[Li HUISHI] an Introduction to Commutative Algebra(Bookfi.Org)

Home , Irreducible component, Valuation (algebra)

An Introduction to Commutative Algebra

From the Viewpoint of Normalization This page intentionally left blank From the Viewpoint of Normalization

I Jiaying University, China

NEW JERSEY * LONDON * SINSAPORE - EElJlNG * SHANGHA' * HONG KONG * TAIPEI CHENNAI Published by World Scientific Publishing Co. Re. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: Suite 202,1060 Main Street, River Edge, NJ 07661 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

British Library Cataloguing-in-PublicationData A catalogue record for this book is available from the British Library

AN INTRODUCTION TO COMMUTATIVE ALGEBRA From the Viewpoint of Normalization Copyright 0 2004 by World Scientific Publishing Co. Re. Ltd. All rights reserved. This book, or parts thereof. may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rasewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

ISBN 981-238-951-2

Printed in Singapore by World Scientific Printers (S)Pte Ltd For Pinpin and Chao This page intentionally left blank Preface

Why normalization? Over the years I had been bothered by selecting material for teaching several one-semester courses. When I taught senior undergraduate students a first course in (algebraic) number theory, or when I taught first-year graduate students an introduction to algebraic geometry, students strongly felt the lack of some preliminaries on commutative algebra; while I taught first-year graduate students a course in commutative algebra by quoting some nontrivial examples from number theory and algebraic geometry, of- ten times I found my students having difficulty understanding. The problem is that in a short semester there is not enough time to go through the sig- nificant background material needed in my course (if you are an instructor of mathematics, can you ask your students to find and read those material themselves and do they usually follow?). Based on my lecture notes on (algebraic) number theory, algebraic geometry, and commutative algebra used at Shaanxi Normal University and Bilkent University, I decided to fuse several things into one - the presentation of this book. As a consequence, the text consists of five chapters that are designed for a (one-semester) common course taken by senior undergraduate students or by first-year graduate students in mathematics. The goal is to introduce to the students the concrete source of commutative

vii ... Vlll Commutative Algebra algebra through the following diagram: DVR -+ PID UFD 1~DVR - + I= I I ED - DD - NND

where DVR = Discrete valuation ring, PID = Principal ideal domain, UFD = Unique factorization domain, ED = Euclidean domain, DD = Dedekind domain, NND- = Normal Noetherian domain, Z[6]= The integral closure of 2[6](or equivalently the integral closure = - of Z) in the number field K Q(d), K[C]= The normalization (or integral closure) of the coordinate ring K[C]of an algebraic curve C in its field of rational functions K(C), - and K[V]= The normalization (or integral closure) of the coordinate ring K[V]of an irreducible algebraic set V in its field of rational functions K (V). Or more precisely, in terms of normal (normalized) structure the lectures demonstrate how-- to get to the center- of commutative algebra by recognizing the roles that 2[6],K[C] and K[V]play in number theory and algebraic geometry, so that, after reading this volume, interested readers may read a course in algebraic number theory at a higher level, or start an advanced course in algebraic geometry with a better background. I assume that the reader has taken the undergraduate courses including Linear Algebra and A First Course in Abstract Algebra (Galois theory may not be included). Moreover, very little about topological space is needed Preface ix to understand the Zariski topology in Chapter 5. Other than these prereq- uisites, the book is self-contained. Exercises are given at the end of each section. Though most of the exercises mainly test the understanding of the text in the usual way, the reader is involved in providing proofs and in working problems that have not been completely solved in the text; and furthermore, students are asked to extend some of the theory that is essential for the subsequent sections. Huishi Li This page intentionally left blank Contents

Preface vii

Chapter 1 Preliminaries 1 0. Conventional Review 1 1. Noetherian Rings 6 2. Factorization of Elements in a Domain 9 3. Field Extensions 18 4. Symmetric Polynomials 28 5. Trace and Norm 33 6. Free Abelian Groups of Finite Rank 40 7. Noetherian Modules 45

Chapter 2 Local Rings, DVRs, and Localization 53 1. SpecR, m-SpecR, and Radicals 53 2. Local Rings and DVRs 58 3. The Ring of Fractions and Localization 67 4. The Module of Fractions 73

Chapter 3 Integral Extensions and Normalization 79 1. Integral Extensions 79 2. Noether Normalization 85 3. Normal Domains and Normalization 90 4. Normal Domains and DVRs 94

xi xii Contents

Chapter 4 The Ring $2~in K = Q(t9) 99 1. dK is Normal and Free of %Rank [K : Q] 102 2. a(&) and Q(u) 108 3. Factorization of Elements in dK 112 4. From dK to Dedekind Domains 117

Chapter 5 Algebraic Geometry 127 1. Finite Field Extension and Nullstellensatz 127 2. Irreducible V and the Prime I(V) 134 3. Point P and the Local Ring Op,v 146 4. Nonsingular Points and DVRs 151 5. Normalization of Algebraic Curves 157 6. Parametrize a Rational Curve via Normalization 162 7. Rational Curves and Diophantine Equations 167 References 171 Index 173 Chapter 1 Preliminaries

This introductory chapter concentrates on basic notions and properties concerning Noetherian rings, factorization of elements in a domain, field extensions, symmetric polynomials, trace and norm, free abelian groups of finite rank, and Noetherian modules, which might not be familiar to some readers. So we include most of necessary proofs for the reader’s convenience.

0. Conventional Review

In this book all rings are commutative associative rings with identity 1, and throughout the text, N = the set of nonnegative integers, Z = the set of integers (ring of integers), Zf = the set of positive integers, Q = the set of rational numbers (field of rational numbers), R = the set of real numbers (field of real numbers), C = the set of complex numbers (field of complex numbers). Let R be a ring and A a subring of R. Then we insist that A has identity 1~ and

And we write

RX = R - (0).

If {Ui}iE~and {Vl,..., Vm}are collections of nonempty subsets of R,

1 2 Commutative Algebra then the sum of {Ui}i,=~and the product of {Vl,..., Vm}are defined as

vl...vm={cvl...v, I}ViEV, , where the sums involved in both C Ui and VI . . . V, are finite sums. So one understands that

0 the sum of given ideals is an ideal; the product of finitely many given subrings (ideals) is a subring (an ideal); and 0 for subrings (ideals) I, J and K, I(J + K)= IJ + IK = JI + KI = (J+ K)I. Let S be a nonempty subset of R and A a subring of R. We set

Z[S]= the subring of R generated by S = {~sz...s:; I si, E S, m E z+,aj EN}

A[S]= the subring of R generated by S over A = { ~a(,,~)s~...s:; 1 a(,,j) E A, si, E S, m E z+,aj E N}

(S) = the ideal of R generated by S

= {crisi I ri E R, si E s}, where the sums involved in Z[S],A[S] and (S) are finite sums. If S = {sl, ..., s,} is finite, we write Z[S]= Z[sl,..., sn],A[S] = A[SI,..., s,], (S)= (s1, ..., sn), and call them a finitely generated subring, a finitely generated subring over A, and a finitely generated ideal of R, respectively. Clearly we n can also write (~1,..., Sn) = C,=lRsi. Let RLR' be a ring homomorphism. Then we insist that

cp is not the zero-homomorphism and cp(1~)= l~,, and we write Kercp, Imcp for the kernel and image of cp, respectively. A very useful consequence of the first isomorphism theorem on ring homomorphism states that Preliminaries 3

if R-%A and R-B* are ring homomorphisms, cp is surjective, and Kercp C Ker$, then there is a ring homomorphism A-%B defined by p(a) = $(r),where cp(r) = a, such that the following diagram commutes:

Kercp L) R A *I JP pocp=$ B if furthermore 1c, is surjective, then p is surjective as well.

Let R be a ring with identity 1 = 1~.If R has no divisors of zero, i.e., a, b E R and ab = 0 implies a = 0 or b = 0, then R is called an integral domain, or simply a domain. If R is a domain, then so is the polynomial ring R[zl, ..., zn]in variables z1 , ..., zn over R. If a, b E R and ab = 1, then a (hence b) is called a unit of R. If every nonzero a E R is a unit, then R is called a field.

0.1. Proposition Every finite domain is a field. Proof Exercise. 17

Thus, if p E Z is a prime number, then the ring Z/(p)of integers modulo p, usually denoted Z,, is a field. Let K be a field. Consider the set of integers o(~)= {m E Z+ I m~ = o for some x E K~ 1. If o(K)= 0, then the characteristic of K, denoted charK, is defined to be zero, i.e., charK = 0; if o(K)# 0, then charK is defined to be the smallest integer p E o(K). In the second case p is a prime number (exercise 3). Every field has a smallest subfield P (with respect to the inclusion relation on subfields), the prime subfield, which is either isomorphic to

Q, if charK = 0, or to

Zp, if charK = p > 0.

Clearly, every finite field F has charF > 0. If a field K has charK = p > 0, then (u + b)P = ap + bP for all a, b E K (exercise 4). 4 Commutative Algebra

If R is a domain, then the field of fractions of R is constructed via the equivalence relation on R x RX: (a,b) - (c,d) if and only if bc = ad. Write f for the equivalence class represented by (a,b), and write Q(R)for the quotient set R x RX/-. Then

Q(R)={%I a,bER, bfo} where the addition and multiplication are defined the same as that for rational numbers. Thus, in Q(R),4 = 0 is the zero of the additive group (Q(R),+), = ~Q(R)is the identity of the multiplicative group (Q(R),.), and if a = f # 0 then a-1 = $. The ring homomorphism

XR : R --+Q(R) r rw - 1 is injective. In the case where R is a field, XR is an isomorphism. So we may view R as a subring of Q(R)and write R Q(R).Consequently, if Q(R)[x] is the polynomial ring in variable x over Q(R),then R[z] Q(R)[x].

If R' is another domain and 9:R --f R' is an injective ring homomorphism, then cp induces an injective ring homomorphism Cp: Q(R)-+ Q(R'), where (f) = 3.Hence Q(R)may be viewed as a subfield of Q(R'), and consequently Q(R)[z]may be viewed as a subring of Q(R')[x].It turns out that if cp is an isomorphism then so is q. In particular, if Q(R)is the field of fractions of the domain R and R c B c Q(R),where B is a subring of Q(R),then Q(B)= Q(R). We assume that the reader is familiar with the basic structural properties of a polynomial ring R[q,..., x,] in variables x1, ..., x, over the ring R, for instance, R is a subring of R[q,..., z,] consisting of constant polynomials, every f E R[Q,..., z,] has a unique expression into the linear combination of monomials: f = r,xy' . . . xEn, where cy = (a1,... ,a,) E N" = { (ai,..., a,) I a1,..., a, E N}, and the degree of f is defined as degf = max { a1 + . . + a, r,xy' . . . xzn # 0 is a term of f 1. If f = 0 then conventionally degf is defined as -m. Thus, for f = C Tax:' . . . xC,an , g = Crp$ ...xk,one knows how to determine the Preliminaries 5 degree of f + g and f . g according to the addition and multiplication of polynomials. In particular, we recall the following important properties of a polynomial ring. Iff E R[z]is a nonconstant monic polynomial, i.e., degf 2 1 and f is of the form

f = rcn + an-lrcn-l +.. . +ao, ai E R, then a division algorithm on g E R[z]by f exists:

g = qf + T, q,r E R[z],degr < degf. Let R be a ring. If Z[sl, ..., s,] is the subring of R generated by s1, .,., s,, then there is an onto ring homomorphism from the polynomial ring Z[zl,..., rc,] to Z[sl, ..., s,]: Z[Zl, ... ,%I - qs1, ..., Sn]

f(z1, .-,4H f(Sl,...,S,) Let A be a subring of R. If A[sl,..., sn] is the subring of R generated by sl,..., s, over A, then there is an onto ring homomorphism from the polynomial ring A[zl,..., rc,] to Ajsl, ..., s,]:

A[~I,.-.,~2] - A[sl,'''7 sn]

Exercises 1. Let A be a subring of the ring R and S C R a nonempty subset of R. Show that Z[S]is the smallest subring of R containing S, that A[S] is the smallest subring of R containing A and S, and that (S) is the smallest ideal of R containing S. (Here the ordering on subrings and ideals is the usual inclusion ordering on subsets.) 2. Complete the proof of Proposition 0.1. (Hint: If R = {al,..., a,} is a finite domain and 0 # ai E R, then aiR = R.) 3. Let K be a field. Show that if charK = p # 0, then p is a prime number. 4. Let K be a field of charK = p # 0. Show that (a + b)P = ap + bP for all a, b E K. (Hint: Check the binomial coefficients of the expansion of (a + b)p.) 6 Commutative Algebra

1. Noetherian Rings

Since Noetherian ring plays a leading role in commutative algebra, we start with this notion. Let R be a ring. R is said to satisfy the maximal condition if every nonempty set of ideals contains a maximal member with respect to the inclusion relation on ideals. R is said to satisfy the ascending chain condition if for every ascending chain of ideals

I1 c I2 5 ..*c In ... there is some k such that Ik = Ij for all j 2 Ic.

