An Introduction to Gröbner Bases

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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 0 1997 by John Wiley & Sons Ltd, Baflins Lane, Chichester, West Sussex P019 IUD, England

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Library ofCongress Cataloging in Publication Data Froberg, Ralf. An introduction to Grobner bases / RalfFroberg. p. — (Pure and applied mathematics) Include:mbibliographical references and index.

ISBN 0 471 97442 0 (alk. paper) 1. Grobner bases. 1. Title. II. Series: Pure and applied mathematics (John Wiley 8; Sons : Unnumbered) QA251.3.F76 1997 512'.24—dc21 97-19668 CIP British Library Cataloguing in hrblicaa'an Data A catalogue record for this book is available from the British Library ISBN

0 471 97442 0

Produced fi'orn camera-ready copy supplied by the author Printed and bound in Great Britain by Biddles Ltd, Guildford and King's Lynn This book is printed on acid-free paper responsibly manufactured fi'om sustainable forestation, for which at least two trees are planted for each one used for paper production.

Preface

1

Rings, Fields, and Ideals 1.1 Definition of a Ring ...................... Exercises

2

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

and Fields ........................ Integral Domains and Fields .............. A Convention ...................... A Finite Ring, Zn ................... Polynomial Rings ....................

1.2

Rings 1.2.1 1.2.2 1.2.3 1.2.4 1.2.5

Zerodivisors .......................

1.3

Ideals 1.3.1 1.3.2 1.3.3 1.3.4 1.3.5

and Rings ........................ Definition of an Ideal .................. The Ideal Generated by a Set ............. Principal Ideals and Euclid’s Algorithm ....... Euclidean Rings .................... Ideals and their Calculus ...............

1.4

Equivalence Relations

1.5

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

1.6

Exercises ............................ Unique Factorization Domains ................ Exercises ............................

1.7

Factor Rings and Homomorphisms ..............

1.8 1.9

Prime Ideals and Maximal Ideals ............... Vector Spaces .......................... Exercises ............................

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

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

{DmOSOSUICtD—ll-l

H.

M

Contents

CONTENTS

vi

2.3

Quotients of Monomial Ideals .................

39

2.4

Prime Ideals ...........................

40

2.5

The Radical of a Monomial Ideal ...............

41

Gr6bner Bases

43

3.1

Monomial Orderings ...................... 3.1.1 A Classification of Orderings ............. Dickson’s Lemma and Some Applications .......... 3.2.1 Dickson’s Lemma ....................

45 46 47 47

3.2.2

3.2

Applications of Dickson’s Lemma ...........

48

3.3 3.4 3.5 3.6 3.7

The Reduction Process ..................... Definition of Grébner Bases .................. Hilbert’s Basis Theorem and Noetherian Rings ....... Gr6bner Bases and Normal Forms .............. Reduced Gr6bner Bases ....................

49 50 51 52 53

3.8

Construction of Gr6bner Bases ................

54

3.9 Free Modules and Syzygies .................. 3.10 Syzygies of Sequences of Monomials .............

55 56

3.11 S-polynomials ..........................

57

3.12 A Criterion for Gr6bner Basis ................. 3.13 The Buchberger Algorithm ..................

57 59

Algebraic Sets

61

4.1 4.2 4.3

61 66 68

Algebraic Sets and Ideals ................... Hilbert’s Nullstellensatz .................... A Dictionary: Algebraic Sets (—) Radical Ideals .......

Primary Decomposition of Ideals

71

Solving Systems of Polynomial Equations

79

6.1

Systems with Only One Solution ...............

79

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

80

6.2

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

80

6.2.1

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

80

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

81 84

Exercises

86

Exercises

6.3 6.4

Decomposition of the Ideal

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

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

87

Exercises

88

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

CONTENTS

vii

7 Applications of Grfibner Bases 7.1

7.2

Membership Problems ..................... 7.1.1 Ideal Membership ................... 7.1.2 Radical Membership ..................

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

7.3

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

7.4

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

7.5

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

8.3

Homogenization versus Dehomogenization ......

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

107 108 111 111 115

Projective Varieties 117 9.1 Projective Closure of an Algebraic Set ............ 121 10 The Associated Graded Ring

123

127 11 Hilbert Series 11.1 Formal Power Series ...................... 127 11.2 Hilbert Series .......................... 129 11.3 Geometric Meaning of Hilbert Series .............

132

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 Griibner Bases

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

CONTENTS

viii

Exercises ..................... 12.4 Differential Grobner 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

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

152 152

14 Software

155

15 Hints to Some Exercises

157

16 Answers to Exercises

163

17 Bibliography

171 171 171

17.1 Books ....................... ....... 17.2 Articles ...................... ....... 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 difliculty 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 19603 there is however also a trend in the opposite direction in these fields. Interest in

constructiveness and concrete calculations is also now increasing. One im— portant 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 efiici— ently, 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

x

PREFACE

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 interested 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 patterned after, and are generalizations of, the complex numbers (C. 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+b6R,anda-bER (b) (commutative) a + b = b + a

2

CHAPTER 1. RINGS, FIELDS, AND IDEALS l

(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

Rsuch that a- 1 = La: a. Neither can webe sure that a-b: b-afor 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 (1 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 (1, 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 at 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 l and l’ 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 + a: = 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+:c) = (—a+a) +1; = (a+(—a))+:c = 0+2; = x, which shows that —a + b is the only solution.

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

1.1. DEFINITION OF A RHVG

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 + a: = b has a. unique solution. To begin with 0 - a = (0 + 0) - a = 0 - a + 0 - a, but it is also true that 0 ~ (1 = 0 - a + 0. It follows that 0 - a = 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 75 1, since if 0 = 1 we get r=r-1=r-0=0foranyr€R,soOistheonlyelementinR.

From the distributive laws (f), the equalities a - b+ (—a) 'b = (a+ (—a)) b=0-b=0, anda-b+a-(—b) =a-(b+(~b)) =a-0=0, we see that —(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) = 1The 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

a’" - a" = a'”+", and (am)" = am". Similarly, in any ring we shall abbreviate expressions like a + a by 2a and more generally for any positive integer nput na=a+a+w+a (n terms). When n < 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

nngs.

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, Oq, OR, 0c, the zero element of Z, Q, R, 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., 13, +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. El EXAMPLE 6. Let R be any ring and n any positive integer. The set of n x n matrices with entries rij E R forms a ring, R", under the usual operations,

i.e. the sum of two matrices C = A +3, B is a matrix with elements c,-j = e.,-n+1; by, and the product 0: A R~,_ B IS a matrix with elements c;,-= Eb, a“, -R bkj. The zero element in Rn is the n x n matrix whith

every entry being equal to 03. We note that if n > 1, and if R has more than one element, then R,I is noncommutative, even though the ring R itself may be commutative. If the R has a unit element, so has R". The unit element 13,, is the matrix (Tij) With 'I‘ij = 03 for '1: 75 j, and 137 = IR for i = j.

El

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. El Exercises

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

{(—22)

a,b€ 1R}

is a commutative ring with unit element with the usual addition and

multiplication of matrices.

1.2. RINGS AND FIELDS

5

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

1.2

Rings and Fields

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

1.2.1

Integral Domains and Fields

DEFINITIONS (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“, 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 = (fl, 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 andb 95 0 we get a = a-l = (1(bb‘1)=(ab)b"1 = 0-b‘1 = 0. El

The most familiar examples Of fields are the rational numbers Q, the real numbers IR, 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 ofelements is a field.

6

CHAPTER 1. RINGS, FIELDS, AND IDEALS

Proof. Let R = {m1,xg,... ,mn} be a finite integral domain and suppose m,- gé 0. We claim that {ah-2:], | k = 1, . . . ,n} contains all elements of R. To show this, it is sufficient to show that may, at xix; if k aé I. This is true, since if mix], = mics; then 35(2), — 2:1) = 0 which gives an, — an = 0 since a: at 0, so an, = an. Since every element in R can be written as mix], for some k, we get that mix), = 1 for some It so 2:; has a multiplicative inverse. El

DEFINITION 13 (SUBFIELD). Suppose k 9 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-kb = a-Kb ifa,b e k. T Then k is called a subfield ofK, 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.

El

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 g 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

[i] + [j] = [k], where k is the remainder when i + j is divided by n, and 0 5 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, 0 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

([1]-mule]. Here [il-Li] = In], Where a = alum. Hence ([z'lmflk] = [r2],

where rlk = (12”! + T2. Thus ijk = kqln + qgn + r2 = (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] + [16]) equal [1'], where r is the remainder when i + j + k is divided by n. Next consider [i]([j] + [16]). Here [j] + [k] = [31], where j + k = qln + 31. Hence [i]([j] + [19]) = [32], where 2'31 = qzn + 32. Thus i(j + k) = iqln + (1211 + 82 = (iql + 42)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 = 111712 is composite, i.e., not a prime. Then [n1][n2] = [0] in Zn, 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 Z1, 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. De-

termine the multiplicative inverses to each element 75 0 in Z7.

8

CHAPTER 1. RINGS, FIELDS, AND IDEALS

1.2.4

Polynomial Rings

The set of polynomials a0 + alz + (12222 + - - - + aux",a¢ e Q, n finite, is denoted Q[m]. With natural addition and multiplication 2 (lift; + Z bimi = 2(0,‘ + bail:i i

1"

i

and

Zuni-2mm” =Z(ao-b,-+a1 -b,-_1 +---+a.--bo):c"' i i i it is easily verified that QM is a ring. In fact, if R is any ring, then the

set R[a:] of polynomials TO + mm + - - - + rum" with coeflicients r,- E R is a ring with natural addition and multiplication. We can identify the ring R

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

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

(R[$])[y]

{fo + fly + - - . + fnynl f:- 6 lb‘l}

= {m + mm + mg + r32:2 + may + r5312 + - - }

{go+.thac+g2.:c2 +---+gmx'" l g.- 6 RM} (REDWSo (R[:v])[y] = (R[y])[a:], and can therefore be denoted by R[a:, y] without any risk for confusion. Likewise, we can define the polynomial ring over R in n variables 21,2:2, . . . ,3”. This polynomial ring is denoted by Rlxlym2a ' ' ' :mnl‘

DEFINITION 16 (DEGREE OF A POLYNOMIAL). The degree of an element m = 2:11 nzf," in a polynomial ring R[:v1, . . . ,mn] is deg(m) = i1 +'- - - +in. The degree of a nonzero polynomial f(a:1,.. . ,zn) = ETi1,....i..$il mzf,"

equals deg(f) = max { deg(a:§1 ~13: I Ti1,_..,i,, 75 0 }. A polynomial of de— gree 0 is called a constant. EXAMPLE 17. The degree of f(:r,y, z) = 2 + 9:2 + myz 6 Z[a:,y, z] is 3.

D

We can consider the ring R as a subring of R[:I:1, . . . ,zn] 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 Ic[2:1, . . . ,mn], where k is a field.

1.2. RINGS AND FIELDS

9

Exercises

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

2. Show that if f,g 6 Man, . . . ,zn]\{0} and f+g 96 0, then deg(f+g) S

maX(deg(f), (1625(9)) and deg(fg) S deg(f)+deg(9)- 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[a:], 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[:c] if R has zerodivisors. 4. Let k: be a field and let R be the subset {ao+a1:v+---+akzk|a1 :0}

of k[:1:]. 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 1' 96 0 in a ring R is called a zerodivisor in R if rs = 0 for some element 3 6 R, s aé 0. An element r 75 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.

El

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

El

EXAMPLE 21. Let R = { (a, b) | a, b E 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 diflicult 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), so (1, 0) is a zerodivisor. El 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,Q{z] and lR[a:1,...,:cm] is O. The characteristic of Zn and Zn[a:1, . . . , mm] is n.

D

10

CHAPTER 1. RINGS, FIELDS, AND IDEALS

Exercises 1. If char(R) = 1) 9E 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’ E 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 a=>a+a’ e a, 2. aEa,rER=>a-r6a.

We note that every ideal is closed under the operation of taking additive inverses, i.e. if a e a, then —a e :1, since —a = a- (—1). It also follows that every ideal contains the element 0, since 0 = a - O. The set consisting of only 0 is an ideal, as is the whole ring R. An ideal a coincides with R if andonlyifle a, sinceiflEa, thenr=1-r6aforeveryreR. Exercises

1. Show that {3n| n E Z} is an ideal in Z.

2. Let k be a field. Show that {f E k[m] | f(1) = 0} is an ideal in k[a:]. 3. Let (c1, . . . , c”) E k", where k is a field. Show that {f E k[2:1,...,a:,,]| f(c1,...,cn) :0} is an ideal in k[a:1, . . .mn]. 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 + ~ - - + rnsnl r; 6 R, s.- 6 S}, is an

ideal denoted by (S). If for an ideal a there is a finite set S = {31, . . . , sn} such that a = (S), then a is said to be finitely generated and denoted (31,...,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—c1,...,:c,.—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 r be an element in a ring R. The set of all multiples of r, {r - r’ | r’ E R}, constitutes an ideal. Such ideals are called principal ideals and r is called a generator for the ideal. The principal ideal generated by r is denoted rR 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 a implies that —a e a. Let n be the smallest positive number in a. We claim that a = (n), i.e., a consists of all multiples of n. Let b be an element in a. Then, b = qn + r for some integers q and r, 0 g r < n, or stated otherwise r = b — qn. Since n 6 a it follows that both b and -qn belong to a, so 1‘ 6 0. But we picked n to be the smallest

positive number in a which forces r to be zero, so b = qn + 0 is a multiple . of n.

Similar reasoning shows that every ideal in Q[a:] is principal. Let a be an ideal in Q[:L'] difi'erent from (0), and let f 96 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 r 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[:c], 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 ¢ 0. Then there are uniquely defined polynomials q,r e k[z] 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 = a,,:1:"+---+ao,an aé 0 and let 9 = bmmm+---+bo,bm 96 0. Ifm < nwe canchooseq = Oandr :9. If m 2 n we see that g = bma;l:r",“"f + 1'1, where deg(rl) < deg(g)

or 1-1 = 0. If n 76 0 and deg(rl) > deg(f), say r1 = ckxk +

+00,

we can continue and write r1 = cka;1:1:"""f + 1'2 with deg(rz) < deg(rl) or r; = 0, so 9 = (bmaglxm'” + cka;1:r"‘")f + r2. 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 + r = qlf + n. Then (q — q1)f = r1 — r. We have deg((q — q1)f) 2 deg(f) if (q — ql) aé 0, which is a contradiction since deg(r1 — r) < deg(f). Thus q = q1. This gives 0 = 0 - f = r1 — r, so r1 = r.

D

In the proof above it is essential that k is a field. There is no division algorithm in Z[:r] as the example 9 = m2 and f = 2a: shows. It is not easy to see a generalization of the division algorithm to Man, . . . , 1:1,]. One can say that the purpose of Grobner 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[:1;]. Let f and g be

nonzero elements in k[:c], k a field. The division algorithm gives q1 and r1

such that g = qlf + T1 with deg(rl) < deg(f) or n = 0. If n 75 0 the division algorithm gives qg and r2 so that f = qzrl + T2 with deg(rg) < deg(rl) or r; = 0. If m 73 0 we get r2 = q3r2 + 1-3 a.s.o. If r3 aé 0 we have

deg(ra) < deg(rg) < deg(rl). Since the degrees decrease, this process is finite, so we must eventually get a remainder which is zero.

9

=

41f + 7‘1

f = qzn + r2 7‘1

=

11372 + 7‘3

TN—2

=

(INTN—1 + TN

rN—I

=

9N+1”'N

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

1.3. IDEALS AND RINGS

13

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

right—hand side. Continuing like this upwards we eventually get the claim that rN divides both f and 9. Thus we have shown that rN is a common

divisor to f and 9. 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 —q2r1, 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 9 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 rN is the greatest common divisor to f and 9, m; = gcd(f, 9).

DEFINITION 26 (GCD). Let f and g be nonzero polynomials in k[:1:]. Then h is a greatest common divisor of f and g, gcd(f, 9), if h divides both f and g, and any other polynomial which divides both f and 9 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 h1 and hg both are greatest common divisors to f and y, then h1 = ch2 for some c E 16. Proof. We only have to prove the last statement. Since h1 is a common

divisor to f and g and h2 is a greatest common divisor to f and y we get that h1 divides h2, h2 = qlhl. Changing the roles of h1 and h2 we get h1 = qghg. Thus h2 = q1q2h2, so q1q2 = 1, hence q1 = (12—1 6 k. El

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 h2 such that h = hl f + hgg. The second equation from the bottom in the Euclidean algorithm shows that m; = 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 m; = -qN - rN_3 + (1 + qN_1)rN_2, hence 1')»; is a linear combination of rN_3 and rN_2. Continuing upwards like this we eventually get the claim that

my 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[:z:] over a field h. Then (f, g) = (gcd(f, 9)). Proof. Let h = gcd(f, y). We have shown that h is a linear combination of f and g which gives h 6 (fly). This gives that (h) g (fig), since

(h) = {rhl r G k[a:] }, and if h E (f,g), then rh 6 (f,g). On the other hand, both f and g are multiples of h, thus f, g 6 (h), which gives (f, g) = {7'1f+’l'29|’l‘1,1'2€k[m]}§(h).

