the result that the set of homogeneous ideals are closed under the usual ideal theoretic operations. 2. Prove Proposition 10 using Lemma 7 in Section 8.1.

9.1

Projective Closure of an Algebraic Set

Let V be an algebraic set in A"((C). Then V can be considered as a subset of P"((C) via the map d) : A"(-1. Then m(a) = (m(a(gl)), . . . ,m(a(gt))). Proof. Since gz- E h(a) we have a(g,-) E a(h(a)) = 0. Thus we have the inclusion (m(a(gl)), . . . ,m(a(gt))) Q m(a). To show the other inclusion, let f e a. We must show that m(f) E (m(a(gl)),...,m(a(gt))). Let f = m(f) + be the decomposition of f in homogeneous components. Then h(f) = m( f):c§+1 + for some d, and because of the ordering we have chosen, we have lm(h(f)) = 935‘,“ lm(m(f)). Since {91, . . . , 9;} is a GrObner basis of h(a), we have that lm(gi) divides lm(h(f)) for some 1'. But lm(gi) = :32“ lm(m(a(g,~))) for some 6. Hence lm(m(f)) = cnlm(m(a(g,-))) for some monomial n and some 0 E Is. Now consider h = f — cna(g,-) E a.

If deg(m(h)) > deg(m(f)), we have m(f) = m(cna(yi)) = cnm(a(gi)) E

(moon), Hammer)». If deg = degmo», we get lm(m(h»
Gfi(R) as follows. If f is homogeneous of degree k, then f Emkand we let ¢(f) = f + 111*“. If f: fo + f1+ -is the decomposition of f in homogeneous components,

we set ¢(f) = ¢(fo) + ¢(f1) +

= (¢(fo),¢(f1),---)- If f E ker(¢), then

all homogeneous components of f belong to ker(¢). If f is homogeneous of degree k: then ¢(f) = f+fik+1, so ¢(f) = 0 if and only if f + g E a for some 9 E Eh“, thus ¢(f) = 0 if and only if f = m(h) for some h E 11. Since ker(¢) = m(a), we have Gfi(R) 2 k[a:1,...,:1:n]/m(a) by the homomorphism theorem, Theorem 61 in Section 1.7, the map ¢ being surjective. I]

DEFINITION 7 (TANGENT CONE). Let V be an algebraic set in A”(k) such that (0,. ..,0) E V. The tangent cone of V at the origin is the algebraic set C(V) = u(m(i(V))). REMARK. Since m(i(V)) is a homogeneous ideal, C(V) consists of lines through the origin. Such a set is called a cone at the origin. In the same

way as a tangent to a. curve is the limit of secant lines, one can Show that C (V) consists of all limits of secant lines in V through the origin. If V is given by equations, one likes to be able to describe C(V) with equations. This is possible if V g A"((C) (we need the Nullstellensatz). THEOREM 8. Let a g (C[:1:1,...,a:,,] with (0,...,0) e 0(a) = V. C(V) = n(m(a)).

Then

126

CHAPTER 10. THE ASSOCIATED GRADED RING

Proof. We have C(V) = n(m(\/E)) g n(m(a)) since m(a) Q m(fi». Suppose that f 6 «H. We must show that m(f) is zero on n(m(a)). We have

f" E a for some k, so m(f") = (m(f))k is zero on n(m(a)), hence m(f) is zero on n(m(a)).

D

Exercises

1. Determine m(a) if a = (a: + 112,911; + 2:3). 2. Show that the multiplication in G'u (R) is well defined and that Gu(R) is a ring.

3. Determine the tangent cone of V = 0((x2 — 3/3)) g A2 (C) at the origin. Draw a picture of (the real part of) the curve V and its tangent cone. 4. V = n((a:y(a: + y — 1))) in A2((C) is the union of three lines. Determine the tangent cone of V (a) at the origin (b) at (0,1)

(C) at (2,0) 5. Determine the tangent cone of V = t:((:1:2 + y2 — 24)) at the origin. Draw a picture of (the real part of) the surface V and its tangent cone.

6. Determine the tangent cone of V = 0((2: + yz, my + 233)) (a) at the origin

(b) at (—1,—1)

Chapter 11

Hilbert Series If R = k[$1, . . . ,zn]/a is a homogeneous algebra we know that R = EBRi, where Ri is the vector space of homogeneous elements of degree 1'. The dimensions dimk Ri have geometrical interpretations. It is practical to save

information about all these dimensions in a generating function. In order to be able to define this function and study it, we need some preliminary work on formal power series.