1.1. Theorem Let R be a ring. The following are equivalent. (i) R satisfies the maximal condition. (ii) Every ideal of R is finitely generated. (iii) R satisfies the ascending chain condition. Proof (i) + (ii) Let I be a nonzero ideal of R. Set

S = {all finitely generated ideals contained in I}.

Then S # 0, and by (i) there is a maximal member in S, say J = C;=, Rai with ai E I. If J # I, then there is some x E I, x 6 J. Thus, J is properly contained in J' = J + Rx and J' E S, contradicting the choice of J. Therefore I = J, a finitely generated ideal. (ii) + (iii) Let

be an ascending chain of ideals in R. Set I = UIi. Then I is an ideal of R and hence finitely generated, say I = CE, Raj with aj E I. Suppose aj E Ii3 with il < i2 < ... < i,. Then aj E Ii,, j = 1, ..., m, and consequently I = Ii,. Let k = i,. Then Ik = Ij for all j 2 k. (iii) + (i) Let S = {Ii} be a nonempty set of ideals in R. If S did not have a maximal member, there would be a strictly ascending chain of ideals out of S, which does not satisfy the chain condition. 0

1.2. Definition A ring R satisfying one of the equivalent conditions of Theorem 1.1 is called a Noetherian ring. Preliminaries 7

Let R be a ring. If every ideal I of R is a principal ideal, i.e., I = (a) = Ra for some a E I, then R is called a principal ideal ring. Principal ideal rings are special Noetherian rings. If a principal ideal ring R is also a domain, then we simply call R a PID. It is a result of the division algorithm in Z and the division algorithm in the polynomial ring K[z],where K is a field, that both Z and k[z]are PIDs (exercise 1). Concerning polynomial rings in finitely many variables over a Noethe- rian ring, we have the following celebrated result.

1.3. Theorem (Hilbert basis theorem) If R is a Noetherian ring then so is the polynomial ring R[z]in variable z over R. Hence, the polynomial ring R[zl,..., z,], in any finitely n variables z1, ..., z,, is Noetherian. Proof We show that if R[x]is not Noetherian then neither is R, by adopting a well-known argumentation (as one may easily find at the site '). Suppose that I is an ideal of R[z]which is not finitely generated. Then a sequence of polynomials from I can be chosen as follows.

fi E I with least degree nl, f2 E I - Rfl with least degree n2, f3 E I - (Rfi+ Rf2) with least degree n3,

k fk+1 E I - IRfiwith least degree nk+l, i=l

where n1 5 722 I 123 5 ' . . I nk+l 5 . . . . Claim Let ai be the leading coefficient of fi. Then

n Ral c Ral + Ra2 c . . * c CRai c . . . i=l is a strictly ascending chain of ideals in R. k If the claim was not true, then Rai = Cfz; Rai for some k, and k this would yield ak+l = riai, ri E R. Note that for i = 1, ..., k, we

http://planetmath.org/encyclopedia/Proofo~ilbertBasisTheorem.html 8 Commutative Algebra have fi = aizna + strictly lower degree terms, ~ifiPk+l-~i= ra.xnk+1 22 + strictly lower degree terms. It follows that

while / k \k

by the choice of fk+l. But clearly degg(s) < degfk+l, contradicting the choice of fk+l. Therefore the claim holds, i.e., R is not Noetherian. 0

The polynomial ring K[zl,..., z,, ...I in infinitely many variables over a field K is non-Noetherian, due to the existence of a strictly ascending chain of ideals:

(51) c (21,52) c . . * c (51,”’, 5,) c . . * . In Chapter 4 we will see that if A is the set of all algebraic integers, i.e., the set of complex zeros of monic polynomials in Z[x],then A forms a ring and it is not Noetherian; while for a finite dimensional field extension Q C K with K a subfield of C, A n K is always Noetherian. Noetherian rings stemming from algebraic geometry are given in Chap- ter 5.

Exercises 1. Show that Z and K[z]are PIDs, where K[z]is the polynomial ring in x over a field K. 2. Let R + R‘ be an onto ring homomorphism. Show that if R is Noethe- rian then so is R’. 3. Let A be a Noetherian subring of the ring R, and let {TI, ...,rs} be a finite subset of R. Show that the subring A[rl,..., T,] of R is Noetherian. 4. Let K be a field which, as a subring, is contained in the ring R. Assume that R is finite dimensional over K. Show that R is Noetherian. Preliminaries 9

5. Let R be a Noetherian ring. The ring of formal power series over R is the associative ring R[[z]]consisting of the formal series

i=O where f(z)= 0 if and only if ri = 0 for all i = 0,1, ..., and the addition and multiplication are defined as for the power series with real coeff- cients in calculus. Show that R[[s]]is Noetherian. (Hint: Define the degree of a series as the lowest power of x.) 6. By Theorem 1.3, Z[x]is Noetherian. Show that the ideal I = (2, x) is not a principal ideal. 7. Let Zz[z,y]be the polynomial ring over the field Z2. Show that in &[x, y]/(z2 + z + y3 + 1) the ideal (TIg) is not a principal ideal.

2. Factorization of Elements in a Domain

Let R be a domain. It is easy to see that the set of units in R, denoted U(R)= {u E R I u is a unit in R 1, forms a group with respect to the multiplication of R. U(R)is called the group of units in R.

2.1. Definition (i) For r E R, u E U(R),the element y = ur = ru is called an associate of r. (ii) Let r,s E R. r is said to be divisible by s, denoted s[r,if r = sz for some z E R, where s (hence t) is called a divisor (or a factor) of r. For u E U(R)and r E R, u and ur are called the trivial divisors of r (note that r = (ur)u-l = (u-lr)u). (iii) For r E R, if r has only trivial divisors in R, then we say that r is irreducible in R; otherwise, r is reducible in R. (So zero is reducible in any domain.) (iv) For r E R, if r is reducible, then r = sz with nontrivial divisors s, t. In this case we say that r has a proper factorization.

Example (i) Let R = Z.Then U(R)= {fl}. (ii) Let R = Z[i]where i = fl.Then U(R)= {fl, f i} (see Chapter 4 10 Commutative Algebra section 3).

(iii) Let R = K[z]be the polynomial ring in z over a field K. Then U(R)= KX. Thus, one easily finds elements in each R that have proper factorization.

2.2. Proposition Let R be a domain, T, s E R. The following hold: (i) r E R is a unit if and only if 7-11. (ii) Any two units are associates to each other, and any associate of a unit is a unit. (iii) r, s are associates to each other if and only if rIs and SIT. (iv) T is irreducible if and only if every divisor of T is either an associate of r or a unit. (v) Any associate of an irreducible element is irreducible. Proof Exercise.

In terms of ideal structure, we may characterize units, divisibility, associates and irreducibility, as follows.

2.3. Proposition Let R be a domain and let r, s be nonzero elements of R. (i) r E U(R)if and only if (r)= R. (ii) TIS if and only if (r)2 (s). (iii) r, s are associates to each other if and only if (r)= (s). (iv) T is irreducible if and only if (r)is maximal among the principal ideals of R (with respect to the inclusion ordering on ideals). Proof Exercise.

2.4. Definition Let R be a domain. We say that factorization into irreducible elements is feasible in R if every nonzero nonunit element may be expressed as a product of finitely many irreducible elements.

2.5. Proposition Factorization into irreducible elements is feasible in a Noetherian domain R. Proof Let R be a Noetherian domain. Suppose that the assertion was not true. Then the set R of nonzero nonunit elements which cannot be factorized into finite products of irreducible elements would be nonempty. Preliminaries 11

Since R is Noetherian, let (y) be a maximal member in

S={(x) IXFQ}.

Then y is reducible because y E R, and y = rs for r, s # U(R). Thus, (y) is properly contained in (r)n (s) (otherwise r or s would be a unit by Proposition 2.3). By the choice of (y) we have

where pi’s are irreducible elements. But then y = PI * *.p,p,+l .-.p,, a product of finitely many irreducible elements. This is a contradiction and hence R = 0.

2.6. Definition Let R be a domain in which factorization into irreducible elements is feasible. For a nonzero nonunit x E R, if any two factorizations

satisfy n = m and (up to the arrangement of divisors) pi = Uiqi, i = 1, ..., n, where ui E U(R),then x is said to have a unique factorization in R. If every nonzero nonunit element of R has a unique factorization in R, we say that R is a UFD (abbreviation of the phrase “unique factorization domain”).

Remark At this stage, it is better to be aware of two facts. (i) There are Noetherian domains which are not UFDs (see exercise 4 of this section and Chapter 4 section 3). (ii) There are UFDs which are not Noetherian (exercise 5).

In order to discuss the uniqueness of factorization into irreducible elements, we introduce the notion of a prime in a domain.

2.7. Definition Let R be a domain, 0 # x E R, and x # U(R).x is said to be a prime if xlab implies xla or xlb for any a, b E R.

2.8. Proposition Let p be a prime in a domain R. The following hold: (i) Any associate of p is a prime in R. (ii) p is irreducible in R. Proof Exercise. 12 Commutative Algebra

2.9. Theorem If factorization into irreducible elements is feasible in a domain R, then R is a UFD if and only if every irreducible element is a prime. Proof Since factorization into irreducible elements is feasible in R, by Proposition 2.2(v), every nonzero nonunit x E R has a factorization

2 = PIf.. pel where pi may be an associate of some irreducible element. First suppose that factorization in R is unique. Let p be an irreducible element and plab where a # 0, b # 0. Then ab = pc for some 0 # c E R.

Consider the unique factorizations: a = pl. . + pn, b = 41. . . qm, c = TI . . .T~, Then

PC = ~(~11**.i rs) = (PI. . *pn)(ql.. ' qm) = ab. By the uniqueness, p divides some pi or some qj. Hence pja or p)b,and this shows that p is a prime. Conversely, suppose every irreducible element is a prime. Consider the factorization into primes

Then pl1q1 (q2 . . . qm). Without loss of generality we may assume plIq1. Then, 41 = ulpl for some u1 E U(R)because q1 has only trivial divisors. Thus, z = ~1.e.p~= (uip1)(~2***qm) and p2***pn= (ulq2)(q3*'.qm)* After repeating this process n times, up to the arrangements of divisors we derive qi = uipi with ui E U(R),and m 5 n. Similarly, n 5 m. So n = m. This shows that factorization is unique in R. CI

2.10. Theorem Every PID is a UFD. Proof Let R be a PID. Then factorization into irreducible elements is feasible in R because R is Noetherian. Let p be an irreducible element in R. Then by Proposition 2.3(iv), (p) is maximal among all ideals. Suppose plab but p /y a. Then (p) is properly contained in the ideal (p,a). By the maximality of (p) we have (p,u) = R. It follows that 1 = ph + aq and b = bph + abq. This yields pjb, showing that p is a prime. By Theorem 2.9, R is a UFD. 0

Remark Recall that before learning a systematic theory on UFDs, in the Preliminaries 13 arithmetic theory on R = Z (or in R = K[x]where K is a field) a prime p is defined as the element which has only the divisors fl (A E K"), fp (Ap). If a, b E R, plab but p /y a, then the Euclidean algorithm output the greatest common divisor gcd(p, a) = 1 in the form

af+pg=L f,gER, that yields plb immediately as in the above proof. That is why we know, without arguing that R is a PID, that R is a UFD. Indeed, there is a class of UFDs that hold a version of Euclidean algorithm, as described below.