U

14

CHAPTER 1. RINGS, FIELDS, AND IDEALS

Let f be a polynomial of positive degree in k[:c] and let c 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 m2 + 1 in 1R[:c]. 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 f, 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 f1f2, then f divides f1 or f divides f2. Proof. Suppose that f does not divide f1. Then gcd(f, f1) = 1 so 1 = h1f + hz f1 for some polynomials h1, h2, which gives f2 = h1ff2 + h2f1f2.

Since f divides both terms on the right-hand side, we get that f divides U

f2-

A polynomial f = cw" + ck_1:r"‘1 + - - - + co is called manic if ck = 1. THEOREM 30. Supposedeg(f) > 0. Thenf = cfl - - - fk, where f,- aremonic irreducible polynomials. This factorization is unique up to the ordering of the fi’s. Proof. We only need to show unicity. Let f = Cl f1 - - - f1, = 6291” ~gm. Being the coefficient of the largest power of a: in f, it is clear that c1 = 02. Since f1 divides 91 gm, we get that f1 divides one of the 95’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 = g]. The equality f1f2 - - . f], =

f192~~gm gives f1(f2"’fk-92“'gm) = 0, so f2"'fk = g2---gm andwe 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” -~-fj‘:," and g = :1.”n with (15,12,- 2 0. It follows from the theorem that f divides 9 if and only if a,- S b.- for all 17.

Thus h = ff‘in(“"b‘)

ffiinwm") divides both 1’ 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(:c — 2)(:z:2 + 1) and g = (:1: — 1)2(a: - 3)(:::2 + 1)(:1:2 + 2), then gcd(f,g) = (a: -— 1)2(:.I:2 + 1). El

1.3. IDEALS AND RHVGS

15

DEFINITION 32 (LCM). Let f and g be nonzero polynomials in k[:r]. 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 = on" + + a0 and c e k then the evaluation of f at c is the element f(c) = akc"+---+ao E k.

THEOREM 33 (THE FACTOR THEOREM). Let It be a field and suppose that

f(z) E k[:c]. Let f (c) e k denote the evaluation of f at c. Then f (c) = 0 if and only if a: — c divides f(1:).

Proof. The division algorithm gives q,r 6 k[:c] such that f = q(.z' — c) + r and deg(r) < deg(:c - c) = 1 or r = 0. Hence 1' e k. Evaluation at c gives f(c) = 0 +r, so f(c) = 0 if and only ifr = 0, hence if and only if :c — c divides f. El

COROLLARY 34. A polynomial of degree n in k[a:] cannot have more than n zeros. Proof. If C1,...,c,. are zeros to 1‘ then (x — c1) ---(:c — cr) is a factor of f D which is impossible if r > deg(f). We also remind the reader of what is usually called the Fundamental

Theorem of Algebra. THEOREM 35 (THE FUNDAMENTAL THEOREM 0F ALGEBRA). Each poly-

nomial f E C[:1:] of degree at least one has a zero in (C. COROLLARY 36. Any polynomial f 6 CM with deg(f) = n > O can be factored in linear factors, f = c(a: — c1) - - - (a: — c"). COROLLARY 37. The polynomials a: — c are the only monic irreducible po-

lynomials in C[.z-]. . If (a: — c)’" divides f(m) but (a: — c)"I+1 does not, we say that c is a root of multiplicity m to the equation f(2:) = 0. COROLLARY 38. An equation f(z) = 0, f e C[:c],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 IDEAIS

16 Exercises

1. Show the following division algorithm in Z. Let m, n e Z, n 96 0. Then there are q,r E Z with r < |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 ii. If m and 72.2 are both greatest common divisors to m and n, then m = :tng. Show also that if

d = gcd(m,n) there are a, b 6 Z such that d = am + bn. . Determine gcd(a:4 + 3x2 + 2, 11:3 — 2:1:2 + a: — 2) in Q[:z:]. Write gcd(m‘ +

3x2 + 2, x3 — 211:2 + a: — 2) as a linear combination of (1:4 + 332 + 2 and 2:3 — 2:):2 + a: — 2. . Let f and g be monic polynomials of positive degree and let f1, . . . , f" be all the monic irreducible factors that occur in the factorizations of

f org. Let f: fflm g," andg=ff1~--1'{," with a.-,b.- 2 0. Show that lcm(f.g) = finmm'bl) ' ' ' 1.1:?w ’b" ). . If f = (:1; — 1)3(:1: — 2)(a;2 + 1) and g = (.1: — 1)2(z — 3)(:z:2 + 1)(:1:2 +2), determine lcm(f, g) .

. If r > 2 we make the following recursive definition of gcd (f1, . . . , fr). We define gcd (f1, . . . , f,) to be gcd (gcd (f1, . . . , f,_1), fr). Show that gcd (f1 ...,f,) divides f.- for z' = l,...,r. Show that if h divides f,- for 'i = 1,...,r then h divides gcd (f1,...,f,). Conclude that gcd (f1, . . . , fr) does not depend on the order of the fg’s. Show that

(f1, - - -,fr) = (gcd(f1,.~,fr))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) S d(a) if a,b 96 0 (2) For each pair a, b of elements, a 51$ 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) = Ial. HR = k[z] we use d(f) = deg(f). Thus Z and k[m] are Euclidean rings. III

In the same way as for Z and k[:z:] we can prove the following theorem. THEOREM 41. In a Euclidean n'ng every ideal is principal. Proof. Let a aé (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 r = b — qa E :1. 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[z'] = {a + ibl a, b e Z }. Put d(a +

ib) = a2 + b2. Show that for a,,6 6 Z[i] we have d(afl) = d(a)d(,6). Show that Z[z'] is a Euclidean ring. Hint: Suppose a = a + ib and ,6 = c+ id, ,6 ¢ 0, are Gaussian integers. Show that a/fl = 7+ q1 +iq2

for some 7 e Z[z'] and some q1,q2 6 Q with |q1| g |a|/2, |q2| S [bl/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,beb} 2. anti

3i a-b={zf=1a.-b,- (“Emu-Eb} 4. a:b={rER|rb6a Vbeb} are all ideals. We prove this for a : b and leave the other as exercises. If r1,r2 6 a: b then (r1 +r2)b = r1b+r2b E a for every b E b since rlb,r2b E b and a is closed under addition. Similarly, if r1 6 a z b and r2 6 R, then

(r2r1)b = r2(r1 b) E a for every b E [1 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 J3 = {r€R|r" e afor somen}, where n is a positive integer.

DEFINITION 43 (RADICAL IDEAL). An ideal in is a radical ideal if x/F = [1.

Note that a 9 fl since 1' E a implies that r” E a for n = 1, and so by definition, 1‘ 6 J5. THEOREM 44. If a is an ideal, then J5 is an ideal which contains a. Moreover, fi is a radical ideal. Proof. We have already shown a g J5, so all that remains is to show \/E is an ideal which is radical. First, we have to prove that J5 is an ideal. Suppose that r1, r2 6 (3. Then, by definition, there are two positive integers m and m such that

1'1”n 6 fi. If we expand the sum (r1 + 1'2)"1"'"2'l with the binomial theorem, we see that every term is a multiple of some 71%;"? with m1 +

m2 = 11.1 + 112 — 1. Since either m1 2 m or m2 2 n2, either 1'?“ or r3” is in a. Therefore, since all its terms are in a, (1'1 + 7'2)"1"'"2‘l E a. We have

proved that r1 + r2 6 J5, but we still have to show that if 1'; 6 x/E and 'r E R, then rm 6 fl. We have 1"; 6 a for some n, hence rnri‘ = (rrl)n e a,

so rm 6 J6, hence J3 is an ideal. We have J5 g \A/E. Let r E \/‘/5. Then 1'" 6 Jr? for some n, hence rm” = (r")'" E a for some m, which gives

1' 6 fl, so

a is radical.

El

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, and c be ideals in aring R. Then H

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

4. agbéfigfi 5. x/a_=¢6 O)

.fim/E=\/\/Em/E

Proof.

1.Letr16ancandr26bnc. Thenrleaandrgebso

r1+r2€a+b.SincerlGcandrgecwegetr1+r26c. 2. LetrEa:c+b:c. Thenr=r1+r2wherer1ceaandr2c€b for every c E c. Thus rc = me + r20 6 (a + b) for every 0 E c, so

1' 6 (0+ [1) : c.

1.3. IDEALS AND RINGS

19

.Letre a:(b+c). Thenr(b+c)€aforeverybeba.nd evercc. In particular rb 6 a for every b 6 [1 since 0 e c. In the same way we seethatrce aforevercc. Hencere azbandr€azcso

re(a:b)n(a:c)Nowsupposere(a:b)n(a:c). ThenrbEafor every b e b and rc 6 a for every c 6 c. Hence r(b+c) = rb+rc e a foreverybebandevercc,sor€a:(b+c).

.IfrEfiJhen‘r”€a§bforsomen,sor6\/b_. 5. Since a" g a we have Va" Q fi. If r 6 J5, then 'r" E a for some n.

Then r""’ = (r")" e a", so 1' 6 Va". . Let r E Vfim/E. Then 1'" E fins/afar some n. Then 1"" e a and 1'” e b for some k, hencer E fiandr e x/b_, 301' E Jam/E. The other inclusion follows from (4). I]

Exercises

Let a, b, and c be ideals in a ring R. Show that

r—Ir—In—I top—I9

PPNPP‘PWE"

1 .a+ b,an b, and a - b are ideals. If a = (S) and b = (T), show that a+b= Pol(k), f I—) ¢f is a homomorphism. If the number of elements in k is infinite then ¢ is an isomorphism; if k is finite then 45 is not an isomorphism.

Proof. That 4) 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, . . . , cu}, then ker(¢) =

((2: -cl)

(av-cu»

'3

CHAPTER 1. RINGS, FIELDS, AND IDEALS

28 Exercises

1. Let k be a field and let c e k. Show that (15: k[:c] —) k,¢(f) = f(c) is a homomorphism. Determine ker(¢). .Letkbeafield andlet c1,...,cn E k. Showthat¢:k[:r1,...,:rn]——i k, ¢(f) = f (cl, . . . , on) is a homomorphism. Determine ker(¢). . Let f : Q(Z) —) Q, [(m,n)] I—> 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 R: has a subfield isomorphic to Q(R). . Show that a field of characteristic 0 has a smallest subfield which is

isomorphic to Q. Let p be a prime number. Show that a field of characteristic p has a smallest subfield which is isomorphic to Z,.

6. Show that R[a:]/ (2:2 + 1) 2 (C. . Show that R = Z2[:v, y] / ($2,311,912) is a ring with 8 elements. Determine a multiplication table for R.

8. Show that Q[a:]/ (1:2 — 2) : (QB/5].

9. Show that R = Z3[a:]/ (m2 + 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 multiplicative 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(( J; 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 : 1R[:c]/ (222). 14. Let ¢ : k[a:1,:1:2] —) k[a:] be defined by ¢(f(:v1,a:2)) = f(2:2,:r3). Show that d) is a homomorphism. Show that the image of ()5 is the ring R in

Exercise 4 in Section 1.2.4. Show that R : k[:vl, $2]/ (2:? — :33). 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 : S/¢(a). 16. Let (1) : Z[.z-] —) Zp[:c], where p is a prime, be defined by ¢(ao + an: + ---+ aux") = [a0] + [a1]:c + - ~ ~ + [an]a:". (a) Show that (15 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 x5 — m2 + 1 is irreducible in {Mr}.

(d) Show that m4 — 32:3 + 39:2 + 7 is irreducible in Q[:r]. 17. This exercise shows that factorization in QM is algorithmic. Let f(:z:) e Z[:1:] be a polynomial of degree n. If f is reducible it has a

factor 9 of degree at most [n/2] (the largest integer S n/2). Take any integers a1, . . . ,a[n/2]+1. We have that g(a.~) divides f(a;), so there are only a finite number of possibilities for g(a,'). A polynomial of degree k is determined by its values in 16+ 1 points. Thus there are only finitely many possible factors 9 to test. Use this to factor 2:4 — 2:3 + 22:2 -— a: + 1 in QL'r].

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 aé R in a ring R is a prime ideal if 1'3 6 p implies that r E p or s 6 p.

LEMMA 64. We have that p is a prime ideal if and only if a1 -- - ak Q p implies that some (ii 9 p.

Proof. Suppose p is a prime ideal. By induction on Is it is clear that it suffices to consider the case when k = 2. Let 0102 g p and suppose (11 Z 33. Take an x e a1\p. For each a 6 (12 we have ma 6 p which gives a e in so a; 9 p. For the converse we note that my 6 p is equivalent to (:5) (y) g )3. Hence if my 6 p then (2:) (y) g p which gives (2:) g p or (y) g p, i.e. a: e p or y E )3.

El

The following property of prime ideals is often used.

LEMMA 65. Let a be an ideal and let p.- be prime ideals for i = 1,. .. ,3. If a g Uf=1p,-, then a Q p,- for some 1'. Proof. We use induction over 3. If s = 2 and a g in U p2,a g p1,a g 132

there exists a1 6 a \ p1,a2 6 a\p2. Then at; + a2 6 a but (11 + a2 9! p1 because a1 ¢ p1 and a2 6 m. In the same way we see that a1 + a2 9.5 132, which gives a contradiction. (Note how in this part we haven’t used that p.- are prime ideals, only that they are ideals.) Now suppose that s > 2 and

that a n p; g Uj9fipj for all i (so all pg’s are really needed). Take for each

30

CHAPTER 1. RINGS, FIELDS, AND IDEALS

i an a,- E (anpi) \U,-¢,p,~. Then we get that a = a1 + a2 ---a, e a. We have a2 - was ¢ p1 (here we use that m is a prime ideal) so 0. ¢ 131 since (11 6 331. Hi > 1 then a1 lip; and az-na, 6p,- soa¢p,-, whichgivesusa contradiction.

El

DEFINITION 66 (MAXIMAL IDEAL). Let R be a ring. An ideal m 75 R in R is a maximal ideal of R if the only ideal of R which strictly contains an is the whole ring R. Prime ideals and maximal ideals can be characterized by means of their

factor rings.

THEOREM 67. An ideal p 95 R in a ring R is a prime ideal if and only if R/p is an integral domain. An ideal m ;é 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 r. In particular the coset a in R/a, which is the zero element in R/a, is denoted 0. Suppose p is an ideal and that 1'3 6 p. In R/p this means that 175— 0. The equation rs— 0 IS equivalent to r— 0 or 5— 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 )3 if and only if R/p is an integral domain. Suppose m is a maximal ideal. If 1' ¢ 111 we get m + (r) = R since m + (7') strictly contains 111. Thus 1 = m + 7's for some m e m and s e R. In R/m this gives I = F5 so any nonzero element 1" in R/m has a multiplicative inverse, so R/m is a

field. Conversely, if R/m is a field and F aé 6 in R/m, then is? = I for some

5 6 R/m, which gives rs +m = 1 for some m E m, hence 111+ (r) = (1) = R for any 1' ¢ 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.

El

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[a:], where k is a field, are (f(m)) where f is irreducible and (0). All these but (0) are maximal.

El

Exercises 1. Let (c1, . . . , c") E k" where k is a field and k” is the set of n—tuples of elements in k. Show that the ideal (2:1 — 01,. . . , z" — cu) is a maximal ideal in k[a;1, . . . ,arn].

2. Show that (at? + 1,502) is a maximal ideal in the ring M351 ,1122].

1.9. VECTOR SPACES

31

3. Show that (x? —a:§) is a prime ideal in k[x1,a:2]. Hint: Exercise Exercise 14 in Section 1.7. 4. Show the following improvement of Lemma 65. Let a g Uf=1ai. If all but at most two of the ai’s are prime ideals, then a g a, for some i.

5. If p is a prime ideal, then J13 = p.

6. Let R be the ring in Exercise 3 in Section 1.1. Show that {(0, b)} is a maximal ideal. 7. Let ¢ : R -) S be a surjective ring homomorphism. Show that, in the correspondence of ideals in R and S (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: 1. v1+v2€Vifvi€V

00:19:???»