11.1

Formal Power Series

Let R be a ring. A formal power series with coefficients in R is a formal expression 2,90 (1a where an E R. Note that we allow infinitely many an to be nonzéro. We define the sum and product of two formal power series just as for polynomials, so

Z (1a + Z M" = Z c" n20

n20

n20

where cn = (1n + bn and

Z (1a - Z M“ = Z a" n20

n20

n20

where dn = (10bn +a1bn_1 +- - -+anbo. The set of such formal series becomes

a ring denoted R[[X]].

In the ring R[[X]] we have (1 — X)(1 + X + X2 + ---) = 1 (which is easily checked). Thus 1 —X has a multiplicative inverse. We can say exactly which elements in R[[X]] have multiplicative inverses. 127

128

CHAPTER 11. HILBERT SERIES

PROPOSITION 1. The formal series F(X) = Zn>0 anX" e R[[X]] has an inverse in R[[X]] if and only if a0 has an inverse in R. Proof. If F(X)G(X) = 1 we get, if G(X) 2 Zn» a", that aobo = 1, so are has an inverse in R. Suppose for the converse that a0 has an inverse

in R. We construct the coefficients for the inverse G(X) = 2,120 a" to F(X) inductively. First we get aobo = 1, so b0 = a0— 1. Suppose we have constructed b0, . . . , bn_1. Next we get bnao + bn_1a1 + - - - + 50% = 0, so bn = —a31(bn_1a1 + ' ' ' + boa”).

I]

Exercises

1. Show that R[[X]] is a ring with unit element 2 a", where b0 = 1 and

bn = 0 if n > 0. 2. If F(X) = Z190 anX" we define the formal derivative of F(X) to be

the series F’(X) = 2,00 naa‘l. (a) Show that (FG)’ = F’G + G’F. (b) Show that (F")’ = nFn‘lF’. (c) Show that G' = 0 if and only if G is a constant.

(d) Show that (G‘1)' = —G'G'2. (e) Show that (G‘")’ = —-nG”G‘"‘1. (f) Show that (1 — X)-" = Ema.) (“g—Ur". 3. This exercise gives an alternative way to determine the series (1 — X)‘”. (a) Let n be a positive integer. Show by induction on m that

("31)+(’:)+~-+(“+:‘1)=0?) (b) Let F(X) = 1+X+X2+- - -. Show that (F(X))" = zmzo (“g—Mm.

11.2. HILBERT SERIES

11.2

129

Hilbert Series

Let R = k[a:1, . . . ,mn]/a be a homogeneous algebra. We have seen that R = ®izoRu where R,- is the k-vector space of homogeneous elements of

degree 1' in R. The dimension of R, is finite, since there are finitely many monomials of degree i, and h(z') = dimk R, is called the Hilbert function of R. As usual it is convenient to gather the information about the h(z')’s (which in general are infinitely many) in a generating function, a formal

power series in Z[[X]]. DEFINITION 2 (HILBERT SERIES). If R = 69520121: is a homogeneous alge— bra, then HR(X) = 21.20 h(i)Xz is called the Hilbert series of R.

EXAMPLE 3. If R = k[a:,y] then R, is generated by mi,zi‘1y,...,yi. These monomials are linearly independent, so dimk R,- = i + 1. Thus HR(X ) =

1+2X+3,’2+---=(1—X)-2.

El

EXAMPLE 4. If R = k[:1:,y]/ (2:2,my) then (images of) the following monomials constitute a k-basis for R: 1, :11,y, y2, y3, y4,. . .. Thus we have h(0) =

1,h(1) = 2, and h(z') = 1 ifz' > 1. Thus HR(X) = 1+2X+X2+X3+- -- = (1+X—X2)/(1—X).

El

We will in the following several times use Theorem 74 in Section 1.9, which states that if f : V ——+ W is a surjective homomorphism of vector

spaces, then dim W = dim V + dim ker(f). We have seen, see Theorem 12 in Section 8.1, that in any homogeneous

algebra, there is either a homogeneous socle element, i.e., an element annihilating every homogeneous element of positive degree, or a homogeneous nonzerodivisor. We will use this to determine what the Hilbert series for

homogeneous algebras look like. LEMMA 5. Suppose that f 6 Man, . . . ,zn] is homogeneous of degree k and that the image of f in the homogeneous algebra R = k[:c1,...,:1:n]/a is a

nonzerodivisor(1'.e., that a = a : (f)). Then HH/(f) (X) = (1 —X’°)HR(X). Proof. The map R; —> Ri+k defined_by fl- I—) ff,- = m is an injective h0-

momorphism of vector spaces since f is a nonzerodivisor, so dimk ( f >2+k = dimk Ri. The map RH], —> RH}, / (f)1+k is a surjective homomorphism of vector spaces, so

hR(i + k) = dim], RH], = dim,c mm + dim,c Ri+k/ (BM = dim], R1 + dimk Si+k = hR(’i) + h3(’t + k).