2.11. Definition A Euclidean domain is a domain R with a function (called a Euclidean function): 4: RX-N that satisfies (i) if a,b E RX and a[bthen $(a) 5 4(b);and (ii) if a, b E RX then there exist q, r E R such that

a = qb + T, where either r = 0 or $(r) < 4(b).

Example (iv) Z is a Euclidean domain with the Euclidean function given by the absolute value function. K[x]is a Euclidean domain with the Euclidean function given by the degree function. (A consequence of applying the division algorithm to both Z and K[x].)

2.12. Theorem Every Euclidean domain R is a PID. Proof Let I be a nonzero ideal of R. If 4 is the associated Euclidean function on R, let us set 4(x*) =min { 4(x)EN O#x E I } . For any 0 # y E I, y = qx* + r with T = 0 or 4(r) < 4(x*). But T = y - qx' E I. By the choice of xi,T = 0. Thus, y = qx'. This shows that I = (x*). 0

2.13. Corollary Every Euclidean domain is a UFD. Proof This follows from Theorems 2.10-2.12. 0 14 Commutative Algebra

Except for Zand K[x],other Euclidean domains will be given in Chapter 4 section 3.

Remark Let K = Q(a).By Theorem 3.4 of Chapter 4, the ring dK of algebraic integers in K is not a Euclidean domain. However, AK is a PID. The reader is referred to http://www.mathreference.com/id,npid.htmlfor a beautiful proof on this fact.

We now proceed to show that the polynomial ring R[x]in variable x over a UFD R is a UFD.

2.14. Lemma (Gauss) Let R be a domain. Then any prime of R is a prime in R[x].

Proof Let p be a prime in R and R = R/(p). Then a direct verification shows that R is a domain, and so is the polynomial ring R[x].If f,g E R[x] and plfg, then fg E (p). For T E R, write F for the image of T in R. Consider the ring homomorphism

R[x] 5 R[x]

Then p(fg) = 7.3= 0. Since R[x]is a domain, it follows that = 0 or 9 = 0, i.e., plf or pjg, as desired. 17

Let R be a UFD. Then for any TI, ..., T, E R, not all zero, the greatest common divisor gcd(a1 , .. ., a,) exists in R (exercise 6).

2.15. Definition Let R be a UFD. If a polynomial T,Z~+T~-IZ~-'+.. .+ TO = f(x) E R[x]has the property that gcd(Tn,rn-l, ..., TO) = d E U(R), then f (x)is called a primitive polynomial.

2.16. Proposition Let R be a UFD. If f,g E R[x]are primitive then so is the product f g. Proof This follows immediately from Gauss lemma.

2.17. Theorem Let R be a UFD with the field of fractions K = Q(R). (i) If f E R[x]and f = gh for some g, h E K[x],then there is a unit Preliminaries 15 a E K[z]such that gal a-lh E R[x]. (ii) Let f,g E R[x],where g is primitive. If 91 f in K[x]then glf in R[z].

Proof (i) Let f = gh be as assumed. Let TO E R be the common denominator of the coefficients of 9. Then rog E R[x].Let d be the greatest common divisor of all coefficients of rag. Then g1 = ag is primitive in R[x],where a = 3 E K, Similarly, there exists /3 E K such that hl = ph is primitive in R[x].Set = f, where a and b have only common divisors which come from U(R).Then

a ,f = aPgh = glhl and af = bglhl.

Now, if a E U(R),then since baP =a, we have ag = 91, a-lh = a-lbph = a-lbhl have coefficients in R, as desired. So it remains to show that a E U(R).If not;, there would be some prime p dividing a. Hence plbglhl but p /y b by the choice of a and b, and p /y 91, p 1hl because both g1 and hl are primitive. This contradicts Gauss lemma. Therefore, a must be a unit. (ii) This follows from part (i). 0

Let R be a UFD and f(x) E R[x]with degf (x)2 1. If d is the greatest common divisor of all coefficients of f(z),then f(z)= dfl(z) where fl(z) is a primitive polynomial. Bearing this fact in mind, Theorem 2.17 enables us to derive immediately the following.

2.18. Proposition Let R be a UFD with the field of fractions K = Q(R), p(z) E R[z]with degp(z) 2 1. Then p(z) is irreducible in R[x]if and only if p(x) is irreducible in K[z]. 17

2.19. Theorem If R is a UFD then so is R[z] Proof Since R is a UFD and R c R[s],by Gauss lemma we need only to consider polynomials of degree 2 1. Let K = Q(R)be the field of fractions of R. Then K[x]is a UFD. Thus every f (x)E R[x]with deg f (x)2 1 is factorized into a product of finitely many irreducible elements in K[z].By Theorem 2.17 and Proposition 2.18, factorization of polynomials of degree 2 1 into irreducible polynomials is feasible in R[z],and irreducible polynomials in R[x]are primes. Hence R[z] is a UFD. 17 16 Commutative Algebra

2.20. Corollary For any field K, the polynomial ring K[zl,..., z,] in finitely many variables 21, ..., z, over K is a UFD. 0

We finish this section by Eisenstein’s criterion concerning the irreducibility of polynomials in R[z],where R is a domain.

2.21. Theorem Let R be a domain and

f(.) = a,zn + . . . + a15 + a0 a polynomial in R[z].Suppose there is a prime p E R such that (a) P X an, (b) plai, i = 0, ...,71 - 1, (c) P2 1.0. Then f(z) is irreducible in R[z].

Proof Suppose f(z)= g(z)h(z)for g(z), h(z)E R[z]where

g(.) = c,zT + . . . + c1z + co

with ci,dj E R and T, s > 1, T + s = n. Then by (b) and (c), p)ao = cod0 and hence p divides co or do but not both. Suppose ~1%. By (a), we may let c, be the first coefficient of g(z) not divisible by p. But note that

This implies p a,, contradicting (b) because m < n. Hence g(5) or h(z) must be a unit of R.

2.22. Corollary If p is a prime number, then the polynomial

f(z) = zp-l + zp-2 + . * * + 5 + 1 is irreducible in Z[z]and hence irreducible in Q[z]. Preliminaries 17

Proof Note that f(z)= 9.If we use the translation z = X + 1, then

(X + 1)” - 1 f(x + + (X + 1) - 1

= 1(xp+ (;) xp-l+ (;) xp-2 +... + ( ) x + 1 - 1) X P-1

= xp-l+ (;) xp-2 + (;) XP-3 + . . . + p.

Now, using p as the prime needed in Theorem 2.21, we conclude that f(z) is irreducible in Z[z] and hence irreducible in Q[x]by Proposition 2.18.

Exercises 1. Complete the proof of Proposition 2.2. 2. Complete the proof of Proposition 2.3. 3. Complete the proof of Proposition 2.8. 4. Let R = K[t2,t3] be the subring generated by t2and t3 in the polynomial ring K[t]over a field K. Show that both t2 and t3 are irreducible in R but none is a prime. However t6 = t2t2t2= t3t3. (See also Chapter 3 (section 2, exercise 2) and Chapter 3 (section 3, Example (iii)).) 5. Show that the polynomial ring R = K[z1,z2,..., 2, ,...] in infinitely many variables over a field K is a UFD. (Hint: Any polynomial in R belongs to a polynomial ring in finitely many variables over K.) 6. Let R be a domain, a,b E R not all zero. Up to a unit multiple, define the greatest common divisor of a and b, denoted gcd(a, b), and the least common multiple of a and b (in case a # 0, 6 # 0), denoted lcm[a, b],as in Z (or as in K[z]with K a field). (In a similar way, for al, ..., a, E R, gcd(a1, ..., a,) and lcm(a1, ..., a,) may be defined.) Show that the following statements are equivalent for a domain R in which factorization into irreducible elements is feasible. (a) R is a UFD. (b) Every irreducible element of R is a prime. (c) For every a, b E R, not all zero, gcd(a, b) (or lcd[a, b] in case a # 0, b # 0) exists. (d) The intersection of two principal ideals of R is another principal ideal. 7. Let R be a UFD, f,g E R[s].Use Theorem 2.17 to show that if f,g 18 Commutative Algebra

do not have common irreducible divisors in R[x]then f,g do not have common irreducible divisors in K[x]either, where K = Q(R)is the field of fractions of R. 8. Let p be a prime number. Show that xn - p is irreducible in Z[x]and hence in Q[x]. 9. Prove that f = llyx8 + 3y7x5+ 9x5 - 7y7 - 21 is irreducible in Z[x,y]. (Hint: Consider f in Z[y][x].)

3. Field Extensions

The study of field extensions stems from the study of zeros of polynomials and the study of irreducibility of polynomials. Let K be a field and f E K[xl,..., x,] a polynomial of degree 2 2. Then the property that f has or does not have a zero in K,and the property that f is reducible or irreducible over K, all depends on the ground field K, for instance, first consider the zeros of x2 - 1, x2 - 2 in Q and the zeros of x2 - 3, x2 + 1 in R, and then consider the zeros of the given polynomials by extending Q to EX, R to C.A full demonstration of this aspect is given in Chapter 4 and Chapter 5. In this section we focus on several fundamental topics concerning field extensions. Let K, L be fields. If K is a subfield of L (including the case where KZLis a ring monomorphism), then we call L an extension field of K, and from now on K C L is referred to a field extension. Let K C L be a field extension and S c L a subset of L. Consider the intersection

of all subfields in L containing S. Then it is an easy exercise to verify that (a) K(S)is the smallest subfield of L containing S, and (b) K(S)= Q(K[S]),the field of fractions of K[S](hence K(S)is also the smallest subfield of L containing K [S]). In view of the above (a)-(b), we call K(S)the subfield of L generated by S over K. If S = {sl, ..., s,} is finite, then we write K(S)= K(s1,..., s,) and call it a finitely generated extension field of K. If S consists of a single element s, then K(s)is called a simple extension field of K. Preliminaries 19

Splitting field 3.1. Definition Let K be a field, and let f(z)be a polynomial in K[z]. If K 2 L is a field extension such that f(x) factors completely into linear factors over L, i.e., f(z)= an(. - ai) in L[z],and f(x) does not factor completely into linear factors over any proper subfield of L containing K, then L is called a splitting field of f(z).

Let K be a field. To see the existence of a splitting field for an arbitrary f(z)E K[z],we start with an irreducible polynomial p(z). Note that the quotient ring

where z is the image of z in L, is a field, for, if p(z) ,Y$(z)- then p(x)h(z) + $(z)g(z) = 1 for some h(z),g(z)E K[z],and hence $(z) is invertible in L. Note that via the natural ring homomorphism K[z]-+ L we may write K c L = K[T].Thus,

K[z]c L[z]and consequently p(l)= 0. It follows from the division algorithm that p(z) is factorized in L[x]as

P(Z) = (. - T)Pl(z), Pl(X) E LM. Now, since K[x]is a UFD, an induction on the degree of polynomials, or a procedure of adding the zeros of each irreducible factor of f(z)successively to the predecessor extension field, yields the following fact.

3.2. Theorem Let K be a field. Every f(z)E K[z]with degf(z) = n > 0 has a splitting field.

Example (i) The field Q(G)serves as a splitting field for both x2 + 3 and z3+ x2 + 3x + 3.

Remark Indeed, any splitting field of f(z)is isomorphic to the one constructed before Theorem 3.2. The reader can refer to any textbook speci- fying field theory for a detailed proof. 20 Commutative Algebra

Repeated zeros and separability

Let K be a field and let f(z) E K[z].In view of Theorem 3.2 we may always talk about the zeros of f(z) in some extension field of K. Furthermore, we explore the following Question When does f(x) have no repeated zeros?

3.3. Proposition f(z) E K[z]has no repeated zeros if and only if f(z) and f'(z) = are coprime, i.e., they do not have nonconstant common divisor. Proof Over a splitting field E of f(z),we have

f(z)= (z - Cyp. . . (z - a,)n" where the ails are distinct. Then it is clear that f(z)and f'(z) have no nonconstant common divisor over E if and only if ni = 1 for i = 1, ..., rn. 0

3.4. Proposition Let E be a splitting field of zn - 1 = f(z) E K[x],where n 2 1. Suppose that charK does not divide n. Then the following hold: (i) f(z)has exactly n distinct zeros (the nth roots of unity over K)in E. (ii) Let

un = {a E E I f(a)= o}. Then U, is a cyclic multiplicative subgroup of EX. Proof (i) By the assumption, this follows from Proposition 3.3. (ii) That U, forms a subgroup of EXis clear. We show that U, contains an element of order n. To this end, let

n = Plel . . .pzs be the factorization of n into primes, and let qz = 2 for i = 1, ..., s. Then, since the polynomial xQ*- 1 has exactly qz zeros in U,, for each i, there is e,-1 at E U, such that a: # 1. Set pt = Then p,". # 1 but p:' = 1. It follows that each pzhas order p:%.Since p;l, ...,pss are pairwise coprime, ,O = . . pS is the desired generator for U,.