. cVickand'ueV

v1+v2 =v2+v1 andvl +(v2+v3) = (121 +v2)+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 + (—1)) = 0 . 1 - 'u = v if v E V and 1 is the multiplicative unit element in k (6162)’U = c1(c2v) if c.- E k and v E V

. c1(1)1 + 122) = c1121 + 011:2 and (01 +62)’U1 = clvl + 6201 if c,- E k and v,- E V

A set of vectors {”3561 is said to be linearly independent if any linear combination c1121 + - - - + one" = 0 of elements in {12,-} implies that

all coefficients c,- = O. A set of vectors {v.«hel is called a generating set for V if every element in V is a finite linear combination of elements in

{21,-}. 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 c.- E k is a vector space with addition (c1, . . . ,cn) + (c’l, . . . ,c’n) = (Cl +c’1, . . . , cn+c’,,) and multiplication c(c1,. . . , cu) = (ccl, . . . , cc"). It" has dimension n since

{(1,0,...,0),(0,1,...,0),...,(0,0,...,1)} isabasis.

EXAMPLE 70. If k g K are fields then K is a vector space over Is. As an example, C is a vector space over R of dimension 2 with basis {1,i}. U

EXAMPLE 71. The ring k[:z:] is a vector space over k with basis {1, 3:, m2, . . ..} More generally, k[:1:1, . . . , 2:7,] is a vector space over It with a basis consisting

of all monomials {mil

a2}: }.

I]

EXAMPLE 72. Let f 6 k[:l;] be of degree d. Then k[:1:]/ (f) is a vector space

over k of dimension d with basis B = {1 + (f) ,a: + (f),...,:z:""1 + (f)}. This follows from the fact that every element in k[a:]/ (f) can be written as 9(3) + (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[a:]/ (f) would yield an element of degree < d which is a multiple of f. D 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 ~ 1122 if 1121 — mg 6 V. The set of equivalence classes {11) + V| w 6 W}

constitute the factor space W/V with addition (wl + V) + (102 + V) = (on + 1122) + 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 (o1 +o2) = f(v1) + f('02) and flea) = cf(o) for all 111,02”) 6 V,c E k. If f : V —) W

is a homomorphism then ker(f) = {v 6 VI 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 VI and Vz are called isomorphic, V1 2 Va if

there exists an isomorphism f : Vl —) 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,...,e,,} for V. We claim that B = {f(e,,+1),...,f(e,,)} is a basis for W. Since f is surjective any element in W has the form

“6181 +' ' '+cnen) = Cif(61)+' ' ‘+Cnf(en) = Ck+1f(€k+1)+' ' '+Cnf(en), thus B is a generating set for W. Suppose ck+1 f (ek+1) +- - -+cnf (6,.) = 0. Then f(ck+1ek+1 + - ~ - +cnen) = 0, so ck+1ek+1 + - ' - + onen E ker(f), thus ck+lek+1 + - - - + onen = clel + - ' - + ckek, so 0161 + - - - +cke,c — ck+1ek+1 -

---—c,,e,, = 0. Since {e1, . . . ,en} is a basis for V we get that c,- = 0 for all El 2', hence {f(ek+1), . . . , f(e,,)} are linearly independent.

DEFINITION 75 (DIRECT PRODUCT). Let V1,V2,... be a sequence (finite

or infinite) of vector spaces over k. Let [I V, = {(111,02, . . .) | v; E V.- }. If we define addition in H V,- and multiplication of an element in H V, with an element in k componentwise, (v1, v2, . . .) +(w1, 102, . . .) = (1)1 + wl, 02 + 102,...) and c('v1,1)2,...) = (cv1,cv2,...), then HV, is a vector space, the direct product of the V,-’s. If we have a finite sequence Vl , . . . , V”, the direct product is often deno-

ted V1 x

x V,,. If furthermore V,- = V for all i then H V,- is denoted

V”. DEFINITION 76 (DIRECT SUM). Let 171,172,... be a sequence (finite or infinite) of vector spaces over It. Let 63V; = {(v1,vg,. . .) | v,- E V,- }, where v; = 0 for all but finitely many i. If we define addition in 69V; and multiplication of an element in 69V,- with an element in k componentwise, (”1,02,”) + (101,1U2,...) = (’01 + 101,02 + £02,...) and 0001,02,...) =

(cv1,cv2, . . .), then 69V; is 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 {6,} is a basis for V and {62} is a basis for W, we claim that {(eg, 0)}U{(0, e9} is a basis for V x W. Suppose X: c,(e.-, 0) +2 c’,(0, e2) = (0,0). The left-hand side equals (2 egeg,c§e§) so Zcie, = 0 in V and 2 die; = 0 in W. Since {e,} are linearly independent in V, we get c,- = 0 for all 2'. Similarly a, = 0 for all i, so {(ei, 0)}U{(0, e;)} is linearly independent. Finally {(e,-, 0)} U {(0,e§)} is a generating set for V x W since an element in V x W is of the form (12,20) with v 6 V and w e W. Since {ei} generates V, we have '0 = E c,-e,- for some 0,- and similarly we have w = Z cgeé for

some cg. Hence (12,111) = (Z cieg,c§e;) = Zei(ei,0) + Ze§(0,e§).

D

34

CHAPTER 1. RINGS, FIELDS, AND IDEALS

The most important property of 69V,- is that the coordinates are unique, if (111,02, . . .) = (101,192, . . .) then 1),- = w,- for all i. Our construction, from

a (finite or infinite) set of vector spaces V,- define a new vector space EBVg, is sometimes called the external direct sum of the Vi’s. Now suppose that we have subspaces V1, Vg, . .. of a vector space V. The sum 2 V,- of‘ the vector spaces Vi consists of all sums 2cm, where v; E Vac; E k and c.- = 0 for

all but finitely many 2'. It is easy to see that 2 V; is the smallest subspace of V that contains all Vi. We now suppose that V = Z V,- and that for all

i, V,- n Shh. V]- = {0}. Then we claim that V : éDVi, and V is called the internal direct sum of the Vi’s. To prove this claim, we consider the map ¢ : 69V,- ——) V defined by ¢((v1,v2,...)) = '01 + 02 + - - -. (Even if there is an infinite number of Vi’s, this sum is a finite sum.) It is clear that (b is a

homomorphism and that ()5 is surjective. Suppose ¢((vl, 02, . . .)) = 0. Then vl+v2+- - - = 0 so '0; = —'vl-v2--~—v,~_1—vi+1—---, s01).- 6 VgfizjafiVj which gives 1),- = 0. This means that d) is also injective. Note that if V is the internal direct sum of {Vi}, then every element 1) E V can be written v=v1 +vz+---,v,- E V,- inauniqueway.

EXAMPLE 78. Let V = lake]. Then V,- = {cXi | c e k} is aone—dimensional vector space with basis X’ and V is the (internal) direct sum of the W8. CI 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; = 213, where v; 6 Vi. Proof. We have already shown “only if” , so suppose each element in V has

a unique expression 2 vi, '1),- 6 Vi. Then ten:#1. Vj = {0} since otherwise, if v,- e V; n 21%,- Vj, v,- # 0, we would have a contradiction to unicity.

I]

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, Q[\/§] (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 16.

1.9. VECTOR SPACES

35

5. Determine a basis for, and the dimension of, the ring k[a:, y]/ (:33, 314) as a vector space over the field k.

6. Show that if V1,V2, . .. are vector spaces over k, then HVi and 69V,a.re vector spaces over k. 7. Show that if V1, 172,. .. are subspaces of a vector space V, then 2 V;

is the smallest subspace of V that contains all V5.

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[:131, . . . ,xn] of the form

zf‘ ”.2191". 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[.'1:1, . . . , xn]) 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 f,for some 12 7e j, then fj is not needed as a generator. For monomial ideals this also gives a suflicient 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 Man, . . . ,xn]. Let f = 205mg,

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

polynomial ring k[a:1,...,a:,.]. If em = cw:1 ---z§;‘,c E k \ {0}, we set mdeg(cm) = (i1,...,in). Let a = (n1,...ns) be a monomial ideal, and suppose f = Ecimi E a. Then f = glm + +9371,i for some 9,- = ZCijmij E k[a:1,...,a:,,]. Let 05m,- be a nonzero term in f. Then am,37

38

CHAPTER 2. MONOMIAL IDEALS

equals the sum of all elements cijmijn, 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 = (a1,...,a,.) and b = (b1,...,b,), the sum is a+b=(a1,...,a,.,b1,...,b,),

and the product is ab = (a1b1,. . .,a1bs, . . . ,arb1,. . . ,arb,) ,

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,a:y,y4) and b = (1132,3312). Thus, for our particular monomial ideals, we get

a + b = ($3,331,114) + (£62,023?) = (m3,zy,y4,w2,zy2)But 2:2 divides 1:3 and my divides $312, so we can remove the generators 2:3, and my2 without changing the ideal. The final answer thus becomes a + b = ($3,$y,y4) + ($213312) = ($3,$y,y4,$ 2113312)

2 ($11,132,314)-

The computation of the product ab is also straightforward. ab

=

($3,341,314) ($2,2y2)

= (x5,m4y2,w3y,z2y3,w2y4,wy6) = ($5,23y,m2y3,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, 71.)).

Thus, (2:211) n (mg/3) = (lcm(m2y,wy3)) = ($2313). For any three ideals, a, b, and c, we have seen that

(a+b)ncI_)(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 0F MONOMIAL IDEALS

39

To prove this, let f = f1 + f2 6 c with f1 6 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 62771 is a term of f2, we can write

f1 + f2 = (fl — 01m) + (f2 + 01m) where f1 - 01m 6 a and f2 +C1m E 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 6 (ant) +(bnc). Consequently, for monomial ideals we get

(m1, . . . ,m,.) n (n1,.. .—,n8) — :2 (mg) n (111-): z: (lcm(m,,n,~)). i=1 j=1

i=1 j=1

In particular, for our monomial ideals a = ($3,131,314), and b = (22,3312), we find that

($3,341,214) n (9112,2312)

(lcm(x3, m2),lcm(:1;3, 2112),” .,lcm(y4, $312))

(x3,m3y2,z2y,wy2,x2y‘,rvy4) (x3,x2y,wy2).

2.3

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):c_3_(a:c)+(b:c), see Proposition 45 in Section 1.3.5. (m1, . . . ,mr) and a monomial n

But for the monomial ideal a =

(man-mar)=(n)=(m1)=(n)+“'+(mr):(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 mu 6 (m1,...,m,.) for every monomial m that occurs in f. If mu 6 (m1, . . . ,m,) then mn is a multiple of some mi, i.e., mn 6 (mi), hence m 6 (mg) : (n). For arbitrary ideals (1, b1, b2, we have seen that a: (51 +52) =(a: bl)n(a: 52), see Proposition 45 in Section 1.3.5, so by induction we get

a: (b1 +---+b5) =n§=1(a:bi).

40

CHAPTER 2. MONOMIAL IDEALS

Let a: (m1,...,mr), b = (n1,...,n,), and b.- = (11;). Then b = b1+---+ 6,. It is easily seen that for any two monomials, m and n, the quotient

ideal (m) 2 (n) = (m/ gcd(m,n)). We conclude that (1:5

=

a:(b1+'--+bs)

a = ((n1)+"'+(ns)) =

(a: (n1))n---n(a: (7%)) ((m1,...,m,) : (n1))I"I---fl((m1,...,m,) : (n,))

((m1)=(n1)+'“+(mr) =(n1))n'“

n ((m1)=(ns)+"'+(mr)t(713)) = ((ml/gcd(m1an1)) + ‘ ' - + (mr/gcd(mnn1))) n - 'n ((ml/gcd(m1,n,)) + ' ' ' + (mr/ gcd(m,.,n,)))

=

(ml/gcd(m1,n1), . . . ,m,/gcd(mr,n1)) n - -fl (ml/gcd(m1,ns), . . . ,m,/gcd(mr,n,)) .

Hence, for our monomial ideals, a = ($3,331, 314) and b = ($2,2y2), the quotient ideal becomes

a: b

=

(273,331,314) : (22,2312)

— (w3/ godfwa. :62), my/ gcd(wy, m2), 314/ soda/4, 902)) n

(163/ gcd(:v3, $312), rcy/ gcd(wy, 2:312), 114/ scd(y4, 95112)) = (ms/zzmy/w, 214/1) n (ma/x,xy/wy,y4/y2) (x,y,y4) n (2:2, 1,312)

= 1 then a is not a prime ideal, since if m = mgm’ then 3,: ¢ a and m’ ¢ 0. but xim’ E a. On the other hand, if all generators are of degree one, so that a is generated by a set of variables, then a is prime since k[:c1 , . . . , mu] / (m1 . . . , an) is a polynomial ring k[zl+1, . . . , run] which is an integral domain. The only monomial ideal which is maximal in k[1:1, . . . ,wfl] is (11:1, . . . ,mn).

2.5. THE RADICAL OF A MONOMIAL IDEAL

2.5

41

The Radical of a Monomial Ideal

Finally we consider the radical of a monomial ideal. But first we need to do some preparation.

DEFINITION 2 (SQUAREFREE MONOMIAL). A monomial m is called squarefree if m = $i1"'$5k,i1 < < ik. A monomial ideal generated by squarefree monomials is called a squarefree monomial ideal. LEMMA 3. A squarefree monomial ideal is radical.

Proof. Let a = (m1,...,mt),m1 =93;1 ---:I:,-,¢. Then a = ($;1,m2,...,mt) n ---n (zik,m2,...,mt) ,

see Section 2.2. If m2 = my, way” then (wivmz, . . . ,mt) = n5n=1(:c,-J.,xjm,m3, . . . ,mt). Continuing like this we see that a is an intersection of ideals of type

(zm1,...,:vm,) = pm, i.e., ideals generated by a set of variables. These ideals are prime according

to Section 2.4. Thus

x/E=x/mn---nmv=x/Hn---m/pzv=mn~-npw=a. Now we turn to the general case.

PROPOSITION 4. Let a = (m1,...,m,) be a monomial ideal with m.- =

23'! ---e§,;’;i,i = 1,...,s and all e“. 76 0. Then «a = (n1,...,ns), where 114' =.’L'i1 ---zik,.,i = 1,...,s.

Proof. Let ci = max{c,-h| k = 1,. . ., k,- }. Then nf‘ 6 a, and hence we have (n1,...,ns)N g awith N 2 c1 +~--+c, —s+1 since (flnl +---+fsn,)N is a sum of terms gm‘i‘I - - mg‘ with d1 + - ~ - +d, = N, so d,- 2 c; for at least one i, therefore every term belongs to a. Since

(n1,...,ns)N (_Z a g (n1,...,n,) we get

I/(n1,...,n.)" 9 fig m but V (n1,...,ns)N = \/(n1,...,n, = (n1,...,ns)

since (n1, . . . ,n,) is squarefree.

El

CHAPTER 2. MONOMIAL IDEALS

42 EXAMPLE 5. 2

2

2

3

_

_

(2?,z2w§,m§x3,x2m4,x4) — ($1,$2$3,$2$3,$2$4:34) -— ($1,$2$3,$4)El

Exercises

1. Let a = (wimgmag) and b = (m§,m3z4) be ideals in k[a:1,22,a:3,z4]. Determine a + b,ab,an b,a: b, and JG. 2. Let a = (2512133) and b = (z1x2,:1:2:n3) be ideals in k[a;1,:cz,:ra]. Determine a + b,ab,an b, a: b, and VS. 3. Write (x1z4,a:2:1:4) as an intersection of prime ideals.

4. Write (x?,x§z§,$§x§,x§z4,mi) as an intersection of ideals of type ,1 , . . ., 2'." u. . (cum 5. Let a be an ideal generated by monomials in k[:c1 , . . . , 27"]. Show that k[a:1, . . . ,xn]/a is a finite dimensional vector space over I: if and only if for each i there is a d,- > 0, such that x?‘ E a.

6. Let a be a monomial ideal such that {2112... ,mfifl} is a part of a minimal system of generators for a, where d; are positive. Show that d1+~-+dn—n+1gdimkk[:r1,...,zn]/agd1d2---d,.

Chapter 3

Grobner Bases Our primary objects of study are factor rings k[a:1, . . . ,wn]/a over a field k. We want to be able to perform calculations in such rings. Elements in k[:r1,...,a:n]/a are cosets f + a where f e k[a:1, . . . ,mn]. The first problem to overcome is that we have no unique way to represent elements in R :-

k[:1:1, . . . ,mn]/a; we have that f+a = g+a in Rifand only if f—g E a. It is clear that a unique way to represent elements in R, a normal form, would

be desirable. When a is generated by monomials we have an ideal situation. If f + a is a coset then f is a linear combination of monomials. We can delete all monomials which belong to a, La, those which are multiples of

generators for a. Thus each element in Man, . . . , 9:1,] /a can be represented as f + a, where f is a linear combination of monomials outside a. This representation is unique, cf.

the introduction to Chapter 2.