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CHAPTER 11. HILBERT SERIES

where S = R/ (f). In Z[[X]] we get

113(1' + k)Xi+k = hR(i).Xi+" + h5(i + k)Xi+k, so

>

.

.

Z hR(i + k)X’+" = Z hR(i)X‘+k + Z: hs(i + k)X‘+’°. i>0

i>0

i>0

Ifj < k we have R ~ Sj, so hR(j)= hs(j). Hence we get

2 hR(1’)X" = X’“ Z hR(i)X" + Z hg(i)X 2'20

1'30

120

i.e.,

HR(X) = XkHR(X) + HS(X). El

LEMMA 6. Let R = k[x1, . . . , mn]/a be a homogeneous algebra and suppose that f is a homogeneous polynomial such that f is a socle element in R,

i.e., such that f E (0) : (fi...,fi) or equivalently f E a : (2:1,...,zn). Then HR/(f>(X) = HR(X) — Xdegf.

Proof. We have dimk(f)1 = 1 ifi = deg(f) and dimk (f) = 0 ifi 96 deg( f) since fg = U for every homogeneous g of positive degree. Hence

dimk(R/(f1): dim;c R, — dim;c (f). equals dimk R — 1 if i— — deg(f) and equals dim,c R otherwise. Thus

HR/(f)(X)= Z dim1(R/ (f))-

Z=dimk R Xi_ Xdeg(f)_

HR(X) — X195”). :1 We are ready to prove our main theorem for Hilbert series of homoge-

neous algebras. THEOREM 7. Let R— — k[z1,...,wn]/a be a homogeneous algebra. Then

HR(X )= pX( )/(1 — X)‘1 for some polynomial p(X) E Z[X] with p(O) = 1 and some d S n.

Proof. Since the set of monomials outside [(a) constitute a vector space basis for R, see Proposition 15 in Section 3.6, we see that HR(X ) = HS(X),

where S = k[:1:1,...,:1:n]/l(a). Then we note that we can assume that k is an infinite field, which is the result of the following reasoning. Let Y be

11.2. HILBERT SERIES

131

a variable and let K = k(Y) be the fraction field of k[Y]. It is clear that K is infinite. Let a = (f1(a:1,...,a:n),...,fk(a:1,...,a:n)) Q k[a:1,...,xn]

and let u’ be generated by the same elements but now as an ideal in K[a:1, . . .,a:n]. Calculating a Grobner basis for a’ is exactly the same as for (1, hence the two ideals have “the same” initial ideals, and we get

HR(X) = HK[ac1.....z:n]/a’ (X). We assume from now on that k is an infinite field. Then we can use the result in Theorem 12 of Section 8.1 that R either contains a nonzerodivisor of degree 1 or a homogeneous socle element. Since 0 : (E, . . . ,fi) = (E, . . . , 37) is a finitely generated homogeneous ideal in R, we see that there exists a homogeneous socle element 5 of maximal degree,

namely of degree max {deg 37}. (We have f5? = 0 for each homogeneous element f of positive degree.) Then HR/(§) (X) = HR(X)—Xdeg” by Lemma 6. Let t be a homogeneous socle element of maximal degree in R/ (3). Since R and R/ (E) coincide in degrees > deg s, we have degt 5 deg s. We conti-

nue like this to get an ideal (s,t, . . .) such that k[x1, . . .,$n]/a + (s,t, . . .) has no socle. (The process is finite since the degrees never increase, and dimk EBigdeg SR,- is finite.)