The last proposition makes the multiplicative structure of a finite field clear. Preliminaries 21

3.5. Theorem Let K be a finite field. Then the multiplicative group KX of K is cyclic. Proof If charK = p > 0, then [K : Z,] = m for some m and hence KXhas n = p" - 1 elements which are all zeros of f(z)= zn - 1 E Z,[zc]. Since p /y n, Proposition 3.4 can be applied to this case. I3

Since K[z] is a UFD, the general discussion may be further reduced to irreducible elements.

3.6. Theorem Let K be a field and let q(z)E K[z] be irreducible. (i) If charK = 0, then q(z)does not have repeated zeros. (ii) If charK = p > 0, then q(z) has repeated zeros if and only if q(z)= g(zP) for some g(z) E K[z]. Proof We apply Proposition 3.3 to both cases. (i) If charK = 0, then since q(z)is irreducible, we have q'(z)# 0 (otherwise p(z) would be a constant), degq'(z) < degq(z), and hence q(z)and q'(z) are coprime. (ii) Suppose charK = p > 0. Let q(z)= a,zn + an-1zn-l+ . . . + alz + a0 with a, # 0. Then q'(z) = na,zn-l + (n - l)a,-l~~-~+ .. + a1 with degq'(z) = n - 1 < degq(z) = n. Thus,

q(z) and q'(z) have a nonconstant common divisor H ra, = 0 * plr, say r = s,p.

Consequently, q(z)has repeated zeros if and only if q(z)= atpxtP + . . . + azpz2P+apxP+ao.Therefore, q(z)= g(zP) where g(y) = ao+a,y+a2py2+ . . . + atpyt,as claimed. 0

3.7. Corollary Let K be a finite field and let q(z)E K[z] be irreducible. Then q(z)has no repeated zeros. Proof Since K is finite, we know that charK = p > 0 for some prime number p. Then Z, is the prime field of K and K is a finite dimensional Z,-vector space, say dimz,K = n. Hence K has pn elements. Thus, the multiplicative group of K, which is KX,has order pn - 1 and XP" = X for all X E K. (We assumed that the reader is familiar with elementary group theory.) It follows that ifg(zp) E K[z],sayg(zP) = ao+alzP+~~~+a,zRP, 22 Commutative Algebra then, after setting a:"-' = bi, i = 0,1, ..., n,

which can never be irreducible. This shows that the irreducible g(x) cannot have repeated zeros by Theorem 3.6. cl

3.8. Definition If a polynomial f(x) E K[x]has no repeated zeros, then f(x) is called a separable polynomial over K, and otherwise an inseparable polynomial over K. (See also Definition 3.11 below.)

Algebraic extension and primitive elements

We now start with a field extension K C_ L and consider CY E L. If there is some f(x) E K[x]such that ~(cY)= 0, then we say that cy is an algebraic element over K;otherwise, we say that a is a transcendental element over K. If every element of L is algebraic over K, then L is called an algebraic extension field of K, and we refer K 2 L to an algebraic field extension. If L contains a transcendental element over K, then K c L is referred to a transcendental field extension. Let K C L be a field extension. Then L is naturally viewed as a K- vector space. In the literature, the dimension dimKL is also called the degree of L over K, denoted [L : K]. Clearly, if a field extension K C L has finite [L: K],then L is algebraic over K. For instance, [C : R] = 2. If L contains a transcendental element over K, then [L : K] = co. It is known that e and 7r are transcendental over Q. So [R : Q] = 00. Another familiar transcendental extension is K c K(z),where K(x)is the field of fractions of the polynomial ring K[x]. Also, not every algebraic field extension is finite dimensional (exercise 4). To understand the structure of a field extension K C L, simple extension plays a key role. Let CY E L. Consider the subring K[cy]C L and the ring Preliminaries 23

homomorphism p: K[z] - K[a]

If a is a transcendental element over K, then

Kerp = (0) and K[z] K[a]. If a is algebraic over K, then kerp # (0) and hence Kerp = (p(z))for some nonconstant p(z)E K[x] because K[z]is a PID. We may assume that p(z) is monic. It is a consequence of the division algorithm in K[z] that p(z) has the smallest positive degree among all polynomials in Kerp. This leads to the following

3.9. Definition For an algebraic element a over K, the monic polynomial p(z),which is the generator of kerp, is called the minimal polynomial of a over K.

3.10. Theorem Let K C L be a field extension and a E L. If a is algebraic over K and p(z) is its minimal polynomial, the following hold: (i) p(z) is irreducible and unique in K[z]. (ii) K[rc]/(p(z)) K[a]is a field containing K. Thus, k[a]= K(a). (iii) If degp(z) = n,then every element p E t(a)has a unique expression

= Xn-1an-’+ Xn-1~n-2 +.. + Xla + Xo, Xi E K.

Thus, {an-’,..., a,1) forms a K-basis for K(a), [K(a) : K] = n. Conse- quently, K(a) is a simple algebraic extension field of K. Proof Using division algorithm by p(z) in K[z], all conclusions are easy exercises. 0

Later in exercise 2 the reader will be asked to show that if a1, ..., a, E L are finitely many algebraic elements over K, then K C K(a1, ..., a,) is an algebraic field extension and [F : K] < 00. When K plays the role as in the case of Theorem 3.6(i) and Corollary 3.7, our next goal is to show that the finitely generated algebraic field extension K(a1, ..., a,) is actually a simple extension. But first, we need the notion of a separable extension.

3.11. Definition (i) Let K C L be a field extension and let a E L be an 24 Commutative Algebra algebraic element over K. If the minimal polynomial p(z) of a over K is separable in the sense of Definition 3.8, then a is said to be separable over K; otherwise a is inseparable over K. (ii) Let K L be an algebraic field extension. If every element of L is separable over K, then L is said to be separable over K; otherwise L is inseparable over K.

By Theorem 3.6 and Corollary 3.7, inseparable field extension is, indeed, quite rare.

3.12. Theorem Let K C F = K(a1,a2, ..., a,) be a finitely generated algebraic field extension. Suppose that 02, ...,am are separable over K. Then F = K(6)for some 6 E F. Proof If K is finite then so is F (by Exercise 2), and the conclusion follows from Theorem 3.5. Suppose that K is infinite. We consider only the case where F = K(a,p) with p separable over K since the general conclusion may be obtained by an induction. Let L be a field over which the minimal polynomial p(z) of a and the minimal polynomial q(z)of P are factorized as

n m

i= 1 j=1 where al,a2 ,..., an,,&,..., Pm E L, and a1 = a, p1 = 0. (The existence of L is guaranteed by Theorem 3.2.) By the assumption, PI, ...,Dm are distinct. Thus, the equations

ai - a1 = &(PI - Pk), k # 1, have only finitely many solutions Xik E K. Hence, there exists c E K such that

ai - a1 # c(p1 - Pk), 1 5 2 5 n, 2 5 k 5 m.

Set 6 = a + cP. Then clearly K(6) C F. Below we show that ,B E K(1Y) and then it follows that F = K(6). Note that a = 6 - cP. We have p(6 - cp) = p(a) = 0. Consider the polynomial r(z) = p(6 - a)E K(d)[z].Then, by the choice of c, ,f? is the only common zero of q(z) and ~(x)in F. This shows that the Preliminaries 25 minimal polynomial of p in K(t9)[x]is of the form t - p for some p E K(t9). Therefore, p = p E K(t9) as expected.

3.13. Definition The element 6 that appears in Theorem 3.12 is called a primitive element of F.

Let F be a field. If every nonconstant polynomial f(z)E F[x]splits in F, i.e., f(z)= ny=lA(z - Ai), A, Xi E F, then F is said to be algebraically closed. Clearly, if F is algebraically closed, then there is no proper algebraic extension of F. For instance, the field C of complex numbers is algebraically closed (this is also known as the content of the fundamental theorem of algebra). Without proof we mention the following theorem (the reader is referred to any textbook specializing field theory for the classical proof given by Emil Artin). Theorem Let K be a field. Then there is an extension field L of K that is algebraically closed.

Liiroth's theorem Within the context of Theorem 3.6(i), Corollary 3.7 and Theorem 3.12, it is easy to see that if K C L = K(t9) is a simple field extension, then any intermediate field extension F of K with K F L is a simple extension. The final part of this section deals with a similar situation on simple transcendental field extension. Let K be a field and x a transcendental element over K. Given coprime polynomials u(x),v(x)E K[x],consider h = - E K(z)and the simple extension K c K(h).Set

where K(h)[t]is the polynomial ring in t over K(h).

3.14. Lemma With notation as above, the following hold: (i) h is transcendental over K. (ii) q(t) is irreducible in K(h)[t]. (iii) [K(z): K(h)]= degq(t) = max{degu(rc), degv(x)}. 26 Commutative Algebra

Proof (i) Exercise. (ii) Note that q(t) is linear with respect to h in the polynomial ring K[h,t] which is a UFD. Hence q(t) is irreducible in K[h,t], for u(z)and ~(z)are coprime by the assumption. It follows from Proposition 2.18 that q(t) is irreducible in K(h)[t]. (iii) By the construction of q(t),q(z) = 0. It follows from part (ii) that q(t) (assuming monic) is the minimal polynomial of z over K(h). Thus, [K(z): K(h)]= degq(t) = max{degu(s), degv(z)}, as desired. I3

3.15. Corollary (i) Let E be any intermediate extension field of K with K E C K(z).Then [K(z): El < co. (ii) Every automorphism of the ring K(z)which is K-linear is given by

Proof Exercise. 0

3.16. Theorem (Luroth) Let K be a field and z a transcendental element over K. Let E be an intermediate extension field of K with K $ E C K(z). Then E = K(y) for some y E K(z)(hence E E K(z))and [K(z): El < 00. Proof By Corollary 3.15, [K(z): E] < co. Let p(t) E E[t]be the minimal polynomial of z over El say

If we multiply p(t)by the least common multiple, say s, of the denominators of ri’s, the obtained polynomial

is primitive in K[z][t]with respect to t (check it!). Write degtf(z,t) for the degree of f(z,t) in t. Then

n = deg,f(z,t) = degp(t) = [K(z): El.

Note that s, = s and all 2 E E. As is transcendental over K, there is at least one E E - K. Set h = # = $ for convenience, where u(z) and u(z) are coprime in K[z].Then

q(t) = hv(t)- u(t) Preliminaries 27 is irreducible in K(h)[t]and

(2) [K(z): K(h)]= degq(t) = max{degu(z), degv(z)} by Lemma 3.14. Since K 2 K(h)& E C K(z),we complete the proof by having the equality

[K(z): El = [K(z): K(h)].

To this end, note that q(z) = 0 and q(t) E E[t].Hence q(t) = p(t)p~(t) with pl(t) E E[t],for p(t) is the minimal polynomial of z over E. Thus, by formula (I),

But f(z,t) is primitive in K[z][t]with respect to t. It follows from Theorem 2.17( ii) that

(3) u(z)v(t)- v(~)u(t)= df(z,t), d E K[z][t].

Suppose deg5f(z,t) = m. Then max{degu(z), degv(z)} 5 m by formula (1). So the above formula (3) implies that

(4) deg,(u(z)v(t) - v(z)u(t))= m and d is a constant. Note that u(z)v(t)- w(z)u(t)is antisymmetric in z and t. Therefore, (2) + (4) yields

[K(z): K(h)]= max{degu(z), degv(z)} = m

= n = [K(z): El, as desired. 28 Commutative Algebra

More results concerning field extensions are given in section 5 and (Chapter 3 Theorems 1.8 and 2.4).