It is this

situation we want to imitate in the general case. It is also easy to achieve a normal form in the case of a single variable. An ideal in k[a:] is generated by a single polynomial, a = (g(:1:)). The division algorithm shows that

any polynomial flat) 6 k[:r] can be written f(a:) = q(x)g(:1:) + r(a:), with

deg(r(a‘)) < deg(g(:c)) (or r(a:) = 0) and that r(m) is uniquely defined by f (:17). Thus each element in k[a:]/ (g(a:)) has a representation f1(a:) + (g(:z:)) with deg(f1) < deg(g), and this representation is unique. We will point out some problems we have to overcome when extending the notion of normal form to the case of several variables, the multivariate case. But before

doing that, we will look at an example of how to get normal forms in the

one—variable case slightly differently. Suppose R = k[a:]/ (2:2 — a: + 1) and that we want to calculate the normal form of 1:4 + 23: + 3. We can view the generator 1:2 — :1: + 1 for the ideal as a rule for rewriting polynomials,

namely since 1:2 ——a:+ 1+ (32—m+1) = 0+ (m2 ——:r+1>, we have 19+ (:1:2 — :1: + 1) = w— 1+ (:32 — a: + 1). Thus, when :32 occurs in a polynomial, 43

44

CHAPTER 3. GROBNER BASES

we can replace it with m— 1. In the same way we can replace any polynomial

m2h(m) which is divisible by m2 with (m — l)h(m) since (:1:2 — m + 1)h(m) + (m2 —— m + 1): 0+ (:1:2 —m+ 1), which gives m2h(m)+ (m2 — m+ 1) = (m— 1)h(m) + (m2 — m + 1). We write this as a rule :1:2 —) m -— 1. Hence we

havem4+2m+3+(m2—m+1)= m2m2+2m+3+(m2—m+1)=(m— 1)m2+2m+3+(m2 —:1:+1)=m3 —m2+2m+3+(m2—m+1)=m2(m— 1)+2m+3+(m2—m+1) = (m—1)(m—1)+2m+3+(m2—m+1) =

m2+4+(m2 —m+1)=m—1+4+(m2—m+1)=m+3+ (m—1)m2+2m+3 = m3 —m2+2m+3 = m2(m— 1)+2m+3-) (m—1)2+2m+3=m2+4—>m—1+4=m+3. We now giveanexample which shows that already in one variable we must be a little cautious when

translating a generator to a rewrite rule. Let R = k[m]/ (m3 — m2). It could seem that we have two choices to make a rule, namely m3 --> m2 or

m2 —) m3. But the second choice would give never-terminating sequences of replacements, e.g., m2 —) m3 —> m4 —) -- -. In one variable we should always replace a monomial m" with a polynomial of lower degree. In several variables we have to order all monomials and make sure that we replace a

monomial with “smaller” terms. This will be carried through in the next

section. Next we consider the example R = k[m, y]/ (m2 — my). Here we have two choices to make a rewrite rule, either m2 —> my or my —) m2. It will turn out that both choices will work, we just have to decide which one to use and then stick to that choice. Which one to use will depend on which kind of questions we want to answer. The first choice will give normal forms

of the type 91 (y) + mgz (y) + (m2 —— my), the second will give normal forms of type g(y) + mf (m) + (m2 —- my). Note that in the first choice the normal forms will be linear combinations of 1, yd and my“—1 for d 2 1; in the second we get linear combinations of 1, yd, m“, d 2 1; so the number of monomials

used in each degree is the same for both choices. We will show that this is no coincidence. The last example will pinpoint a more serious difiiculty we

have to overcome. Let R = k[m,y, z]/ (m2 — mz, my — 22). Suppose that we

decide to use the rules m2 —> mz and my —> 22. If we try to get a normal form of m2y we could use the first rule to get m2y —) myz and then the second rule to get myz —) 23. On the other hand we could start with the second

rule to get mzy —) mz2. Neither 23 nor mz2 can be rewritten by means of the rules. This will be resolved by adding a new rule. As soon as we have an ambiguity like this, in a systematic way we will add a new rule which

gives a common answer. In the example we add the rule mz2 —) 23. This is okay since mz2 — 23 = y(m2 — mz) + (z — m)(my — 22) E (m2 — mz,my — 22), so mz2 + (m2 — mz,my — 22) = 23 + (m2 - mz,my — 22). One way of looking at Grobner bases is as a sufliciently large set of rules to get unique normal forms of elements. This will be systematically developed later on.

3.1. M0NOMIAL ORDERINGS

3.1

45

Monomial Orderings

Let k be a field and A = k[a:1, .

,zn]. An element 2:? - - - mi: in A is called

a monomial and an element cm;1 mmf; with c e k \ {O} is called a term. (This terminology is not universally agreed on.) Let 9)? denote the set of all monomials in A. We will introduce an order relation -< on :m. We define 4 to be an admissible order relation on 9)! if it is a total order which is

compatible with the multiplication of monomials, i.e., o For any pair ofmonomials m,n we havem p1 >- p2 >is finite according to Lemma 5, hence after a finite number of steps no element in supp(fN) is divisible by m. We have fN = f — hgl for some polynomial h, and we define fN to be the remainder of f with respect to 91. In the case of one variable we have the usual remainder in the division algorithm.

EXAMPLE 7. Let again f = m? +a:§a:2 +5123, g1 = if —a:1:c2 and let < be Lex. Then f1 = 3mfx2 +103 by the example above, so f2 = fl — 3w2(a;¥ -:r1:1:2) =

32:1 10% +253, which is the remainder of f with respect to 91 since no monomial I] in f2 is divisible with mi. We will now formulate the process we have just described in a way

which makes it easy to generalize to the case of remainder with respect to an ordered sequence of polynomials (gl, . . . , 9,). The remainder can recursively be described as follows. If lm(gl) divides lm(f), let rem(f, (91)) = rem(f — um, (91)), where n is a term chosen such that lt(f) = lt(ngl). If lm(gl) does not divide lm(f), let rem(f, (91)) = lt(f) +rem(f — lt(f), (91)). This gives a recursive definition of rem(f) with respect to 91. Now let (91,... , 9;) be an ordered sequence of nonzero polynomials and f a polynomial. We recursively define rem(f, (91 , . . . , 93)) = rem(f —ngk, (gl, . . . , 95)), where k is the smallest index such that lm(gk) divides lm(f) and n is a. term chosen so that lt(f) = lt(ngk). If no lm(gi) divides lm(f) we define rem(f: (91: - - - 198)) = (lt(f) + rem(f — lt(f)1(gla ' - ' a 98))'

finite since in both cases the leading monomial drops.

3.4

The process is

Definition of Grobner Bases

For a general sequence (g1, . . . , 9,) the remainder is not very well behaved. For instance the result depends on the order of the elements in the sequence. This is seen by the following example.

EXAMPLE 8. Let f = 2¥$2,(91,gz) = (3%,:51102 — 9:3) and let -< be Lex. Then rem(f, (gl,gz)) = 0 but

rem(f, (92,111))

= rem(zfzz — 21(11132 — $3),(92,91))

3.5. HILBERT’S BASIS THEOREM AND NOETHERIAN RINGS

51

rem($1$§: (92, 91)) rem($1$§ —' 502001932 — $3),(92,91)) T311433, (92,90) = 1‘3C]

We have already noted the following fact, but since it will be used several times we formulate it as a lemma. LEMMA9. Let f,gl,...,gs E k[1:1,...,:1:,,]. Then f—rem(f, (91,...gs)) E (91,...93). In particular, ifrem(f, (91,...gs)) = 0 then f E (91,...98).

Our next aim is now to choose a generating set 91, . . . , gjv for a = (91 , . . . 9,) so that the remainder of a polynomial with respect to (g1, . . . , gjv) is unique in the sense that it only depends on the ideal, so we want to achieve that rem(f1, (91,. . .,g}v)) = rem(f2, (gi, . . . ,gfv» ifand only if f1 — f2 6 a. DEFINITION 10. Let a be an ideal in k[z1,...,a:,,].

A set {gl,...gs} of

elements in a such that (lm(gl), . . . , lm(g,)) = 1(a) is called a Grb’bner basis for a.

First we note that a Grobner basis for an ideal is a generating set for the ideal. LEMMA 11. If {91, . . .95} is a Gr6bner basis for a, then (gl, . . .g,) = a. Proof. Clearly (91,...93) g; a since 9,- E a for all i. Let f e a. Then lm(f) E (lm(gl), . . . ,lm(g,)), hence lm(f — ngk) < lm(f) for some 9], and some term n. Since f —ng;c E a, we get by recursiveness that f E (91,. . . 9,). III

3.5

Hilbert’s Basis Theorem and Noetherian

Rings We get as a corollary of Lemma 11 that every ideal a in a polynomial ring is finitely generated. This is a classical and important result by the German mathematician David Hilbert from 1891. COROLLARY 12 (HILBERT’S BASIS THEOREM). Everyideal in a polynomial ring over a field is finitely generated. Proof. Being a monomial ideal, we know that l (a) is finitely generated by Dickson’s Lemma. Let 1(a) = (m1 ,mn), and choose f,- e a such that lm(fi) = mai = 1,...,N. Then a = (f1,...,fN) by Lemma 11. El

52

CHAPTER 3. GROBNER BASES

A ring in which every ideal is finitely generated is called Noethen'an after the German mathematician Emmy Noether. There are different ways to state the condition of being Noetherian. PROPOSITION 13. Let A be a ring. The following conditions are equivalent. Every ideal in A is finitely generated.

Every strictly ascending sequence of ideals a1 C a2 C --- is finite. Every ascending sequence of ideals a1 g a2 9 -- - is eventually constant, i.e., there is an n such that an = an+1 = - - -.

0 Every nonempty set of ideals M = {ad} has a maximal element, i.e., there is an an“, E M such that no other ideal in M strictly contains

0a., . REMARK. The second (or third) condition is called the ascending chain condition for ideals, the ACC. The last condition is called the maximum condition for ideals.

Proof of Proposition 13. Suppose that every ideal in A is finitely generated and that a1 C (12 C - - - is a strictly ascending sequence of ideals in A. Then

a = Ua; is an ideal. Since a is finitely generated the chain must stop at an as soon as all generators of a are contained in an. That the second

and third conditions are equivalent is trivial. If some nonempty set M of ideals doesn’t have a maximal element, we can construct an infinite strictly ascending chain of ideals in M. Thus the second condition implies the fourth. Assume that the maximum condition holds and that a1 C (:2 C - --

is a strictly ascending sequence of ideals in A. Let an be a maximal element in M = {a}. Then the sequence stops with an, so the maximum condition

implies ACC. If some ideal a in A can’t be generated by finitely many elements, we could construct an infinite strictly increasing sequence of ideals

in A by adding new generators one at a time, so ACC implies that every ideal is finitely generated. El

3.6

Grfibner Bases and Normal Forms

The importance of a Grobner basis for an ideal comes from the fact that for every polynomial in k[a:1, . . . ,xn] one can calculate a unique normal form

with respect to the ideal with the aid of the Grobner basis.

PROPOSITION 14. Let {gl,...g,} be a Grobner basis for the ideal a = (91,...95). Then rem(f1,(gl,...g,)) = rem(f2,(g1,...gs)) if and only if

f1 — f2 6 a. In particular rem(f, (gl, . . . ,g,)) = 0 ifand only iff e a.

3. 7. REDUCED GRO'BNER BASES

53

Proof. Suppose rem(f1,G) = rem(f2,G), where G = (91,...gs). Since f,- — rem(f,-,G) E a,i = 1,2, cf. Lemma 9 in Section 3.4, we get f1 — rem(f1,G)—(f2—rem(f2,G)) = f1—f2 e (1. Now suppose fl—fz E a. Then

rem(f1,G)—rem(f2,G) = (f2—I‘em(f2,GD—(fl—I'emUl,G))+(f1-f2) e a since f; — rem(f,,G) E a,i = 1,2. But rem(f,,G),z' = 1,2, is a linear combination of monomials outside 1(a). If rem(f1,G) ¢ rem(f2,G) we would have lm(rem(f1,G)-rem(f2,G)) ¢ a. But rem(f1,G)—rem(f2,G) e

a gives lm(rem(f1, G) — rem(f2,G)) e a, a contradiction.

D

We will formulate what we have shown in a somewhat different way. Since f —rem(f, G) 6 a we could represent each element f in k[:c1 , . . . , wn]/a with a. polynomial rem(f, G) which is a linear combination of monomials outside 1(a), so these monomials generate k[a:1, . . . , mn]/a as a vector space over It. The last argument in the proof shows that these monomials are

linearly independent modulo (1, hence the monomials in k[a:1, . . . , mu] \ l(a) constitute a vector space basis for Man, . . . , mn]/a. We have derived a way to obtain a unique representation of the elements in k[:1:1, . . . , zn]/a (if we have a Gr6bner basis). For reference we formulate a proposition.

PROPOSITION 15. Let a be an ideal in k[a:1, . . . , run]. The monomials outside 1(a) constitute a k-basis of Man, . . . ,mn]/a. It follows from the definition of Gr6bner basis that any permutation of a Grobner basis is a Gr6bner basis. Since for any f E k[a:1, . . . , 3"] we have

that rem(f, G) is a linear combination of monomials outside l(a), and these monomials are linearly independent modulo a, we get that rem(f, G) = rem(f, G’) if G and G’ are two Grébner bases of a (for the same ordering). In particular the remainder does not depend on the order of the elements in

a Griibner basis. Thus, given a fixed ordering, rem (f, G) will be uniquely defined and does not depend on the Grébner basis G of a. We will call this unique element the normal form of f and denote it by NF(f). For practical reasons it is convenient to have a Gréibner basis as small as possible. We will see that there is a natural choice. We will give conditions on the Gréibner basis which will force the Gr6bner basis to be unique.

3.7

Reduced Gréibner Bases

Gr6bner bases are not unique. For any Grébner basis G of a and any element f E a we can extend G with f and get a new Gr6bner basis for a. We will now introduce the concept of reduced Grb‘bner basis and prove that

reduced Griibner bases are unique. If a = (91, . . . , 9,) and G = {91, . . . , 9,} is a Grobner basis, we define G to be reduced if

54

CHAPTER 3. GROBNER BASES' o {lm(gl), . . . , lm(g,)} constitutes a minimal set of generators for 1(a) o g,- are monic, i.e., lc(g,-) = l,i = l, . . .,s o no lm(gi) divides any monomial in supp(gj),i ¢ j Given a Grobner basis G it is easy to construct a reduced Grobner basis.

First we pick a subset G’ = {g;,, . . . , 9,3} of G so that the first condition in the definition is fulfilled. Then we multiply each 9;, with lc(g,',)‘l so that we get monic elements. To finish, we take the remainder of each 9,,

with respect to G’ \ {g,-,}. Thus reduced Grobner bases exist. They are also unique since if {91, . . . , 9,} and {h1, . . . , hs} are two reduced Grébner

bases for a with 1m(g,:) = 1m(h.-), we get that g,- — h,- E a. If g,- — h; at 0 we

would have lm(gi - hi) 6 l(a), but the leading terms in g,- and hi are equal, hence cancel in g,- — hi. But gi — h,- is a linear combination of monomials

outside l(a), which gives a contradiction. EXAMPLE 16. If G is the reduced Grobner basis for a in any ordering, then

a=k[1:1,...,a:n]ifandonlyifG={l}.

El

EXAMPLE 17. Different orderings of course give different Grobner bases.

Let a = (53%,231173 — 2:3). Then {xizg — m1x3} is the reduced Grobner basis for a in Degrevlex. In Lex or Deglex, however, the reduced Grobner

basis is {wins-$21M -- 2:333}.

El

Exercise 1.

{3? + :52 + 31:53 + 21m2,x1x§ + 21,3:3 + m2,m¥m§ — x2,x1mg+ 2:132}

is a Grobner basis for the ideal a = (x? + $2, :6l + 2:1) in Lex. Construct the reduced Grobner basis for a.

3.8

Construction of Griibner Bases

A Grobner basis for an ideal 0 gives us normal forms for elements modulo a. This will help us to solve lots of problems. We have seen that Grobner bases do exist, but this does not help us much if we don’t know how to calculate them. The remaining parts of this chapter will be dedicated to

how we construct a Grobner basis for an ideal a = (fl, . . . , fk) given by generators. However, we require some new concepts and some theory, so a

little patience may be needed.