Hence we eventually reach a ring R1 = R/s

without socle and HR/5(X) = HR(X) — q(X), where q(X) E Z[X] has degree deg 3. Then R1 contains a nonzerodivisor of degree 1, which after a. linear change of coordinates, we can assume to be one of the 227’s, and

HR, /(r.-)(X) = (1 — X)HR1 (X). We now use induction over the number of variables n. If n = 0 we have R = k so HR(X) = 1. Suppose that the statement is proved for n — 1 variables, so H31 /(m_.-) (X) = p1 (X) / (1 — X)e with p1(X) E Z[X] and e g n — 1. Then (1 — X)(HR(X) — q(X)) = p1(X)/(1 — X)e, so HR(X) = (p1(X) + (1 - X)e+1q(X))/(1 — X)e+1. D For large 1', hR(i) is a polynomial in i, the Hilbert polynomial of R. COROLLARY 8. Let R = k[:1:1,...,mn]/a be a homogeneous algebra with Hilbert series p(X) / (1 — X)d, where 11(1) 76 0. Then the Hilbert function of R, hR(i) = dimk R, is a polynomial of degree d — 1 in i with highest

coefficient p(1)/(d — 1)! for large 1'. (Ifd = 0, hR(i) = 0 for large 2'.) Proof. Suppose that p(X) = a0 + a1X +

+ ak. Then HR(X) =

1/(1 — X)d + alX/(l — X)d + - - - + ak/(l - X)d. The coefficient of X1 in aj/(l — X)d equals a,» times the coefficient of Xi‘j in 1/(1 — X)‘i which is (”C—{11:14) which, if i 2 j, is a polynomial of degree d —— 1 in i with highest coefficient aj/(d — 1)!. Hence the coefficient h¢(R) of Xi in HR(X ) equals a polynomial in i of degree d — 1 with highest coefficient (1+a1+---+ak)/(d—1)!=p(1)/(d—1)!ifi2k. I] EXAMPLE 9. Let R = If:[ai:,y,z]/ (2:2,:zyz): Ifi 2 3 the nonzero monomials

of degree 72 in R are mfg/"3,0 Sj S 1,3121‘J,0 g j g 1, and yjzi'j,1 S

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CHAPTER 11. HILBERT SERIES

j g z' — 1. They are i + 3 in number, hence the Hilbert polynomial of R is i + 3. El

Exercises 1. Show that the number of monomials of degreei in k[:1:1, . . . , run] is ("+1.”). 2. Show that Hk[21,...,x,.](X) = (1 — X)—".

3. Determine the Hilbert series HR(X) and the Hilbert polynomials hR(i) of

(a) k[$,y,Z]/(wy,$z,312) (b) k[$,y,2l/($2»y2,22) (c) k[:1:, y, z]/ (39,2314, 1122)

11.3

Geometric Meaning of Hilbert Series

Let b be a homogeneos ideal in k[:1:1,...,a:n,a:n+1]. In this section we will show that the Hilbert series of Man, . . .,:1:n+1] / [1 contains geometric

information on Q3(b) (or on u(a) if b = h(a)). We start with a purely algebraic definition. DEFINITION 10 (DIMENSION OF AN ALGEBRA). Let R be a homogeneous algebra with HR(X) = p(X)/(1 — X)d, where 12(1) gé 0. Then we define the dimension of R to be (1. EXAMPLE 11. Using Corollary 8 and Example 9, we see that the dimension

of k[x,y,z]/ (9:2,zyz) is 2, since degz’ + 3 = 1.

D

The following lemma is an immediate consequence of the definition. LEMMA 12. If f is a homogeneous nonzerodivisor in the homogeneous a1-

gebra R, then dim R/ (f) = dimR — 1.

Proof. If HR(X) = p(X)/(1 — X)d with 11(1) 96 0, then HR/(fl(X) =

(1 — X’“)p(X)/(1 — X)"1 f (1 + X +

+ Xk-1)p(X)/(1 — X)d‘1, and

kp(1) 96 0, where k = degf.

The Hilbert series has geometric meaning.

CI

11.3. GEOMETRIC MEANING OF HILBERT SERIES

133

DEFINITION 13 (DIMENSION OF AN ALGEBRAIC SET). Let a be an ideal

in (C[271 , . . .,:rn] and let V = 0(a). The dimension of V is defined to be diIIIl' = dim((C[a:1, . . . ,mn+1]/h(u)) -— 1. In order to show that this is a good definition, we must prove that the dimension does not depend on the ideal (I used to define V. It is enough to

prove that dim (C[.1:1 , . . .,:vn]/a = dim (C[a:1 , . . . find/JCT.