Exercises 1. Let K L be a field extension and a1, ...,a, E L. Show that if a1, ..., a, are algebraic over K, then K 2 F = K(cr1, ...,am) is an algebraic field extension and [F : K] is finite. (Hint: Note that F =

K(al)(a2).. + (a,) and use Theorem 3.10(iii).) 2. Let K L E be a tower of algebraic field extension, i.e., L is algebraic over K and E is algebraic over L. Use exercise 1 to show that E is also algebraic over K. Moreover, show that if [L: K]< co and [E: L] < 00, then [E: K]= [L: K][E: L]. (Hint: If a E E and Anan + .. . + Xla + XO = 0 for Xi E L, then consider K C: K(X,, ..., XO) C K(X,, ..., Xo)(a).) 3. Use Theorem 3.10(iii) to show that if K C: L is a field extension, then all elements of L which are algebraic over K form a subfield of L containing K. 2 is called the algebraic closure of K in L. (Hint: For alpE L, algebraic over K, consider K & K[a] k[a][p].) 4. Let F be the subfield of C consisting of all algebraic elements over Q. Use (section 2, exercise 8) to show that [F : Q]= co. 5. Show that if K is an algebraically closed field, then K is infinite (or equivalently, that a finite field cannot be algebraically closed). (Hint: If F = {ul,..., un} is a finite field, consider the polynomial p(x) = ny="=,x- ui) + 1 in F[x].) 6. Let d E Z be square-free. Then every element a E a(&) is of the form a = r + sd,where r, s E Q. Show that a has the minimal polynomial

pa(.) = x2 - 2rx + (r2- s2d).

7. Let F = Q(a,8). Find a primitive element for F. 8. Prove Lemma 3.14(i). 9. Complete the proof of Corollary 3.15.

4. Symmetric Polynomials

Let R be a ring and R[xl,..., x,] the polynomial ring in variables 51, ...,x, over R. Put N" = {a = (a1,..., a,) I>ai EN . Preliminaries 29

Then every element f(z1,..., 2,) E R[q,..., z,] has a unique expression

Let S, denote the permutation group of {1,2, ..., n}. A polynomial f(sl,..., 2,) E R[zl,..., z,] is said to be symmetric if

For example, 5: + 5; + xi, (21 + 22 + 53 + ~)(~11~2~3a)~. Important symmetric polynomials are those elementary symmetric polynomials:

Let R[sl,..., s,] be the subring of R[xl,..., xn] generated by R and {s1,s2 ,...,s,}. Then it is clear that every g(s1,s2,..., s,) E R[sl,..., s,] is a symmetric polynomial. The next theorem, due to Newton, shows that the converse is also true. 30 Commutative Algebra

4.1. Theorem If f = f(z1,..., 2,) is a symmetric polynomial in R[zl,...,z,], then f(21,...,2n) E R[si,..., sn]. Proof To reduce f = f(sl,..., 2,) = c,c,z~' . . . zEn into a polynomial in elementary symmetric polynomials, in view of previous (*) we order the set of monomials

{2yiZ;z..*2,:- I (a1,...,a,) ENn by the lexicographic ordering:

2~1. . . 2:n +lez 271 . . . 25

if and only if

a1 = P1 , a2 = Pz, ..., a,-1 = a3-1 while a, < p, for some s 5 n.

Thus, the terms of f are ordered lexicographically (note that +lez is a total ordering), and we may assume that the leading monomial of f is xy'x;' . . . x;n. Since f is symmetric, zztl)zzi2,. . . x;;,, occurs in f for every 7r E S,. It follows that the leading monomial of f has the property that 011 2 a2 2 2 a,. For example, the leading monomial of

ki k2.. .sk- (21 + . . . + 2,)kl . . . (21 . . . 2,)kn $1 s2 is

++"'

By choosing kl = a1 -aZ, ..., k,-1 = a,-1 -a,, k, = a,, we can make this the same as the leading monomial off. Suppose that the leading coefficient of f is c, then f - csf's~. . . s? has a lexicographic leading term

d$ xp . . . xk , P1 2 P2 2 ... 2 Pn which comes after c~y'x~~. . . x:n in the ordering. Since only a finite number of monomials x~'x~. . . 22 in f satisfying y1 2 72 2 . . . 2 7, follow ~~'IC;~. . . xgn lexicographically, a finite number of repetitions of the above process reduce f to a polynomial in s1, .. ., s, . 0

Example (i) The symmetric polynomial Preliminaries 31 is written lexicographically. And by the method given in the proof we may derive that f = sls2+3s3. Similarly, (IC1+22)(zl+x3)(22+23) = s1s2-s3.

An application of symmetric polynomials to field extension is given as follows. If K C L is a field extension, a,xn + a,-~x"-~+. . . + ao = f(x) E K[x] with degf(z) = n, and f(~i)= 0 with TI, ..., T~ E L, then, f(x) factors in L[x]as

f(x) = a,(z - T1)(Z - TZ). (x - r,)

= an(xn + c1xn-l + C2xn-2 + . . . + cn) where ci = (-~)ZS~(T~,T~, .,., T"), i = 1, ..., n. After comparing coefficients of both sides, we have

(-1)iUnSi(T1,T2, ..., T,) = a,-< E K, 2 = 1,..., n. 4.2. Corollary Let K C L be a field extension, a,xn + an-lxn-l + ... + ao = f E K[x]with degf = n, and f(~i)= 0 with TI, ..., T, E L. If h(x1, ...,2,) E K[q,..., x,] is a symmetric polynomial, then h(rl,~2,..., T,) E K, i.e., {TI, ..., T,} defines a function K[s1, ..., s,] - K

Example (ii) Suppose that TI, 7-2, 7-3 are the zeros of f(z)= s3+x2-x+1 in C. Find T: + T; + T: and T: + T; + T:.

Solution Since f(x) = (x - T~)(z- ~2)(x- ~3),it follows that

7-1 + 7.2 + T3 = -1, 32 Commutative Algebra

More generally, the following recurrence relations, called Newton's for- .. mulas, can be used to establish formulas for pi = (-1)%(z2,+ xi + . . . + zk), i 2 1, in terms of s1,s2, ..., sn.

Pl + s1 = 0,

Pz + SlPl + 2sz = 0,

...

Pn + SlPn-1~2pn-2 +.. . + sn-lpi+ TLS, = 0. We close with an application to polynomial building,

Example (iii) Let r1,rz,r3 be the zeros of f(x) = x3 - z + 2 in C. Find the polynomial g(x) that has zeros rf ,r;, rg . Solution Suppose the desired polynomial is of the form g(z) = x3 + Ax2 + Ba: + C. Then

A=-(rf+r; +rg) =-~2(rl,r2,~3)

= -s1(r1,r2,r3)2 + 2sz(r1,r2,r3)

= 0 + 2(-1) = -2,

= (-1)2 - 2(0)(-2) = 1,

C = -T~T;T~= -~3(~1,~2, ~3)~ = -4.

Hence g(z) = x3 - 2rc2 + z - 4.

Exercises 1. Express the product (xf + x;)(z? + zi)(zz + xg) in terms of sl,s2, s3. 2. Let T1,TZlT3be the zeros of f(z) = x3 - 6z + 11 - 6 in C. Determine the polynomial g(a:) that has zeros r: + r;, T: + ri, r; + T:. 3. Let TI,r2,r3, 7-4 be the zeros of f(z)= adz4 + a3x3 + 1322' + alx + a0 in Preliminaries 33

@, where ai E Q. Suppose a4 = -5 and the elementary polynomials in 3 1 TI,TZ,T~,~~are s1 = %, s2 = 16, s3 = -8, s4 = -E. Find a3,a2,a1,ao. 4. Let R be a ring. A polynomial belonging to R[xl,..,xn]is said to be antisymmetric if it is invariant under even permutations of the variables, but changes sign under odd permutations. Let

Show that (a) A is antisymmetric, and (b) if 2r = 0 implies T = 0 for r E R, then any antisymmetric polynomial f is expressible as a polynomial in the elementary symmetric polynomials, together with A. (Hint: Note that f(x11x21x3,...,xn) = -f(x2,xl,x3, ... ,xn)r 2f(xl,xl,x3,...,%)= 0. Thus, f vanishes when 21 = 52. So a division on f by x1 - x2 in R[Q, ..., xn][xl]yields (21 -xz)lf. Similarly, (z1-xi)lf, i = 3, ..., n. Writing f = niZ1(x1- xi)fi, where fl E R[x2,..., x,] and is antisymmetric. Now an induction on n finishes the proof.)

5. Trace and Norm

Throughout this section we let K s L be a simple algebraic field extension, that is, L = K(6),6 E L. If p(x) E K[x]is the minimal polynomial of 6 over K, we may set a tower of field extensions

KcLcE such that E contains all distinct zeros of p(x),say dl = d,82, ..., 6,, where m 5 n = [L: K] = degp(x), that is, E contains the splitting field of p(x).

5.1. Proposition With notation as above, there are exactly m distinct ring monomorphisms L -+ E that are K-linear. Moreover each K-linear monomorphism L --f E is given by 6 4 6i, 1 5 i 5 m.

Proof If a: L t E is a monomorphism as described, then 0 = a(p(6))= p(a(d)),i.e., ~(19)is a zero of p(x). Note that the elements of L are of the form EX,@. It follows that if LZEand L"2.E are two K-linear ring 34 Commutative Algebra monomorphisms such that a1(6) = a2(6),then 01 = 02. Conversely, each 6i defines a desired monomorphism ai: L - E

CXjSj H cxj79; because all t9i's have the same minimal polynomial p(z). 0

With the help of Proposition 5.1 we may determine, for every a E L, the minimal polynomial pa(.) of a over K and the splitting field of pa(%).

To see this, let a1, ..., em be all distinct monomorphisms L 4 E defined by ai(6)= t9i, a = 1, ..., rn. Suppose that each 6i has multiplicity ei 2 1, that is, p(z) = nzl(z- 6i)ei in E[z].Then el + e2 + . . . + em = n = degp(s), and each a E L = K(8)is associated to a monic polynomial in E[z],that is,

For convenience, we call fa(z)the total polynomial of a.

5.2. Proposition Let K L = K(6)c E be as above. For any a E L = K(6),the following hold: (9 fa(z)E Wzl. (ii) Let pa(.) E K[z]be the minimal polynomial of a over K. Then fa(z)= pa(z)' for some s 2 1. (iii) E contains the splitting field of the minimal polynomial pa(.) of a over K.

Proof (i) Since a = ~(29)= Cyi: A@, where T(X) = CyL; Xixi E K[z], we have

i= 1 Preliminaries 35

Note that all Xi E K. After expanding the latter product we see that the coefficients of fa(z)are given by symmetric polynomials in the n zeros of p(z). By Corollary 4.2, fa(z)E K[z]. (ii) By part (i), fa(.) E K[z]and fa(a)= 0. It follows that fa(z)= pa(s)sh(z),where h(z)E K[z]and pa(z), h(z) are coprime and both are monic. If h(z) is not a constant, then some C~(CY)is a zero of h(z). Let CY = r(6) = Eyz: Xi62 with r(z) = CyI; Xizz E K[z].Then ai(a)= r(6i) and h(ai(a))= h(r(6i))= 0. Set g(z) = h(r(z))E K[z].Then g(Si) = h(r(6i))= 0 implies p(z)lg(z),for p(z) is the minimal polynomial of 6 and hence the minimal polynomial of each 6i. It follows that 0 = g(6) = h(r(6))= h(a)and pa(z)lh(z),a contradiction. This shows that h(z)is a constant and h(s) = 1 because it is monic. Thus, fa(z)= pa(z)'. (iii) By parts (i) and (ii), fa(.) E K[z]and fa(z)= nEl(z - ai(~~))"~= p,(~)~for some s 2 1. So pa(.) factors into linear divisors over E.