3.9. FREE MODULES AND SYZYGIES

3.9

55

Free Modules and Syzygies

Let A be a ring and let A" = { (a1, . . . ,an) I ai E A}. We define an addition ofelements in A" by (a1,. . . ,an) +(b1,.. .,bn) = (a1 +b1, . . . ,an+bn) and a multiplication of elements in A" with elements in A by a(a1, . . . , an) = (M1,..., aan). With these operations A” is called a free module. A" has a basis e1 = (1,0,...,0),e2 = (0,1,...,0)...,en = (0,0,...,1) in the sense that every element in A" is a linear combination, with coefficients in A, of e1, . . . , en and furthermore that e1, . . . , en are linearly independent, i.e., if a1e1 +

+a,.en = (0,0,...,0) then a; = O for i = 1,...,n. All rules

for addition and multiplication which are true for vector spaces are also true for free modules. In fact, if A is a field then A" is nothing but an n-dimensional vector space. On the other hand, if n = 1 we can identify A1 with A. Thus the concept of free module is a common generalization of the concepts of vector space and ring. A nonempty subset M of A" is called a submodule of A" if M is closed under addition and multiplication with elements in A. Thus M g A" is a submodule of A" if M 76 0 and the following two conditions apply: 0 (a1,...,a,,),(b1,...,b,,) 6 M implies that (al +b1,...,an+b,,) e M o (a1,...,an) E M,a 6 A implies that (aa1,...,aa,,) E M In case A is a field, a submodule is just a subspace of A". In case n = 1, a submodule is just an ideal in A.

Let F = (f1...,f,,) e A". An element 3 = (31...,sn) in A" such + snfn = 0 is called a syzygy of F. The set of all syzygies that .91 f1 +

of F constitute a submodule of A”. From now on let A = k[m1, . . . ,zn]. Let F = (fl, . . . , fk) be a fixed element of A". We say that an element

H = (h1,... ,hk) e A" represents f 6 A with respect to F, if h1f1 + - - - + fkhk = f. An element 5 e A" representing 0 with respect to F is just a syzygy of F. If H and H’ both represent f with respect to F, then their difference H — H’ is a syzygy of F. If H represents f and S is a syzygy of F, then H + S represents f with respect to F. Exercises

1. Let F e A”. Prove that the set of syzygies of F constitute a submodule of A". 2. Let A = k[x1,a:2] and let F = (2:1, 2:2). Determine all syzygies of F.

3. Let F = (fl, . . . ,fk) be a fixed element of A". Show that if H and H’ both represent f with respect to F, then their difference H — H’ is a

56

CHAPTER 3. GROBNER BASES syzygy of F. Show that if H represents f with respect to F and S is a syzygy of F, then H + .5' represents f with respect to F.

3.10

Syzygies of Sequences of Monomials

In general it is not a trivial task to determine the syzygies of a sequence of

elements F = (fl, . . . , fk) E A". If the fi’s are monomials the task is much easier. We begin with an example. Let F = (m1, m2) = (wfwgmg, 23.2%) and suppose (a, b) is a syzygy of F. We have aml +bm2 = 0, so am; = —bm2 E

(m2), hence a 6 (m2) : (m1) = (mg/gcd(m1,m2)) =. (flaws/$233) = (2?), so a, = mfal. In the same way we see that b 6 (m1) : (m2) = (ml/gcd(m1, m2)) = (mgzg/mgws) = (2:25:23), so b = 2:22:3b1. We get aml + bmz = zgalml + z2n1m2 = fixgxgal + x§z§z§bl = zfzgmflal + b1) = 0, so a1 + bl = 0. Hence every syzygy is a multiple of (m2/ gcd(m1,m2), —m1/gcd(m1,m2)) = (17%: “$233)-

If F = (m1 , m2, m3) is a sequence of three monomials it is easy to find three syzygies, namely

512

= =

("12/ gcd(m1,m2), -m1/ gcd(m1,m2), 0) (mz/ 80d(m1,m2))€1 — (ml/ n(m1,m2))€2,

813 = ("13/ n(m1:m3)) 0, -m1/ 80d(m1sm3)) 323

(m3/ n(m1,m3))€1 — (ml/ gcd(m1,m3))e3, and (0, m3 / gcd(m2, m3), —m2/ gcd(m2, m3)) (m3/ gcd(m2,m3))e2 — (mg/ gcd(m2, m3))e3.

If for example F = (wfwgw3,m§m3,x¥m§) then (32, —:c¥,0), (1:3,O, —a:2) and (0,2:fiwg, —a:§) are syzygies. The next proposition shows that for sequences with three or more monomials it suffices to calculate these pairwise

syzygies, since every syzygy is a linear combination of them. Note that n/ gcd(m, n) = lcm(m, n) /m. PROPOSITION 18. Let m1, . . . ,mk be monomials. IfS = (.91, . . . ,sk) is a syzygy of (m1, . . . ,mk) then S is a linear combination of the syzygjes Sij =

lcm(mi’ mi) ,,

vi _

mi

wherelSi lm(f) or max {lm(qifi)} =1m(f). If max {lm(qif¢)} = lm(f) we are ready since then lm(f) = lm(qi) lm(fi) for some i, so lm(f) E (lm(fl), . . . , lm(fk». Now assume that m = max {lm(qif;)} > lm(f) and then we denote the set

{i| lm(qifi) = max {lm(qifi)}} = {z}, . . . ,it} = I. The idea is now to add a syzygy Z to Q = (q1,.. .,qk) to get another representation Q’ = Q + Z for f with max {lm(qfiffl} -< max {lm(qifo}. Since there is no infinite decreasing sequence of monomials, see Lemma 5, we will after a finite

number of steps come to the first case max {lm(q;f,-)} = lm(f), and we can conclude the result. If max {lm(qif,)} > lm(f) we must have that the leading terms in Ziel (15f; cancel. But then SQ = (31, . . . ,sk), where s, = lt(qi) ifi e I and 3, = 0 otherwise, is a syzygy for (lm(fl), . . . , lm(fk)). Hence SQ is a linear combination of 3ij ’8 according to Proposition 18, say

SQ: 2131C[a:1,...,:u,,]/q1 x

xC[a:1,...,a:,.]/qs

by ¢(f) = (f + m, . . . ,f + q,). It is easy to see that (15 is a homomorphism. The kernel equals {flf+cli=Qiyi=law-,3}={flf6‘li:7;=1,---,3}=ng=1Qi=ai

hence d) gives an isomorphism 2,0 : C[2:1, . .. ,zn]/a —) im(¢), defined by ¢(f + a) = (f + q1,. ..,f + q,)), according to Theorem 61 in Section 1.7. It remains to show that 45 is surjective. We show this by induction on 3. If

s = 2 there are q,- E qi,z' = 1,2, such that q1 + q2 = 1 since q1 + qz = (1).

Hence ¢(q1) = (0+q1,1+q2) and ¢(q2) = (1+q1,0+q2) so ¢(f2q1 +fiqz) = (fl + m, f2 + qz) and 45 is surjective. By induction we have C[:1:1,...,a:n]/ 0:11 q; : C[a:1,...,':vn]/q1 x

x C[:1;1,...,a:n]/q,_1,

so for each (s — 1)-tuple (f1,...,f,_1) there is an f such that f + q.- = fi+q,-,i = 1,...,s— 1. By Lemma3, qln---nq,_1 +qs = (1), so there exist a e qln- - -nq,-1 and b 6 q, such that a+b = 1. Let (f1, . . . ,f,) be any s-tuple of polynomials. By induction there is an f such that f + q.- = f.- +q.forz': 1,...,s—1.Sincea€ qi forz' - yl > 112 > ya. The result is {$3 - 311,32311 — y2,a:y3 — yiyzwyf — 313,113 — gill/2,2412% - 111‘,

111113 — 313.113 - 21?}Hence

k[a:3,a:5, m7] 2* k[y1,y2,ys]/ (1153‘ - yfyzwzya — yiyiys — 113.113- 1/?)-

96

CHAPTER 7. APPLICATIONS OF GROBNER BASES

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

21% - y? = -y2(y1y3 — 31%) + y1(y2ya — 31?), so yg — y? is not needed as a generator.

El

Exercise

1. Write k[a:3, 2:7, ms] 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,...,f,) and b = (91,...,g3) 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: (1+ [1 = (f1,...,f,,g1,...,g,) and a - b = (f1 91, . . . , 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 Proposi-

tion 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,...,g,) be ideals in k[a:1, . . . , mn]. Consider the ideal

c=(yf1,.--,yfn(1-y)91,---,(l-y)gs) in k[x1,...,a:n,y]. Then anb=c0k[:c1,...,a:n].

Proof. Suppose f E anb. 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 cnk[a:1,...,a:n]. Then f: h1($12- - -1$nvy)yf1(a71v "’xn) +

+hr(xl:"'azn:y)yff(mlv ..,.’En)+

hr+1(m1,...,a:my)(1 —y)gl($1,---,$n) + "'+

7.4. IDEAL OPERATIONS

97

flu-+8051, - - - filmyxl — y)gs($1, - - - ’98:)-

Since f E k[a:1, . . . ,mn] we get f=hr+1($1)'“,$n)0)gl(mla'--ymn)+"'+

hr+s($ls - ' - ,2)", 0)g,(-’I71, - - ' #6106 b

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

h,(a;1,...,a;n,1)f,.(a:1,...,:cn) 6 a. U

EXAMPLE 11. Let a = (11:2 +y2,a:y) and b = (2:2 — 112). Let c = (2:2t + y2t,a:yt, L732 — ty2 — 11:2 + 3/2) . A Gr6bner basis for c in Lex with t > :1: > y is

{2m2t — :32 — y2,a:yt, 2y2t + 2:2 + 212,1:3 — $312,223; — gs}. Hence an I) = ($3 — :vy2,a:2y — 313).

El

Exercise

1. Determine (2:3 + 11:,xy + 312) n (3/3 + 11,231 + 2:2) in k[:c,y]. 7.4.2

Ideal Quotient

Let a be an ideal and b = (91, . . . , 9;); see Proposition 45.3 in Section 1.3.5.

Then a : b = nf=1a : (9;). Hence it is sufficient to consider the case when E: is principal, since we know how to perform intersections.

PROPOSITION 12. Huh (9) = (h1,...,ht) then a:(g)=(h1/g,...,ht/g). Proof. Let f e a : (9). Then fg 6 an (g). Let h’ e an (9). Then h’ = gh for some hand h E a: (9). Thus h’ 6 an(g) ifand onlyifh’/g e a: (9). El

EXAMPLE 13. Let a = ($2+y2,xy) and b = (2:2 —y2). Since an [1 = (x3 — $112,291; — 113) we get a: b = (x,y). El Exercise

1. Determine (m3 + m,a:y + yz) : (3/3 + y,my + 1:2) in k[a:,y].

CHAPTER 7. APPLICATIONS OF GRéBNER BASES

98

7.5

Supplementary Exercises

1. Show that xsy — 2:341:12 6 (mg/3 - 2233/ + 32:2y2, $3342 + 52:21:13). 2. Show that 11:10 + 1:16 E Z2[a:4,a:6 + 37,110]. 3. Show that

\/(m2—m+mz—z,a:y—y+y2,y+yz+z2+2z+1) = (z—1,y,z+1). Let b = (my - y,a:z — 1:,yz — 2). Determine b n k[a:, y] and b n k[:c].

Let R = k[a:1,...,a:n]/ (91,...,g,) = k[:v1,...,xn]/a. Define a homo— morphism ¢ : k[y1, . . . ,y,] —+ R by ¢(y.-) = f,- +a. Show that ker(¢) =

((1/1 — f1,---,yr —fr)+a)nk[y1)"')y1‘]‘ Show that R = Q[a:]/ (9:3 + 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 polynomial equation of degree at most 3 with coeflicients in Q. (Hint: Since dimQ R = 3, we get that l, a, a2, a3 are linearly dependent.) Determine an equation for (the image of) a: + 1.

Write k[a,-4 + 22,33] as k[y1,y2]/a. Write k[z4,a:3y,my3,y4] as a factor ring of fizz/1,112,313,314]. Show that (3/2 — xz,yz — a: 3,z2 —x2y> is a prime ideal in k[:c,y,z]-

Hint: Write k[t3, t4, t5] as a factor ring of Min, 9, 2]-

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[:v1, . . . ,mn]. 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 23 +224;2 is homogeneous but 2:3 +2.1:y2 -a:y is not. The homogeneous elements of degree 1' in A constitute together with {O} a vector space A,over k with the set of monomials of degree 2‘ 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 i + j. Every element f E A \ {0}

can be uniquely written as f = fo + f1 + ~ - ~ + fk, where f; 6 Ai. The fi’s are called the homogeneous components of 1‘. Another way to express this is to say that A is the direct sum of the Ai’s, A = 6920A“ 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

El

100

CHAPTER 8. HOMOGENEOUS ALGEBRAS

LEMMA 2. An ideal a is homogeneous if and only if the following condition holds. If F 6 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 11. Then F = g1f1+ -+ gkfk Let F- be the homogeneous component of degree j of F. Then F— - 90“?leg f1)f1 +- -+ 9;"._deg f") f], where 9,“ deg f‘) is the homogeneous component of degree (j — deg fg) of gi. Hence .Fj E a. For the converse let a be generated by M, . . . , hk. Each

homogeneous component h?) of each h.- belongs to a. Then a is generated

by the 119’s.

1:

REMARK. Another way to express the statement of the lemma is the following. Let a be a homogeneous ideal and let a5 be the k-vector space of homogeneous polynomials of degree 2’ in a. Then a = flaizoai.

DEFINITION 3 (HOMOGENEOUS ALGEBRA). A ring R = k[a:1, . . .,:c,.]/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. LEMMA4. If R— — k[a:1,.. .,;'rn]/a is a homogeneous algebra then any ele-

ment in R can be written f: fo +-

+ fk, where f, is homogeneous of

degree 2 or zero, and where f.- is unique. Equivalently, R: 69R.- where R,-

is the image of A; in the natural map A —) A/a = R. On the other hand, ifR = k[rl, . . . ,xn]/a and R = 69R,- where R; is the image on, then a is homogeneous. Proof. Since any f E A can be written f= fo + + fk with f,- 6 A5, then any element_f 6 R can be written 1‘: fo + + fl, f.- 6 __S__uppose Ag.

f= fo+

+fk=90+

+9k=9,50fo-90+

This means that f— g— — (f0 — go) +

+fk-gk=0-

+ (f1. — gk) 6 01. But f,- —

'

the homogeneous comp—onent of degree i of f— 9, so f,-— g,- e a since a is homogeneous. Hence f_,-- gi— 0, so fi— m. Now suppose that R: 69R. where R5: im A5. This means that fo + + fk —- fo +- -+ fk— 0 if and only iff, =0 for everyi,i..,e fo+c -+fk E aifandonly iff; E aforall 2', Le, if and only if a is homogeneous. El COROLLARY 5. Let a g b be homogeneous ideals in Man, . . . ,mn] and let R = k[a:1,...,mn]/a = EBizoRi,S = k[:v1,...,2:n]/b = 635205;. Then S,- : ¢(R,-), where (15 : 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 f1 — f2 6 a g b, so

f1 + b = f2 + 5. Hence ()5 is well defined, and it is easy to check that q) is a homomorphism.

[j

The set of homogeneous ideals behave well under ideal operations. LEMMA 6. If a and b are homogeneous ideals in k[a:1,.. . ,zn] then a + b, ab, a 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 a n b 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) = (g1,. ..,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 = Fil + + Fim 6 J5,n G Aij,z'1 < - - - < im. Then F" E a for some n. The homogeneous component of lowest degree in F" is Fi'l'. Since a is homogeneous we have F}: 6 a, so F,-1 6 J5. Hence F — F11 = Fig + - - - + Fin. 6 x/E and we are ready to use induction on m.

III

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, 0. ¢ a implies that b E a for homogeneous elements a, b, then a is a prime ideal.

2. If ab 6 a,a. ¢ a implies that b" 6 a for some n for homogeneous elements a, b, then a is a primary ideal.

Proof. 1. Suppose that fg e awhere f = fo+- - ~+fk andg = go+- - -+gm with f5, 9,- 6 Ai. Suppose that f 93 a. Then f; g! a for some 2', let to be the smallest such t. The homogeneous component of degree 2'0 of fg is fog;o + - ' - + fie go. Since fg e a and a is homogeneous we get that this component belongs to a. Since all terms except possibly fiogo belong to a, we get that fiogo, and thus go, belongs to a. Then f(g — go) 6 a and we can repeat the same argument to get g1 G a. We are ready to use induction on m.