DEFINITION 14 (DIMENSION OF A PROJECTIVE SET). Let a be a homogeneous ideal in (C[x1,. . .,a:n+1] and let V = 811(a). The projective dimension of V is defined to be dimV = dim((C[:r1,.. . ,zn+1]/a) — 1. We must also in this case show that dimV is well defined. We need

some lemmas, but first we give an example.

EXAMPLE 15. Let a = (:L') n (y) n (a: —y) = (2:21; —:1:y2). As an affine algebraic set in A2(C), 0(a) is the union of three lines through the origin. As a projective set in P1((C), 11((1) consists of the three points (0 : 1), (1 : 0),

and (1 : 1). We have h(a) = ($23; —— 333,12) 9 (C[z,y,z], so

q,y,zI/h(u) (X) = (1 - X3)/(1 — X)3 = (1 + X + X2)/(1 — X)2. The dimension of the affine algebraic subset 0(a) of A2 (C) is thus one, as it should be. Since qyyl/(wzyflyzfiX) = (1 — X3)/(1 — X): (1 + X +X2)/(1 —- X),

the dimension of the projective algebraic subset iU(a) of P1(C) is zero.

El

LEMMA 16. Let R and S be homogeneous algebras. If hR(i) S h5(i) for all large 2', then dimR S dim 5'.

Proof. Both in; (2') and hs(i) are polynomials p(z') and q(i), respectively, for large 2'. If hR(i) g hs(i) for all large 1', we must have deg 10 5 deg q, hence dim R g dim S, by Corollary 8.

D

LEMMA 17. Let a g b be homogeneous ideals in A = k[:c1, . . . ,zn]. Then

dimk[:r1,...,a:n]/b g dimk[a:1,...,:rn]/a. Proof. We have a,- g b,- so dimk(ai) g dimk(bi). Hence ILA/ha) = dimk(A/b),- = dimk A,- — dimk bi S dimk Ai — dimk a, = dimk(A/a),- = hug/“(’13.

Thus the conclusion follows from Lemma 17.

D

134

CHAPTER 11. HILBERT SERIES

LEMMA 18. Ifa is a monomia] ideal in A = k[:l:1, . . . ,xn] then dim A/a = dimA/fi. Proof. Letebemax{klzi1---:rik ¢ J6} forsome 1 Sil
2 : A3 ——) A3 defined by 2((r1,r2,r3)) = r1(y, —:L',0)+7'2(2,0, —:I:)+r3(0, —2,y). We denote the sequence ((2, —y,:r)) (consisting of one element) by syz2(s). A third syzygy would be an element 1' in A such that r(2, —y, m) = (0,0,0). This gives 1' = 0. Hence ker(‘I>3) = 0, if (1)3 : A ——> A3, 3(r) = r(2, —y,:c). The third module of syzygies is the zero submodule (i.e., the zero ideal) of A. El

12.2. MORE GENERAL ORDERINGS WITHIN MORE GENERAL RINGS 141 The construction of the first syzygy of a sequence of elements is the submodule analogue of calculating (0) : a for an ideal a, and similarly for higher syzygies. This can be done with Grobner bases for modules. The Syzygy Theorem states the following. Let f = (fl, . . . , f5) be a sequence of homogeneous elements in A = k[a:1, . . .,2:n]. Let syz1 (f) = (g1, . . . , gt) be a sequence of homogeneous elements that minimally generate ker(1), where (1)1 :As —) A is defined by Q1((r1,...,rs))= rlfl + - - - +1'sfs. Let

syz2(f) be a sequence minimally generating ker(2), where {>2 : A‘ ———) As is defined by ‘I>2((r1,...,rt)) = mm + . . - + rtgt and so on. Then ker(n)

is the zero submodule. It can be shown that different choices of minimal generators for the kernels give isomorphic syzygy modules.

By using a suitable ordering, it is even possible to give a proof of this theorem using Grobner bases. This was done in an unpublished Diplom-

arbeit by F.-O. Schreyer. We refer the interested reader to the book of D. Eisenbud. Further Reading

D. Eisenbud, Commutative algebra with a view toward algebraic geometry,

Springer 1995. F.-O. Schreyer, Die Berechmmg von Syzygien mit dem verallgemeintern Weierstrass ’schen Divisionssatz, Diplomarbeit, Hamburg 1980. Exercise

1. Let A = k[:1:,y, z] and s = (112,933,, (132,112,312, 22). Show that the submod— ule of (first) syzygies is a submodule of A6 generated by 8 elements. Show that the submodule of second syzygies is a submodule of A8 generated by 3 elements. Show that the third module of syzygies is the zero submodule

of A3.