We now introduce two functions on L that will play important roles in Chapter 3 section 3 and throughout Chapter 4. Let Q E L = K(6). By Proposition 5.2(i), the total polynomial fa(z) of CY belongs to K[z].By the definition, fa(z)= nEl(z- o~(cY))"~,where el + .. . + em = n = [L : K] = degp(z). If we check the expanded expression of fa(z),then fa(.) = zn + &-1zn-' + ... + c1z + co with m c++l = - eiai(a) and co = nzlo~(Q)~+. Thus, we have obtained two well-defined functions:

TLIK: L + K

i= 1

i= I

5.3. Definition For Q E L, TL/K(Q)is called the trace of Q in K and NL,K(Q)is called the norm of CY in K.

5.4. Proposition For Q,P E L, X,p E K, the following hold: (i) TL/K(XCY+ PP) = XTL/K(CY)+ /-LTL/K(P). 36 Commutative Algebra

Knowledge on bilinear forms needed by the next theorem and Chapter 3 Theorem 3.2 is given as an appendix at the end of this section. Viewing L as an n-dimensional K-vector space, Proposition 5.4 enables us to define a symmetric bilinear form on L: LxL- K

TL/K(QP) 5.5. Theorem For K C L = K(6) with [L : K] = n, if the minimal polynomial p(z) of 6 over K has n distinct zeros 61 = 6,192, ..., 8,, then the bilinear form defined above is nondegenerate. Proof Set r(k)= (6!,..., di), Ic = 0, ..., n - 1, and write V for the Vander- monde matrix r(0) 1 1 ... 1 V= [ ;;;; :; :; ::: :: ... I=[ 6:-' q-1 I* r(n - 1) . . . 6;-1

Let ~1,..., on be all the n distinct K-linear monomorphisms from L to E as described in Proposition 5.1 such that ai(6) = 6i,i = l,...,n. If we consider the standard K-basis {1,6,..., of L, then since n n T,/,(6k@) = cCi(6k6j) = cspt = d(Ic)(6(j))t, i= 1 i=l the matrix of the bilinear form is given by

(TL/K(dW))= VVt and hence det (T~/~(s~dj))= (det(V))2.

Since all the 6i are distinct, det(V) = n,,,(6i - 6j) # 0. This shows that the bilinear form is nondegenerate. 0 Preliminaries 37

5.6. Corollary If K & L is a finite dimensional separable field extension, then Theorem 5.5 hold. cl

Appendix. Bilinear forms Let U and V be two vector spaces over a field K, U x V = {(u,v)[ u E U, v E V)the Cartesian product of U and V. A bilinear form on U x V is a mapping

<, >: UXV- K

(u,v) ++ < u,v > satisfying

< xu1 +pu2,v> = x < u1,v > 3-p < U2,V >

< u,xv1 +pv2 > = x < u,v1 > -I-p < u,v2 > for all u1,u2 E U,211,212 E V and A,p E K. A bilinear form < , > on V x V is called a bilinear form on V. A bilinear form < , > on V is said to be symmetric if < x,y >=< y,x > for all z,y E V. If < , > is a bilinear form on U x V which satisfies

< u,v’ >= 0 for all u E U implies v’ = 0, and

< u’,v >= 0 for all ‘u E V implies u’ = 0, then < , > is called a nondegenerate bilinear form. Let U and V be finite dimensional K-spaces. Given a basis {ul,..., um} of U and a basis {VI, .., vn} of V,if < , > is any bilinear form on U x V, then there is an associated m x n matrix A = (aij) with

aij =< ui, vj >, i = 1, ..., m, j = 1, ..., n, and < , > is completely determined by A, that is, given

i= 1 j=1 38 Commutative Algebra itf follows from the bilinear property of < , > that

< U,Y > = < cxiui,cpjvj >

Conversely, any m x n matrix over K yields a bilinear form on U x V in this way. Now, suppose that {ui,..., u&} and {Y;, ..., vi} are new bases for U and V respectively, and that

m. n

i=1 j=1

Then it follows from a change of bases and (I) that

(Al, ..., A,) = (A;, ..., A&)P,

(PI, '"7 pn) 1(pi, ...t~i)Q for some invertible matrices P = PmX,, Q = Qnxn. Thus,

< u,Y >= (A;, ..., A~)PAQ~ and consequently, the matrix referred to the new bases is PAQt.

5.7. Theorem (i) Let U and V be finite dimensional vector spaces over a field K, where dimKU = m and dimKV = n. If < , > is any nondegenerate bilinear form on U x V, then m = n, and for any basis {ul,..., un}of U there exists a unique basis {vl, ..., vn} of V such that

0, if i # j, < Ui,Yj >= 6.. - 1,if i = j. Preliminaries 39

(ii) A bilinear form < , > on a finite n-dimensional K-space V is nondegenerate if and only if the associated matrix A = (aij),where aij =< Vi,vj >, is invertible for any basis {q,... ,v,} of V. Proof (i) Let {u~,..., u,} be any basis of U. Consider the linear mapping induced by < , >

0 : V - Km = {(Ail ..., A,) I Xi E K}

Then 0 is injective because < , > is nondegenerate. Thus, n = dimKV 5 dimKU = m. Similarly we also have m 5 n. Hence m = n. Note that 0 is now an isomorphism. If we use the standard basis {el, ... ,em} of Km,where

ej = (0,...,0,1,0,...,0),+ j = l,...,m, j-1 and write vj for the inverse image of ej under 0,then it is clear that {w~,...,~~}is a basis for V and < uilvj >= Sij, i,j = l,...,m. If {w;,... ,&} is another basis of V with this property, then a(vi - v:) = 0 implies zli = v:, i = 1, ..., m, because a is injective. (ii) This follows from part (i) and previous discussion on the associated matrices of < , > with respect to given bases.

Exercises 1. Complete the proof of Proposition 5.4. 2. Let d E Z be square-free, K = Q(&). Find all Q-linear ring monomorphisms K 4 C. 3. Let K = a(&) be as in exercise 2 above, a = r + s&. Show that TK,Q(~)= 2r and NKIQ(~)= r2 - s2d. (Compare with section 3, exercise 6.) 4. Let 6 = 4 + a,K = Q(6). Find the minimal polynomial p(z) of 6. (Hint: Note that ()(A),a(&) c K. Any Q-linear ring monomorphism

u: K -+ C induces a Q-linear ring monomorphism GI: Q(a)--f Q1 and

a Q-linear ring monomorphism 02: a(&) -+ Q1, such that ~(6)= al(4)+ 02(&). The answer is p(z) = x4 - lox2 + 1.) Can you generalize this result to 6 = fi + 4 for arbitrary square- free p # q? 40 Commutative Algebra

6. Free Abelian Groups of Finite Rank

Let G be an abelian group with the binary additive operation + and the identity element 0. For g E G, we write Zg for the cyclic subgroup of G generated by g, and consequently, we write CgEnZg for the subgroup of G generated by a nonempty subset R C G.

A subset S2 = {gi}iE J of G is said to be Z-linearly independent if for any finitely many gill gi2, ...,gin E 0, there do not exist s1, ..., sn E Z, not all zero, such that slgil + s2gi2 + . . . + sngin = 0. If R is not Zlinearly independent, then it is Zlinearly dependent.

6.1. Definition Let = {gi}iGJ C G. If G = CgiEnZgi and R is Z- linearly independent, then G is called a free abelian group and $2 is called a Zbasis of G, or just a basis of G.

Below we focus on free abelian groups with finite Zbasis.

6.2. Proposition If a free abelian group G has two bases (91, ...,gn} and {hl, ..., hm},then m = n. Proof Suppose m < n. Then, as dealing with vector bases over a field in linear algebra, after expressing each gi as a Zlinear combination of hi's, we may derive, by passing to Q, that (91, ..., gn} is Zlinearly dependent, a contradiction. Hence, m 2 n. By symmetry, n 2 m. Thus, m = n as desired. 0

6.3. Definition An abelian group G with a basis of n elements is called a free abelian group of Z-rank n, or just a free abelian group of rank n.

Z is a free abelian group of rank 1 and { 1) is a basis for Z. The direct sum

of n copies of Z, where

(kl,.., kn) + (el,..., en) = (kl + e,, ..., k, + en>,

m(k1, ..., kn) = (mkl,..., mkn), m E Z, Preliminaries 41

ei=(O ,..., O,I,O,..., 0) i=l,..., n

6.4. Proposition Any finitely generated abelian group G = Cy=lZgi is a homomorphic image of some free abelian group of rank n. If G is free and (91, ..., gn} is a basis of G, then G E Z". Proof Exercise.

In view of Proposition 6.4, from now on we write

i= 1 for the free abelian group G with basis (gl, ...,gn}. Let M,(Z) be the set of all n x n matrices over Z. If A E Mn(Z) and det(A) = fl,then we say that A is unimodular. If A E Mn(Z) is unimodular, then A is invertible and

A-l = -A* =&A* det(A) where A* is the adjoint matrix of A. Clearly, the construction of A* implies A* E M,(Z). It follows that A-' E M,(Z).

6.5. Lemma If {ul,.,., u,} is a basis for the free abelian group G, then (~11,..., vn}, where

n vi = Caijuj, aij E Z,i = 1,..., n, j is a basis for G if and only if A = (aij)is unimodular. Proof With the help of the above remark, this is an easy exercise.

6.6. Theorem Let G be a free abelian group of rank n and H a nonzero subgroup of G. Then the following hold: (i) H is free of rank s I n. 42 Commutative Algebra

(ii) There exist a basis (91, ...,gn} for G and integers [I, ..., ls E Zf such that {elgl, ...,e,gs} is a basis for H. Proof We prove the theorem by induction on the rank of G. If rankG = 1, then G 2 Z and the conclusions (i) and (ii) are clear. Suppose the assertions (i) and (ii) are true for any free abelian group of rank < n. Let G be a free abelian group of rank n and let H be a nonzero subgroup of G. Then, with respect to a fixed basis {el, ..., en} of G, H contains elements

(1) h = klel + . . . + knen, some ki's are positive.

Choose a basis {XI, ..., xn} of G such that !I is the smallest positive coefficient with respect to (l),and rearrange the members of this basis (if necessary) so that H contains an element of the form

On division by el, write

If we define

92 = 52

then

[f] = [;::::;][:)Xn Sn 0 0 ...

where the square matrix is clearly unimodular. By Lemma 6.5, (91,x2, ..., x,} Preliminaries 43 is a basis of G. With respect to this new basis and previous formula (2),

fl = Clxl+ 7712x2 + . . . + mnx,

= Clgli 7-252 + . . . + r,x,.

By the choice of Cl, we must have r2 = . . . = r, = 0. It follows that

(3) fl =am.

Now, with respect to the new basis {g1,22, ..., xn}, each h E H has the expression

h = clgl+ ~2x2+ . . . + GX,, ci E Z.

Write c1 = C1q + r with q,r E Z, 0 5 r < el. Then by the above (3), H contains

h - qfl = Ciqgi + rgl - C1qg1+ ~2x2+ . . . + GZ,

h ++ ~2x2+-..+Gz, defines a group homomorphism and H Zf1 + p(H). Note that p(H) is a subgroup of G* that is free of rank n - 1. By the induction hypothesis, there exist basis (92, ...,gn} of G' and integers &, ...,C, E Z+, where s 5 n - 1, such that {&g2,...,Cngs} forms a basis for cp(H). Thus, {Clgl = f1,[2g2, ..., !,gS} forms a basis for H, as desired. 0

6.7. Theorem Let G be a free abelian group of rank n and H a subgroup of G. The following statements hold: 44 Commutative Algebra

(i) GIH is finite if and only if rankG = rankH. (ii) If rankG = rankH = n, (51, ..., xn} is a basis for G, {yl,..., yn} is a basis for H, and

then the number of elements of GIH is equal to Idet(A)I, where A = (aij). Proof (i) By Theorem 6.6, choose a basis (91, ...,gn} of G and a basis { fl,..., fs} of H with fi = eigi and ti E Z+, i = 1, ..., s 5 n. Thus

and we have the group isomorphism

It follows that GIH is finite if and only if n = s. (ii) By the proof of part (i), if GIH is finite then it has exactly ell2 .-.en elements. Employing the chosen bases in part (i), we have

j=1

j=1 Preliminaries 45

Then (bij) = B and (dij) = D are unimodular, and

c = (c.23.) = [!!;!/! .. 0 0 ... en

Taking the matrix A = (aij) from the assumption of part (ii) into ac- count, we get A = BCD and det(A) = det(B)det(C)det(D). Therefore, Idet(A)I = ll&...l,.