102

CHAPTER 8. HOMOGENEOUS ALGEBRAS

2. Suppose fg E a where f = fi, +---+fi.. and g = gjl + -~-+gj, with fi,-:9j.- homogeneous components. Suppose f ¢ a. If ft, 6 a then (f — fi1)g E a, f — fil ¢ a, so we can just as well assume that fgl gt a.

If we can show that for each i there is an m,- such that 9'? E a if m = max{mi}. The homogeneous component of lowest degree in fg is f11 9,1 Since fg e a and a is homogeneous we have fixgj1 E a. Since fil 9! a we have 9'"1 E a for some m1. Now suppose we have proved

91-": ,.

, 91"" E a. LetM: max{mj1,...,mj,}. Consider

F = f(gj.+1+---+gj.)" = f(g - (91-1 +---+gj.))" =

1)n_'9i(91'1 +"'+gj.)"“ = fZ i=0 C )(

(-1)"f(gj.+-- my”:(’f)(—1)"-‘g‘ 1wehaveq;:(g)=(1)ifi>l,soa:(g)=q1:(g) whichisimprimary. We consider the set of ideals a : (g) with g homogeneous such that a : (g) is {DI-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 idea], it suflices to prove that if a: and y are homogeneous elements such that my 6 a : (f) ,2: g! a: (f), then 3; e a : (f); see Lemma 7.1. Suppose my 6 a:(f),:1: ¢ :1 : (f) and 3,3] homogeneous. Then y E a : (fr) and a : (far) 96 (1) since fa: g! a. Since a : (fr) is homogeneous (Lemma 6), we have a : (fro) g 932. Since

mtgx/awxmx/azuzmx/fiwt, so 1/a : (fx) = am, so a : (fzt) is flit-primary by Proposition 17 in Chapter 5. But a : (fr) = a: (f) since a : (f) g a : (fr), fa: is homogeneous, a : (far) is DIR-primary, and a: (f) 18 maximal. Thus 3/ E a : (f) and a: (f) is aprime ideal. But then a: (f)_= Va: (f) =9fl, which gives f9)! 9 a. For the factor ring R this means f--9Jt= 0. Now suppose that 931 does not belong to a. By

Theorem 15 in Chapter 5 it suflices to show that there is a homogeneous element in 9)! which does not belong to m U - - - U 11,. Suppose the contrary, that 93! g p1 U ~ 0 - U 31,. By deleting some pi if necessary, we can assume that M g Ui¢jpi for every j. Take for each j a homogeneous element 2.3 e 932, 9:1 65dUi¢i Let degxl = d1 and degmzu = d2. Consider the element 11:12 + (1'2" 2:8)‘11. This 1s a homogeneousmelement (of degree dldz). Since 331 g! U191”); and all p,’s are prime ideals, we have 21 ¢ U4¢1p.-. Since 2:1 e Uf_1p,- we must have :121 6 p1, and thus $111 6 p1. As above

x2 6 p2,“ ,m, 6 33,. Thus 2:2

4:3 6 p, ifz' > 1, so (awn-11:3)"ll E )3; if

13> 1. As above 2:2,...,:r, ¢ p1, so ($2 ---z,)dl ¢ p1 and 52:2 Mars)“ 6 )1.if i > 1. We claim that the homogeneous element y = 2:12 + ($2 "-$8)d1

8.2. HOMOGENIZING AND DEHOMOGENIZING

105

does not belong to any pi, since ify 6 p1 then y—z‘f’ = (:1.'2---a:,.,)“ll 6 m, a contradiction, and if y e p,- for somez' > 1 then y — (1:2 - - -:z:,)‘11 = $112 6 33;, a contradiction. Finally suppose that k is an infinite field. For each i we have that the elements of p; of degree 1 constitute a k-subspace V,- of the space V of all elements of degree 1, and that V,- aé V. If all elements of degree 1 belong to some p;, we get V = U§=1V,-. But a vector space over an

infinite field cannot be a union of a finite number of proper subspaces.

I]

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[a:,y]/ (32y +3312). 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 Man, . . . , 23"] there is a natural way that to construct a homogeneous polynomial (or homogeneous ideal) in a polynomial ring k[:1:1, . . . , mmy] with one more variable.

8.2.1

Homogenizing Polynomials

Let f = fo +

+ fk E k[a:1,...,:rn] = A, f,- homogeneous of degree 2',

fl: 75 0DEFINITION 13 (HOMOGENIZATION OF A POLYNOMIAL). The homogenization of f with respect to y is

h(f) = fol/k + f1yk_1 + - - - + fk—ly + fl: 6 k[$1,- - - 411,11]Thus for example h(1 + 312:2 + $3) = y3 + 2:12:23; + :33. It is practical to have a. formula for homogenizing.

LEMMA 14. We have h(f(a:1,...,a:,,)) = ydegff(:rl/y,...,a:,./y). Proof. Iff = fo +~--+fk,fk 9E 0, we have

h(f): foy" +"'+fk =yk(f0+f1/y+‘“+fk/yk) and it is easily checked that fi/y‘ = f,-(z1/y. . . . ,mn/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) = degf + deg 9, so

h(fg) = ydegf+deggfg(x1/y, - - - ,xn/y) = ydes f+deggf(xl/y, . . . , mn/y)g($1/3/, - - , «Tn/y) =

yde"f(m1/y,- - -,=vn/y)yde“g(m1/y,~ - ,wn/y) = h(f)h(9)Homogenization works less well with sums, unless deg f = degg

deg(f + 9)EXAMPLE 16. If f = 221,9 = 27% then h(f) = f, h(g) = g, and h(f +g) = any + 2:? aé h(f) + h(g). If f = 1 +m1w2,g = 11:1 —a:1a:2, we have h(f) =

3/2 + z1$2ah(9) = $1.11 — 101-732, h(f + 9) = y + 1'1 ¢ h(f) + h(9)-

D

LEMMA 17. We have h(f + g) = ydez(f+y)-degfh(f) + ydeg(f+9)-deggh(g).

In particular, ifdegf = degg = deg(f+g), wehave h(f+g) = h(f) +h(g). Proof. We have

h(f + g) = yde‘(’+9)(f(m1/y,---,a:n/y) + 9(31/9, - - - .wn/y» = ydeg"+9)‘d°‘ f31“"flan/y, - - - ,wn/y)+ yd°‘(’+”)“'°‘”yde‘ggcvi/y, . - - ,xn/y) = ydes(f+9)-des fh(f) + ydes(f+9)—deg 911(9).

Exercise

1. Let f(:r) E k[a;]. 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[:z;1, . . . ,mn]. Then h(a) 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. If a = (f) then h(a) = (h(f)). Proof. It is clear that (h(f)) Q h((f)). If g e (f) then 9 = fgl for some

all(13;! h(g) = h(fgi) = h(f)h(91) 6 (MB) by Lemma 15, hence MM) 9 .

El

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

might be strict inequality. If a = (311:2 — 1,27% — 22) then :53 — any 6 h(a) but 2:3 — 21y ¢(:1:1:1:2 —y2,z¥ — 1,132).

El

Exercises

1. Show that if a = (:1:l — 1,3} — 272), then :63 — any 6 h(a) but 2:3 — 11:11! ¢ ($1532 — yzix¥ — 3/332)-

8.2.3

Dehomogenizing Polynomials

There is an Opposite operation to homogenization: for each homogeneous

polynomial we can define a polynomial in a polynomial ring with one less variable. In fact there are just as many ways to dehomogenize a homogeneous polynomial as there are variables.

DEFINITION 21 (DEHOMOGENIZATION OF A POLYNOMIAL). If the polynomial f (9:1 , . . . , :13", y) e k[a:1 , . . . , a3", y] is homogeneous, then the polynomial a(f) = f (:31, . . . ,xn, 1) is the dehomogenization of f with respect to 3/. If f and g are homogeneous of the same degree, f + g is homogeneous

(or 0) and a(f + g) = a(f) + (1(9), and for all homogeneous f, 9 we have that fg is homogeneous and a(fg) = a(f)a(g).

8.2.4

Dehomogenizing Ideals

We can define the operation of dehomogenization not only on homogeneous polynomials, but also on homogeneous ideals.

108

CHAPTER 8. HOMOGENEOUS ALGEBRAS

DEFINITION 22 (DEHOMOGENIZATION OF AN IDEAL). If the ideal a in the ring k[:1:1, . . .,:1:n, y] is homogeneous, then a(a) is the ideal in Man, . . . ,mn], generated by the dehomogenizations of all homogeneous elements in a. The map d:k[:1:1,...,a:,,,y] —) k[a:1,...,a:,.]

defined by d(f(m1, . . . ,zn,y)) = f(a:1, . . . ,2”, 1) is a homomorphism, so if

a is an ideal in k[m1,...,zn,y] then d(a) = {d(f)| f E a} is an ideal in k[a:1, . . . , 17“]. We claim that if a is homogeneous then d(a) is generated by the images of homogeneous elements in a. Let f = fo + - - - + fk E a, where fi is homogeneous of degree 1. Then d(f) = d(fo) + d(f1) + + d(fk) and f.- E a since a is homogeneous, so d(fg) E a(a). Thus it is clear that d(a) = a(a).

8.2.5

Homogenization versus Dehomogenization

The operations of homogenization and dehomogenization are almost inverse

to each other. More exactly, if f e k[x1,... ,mn] = A then a(h(f)) = f, but if g is a homogeneous polynomial in A[y] then h(a(f)) = 31""f where y'" is the largest power of 3/ which divides f. In particular, if y does not divide f then h(a(f)) = f. For an ideal a in A we get a(h(a)) = a. LEMMA 23. Let a be an ideal in A and let f be a. homogeneous element in

A[y]. Then f e h(a) if and only if f = y'”h(g) for some 9 e a and some m. Proof. If f = ymh(g) then obviously f E h(a). If f E h(a) then f = Egih(f,-) for some homogeneous g.- 6 A[y] and some f,- E a.

We can

suppose that deg(g.-h(1%)) = deg(f) for all 2', because all other terms must cancel. Let y’"" be the largest power of y that divides g,- and let 9,- = y’"‘ 9;. Then

29mm) = Zymigéhm) = Zym‘h(a(g£))h(fe) = 211"" h(a(g£)f.')If min {m.} = m this sum equals 31'" z ymi—mh(a(g£)f,-) = y’" E h(a.(g£)f,-) and a(g£)f; e a. El LEMMA 24. If a is an ideal in k[:1:1, . . . ,zn] then h(a) : (gm) = h(a) for each

m. Proof. That h(a) Q h(a) : (ym) is clear. If y'"f e h(a) for some homogeneous f 6 Han, . . . ,zn] then ymf = y"h(g) for some 9 E a and n 2 m since y does not divide h(g). Thus y’"'"f = h(g) and y’"‘"f 6 h(a), so

f = y"‘"‘(y"“"f) E h(a)

'3

8.2. HOMOGENIZLNG AND DEHOMOGENIZING

109

For ideals we have:

LEMMA 25. If a is a homogeneous ideal in k[z1, . . . ,a:my] then a g h(a(a)) and y’"h(a(a)) g a for some positive integer m. Proof. Suppose f is a homogeneous element in (1. Then a(f) E a(a), hence

y“"f = h(a(f)) e h(a(a)) for some k, so f = y"h(a(f)) 6 h(a(a)) for some k.

Thus a g h(a(a)). Now suppose a = (f1,..., fr).

h(a(f;)) = y—m" f5. Let m = max{m,-}. Then y'"h(a(a)) g a.

We have

I]

LEMMA 26. Let a and b be two homogeneous ideals in k[a:1 , . . . , mm y]. Then

a(a) = 0(6) if and only ifa : (ym) = b : (ym) for some integer m. Proof. We have for each m 2 1 that a(a : (ym)) = a(a) : a((ym)) = a(a) : (1) = a(a). (Here we have used that a(a : b) = a(a) : a(b); see the exercises below.) If a : (3,“) = b : (34“) we get

0(a) = “(‘0 = (1) = (1(0) = “((yml) = a(a = (11m)) = “(b I (31ml) = 0(5) 20((3lmll = 0(5) 1 (1) = 0(5)Now assume that a(a) = a(b). Since a : (y‘) g a : (y‘H) we have by Proposition 13 in Section 3.5 that a : (31“) = a : (ym‘H) = - -- for some m, and we can assume also that b : (ym) = b : (ym+1) = by choosing m large enough. Since

a(fi= (ym)) = 0(a) = “(5) = “(VJ = (11"?) we have h(a(a : (ym))) = h(a(b : (ym))). By Lemma 25 we have that y'"(h(a(a : (ym)))) g a : (31“) for 171. large enough. Thus it follows that h(a(a : (ym))) g a : (ym). Since for any ideal b it is true that b g h(a(b)), we have a : (gm) 9 h(a(a : (ym))), so we get h(a(a : (ym))) = a : (y’") and therefore

a = (11”) = h(a(a = (y”))) = a(0 = (11"?) = 4(a) = a“) = “(b = (31"?) = h(a(b = ~1 of > with the ordering of the monomials in

y, i.e., mil “wiry" >-1 xf‘ Hang-3H if mi“ ”41:5: >- xf‘ nab-f;- or if 2:? wars," = :t1 -- 03!," and e > f, is a good extension of >. (b) Show that Degrevlex on A[y] with 2:1 > > zen > y is a good extension of Degrevlex on A with 2:1 >- - - - > 1:“. 2. Prove Proposition 32. 3. Determine the reduced Grobner basis in Degrevlex with :51 >- 22 >- 933 of (2:132 — 1, mm — 1:1, 2:3 - :02 —- 1) by determining the reduced Grobner basis in Degrevlex with x1 >- .732 > 233 > y of

($1232 — y 2 ,wzxs — 5319,332 - wzy - y 2 )4. Show that if >—1 is a good extension of >- then the grading >1,g of >1 is a good extension of the grading >-g of >. In particular, if >- is a degree ordering and >1 is a good extension of >-, then >1,g is also a good extension of >. Conclude that Deglex with 11:1 > -- - > 27,. >- y is

a good extension of Deglex with 11:1 > ~ - - >- (an. 5. Determine the reduced Grobner basis in Deglex with a: >- y >- z of (my — z,:z:z — y,yz — :3) by determining the reduced Grobner basis in

Deglex with a: >- y >- z >- t of (my — zt,a:z — yt,ya: — wt). 6. Let a = (f1,...,fk) be an ideal in k[a:1,...,:rn]. Show that if G = {91, . . . , 9,} is a Grobner basis of (h(f1), . . . ,h(fk)) in Degrevlex with 2:1 >>- 3,, >- y, then {h(a(gl)), . . . ,h(a(g,))} 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 (afline 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 afiine space k" points “at infinity”. We start by describing

the projective line P1 (R), which is obtained by adding one point 00 to R. We give the elements in P1 (R), which we call points, coordinates in the

following way. Define an equivalence relation on the set R2 \ {(0,0)} by (z, y) ~ (whyl) if (2,1,1) = (twl,ty1) for some t 96 0. Then P1(1R) is the set of equivalence classes. We will write points in P1 (R) as (a: : y), where (z,y) is a representative for the class (a: z 3/). Thus (1 : 2) = (3 : 6). If one representative (z, y) for a class has y 75 0, then every representative for this class has second coordinate 76 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 z 0), 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 k"+1\{(0, 0,. . . , 0)}, where the equivalence is defined by

($1,---,93n+1)~(y1,---,yn+1)if (9:1,...,:1:n+1)=(ty1,...,tyn+1) for some t 96 0. We write the equivalence classes (x1 : - - - : a) where (9:1,...,:1:n+1)

is a representative for the class. We identify (:51, . . . ,mn) E k“ with (2:1 : 117

118

CHAPTER 9. PROJECTIVE VARIETIES : 9:." : 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) 9é (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(:c, y) E R[z,y] and consider the affine curve u(f) =

{(a,b) 6 1R2;f(a,b) = 0}- L6t h(f) = zdegfflm/zw/Z) 6 Mas/,2] be the

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

h(f)(ta, tb, to) = tde‘fh(f)(a, b, C), so if h(f) is zero on a representative ((1, b, c) for a point (a : b : c) in P2 (R), then h(f) is zero on every representative for (a : b : c). 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

U(f)-

n

EXAMPLE 3. A line 1 in R2 has an equation Ax+By+C = 0 with (A, B) yé (0, 0). The projectivization of l is the set of points in P2 (R) satisfying the equation A2: + By + 02 = 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 + 02 = 0 with (A, B, C) aé (0,0,0). Besides the projectivizations of affine 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 A12: + Bly + 01 = 0 are two nonparallel affine lines intersecting in the point (a, b), then the projective lines Am+By+Cz = 0 and A1m+Bly+Clz = 0 intersect in the projective

point (a : b: 1). IaJ+By+C = 0 and Az+By+01 = 0 are two parallel lines, and thus do not intersect, then the projective lines A9: + By + Oz = 0 and A9: + By + 012 = 0 intersect in the point (B : —A : 0) at infinity. The intersection between A2: + By + 02 = 0 and z = 0 is also (B : —A : 0) (if (A, B) 96 (0, 0)). Thus any pair of projective lines intersect in one projective point.