12.2

More General Orderings within More General Rings

We cannot expect a direct generalization of the concept of Grobner bases

to ideals in a power series ring k[[a:1, . . . ,xnfl, since in general a power series does not have a leading term. There are also other rings which are of importance in commutative algebra and algebraic geometry. We will mention an important example. If V is an algebraic subset of C" containing the origin, and one wants to study the behaviour of V in a neigh—

bourhood of the origin, the natural ring to study is not C[:r1, . . . ,wn] but

142

CHAPTER 12. VARIATIONS 0F GROBNER BASES

its “localization at the origin”, denoted (C[a:1,. ..,:1:n]($1,m,xn). This ring, which lies between (C[a:1,. . . ,mn] and the fraction field of C[m1,. . . ,zn], con-

sists of {f/gl f,g E C[z1,...,a:n],g(0,...,0) 760}. This is a local ring, i.e., there is a unique maximal ideal, namely (51:1,...,a:n). Every ideal in (C[m1,. . . ,mn]($hwzn) can be generated by polynomials in C[:r1,...,:c,.], since a generator f /9 can be replaced by f, g being invertible. If one wants to study ideals in Cl$1,~-,$nl(m,...,zn) by means of Grobner ba-

ses, it turns out that one has to consider new kinds of orderings; we must demand that am < 1 for every i. This implies that one has to be more careful to get a terminating algorithm; it is not possible just to copy the Buchberger algorithm. One important concept to introduce is the ecart of a polynomial f, which is the difference between the highest and the lowest

degree of a monomial occurring in f. When one has a choice when making a reduction, one has to minimize the ecart. The first to give a standard basis algorithm for these kinds of rings was Mora. He was interested in getting an algorithm to compute the tangent cone of an algebraic set. If (f1, . . .,fr) g m = (2:1, . . . ,mn) ; (C[a:1,...,mn] then the associated graded

ring of C[m1, . . . ,xn]/ (f1, . . . , f,) with respect to E coincides with the associated graded ring of C[x1, . . . ,mn](zlw,mn)/ (f1, . . .,f,~) with respect to E. His results have subsequently been simplified and generalized by Grabe

and the Singular team. There is now a theory of standard bases for general orderings where some of the variables are > 1 and some are -< 1. Standard bases have the same applications as Grobner bases and there are also algorithms to make ideal operations, calculating syzygies and so on in this

more general setting. Further Reading H.-G. Grabe, The tangent cone algorithm and homogenization, J. Pure

Appl. Algebra 97 (1994), 303—312. H. Grassmann, G.M. Greuel, B. Martin, W. Neumann, G. Pfister, W. Pohl, H. Schonemann, and T. Siebert, Standard bases, syzygies and their implementation in SING ULAR, Beitrage zur angewandten Analysis und Informatik, pp. 69—96, Shaker Aachen 1994. T. Mora, An algorithm to compute the equations of tangent cones, Proc.

EUROCAM 82, Springer Lecture Notes in Computer Science (1982).

12.3

Grobner Bases for Noncommutative Rings

In this section we will study ideals in the noncommutative polynomial ring

over a field h (or the free associative algebra over k). The noncommutative

12.3. GROBNER BASES FOR NONCOMMUTATIVE RINGS

143

polynomial ring k < x1, . . . ,3», > consists of linear combinations of monomials (or words) xi, - - - 33,-, , where the variables do not commute. There are for example nine monomials of degree 2 in k, namely an: = x2,xy,a:z,yw,yy = y2,yz,z:r,,zy,zz = 22' An ideal in B = k < 221, . . . , 93,, > (or a two-sided ideal) is a nonempty subset a of B such that

o a,b€a=>a+b€a o a6a=>ras€aforallr,sEB

The foundations for noncommutative Grobner bases were laid down by Bergman. A monomial ordering in B is a total ordering of the monomials in B compatible with multiplication, i.e., o For any monomials m,n we have m -< n or n -< m or m = n

o 1 < m for every monomial m gé 1 0 ml < m2 implies mmln < mmzn for any monomials m,n There are lots of monomial orderings, but we will in this section stick to one

choice, namely Deglex. Here mi, mam-k >~ wj, "‘$jm if k > m or if k = m and i1 = j1,...,i3_1 = js_1 and is < js for some 3. Thus we have made the choice that $1 > > 11:”. In k < wl,x2,a:3 > the smallest monomials are 1