Exercises 1. Let i = G.Show that Z[i] = {a + bi [ a, b E Z} Z Z@ Z. 2. Complete the proof of Proposition 6.4. 3. Complete the proof of Lemma 6.5. 4. An abelian group G is said to be torsion-free if G does not have finite order nonzero element. Show that a finitely generated torsion-free abelian group is free of finite rank. (Hint: Use Proposition 6.4 and refer to the proof of Theorem 6.7.) 5. Show that a finitely generated abelian group G is either finite or isomorphic to the direct sum of a free abelian group of finite rank and a finite abelian group.

7. Noetherian Modules

Let R be a ring.

7.1. Definition Let M be an abelian group with the binary additive operation + and the identity element 0. We say that M is an R-module if there is a mapping a: RxMd M

(r,m) H a(~,m)= rm (called the R-action on M)satisfying (Ml) (r + s)m = rm + sm, (M2) r(m+ m’) = rm + rm’, (M3) r(sm)= (rs)m, 46 Commutative Algebra

(M4) lm=m for all r, s E R and m,m' E M.

By definition, a Z-module is nothing but an abelian group M (binary operation is written additively). Conversely, given an (additive) abelian grooup M, M can be made into a Z-module by defining

Om = 0 and lm = m for m E M, then inductively

(n+ l)m = nm + m for n E Z', m E M, and

(-n)m = -nm for n E Z', m E M.

If R = K is a field, then an R-module is nothing but a K-vector space. In this sense we may view an R-module as the generalization of a vector space. However, since not every nonzero element in an arbitrary ring R is a unit, many of the techniques developed in vector space theory cannot be performed directly to deal with R-modules. From the definition it is clear that if M is an R-module then every r E R defines an endomorphism of the abelian group M,that is, pr: M + M with p,(m) = rm. One easily checks that this yields a ring homomorphism 0: R + EndzM with u(r)= pr, where EndzM is the ring of endomorphisms of M. Conversely, if M is an abelian group then any ring homomorphism 9: R 4 EndzM induces an R-module structure: rm = p(r)(rn). This is the idea of modern representation theory of rings and algebras. Two special kinds of module will be used frequently in the follow-up chapters:

If R is a subring of a ring S (note that 1~ = 1s by our convention made on rings), then S is an R-module with the action given by the ring multiplication. If I is an ideal of the ring R. Then I is an R-module with the action given by the ring multiplication. Let M be an R-module and N an (additive) subgroup of M. If rz E N for all r E R and x E N, then N is called an R-submodule of M. Given a family {Ni}i,=r of R-submodules of M, the sum xiGINi of subgroups forms an R-submodule in a natural way; and the intersection Preliminaries 47 ni, I Ni is an R-submodule. Given an R-submodule of a module M, an R-action on the quotient group M/N is defined as

riTi = i%i, r E R, m E MIN. With the R-action defined above, M/N is called the quotient R-module determined by N. Let M and N be R-modules. An R-module homomorphism from M to N is a homomorphism of abelian groups $: M + N satisfying $(rm) = r$(m) for all r E R and m E M. It can be verified directly that Kerll, is an R-submodule of M, that Im$ is a submodule of N, and furthermore, that the following R-module isomorphism theorems hold: (a) M/Ker$ Im$. (b) Let A, B be submodules of the R-module M. Then, (A+ B)/B 2 A/(An B). (c) Let A, B be submodules of the R-module M. If A B then (M/A)/(B/A)E M/B. (d) Let N be a submodule of the R-module M. Then there is a bijection between the submodules of M which contain N and the submodules of M/N: a: A-(A+N)/N

such that a(A+ B) = a(A)+ a(B)and a(An B) = a(A)n a(B)for all submodules A, B of M containing N. Let S C M, where M is an R-module, and T G R. Put TS = { finite sums Crimi I ri E T, mi E s 1 With notation as above, the reader is also asked to check the following statements.

(e) If T = R, then RS forms an R-submodule of M; moreover, RS = Rmi and it is the smallest R-submodule of M containing S. (f) If T is an ideal of R, then TM forms an R-submodule.

The R-submodule N = RS obtained in part (e) above is called an R- submodule of M generated by S, where S is called a set of generators of N. If M = RS with a finite set of generators S = {rnl,..., ms}, then M = C:==,Rmi and is called a finitely generated R-module. 48 Commutative Algebra

Given a family of R-modules { Mi}iEJ, the direct sum of abelian groups

@Mi = { (mi)icj 1 0 # mi E Mi for only finitely many mi iEJ is an R-module, where

and is called the direct sum of {Mi}iE~. For the direct sum &jMi of given R-modules defined above, it is not hard to see that there is an injective R-module homomorphism Mi - @iGjMi with m, k (X~)%~J,where xi = mi and xj = 0 for j # i. Hence Mi is isomorphic to a submodule of @i€~Mi.Conversely, let {N~}~EJbe a family of submodules of some R-module M, and let N = CiE Ni. Then

defines an R-module homomorphism. If 4 is an isomorphism then N is said to be the direct sum of its submodules Ni, i E J, and we also write N = @iEJNZ. Let M be an R-module and suppose that IM = @iEJMifor some submodules Mi c M,i E J. Then it is clear that every element m € A4 has a unique expression m = C mi, i.e., C mi = 0 if and only if mi = 0.

7.2. Definition An R-module M is said to be free if there are E M, i E J, such that M = @iE JRJ~,where {

Example (i) Any vector space V over a field K is a free K-module. Any free abelian group (as defined in section 6) is a free Z-module.

(ii) For any set J of indices, F = @iEjRi with Ri R (as R-modules) is a free R-module.

7.3. Proposition Any R-module M is the homomorphic image of some free R-module. Preliminaries 49

Proof The R-module homomorphism C%mcMRm -+ M = zmE~Rm defined by CrmH zr,m does the job.

An R-module M is said to be Noetherian if every R-submodule of M is finitely generated.

7.4. Theorem For an R-module M, the following statements are equivalent. (i) M is Noetherian. (ii) For any ascending chain

Mi M2 C_ c Mn c *** of R-submodules in M, there is some k such that Mk = Mj for all j 2 k. (iii) Every nonempty set of R-submodules has a maximal element with respect to G. Proof Exercise (see the proof of Theorem 1.1). 0

7.5. Theorem (i) Let p: M -+ H be an onto R-module homomorphism. If R is Noetherian then so are Kercp and H. (ii) Let N be an R-submodule of the R-module M. Then M is Noetherian if and only if N and M/N are Noetherian. Proof To better understand the argumentation, the reader is reminded to bear the foregoing R-isomorphism theorems (a)-(d) in mind. (i) This follows from the fact that ascending chains of R-submodules in Kerp and H correspond to some ascending chains of submodules in M. (ii) If M is Noetherian then so are N and M/N by part (i). Now let Mi c Mz c ... c M, c ... be an ascending chain of R-submodules of M. Then NnM, c NnM2c...cNnMnc... is a chain of R-submodules in N and for some l 2 1

(1) N n Me = N n Me+i, i = 1,2, .... On the other hand, we also have a chain of R-submodules in M/N Mi+N M2+N Mn+N ... c ~ c '.. c c N N N 50 Commutative Algebra and (without loss of generality) for Z 2 1

Me+N Me+i+N , -- - , 2 = 1,2, ... N N Thus, for m E Me+i, formula (2) implies m = m' + x for some m' E Me and x E N. But then

x = m - m' c Me+in N = Men N by (1) above. It follows that m - m' = m" with m" E Me n N, and consequently m = m' + m" E Me. This shows that Mj = Me for j 2 e, that is, M is Noetherian.

7.6. Theorem (i) Given finitely many Noetherian R-modules MI,..., M,, the direct sum is a Noetherian R-module. (ii) If R is a Noetherian ring and M is a finitely generated R-module, then every submodule of M is Noetherian, in particular, M is Noetherian.

Proof (i) Set M = M1@M2. Then MI and M/M1 M2 are Noetherian by the assumption. It follows from Proposition 7.5(ii) that M is Noetherian. Now an induction on s shows that @:==,Mi is Noetherian. (ii) Suppose M = C&,RJi, Ji E M. Then there is an onto R-module homomorphism

R@R@-@R

So the conclusion now follows from part (i) and Theorem 7.5. 0

We complete this chapter with the celebrated Krull's intersection theorem.

7.7. Theorem (Krull) Let R be a Noetherian ring and I an ideal of R. Given a finitely generated R-module M, let

00 U= n IW. n= 1

Then IU = U. Preliminaries 51

Proof First note that every InM, hence U and IU are R-submodules. So it is clear that we need only to show U g IU. For this purpose, noticing U n IU = IU, let us consider R= {IS SasubmoduleofM,SnU=IU } . Then R has a maximal member, say S, with respect to C on submodules, for M is Noetherian by Theorem 7.6. Claim For the maximal S obtained above, there is some n such that I"M C S, and consequently, U = InM n U S n U = IU. To find the above claimed n, let I = xi==,R&, & E I. If we can find, for each ti, some ni such that (*I ("M 2 s, then there will be some n, large enough, such that I"M C S. As a matter of fact, we may reach the above mentioned property (*) for any a E I. To see this, define, for each k 2 1, the R-submodule Mk={mEMIakmES 1. Then we obtain an ascending chain

Mi E M2 5 ... Mq 5 ... and there is some z such that M, = Mj for all j 2 z. For this fixed z, obviously IU 2 (aZM+ S)n U. On the other hand, if u E (aZM+ S)n U, then u = azm + u with m E M and u E S. Hence au E aU 2 IU C_ S, and az+'rn E S. This shows that m E M,+1 = M,, and it follows that azm E S. But this yields u E S n U = IU, and consequently IU = (aZM+ S)n U. If azM+ S = MI then U = IU; otherwise, by the maximality of S in R, a"M 2 S, as desired. 0

7.8. Corollary Let R be a Noetherian domain, and let M be a finitely generated torsion-free R-module, i.e., for r E R and m E MIrm = 0 implies r = 0 or m = 0. Then

n=l Proof This follows from Theorem 7.7 and later exercise 6. 52 Commutative Algebra

Exercises 1. Find all Zsubmodules of Z and all Zmodule homomorphisms Z -+ Z. 2. Complete the proof of Proposition 7.4. 3. Let R be a domain with the field of fractions K. Let X E R be nonzero and nonunit. Show that R[i], the subring of K generated by over R, is not a finitely generated R-module. (Hint: If there was a finite set of generators, then 1, i,+, ..., 1 would be a set of generators for some s > 0. After expressing as an R-linear combination of the foregoing generators, see what happens.)

4. Show that there is no nonzero Zmodule homomorphism Q --f Z. 5. Let R be a ring and let Ill... ,Is be finitely many ideals of R. Suppose that R/Ij is Noetherian, j = 1, ..., s, and that nglIj = (0). Show that R is Noetherian. (Hint: Consider the R-module homomorphism

R ---f @j=,(R/Ij) with r H Cxj, where xj = F E R/Ij, j = 1, ..., s.) 6. For any ring R, one may also define matrices (rij)mxnof finite order with entries rij E R, define addition and multiplication of matrices, and define the determinant, adjoint and inverse of a square matrix, as in classical linear algebra. Let M = ct=,R& be a finitely generated R-module, where & E M, i = 1, ..., s, and let I be an ideal of R. Show that if IM = M then there is some r E R such that rM = (0) and 1 - T E I. (Hint: Note that IM = M implies & = C,"=,aijtj, i = 1, ..., S, aij E I. Thus

(al:~La:l:l:;: ::: l) (i)= is)

asi as2 ... ass -

Multiplying by the adjoint (aij)* of (aij), it follows that det(aij)M = {0}, where det(aij) = 1 - a for some a E I.) 7. Let R and K be as in exercise 3 above. Use problem 6 to show that K is not a finitely generated R-module. (Hint: Take a nonzero nonunit X E R and note that XK = K.) 8. Let R = A[xl, ..., xn] be the polynomial ring in x~,...,x,over a ring A. Let Ri = Ccrl+...+a,=iAx:' . . . xzn , i E N,which is called the ith homogeneous part of R. Show that, as A-modules, R = @iE~Ri,and that, as subsets, RiRj = Ri+j, i,j E N. Chapter 2 Local Rings, DVRs, and Localization

Commutative algebra studies various problems related to “factorizations into irreducible components” and “zeros of polynomials” by means of rings and their modules. In this chapter and the next, we introduce the fundamental structures and methods that are essential in demonstrating the principles of commutative algebra such as “singularities versus normalization” and “global concern versus local solutions”.