El

EXAMPLE 4. The affine hyperbola my = 1 in R2 corresponds to the projective curve my = 22. We have added the two points (0 : 1 : 0) and (1 : 0 : 0) corresponding to the two directions of the asymptotes to the affine points. The affine parabola y = :32 corresponds to the projective

curve yz = 1:2, which has one point (0 : 1 z 0) at infinity. The affine

circle :02 + y2 = 1 corresponds to the projective curve 2:2 + 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 affine curves just describe different aspects of the same projective curve.

El

The theory of projective sets runs in parallel to the theory of affine algebraic sets. Let P = (cl : - - - : on“) E P"(k). If f is a homogeneous polynomial such that f (c1, . . . , on“) = 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 Q 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 polynomial ring k[:z:1, . . . , zn+1], then the set of all projective points that are zeros of all polynomials in a is denoted 23(a) and is called the projective set or variety defined by a.

EXAMPLE 7. If a = (x1, . . . ,az,,+1) then 58(a) = (0. Therefore (2:1, . . . ,$n+1) is often called the irrelevant maximal ideal. More generally, if a is a homogeneous ideal with J5 = (2:1,...,mn+1) then 2101) = (0, since for each i we have 2:2" 6 a for some positive m, hence (0, 0, . . . ,0) is the only common zero, and (0,0, . . . ,0) does not give a point in P”(k). D PROPOSITION 8. (a) Il 9 X2 g P"(k) then 3(X2) g 3(X1). Ifa g b are homogeneous ideals in k[:vl, . . . ,$n+1] then Q16!) g SU(a). (b) For any subset X g P”(k) we have X g ‘13(3(X)), with equality if and only if X is a projective set.

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

=

’11(a)Um(b),

mat) momma), ‘B(a+b) = momma). (e) Ifa is a homogeneous ideal in k[a:1,.. .,:1:n+1] then a g 3(m(a)) and the inclusion may be strict.

Proof. Parallel to the proof of Proposition 6 in Section 4.1.

I]

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

EI-

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 {Xa} of projective subsets of P”(k) has a minimal element, i.e., an element X0,o not containing some other Xa. ' Proof. Parallel to the proof of Theorem 9 in Section 4.1.

El

THEOREM 12. Every projective set in P"(lc) 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 76 i, then the irreducible

components are unique.

Proof. Parallel to the proof of Theorem 10 in Section 4.1.

D

THEOREM 13 (HILBERT’S PROJECTIVE NULLSTELLENSATZ, WEAK FORM). Let a be a homogeneous ideal in C[a:1 , . . . , mn+1], with \fd strictly contained in (:31, . . . ,mn+1). Then ”(a) 95 0. Proof. Consider the affine algebraic set 0(a) in A"+1(C). As an affine algebraic set, 0(a) is called the afi'ine cone of the projective set Q'Ka). The affine Nullstellensatz, Theorem 11 in Section 4.2, shows that 0(a) is nonempty. Since J5 aé ($1,...,a:n+1), 0(a) does not consist of only (0,0,...,0), so D SU(a) is nonempty.

COROLLARY 14 (STRONG PROJECTIVE NULLSTELLENSATZ). For any homogeneous ideal a g C[a:1,...,a:n+1] with J5 strictly contained in the

ideal (2:1, . . . , 2"“) we have 3(’Il(a)) = J5. Proof. As in the proof of the affine case, Corollary 13 in Section 4.2, we

have \/E g Balk/3)) = 3(‘13(a)). Suppose that f = fo+- - -+fs 6 3(m(a)), f,- homogeneous of degree 1'. Then f(zl, . . . ,a) = 0 for every (9:1 :

:

$714.1) 6 j(Qi(c"))a SO f(t$1,- ' “:t$n+1) = f0 + tf1($17- ' - 1$n+1) + H ' +

t’f,(:1;1, . . . , $n+1) = 0 for every t e (3. Since C is infinite, this polynomial in t must equal the zero polynomial, hence f,-(:cl, . . . ,zn+1) = 0 for every i. Thus 3(m(a)) = i(n(a)) and the claim follows from the affine case. El

9.1. PROJECTIVE CLOSURE OF AN ALGEBRAIC SET

121

COROLLARY 15. If X is a projective set in P"(C) then Q3(’J(X)) = X. If

a = J5 and is strictly contained in (2:1, ...:cn+1) then 30130:» = a. Proof. Parallel to the proof of Corollary 14 in Section 4.2.

E]

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 ()5 : A"((C) —) P”(C) defined by ¢(c1,. .. ,cn) = (c1 : : cn : 1). DEFINITION 16 (PROJECTIVE CLOSURE). Let V be an algebraic subset of A"(C). The smallest projective set h(V) in P"(C) containing V (i.e., con— taining ¢(V)) is called the projective closure of V. On the other hand, if V is a projective set in P"((C), we set a(V) = {(a1,...,an)| (a1 : ---:a,, :1) 6V}. DEFINITION 17 (AFFINE RESTRICTION). If V is a projective set in P“(C) then a(V) is called the affine restriction of V. PROPOSITION 18. If V = $01) is a projective set then a(V) = u(a(a)), i.e., the afline restriction of V is the zeroset of the dehomogenization of a. In particular a(V) is an algebraic set. Proof. This follows directly from the definitions.

El

It is not equally easy to describe the projective closure with equations. If the projective set V is described with equations, i.e., as a zeroset of an ideal given by its generators, we need the Nullstellensatz to be able to describe the projective closure of V. This is the reason for restricting ourselves to algebraic sets in A"(C) here. PROPOSITION 19. Let a be an ideal in C[xl,...,z,,]. Then h(o(a)) = SU(h(a)). In words, the projective closure of 0(a) is the zeroset of the homogenization of a.

122

CHAPTER 9. PROJECTIVE VARIETIES

Proof. We shall show that ‘1](h(a)) is the smallest projective set containing 0(a). We first show that m(h(a)) contains 0(a). Suppose that ((11 , . . . , an) e h(a). We must show that (al : : an : l) e 23(h(a)). Let g 6 h(a). Since a = a(h(a)), we have (1(9) 6 a. But 9 = 1v$+1h(a(g)) for some m. Hence

g(a1,...,a,,,1) = a(g)(a1,...,an) = 0, so 0(a) is contained in ‘B(h(a)). For the second part of the proof suppose W = m((f1, . . . , 1%)) is a projective set containing 0(a). We shall show that W contains m(h(a)). Since

f;(a1, . . . ,an, 1): 0forevery (a1, . . . ,an) E 0(a), we get a(fi)(a:1, . . . ,zcn) E i(n(a)) = fl, so (a(f¢))'" e a for some m. This implies that h((a(f;))”‘) = (h(a(f,-)))’" E h(a). Thus (h(a(f,-)))’" is zero on %(h(a)), hence h(a(f,-)) is zero on ZU(h(a)). But h(a(f;)) = z;_|lilfi for some k 2 0, and hence f.- is zero on m(h(a)), so Q3(h(a)) Q Q}((f1, . . . ,f,)). D COROLLARY 20. Let a be an ideal in C[a:1,...,a:n]. If {gl,...,gt} is a Grdbner basis ofa then m(h(gl), . . . , h(gt)) is the projective closure of 0(a). Proof. This follows from Proposition 33 in Section 8.3.

El

REMARK. One could now rather directly use the results of the previous chapter to draw the following geometric consequences. Let V be an algebraic set in A"((C). If V is irreducible then the projective closure is irreducible. If V = UV,- is the irredundant decomposition in irreducible

components, then h(V) = Uh(V,-) is the decomposition in irreducible components, and no h(Vi) is contained in the “hyperplane at infinity” defined

by mn+1 = 0. We have that a(h(V)) = V. If W is an irreducible projective set which is not contained in the hyperplane at infinity, then a(W) is irreducible. If W = UWi is the irredundant decomposition of W in irreducible

components, then a(W) = U;eIa(Wg) is the decomposition in irreducible components, where the union is taken over the set I for which W; is not contained in the hyperplane at infinity for i e I. Exercise

1. Determine the projective closure in P2((C) of the algebraic subset of

A2(C) given by “(1:132 — 1,50% -— m2».

Chapter 10

The Associated Graded

Ring We will in this section consider an algebraic construction used to study an algebraic set V in k" near a point P E V. If P = (a1,...,an), we can make a translation X1 = 2:1 — ab... ,Xn = m" — an, and hence we may

assume (as we do from now on) that P = (0,...,0). If f(a:,y) = 0 is a curve passing thro§h the origin, we have a well-defined tangent at the origin if (%£(0,0), 3y (0,0)) 75 (0,0). In that case we say that (0,0) is a regular point on the curve. If f(0,0) = %(0,0) = %(0,0) = 0, we say that (0,0) is a singular point on the curve. If f(:1:,y) = fo + f1 (93,11) + f2(:1:,y) + ---, where f,(:c,y) is the homogeneous component of degree 12 of f, then f (0, 0) = 0 means that f0 = 0 and (0,0) is a singular point if and

only if fo = f1(w,y) = 0- If fizzy) = fm(x,y) + fm+1(w,y) +

where

fm (11:, y) 51$ 0, then fm(a:,y) = 0 approximates the behaviour of f (x, y) = 0 near the origin as we are used to from the theory of Taylor expansions. We

call the ideal (fm(m, y)) the cone of the ideal (f(a3,y)). We now turn to the general situation. DEFINITION 1 (CONE OF AN IDEAL). If f(a;1,...,a:n) = fm(a:1,...,x,.) + fm+1(a:1,...,a:,.) + with fm(a:1,...,zn) 75 0, we set m(f) = fm. Let a g k[:1:1, . . . ,zn] with (0,. . . ,0) E 0(a). The cone of a is

m(a) = ({m(f)| f E a}). LEMMA 2. We have m(fg) = m(f)m(g). I3

Proof. This is trivial.

123

124

CHAPTER 10. THE ASSOCIATED GRADED RING i If a is principal, it is easy to determine m(a).

PROPOSITION 3. If a = (f) then m(a) = (m(f)).

Proof. If g e (f) then 9 = f91 for some 91 and m(g) = m(fgl) = m(f)m(gl) e m((f)). Hence m((f)) Q (m(f)). The other inclusion is trivial. El

If a = (fl, . . . , fk) is not principal then (m(fl), . . . ,m(fk)) g m(a), but there might be inequality.

EXAMPLE 4. Ifa = (a: + 312,501; + 2;") then (m(a: + y2),m(a:y + 1173)) = (3,39): (m), but 3/3 —:c3 = y(z+y2) — (mg/+223) e a, so y3 —:t3 = m(y3 —a:3) E m(a) and 113—173 ¢ (11:). El THEOREM 5. Let a be an ideal in A = k[zl, . . . ,xn]. Let -< be Degrevlex on A and extend -< on k[m1, . . . ,$n+1] to the product ordering -1 m2w3+1 if d > e or if

d = e and m1 > m2, where the mi’s are monomials in A). Let {91, . . . ,g;} be a Grobner basis ofh(a) in >1. Then m(a) = (m(a(g1)), . . . ,m(a(gt))). Proof. Since 9,- E h(a) we have a(g,-) e a(h(a)) = a. 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(fizz+1 +

for some d, and because of the ordering

we have chosen, we have lm(h(f)) = sci“ lm(m(f)). Since {91, . . . ,gt} is a Gréibner basis of h(a), we have that lm(gg) divides lm(h(f)) for some 1?. But lm(gi) = $5,“ lm(m(a(g,~))) for some e. Hence lm(m(f)) = cnlm(m(a(g,-))) for some monomial n and some c e It. Now consider h = f — ena(g;) e a.

If deg(m(h)) > deg(m(f)), we have m(f) = m(cna(9i)) = mm(a(9«')) E (m(a(91)),-~,m(a(gt)))- If deg(m(h)) = desm(f)), we set lm(m(h»
0.

2. If F(X) = 2,20 anX” we define the formal derivative of F(X) to be

the series F’ (X) = 2,120 naa—l. (a) Show that (FG)’ = F'G + G’F.

(b) Show that (F")’ = nF”‘1F’. (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)-n = no ("+3-1)Xm. 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

(11—1) + (n) +...+ (n+m—l) = (n+m) 0

1

m

m

(b) Let F(X) = 1+X+X2+- - -. Show that (F(X))” = Ema, ("+'"‘1)X"'. m

11.2. HILBERT SERIES

11.2

129

Hilbert Series

Let R = k[:t1, . . . ,xn]/a be a homogeneous algebra. We have seen that R = EBQORi, 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 1', and h(i) = dimkRg 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 = €9.20R.‘ is a homogeneous alge-

bra, then HR(X) = 21.20 h(z')X’ is called the Hilbert series of R. EXAMPLE 3. If R = k[:r,y] then R,- is generated by maxi—131,. . . ,y‘. These monomials are linearly independent, so dimk R; = i + 1. Thus HR (X) =

1+2X+3X2+---=(1—X)-2.

EI

EXAMPLE 4. If R = k[a:, y]/ ($2,273)) then (images of) the following monomials constitute a k-basis for R: 1, z, y, y2, y3, y4, . . .. Thus we have h(0) =

l,h(1) = 2, and h(i) = 1 ifi > 1. Thus HR(X) = 1+2X+X2+X3+~~= D (1+X—X2)/(1—X). 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 Han, . . . ,xn] is homogeneous of degree It and that the image of f in the homogeneous algebra R = k[a:l,.. .,:1:,,]/a is a.

nonzerodivisor (i.e., that a = a : (f)). Then HR/(f) (X) = (1 — X")HR(X).

Proof. The map R,- —) RH], defined_by f,- I—) ff,- = ff,- is an injective homomorphism of vector spaces since f is a nonzerodivisor, so dim], (flHk = dim], Ri. The map Ri+k —+ R,-+k/ (fli+k is a surjective homomorphism of vector spaces, so

mm + k) = dim], R,” = dim], (BM + dim], Ri+k/ (BM = dim], R; + dim], 35+]; = hR(i) + hs(’i + k).

130

CHAPTER 11. HILBERT SERIES

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

hR(i + k)X‘+" = hR(i)X"+" + hs(i + k)X‘+", SO

2 hR(i + k)X‘+" = Z hR(z-)X‘+k + Z hs(z' + k)X‘+". 1'20

1'20

£20

Ifj < k we have Rj 2' Sj, so hR(j) = hs(j). Hence we get

2 hR(z')X" = X" Z hR(zf)X‘ + Z hs(i)X‘, .20

£20

£30

i.e.,

HR(X) = X"H3(X) + HS(X). El

LEMMA 6. Let R = k[a:1, . . . , xn]/a be a homogeneous algebra and suppose that f is a hom_ogeneous polynomial such that f is a socle element in R, i.e., such that f E (0) : (fi...,:c_,{) or equivalently f E a : (11:1,...,:1:,.). Then HR/(f)(X) = HR(X) — Xdegf.

Proof. We have dimk (f1. = 1 ifi = deg(f) and dimk(f) = 0 ifi ¢ deg(f) since in = U for every homogeneous g of positive degree. Hence dimk(R/ (f)z = dimk R,- — dimk (f): equals dimk R; — 1 ifz' = deg(f) and equals dim;c R,- otherwise. Thus

U) =

RiXi _ Xdes HR/(f)(X) = EdimAR/ (BL-X" = Zdimk HR(X) - Xdew).

D

We are ready to prove our main theorem for Hilbert series of homoge-

neous algebras.

THEOREM 7. Let R = k[:1:1, . . . ,zn]/a be a homogeneous algebra. Then HR(X) = p(X)/(1 — X)d for some polynomial p(X) 6 Z[X] with p(0) = 1 and some d g n.

Proof. Since the set of monomials outside 1(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[a:1, . . . ,mn]/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(2:1,...,2:,,),...,fk(2'1,...,2:,,)) g k[2:1,...,:rn] and let a’ be generated by the same elements but now as an ideal in K[.1:1, . . . ,2:,,]. Calculating a Grobner basis for a’ is exactly the same as for a, hence the two ideals have “the same” initial ideals, and we get HR(X) = HKIzl....,wn] /,,I (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 : (ff, . . . , fl) = (E, . . . ,3?) is a finitely generated homogeneous ideal in R, we see that there exists a homogeneous socle element s of maximal degree, namely of degree max {deg 57,}. (We have f3?- = 0 for each homogeneous element f of positive degree.) Then HR/(g) (X) = HR(X)—Xdeg 3 by Lemma 6. Let f be a homogeneous socle element of maximal degree in R/ (s). Since

R and R/ (§) coincide in degrees > deg s, we have degt 5 deg s. We continue like this to get an ideal (s, t, . . .) such that k[2:1, . . . ,mn]/a + (s,t, . . .) has no socle. (The process is finite since the degrees never increase, and dim;c eaisdeg 8R,- is finite.) Hence we eventually reach a ring R1 = R/s

without socle and HR/,,(X) = HR(X) - q(X), where q(X) E Z[X] has degree degs. Then R1 contains a nonzerodivisor of degree 1, which after a. linear change of coordinates, we can assume to be one of them— 2:-’s, and

HR, /(:,_)(X) (1 — X)HR1 (X) We now use induction over the number of variables n. If n = 0 we have R = It so H3(X) = 1. Suppose that the statement is proved for n — 1 variables, so HR1 ME) (X) = 1.01 (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 - XV, SO 171200 = @100 + (1 - X)‘iJ'1¢1(X))/(1 - X)"’“-

D

For large 2', h3(1') is a polynomial in i, the Hilbert polynomial of R.

COROLLARY 8. Let R = k[2:1,. . . ,2:,.]/a be a homogeneous algebra with Hilbert series p(X) / (1 —— X)‘, Where p(l) yé 0. Then the Hilbert function of R, hR(z') = dim], R; is a polynomial of degree d — 1 in i with highest coefficient p(1)/(d — 1)! for large 2'. (It'd = 0, hR(i) = 0 for large 2'.)

Proof. Suppose that p(X) = a0 + a1X + + akX". Then HR(X) = 1/(1 — X)‘1 + alX/(l — X)d + "'+‘1k/(1 — X)d. The coefficient of Xi in aj/(l — X)d equals (1,- times the coefficient of X“j in 1/(1 — X)d which is aj(i—fl1_"{'1) which, if i 2 j, is a polynomial of degree d — 1 in z' with highest coeflicient aj / (d — 1)!. Hence the coefiicient hi(R) of X‘ in HR(X) equals a polynomial in i of degree d — 1 with highest coefficient

(1+a1+---+ak)/(d—1)!=p(1)/(d—-1)!ifi>k.

El

EXAMPLEQ. Let R: k[2:,y,z]/ (2:2 ,.2:yz) If t > 3 the nonzero monomials ofdegree i in R are mJy‘J ,0 < j < 1, s“j, 0 < j_ < 1, and yjz“j, 1
A3 defined by {>2((7‘1 , 1‘2, 13)) = n (y, —:E, 0)+r2 (2, 0, -:L‘) +r3(0, —2, y). We denote the sequence ((2, —y, 3)) (consisting of one element) by syzz (s). A third syzygy would be an element 1' in A such that r(z, —y, z) = (0, 0, 0). This gives 1' = 0. Hence ker(i>3) = 0, if {>3 : A —) A3, 3(r) = r(2, —y,a:). 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, . . . , f,) be a sequence

of homogeneous elements in A = k[:c1, . . . ,mn]. Let syzl (f) = (91,. .. ,gt) be a sequence of homogeneous elements that minimally generate ker(1), where in :A’ ——-) A is defined by 1((r1, . . . ,rs)) = r1f1 + - - - +rafs. Let

syz2 (f) be a sequence minimally generating ker(2), where 2 : A‘ —> A” is defined by (P2((r1, . . . ,rt)) = mm + ' ~ - + mg 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 Diplomarbeit 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 verallgemeintem Weierstrass ’schen Divisionssatz, Diplomarbeit, Hamburg 1980. Exercise

1. Let A = k[a:, y, z] and s = (22, my,:vz,y2,yz,z2). Show that the submodule 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, . . . ,mn]], 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” contain-

ing the origin, and one wants to study the behaviour of V in a neighbourhood of the origin, the natural ring to study is not C[a:1, . . . , 23”] but

142

CHAPTER 12. VARIATIONS OF GROBNER BASES 9

its “localization at the origin”, denoted C{a;1, . --,$nl(z1,...,z,.)- This ring, which lies between C[:L'1 , . . . , an] and the fraction field of C[:z:1 , . . . , 21“], consists of {f/gl f,g e C[a:1,...,a:n],g(0,...,0) 75 0}. This is a local ring, i.e., there is a unique maximal ideal, namely (2:1,...,a:n). Every ideal in C[a:1, . . . :znl(z1,....wn) can be generated by polynomials in (C[zl , . . . ,mn], since a generator f/9 can be replaced by f, 9 being invertible. If one wants to study ideals in (C[:c1,.. .,mn](,,l,m,,n) by means of Grfibner bases, it turns out that one has to consider new kinds of orderings; we must demand that to, -< 1 for every 2‘. 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 court 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,...,f,) g m = (21,...,a:,,) g (C[a:1,...,a:n] then the associated graded ring of C[a:1, . . . ,zrn]/ (f1, . . . , f,) with respect to E coincides with the associated graded ring of C[a:1, . . . ixn](21,...,zn)/ (f1, . . . , fr) with respect to '5. His results have subsequently been simplified and generalized by Grabs

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 al-

gorithms 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 imple-

mentation 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

Griibner Bases for Noncommutative Rings

In this section we will study ideals in the noncommutative polynomial ring

over a field k (or the free associative algebra over It). The noncommutative

12.3. GRéBNER BASES FOR NONCOMMUTATIVE RINGS

143

polynomial ring k < 1:1,. . . ,zn > consists of linear combinations of monomials (or words) .7251 - - - 2.3-, , where the variables do not commute. There are for example nine monomials of degree 2 in k , namely

m = mzwywzwwwy = y2,yz,zm,zy,zz = 2’An ideal in B = k < x1 , . . . , 2:" > (or a two-sided ideal) is a nonempty subset a of B such that o a,b€a=>a+b€a o aea=>rasEaforallr,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 75 1 0 m1 < m2 implies mmln ~< mmgn for any monomials m, 11.

There are lots of monomial orderings, but we will in this section stick to one choice, namely Deglex. Here mg, - . 4:1,. > mjl ~ - ~mjm if k > m or if k = m and i1 = 3], . . . ,i,_1 = j,_1 and is < j, for some 8. Thus we have made the choice that $1 >- - n >- a)". In k < $1,532,113 > the smallest monomials

are 1 0 13 a minimal generator for a monomial ideal a, show that a is the intersection of a + ($211) and a + (:3??- $73k an]; 5. Show that (the 1mage of)2:1: is a lnonzerodivisorz'1n k[:z:, y, z] /p but that no power of :1: belongs to p2. To prove the last statement, one can e.g. calculate the Grobner basis of p2 in Lex with z >- y > w. 6. An irreducible ideal is primary. Use Exercise 1.

8. Write the monomial ideal as an intersection of irreducible ideals. Show that the intersection of irreducible monomial ideals with the same radical is of the given form. Section 6.3

1. The Grobner basis in Lex with a: > y is {$2 — 1,:cy — a: — y + 1,112 — 1}. 2. The Grobner basis in Lex with :1: >- y >- z is {x—yz,y2—zz,yzz-y,23 —z}. 3. The Grobner basis in Lex with a: > y >- z is (a: — 1,y,z — 1). 4. Determine the dimension of the vector space k[a, b, c, d]/l (a). 5. Calculate dim;c k[a:1,a:2,a:3]/l(a).

160

CHAPTER 15. HINTS TO SOME EXERCISES

Section 6.4

lb. Minimize 2:2 + y2 + 22 when :53] + zz + yz — 1 = 0. It is easiest first to determine A.

Section 7.1.1 1. Calculate a Grobner basis for a in Lex with a: > y > 2.

Section 7.1.2 1. Show that z,y,z E x/E. Section 7.5

3. Show that a: — 1,y,z+ 1 6 fl and that 1 9! JG. 6. To show that R is a field, it suffices to show that m3 +:c+ 1 is irreducible. If 2:3 + a: + 1 were reducible the equation would have a rational root.

9. Show that k[t3,t4, t5] = k[a:,y,z]/ (112 —- zz,yz — 23,22 — 1:231) .

k[t3, t4, t5] is an integral domain. Section 8.1

1. Show by induction on n that V is not the union of n proper subspaces.

For the induction step suppose V = U?V.- and Uf‘l Z V” and Vn g Uf‘lVi. Take a: E UTIV; \ V" and y 6 12,, \ U’f'lVi. Consider a: + y. Then show that Z2 x Z2 is the union of three proper subspaces.

2. 2:, y, and a: + y are the only homogeneous elements of degree 1. The set of zerodivisors is the union of (2:) , (y), and (:1: + y). Section 8.3 3. The reduced Grobner basis for the homogeneous ideal is {$1972 — 2112,972173 — $131,933 -‘ $211 — 212, xi?! — 933312,??33/ — 517110311 + 1321/2}-

5. The reduced Gr6bner basis for the homogeneous ideal in Deglex with z>y>z>tm

{11:31 — zt,a:z — yt,a:t — yz,yzz - zt2,y2t — 24213312:2 — yt2,z3t — zts}.

161 Section 11.1

23. Use the definitions of multiplication and derivation and calculate both sides.

2b. Use Exercise 2a and induction.

2c. 1 = G - G‘l. Use 2a. 2e. G‘" = (0—1)”.

Section 11.2 2. Use Exercise 1 in this section and Exercise 2f in Section 11.1.

Section 11.3 lb. For each i > n” there are just as many such monomials as there are

monomials 2:3" -- 411"" with m; < N, i.e., (n — 1)”. 1c. There are n(n — 1)” such monomials.

1d. Let c,- be the number of monomials mfl'f“ we?" of degree j with m.- < N for all i. The number of monomials 21'“ -- mg"- of degree i with

m,- 2 N for 1 g i 5 k and deg(z;"4'_°f1m:v;"“)=j, mj < Nfor k+1 g i g n then equals Cj times the number of monomials :ri’” - - - 2,?" of degree i—kN — j, which is a polynomial p,- (i) in i of degree [6— 1. The total number of monomials asked for is Egg-k)" ij (i), which is a sum of polynomials of degree k — 1, and thus a polynomial of degree k — 1. 1g. This follows from Exercise 1e and Corollary 8. 2. Show that (C[a:1 , . . . ,mn]/a : C[:z:k+1, . . . ,ztn]. 3. Use Exercise 2 in Section 11.2. 4. Use Lemma 5. 5. Use Lemma 5 several times.

Section 12.3 3. Show that main 6 1(a) if and only if q > 1' — 2.

Chapter 16

Answers to Exercises Section 1.2.3

Table 16.1: Addition in Z6.

+ [0] [1] [2] [3] [4] [5]

| [0] [0] [1] [2] [3] [4] [5]

[1] [1] [2] [3] [4] [5] [0]

[2] [2] [3] [4] [5] [0] [1]

[3] [3] [4] [5] [0] [1] [2]

[4] [4] [5] [0] [1] [2] [3]

[5] [5] [0] [1] [2] [3] [4]

Table 16.2: Multiplication in Z6.

‘ [0] [1] [2] [3] [4] [5]

| [0] [0] [0] [0] [0] [0] [0]

[1] [0] [1] [2] [3] [4] [5]

[2] [0] [2] [4] [0] [2] [4]

[3] [0] [3] [0] [3] [0] [3]

[4] [0] [4] [2] [0] [4] [2]

[5] [0] [5] [4] [3] [2] [1] 163

164

CHAPTER 16. ANSWERS TO EXERCISES

Table 16.3: Addition in Z7.

+ [0] [1] [2] [3] [4] [5] [6]

| [0] [0] [1] [2] [3] [4] [5] [6]

[1] [1] [2] [3] [4] [5] [6] [0]

[2] [2] [3] [41 [5] [6] [0] [1]

[3] [3] [4] [5] [6] [0] [1] [2]

[4] [4] [5] [6] [0] [1] [2] [3]

[5] [5] [6] [0] [1] [2] [3] [4]

[6] [6] [0] [1] [2] [3] [4] [5]

Table 16.4: Multiplication in Z7.

' [0] {1] [2] [3] [4] [5] [6]

| [0] [0] [0] [0] [0] [0] [0] [0]

[1] [0] [1] [2] [3] [41 [5] [6]

[2] [0] [2] [4] [6] [1] [3] [5]

[3] [0] [3] [6] [2] [5] [1] [4]

[4] [0] [4] [1] [5] [2] [6] [3]

[5] [0] [5] [3] [1] [6] [4] [2]

[6] [0] [6] [5] [4] [3] [2] [1]

In Z7 we have [11‘1 = [11,[21‘1 = [4],[?»]"1 = [5114]"1 = [21,[51‘1 = [3],

and [61-1 = [6].

Section 1.3.1

4- {[0]} , {[0], [2], [4]} , {[0], [3]} ,Ze. Section 1.3.3

5. gcd(:1:4 +3.1:2 +2,:z:3 —2:1:2 +x—2) :29 +1. 9:2 + 1 = (2:4 +3.7:2 + 2)/6 — (2:3 — 2:1:2 +2: — 2)(x +2)/6. 7. lcm(f,g) = (:1: — 1)3(m — 2)(:1: — 3)(.’B2 + 1)(:z:2 + 2). Section 1.4

11. (6).

165

Section 1.5 4. {a+bi'| a,bEQ}.

Section 1.6

5. In Z2[3]: 3,3+1,32+$+1,33+32+1,33 +3+1. In Z3[3]: 3,3+1,3+2,32+1,32+23+2. Section 1.7 1. ker(¢) = (3 -— c). 2. ker(¢) = (31 — c1,. ”,3" — c"). 7. In the table we use a representative a for the class a + (32, 3g, yz). We have omitted 0 and 1.

Table 16.5: Multiplication in Z2[3,y]/ (32,3y,y2). 3

|3

y

3+1

y+1

3+y

3+y+1

0

0

3

3

0

3

y

0-

0

y

y

0

:1

3+1 y+1 3+y

3 3 0

y y 0

1 3+y+1 3+y

3+y+1 1 3+y

3+y 3+y 0

y+1 3+1 3+y

3+y+1

3

y

y+1

3+1

3+y

1

9. In the table we use a representative a for the class 0+ (32 + 1) and write the elements in Z3 as 0,1, and 2. We have omitted 0 and 1.

Table 16.6: Multiplication in Z3[3]/(11:2 + 1). 2 3 3+1 3+2 23 23+1 23+2

2 1 23 23+2 23+1 3 3+2 3+1

3 23 2 3+2 23+2 1 3+1 23+1

3+1 23+2 3+2 23 1 23+1 2 3

3+2 23+1 23+2 1 3 3+1 23 2

23 3 1 23+1 3+1 2 23+2 3+2

17. 34—33+232-3+1=(32—3+1)(32+1).

23+1 3+2 3+1 2 23 23+2 3 1

23+2 3+1 23+1 3 2 3+2 1 23

166

CHAPTER 16. ANSWERS TO EXERCISES

Section 1.9 3. A basis is {1, W}. The dimension is 2.

4. A basis is {1+a,a:+a,y+a}, where a = ($2,331,312) (and 1 = [1]). The dimension is 3.

5. A basis is {1,3,1},x2,my,y2,m2y,xy2,y3,$2112,my3,a:2y2}. The dimension is 12.

Chapter 2 1. (1+ [1 = (m§,m§,m224,x§,m3x4) ab = (fiwgwfifgm,wfimwwwimfimgmgm)

a n b = (zi’wg, £13134,mgmflgmgmwgzgzgm) a: b = (x§,w2m4,x3x4,x§)

x/5 = (161,912-104,273)2. a + b = (z§,m1z2,$2$3)

ab = (wi’zg, mfmgmsmgzgxlxgze)

a n b = (mgmzmlmg) a: b = (mf,x1z2)

x/5 = ($1) 3- ($1,972)0 (934)-

4. ($133,182)n(m¥,zg,m§,m4)n(z¥,m§,m4). Section 3.1.1

2. (1,1,...,1),(1,1,...,1,1—n),(1,1,...,2-—n,0),...,(1,—1,0,...,0). 3. w‘l‘zgxg >~mfm§m§ ifa+b> d+eorifa+b=d+eanda>dorif a=d,b=eandc>f. 4.1