1. SpecR, m-SpecR, and Radicals

Let R be a ring.

1.1. Definition (i) An ideal P 5 R is called a prime ideal if P has the property:

for a, b E R, ab E P implies a E P or b E P.

Write SpecR for the set of all prime ideals of R and call it the prime spectrum of R. (ii) An ideal M R is said to be maximal if for any ideal I of R,

M 2 I implies A4 = I or I = R.

Write m-SpecR for the set of all maximal ideals of R and call it the maximal spectrum of R.

A subset S c R is called a multiplicative set if 1 E S and a, b E S implies ab E S.

53 54 Commutative Algebra

1.2. Proposition (i) An ideal P of R is prime if and only if RIP is a domain, or equivalently, if and only if S = R - P is a multiplicative set. (ii) An ideal M of R is maximal if and only if RIM is a field. Consequently, R is a field if and only if (0) is maximal in R. Proof By definition, both parts (i) and (ii) may be verified directly. 13

Example 1 (i) If p R 4 B is a nonzero ring homomorphism from R to a domain B, then Kerp = P is a prime ideal of R by Proposition 1.2(ii).

(ii) Let R be a domain. Then (0) is a prime ideal of R. If 0 # a E R, then the principal ideal (a) is a prime ideal if and only if a is a prime. Thus, a PID R has SpecR = ((0)) U {(p) I p E R a prime}. In particular, SpecZ = { (0)) U { (p) 1 p E Z a prime number}; and if K[z]is the polynomial ring in 3: over a field K, then SpecK[z] = { (0))U { (f)1 f is irreducible in K[z]} (in case K is algebraically closed, SpecK[z]= {(z - A) I X E K)U ((0))).

(iii) In the polynomial ring K[zl,..., z,] in ~1,...,z, over a field K,

(0) c (21) c (zl,z2) c (zl,z2,23)c "' c (zlr.*.r%) is a chain of prime ideals. For any n-tuple P = (al,..., a,) E K", define the function

fbl,..., 4 i-+ f(P)= f(a1,... ,a,) where iff = c E K is a constant then f(P)= c, and, for f,g E K[zl,..., z,], define (f+g)(P)= f(P)+g(P),(fg)(P) = f(P)g(P).Then $JPis an onto ring homomorphism with KerQp = (21 - al, ..., z, - a,) (check it!). By Proposition 1.2(ii), (21 - al, ...,2, -a,) is a maximal ideal of K[zl,..., z,]. In Chapter 5 section 1 we will see that if K is algebraically closed then every maximal ideal of K[zl,..., z,] is of the form (51 - bl, ..., z, - b,) for some (bl, ..., b,) E Kn. (iv) If p is a prime number in Z,then (0), (p),(z) and (p,z) are prime ideals of Zjz], but only the last one is maximal. Indeed, the next proposition states a more general result.

1.3. Proposition Let R be a PID, K its field of fractions, and R[z] the polynomial ring in z over R (hence a UFD). If P E SpecR[z], then either Local Rings, DVRs, and Localization 55

P = {0}, or P = (f)for some irreducible element f E R[z],or P = (p,f), where p E R is an irreducible element (hence a prime) and f E R[z]is a polynomial such that 7 is irreducible (hence a prime) in the polynomial ring (Rl(p))[zI. In the case where P = (p,f), P E m-SpecR[z].

Proof If P = (0) or P = (f) for some 0 # f E R[z],then it is done by Example (ii) above. Suppose P is not principal. Then there exist fl,f2 E P which do not have common divisor in R[z].Note that R is a UFD. By Chapter 1 (section 2, exercise 7), fl,f2 do not have common divisor in K[z],that is gcd(f1, f2) = 1. Thus, gfl + hf2 = 1 for some g, h E K[z].Multiplying by the common denominator, say u,of all coefficients in g and h, we have

This shows that Rn P # (0). But R is a PID. Hence R n P = (p) for some irreducible p E R. By later exercise 2, the polynomial ring g[z]is a PID. Since p E P, it follows from exercise 2 and the onto ring homomorphism

that if we pull the image of P in $[z] back into R[z],then P = (p,f) as desired.

Except for SpecZ[z], another typical case of Proposition 1.3 is SpecK[z,y] where K is a field. In particular, it follows from the building of SpecK[z,y] that all irreducible plane curves over K are established (see Chapter 5 section 2). To indicate the existence of prime ideals in an arbitrary ring, we need Zorn’s lemma that is equivalent to the axiom of choice in set theory.

1.4. Lemma (Zorn) Let R be a nonempty partially ordered set with the partial ordering k.If any totally ordered subset U c R has an upper bound in 0, then R contains a maximal element.

1.5. Proposition Let R be a ring. (i) Any ideal I # R is contained in a maximal ideal. 56 Commutative Algebra

Proof (i) Set R = {proper ideals of R containing I} and let C be the inclusion ordering on 0. Then I E R and c is a partial ordering on R. If V = {Ji}i€~is any totally ordered subset of R, then J* = uiE~Jiis an upper bound of V in 52 because 1 # J*. By Zorn’s lemma, R contains a maximal element M. By the definition of R, M is also maximal among all ideals of R. (ii) This follows from part (i) immediately. 0

1.6. Proposition Let R be a ring. (i) If S is a multiplicative set and I is an ideal of R with I n S = 0, then there exists a prime ideal P such that

P>Iand PnS=0.

(ii) Every prime ideal contains a nonzero minimal prime ideal (with respect to G on SpecR). Proof (i) Consider the partially ordered set R = {ideals J with J 2 I and J n S = 0) with the inclusion ordering 2. By Zorn’s lemma, R contains a maximal element P. We claim that P E SpecR. For if a, b E R, a, b # P, then

(aR+ P)nS # 0, (bR+ P)nS # 0, because P is properly contained in both aR + P and bR + P. Thus, there areas+pl E (aR+P)nS, by+p2 E (bR+P)nS, and (as+pl)(by+p2) = p’+abxy E S since S is a multiplicative set, where p’ E P. This shows that ab # P by the choice of P. (ii) Exercise.

Let R be a ring and a E R. If an = 0 for some n 2 1, then a is called a nilpotent element. Set r(~)= {a E R I a is nilpotent 1 and call r(R) the nilradical of R. R is said to be reduced if r(R)= (0)

1.7. Theorem (i) r(R) = npESpecRP. Consequently r(R)is an ideal of R. Local Rings, DVRs, and Localization 57

(ii) r(R) = nQ where Q runs over all nonzero minimal prime ideals of R. Proof (i) The inclusion r(R) C nP is clear. The inclusion 2 follows from the fact that if a E R is not nilpotent then there is some prime P not containing a. To see this, consider the multiplicative set S = (1, a,a2,..., an, ...}. Then 0 @ S, for a is not nilpotent. By Propo- sition 1.6(i) (taking I = {0}), there is a prime ideal P with P n S = 8, as desired. (ii) This follows from the proof of part (i) and Proposition 1.6(ii). 0

For a ring R, we also set the intersection J(R)= n M ME m-SpecR and call J(R) the Jacobson radical of R.

1.8. Theorem With notation as above, J(R) = {r E R I 1 -yr E U(R) for ally E R } .

Proof By Proposition 1.5(i), 1 - yr is a unit if and only if 1 - yr is not contained in any maximal ideal of R. Now the assertion is clear.

1.9. Theorem Let R be a Noetherian ring and J(R) its Jacobson radical. If I is an ideal of R and I C_ J(R), then np==,InM= (0) holds for any finitely generated R-module M. Proof This follows from Chapter 1 (section 7, Theorem 7.7 and exercise 6).

Exercises 1. Complete the proof of Proposition 1.6(ii). (Hint: Apply Zorn’s lemma to the prime ideals contained in a prime ideal P by defining P2 < PI if Pl c P2.) 2. Let R be a PID. Show that SpecR = m-SpecR. 3. Let R be a UFD. (a) Show that every minimal nonzero prime ideal of R is principal. (b) Show that, without counting associates, there is a one-to-one and onto correspondence between prime elements of R and minimal nonzero prime ideals of R. 58 Commutative Algebra

4. Let R be a ring and I an ideal of R. Define fi={a E R I an E I for somen 2 1 1 and call 4 the radical of I in R. Show that

J7= n p. PEspeCR, P>I

5. Let I and J be ideals of a ring R. Show that = fi n 6. 6. Let R be a ring. Two ideals I, J of R are said to be comaximal if I + J = R (for instance, if one of them is maximal). (a) Show that if I and J are comaximal then I.J = InJ, I + J2 = R, and Im + J” = R for all integers m, n 2 1. (b) If I1, ..., IN are ideals of R, and Ii and Ji = njfiIjare comaximal for alli = l,...,N, show that Ipn...nI; = (Il...IN)n= (IIn . . . n IN)^ for all integers n 2 1. 7. Show that in Z[&] the ideal P = (2,&) is a maximal ideal but not a principal ideal. (Hint: Z[&]/P E Z2) 8. For f = 1 + x2,g = y2(x + x3) + (y - 1)x2 + y + 2 in the polynomial ring R[x,y], is the ideal (f,g)a prime ideal or a maximal ideal?

2. Local Rings and DVRs

2.1. Definition Let R be a ring. If R has only one maximal ideal, then R is called a local ring.

By definition, all fields are local rings, for (0) is the only maximal ideal in these rings.

2.2. Theorem The following statements are equivalent for a ring R. (i) R is a local ring. (ii) All nonunits of R form an ideal. Proof This follows from Proposition 1.5. I?

Example The following examples, which are from different aspects of mathematics, may perhaps help to qualify the name “local ring”. Local Rings, DVRs, and Localization 59

(i) (Foundation of modern scheme theory) Let V be the real line (or a topological space, or a differentiable manifold), and P E V a point. Consider the set E of real-valued continuous functions (differentiable functions in case V is a differentiable manifold) on some open interval (open neighborhood) around P. Then, two functions f ,g E E may be “locally” identified by the equivalence relation: f - g if and only if they agree on some open neighborhood of P. The quotient set Ep = (E x E)/ - forms a ring with the addition and multiplication induced by that on functions. Elements of Ep are called function germs at P. It is easy to see that the subset

mp = {If~~pf(p)=~} forms an ideal of Ep. If g @ m, then there is an open neighborhood U of P on which g is nonzero. So h = is defined on U and gh = 1 on U. By Theorem 2.2, Ep is a local ring.

(ii) Let R[z] be the polynomial ring in z over R, and R(x) its field of fractions. For any a E R,

is a local ring with the unique maximal ideal ma = { R(z) I g(a) # fo9(X) E 0, f(a) = 0). R[z],is called the local ring of the point a, due to the fact that each rational function in R[z], is “locally defined” on some Zariski open neighborhood of a. (See Chapter 5 section 3 about this topic in a more general setting.) (iii) Let p be a prime in Z. Then

is a local ring with the unique maximal ideal mp = { E Z(p)[ Ic E Z}. (iv) Let K be a field and R the ring of formal series in one variable, i.e.,

If f = a0 + ala: + u2z2 + ... with a0 # 0, then f = ao(1 + zg) for some g E R and

ao(l+zg)’a;l(l-xg+x2g2-~-)= 1. 60 Commutative Algebra

Conversely, if fh = 1 for some h E R, then a0 # 0. Thus, R is a local ring with the unique maximal ideal m = XR = (x).

The next easy but powerful lemma enables us to see the link between finitely generated modules over a local ring R and finite dimensional vector spaces over the field R/m, where m is the unique maximal ideal of R.

2.3. Lemma (Nakayama) Let R be a ring and I an ideal contained in the Jacobson radical J(R) of R (see section 1). If M is a finitely generated R-module and IM = M, then M = (0).

Proof If IM = M and M # {0}, then let M = CB1R

2.4. Corollary Let R be a local ring with the unique maximal ideal m. Let M be an R-module and N M a submodule. Suppose that M/N is a finitely generated R-module and M = N + mM. Then N = M. In particular, let M be a finitely generated R-module. If 51, ..., Es E M are such that their images span the vector space = M/mM over the field Rlm, then (1, ..., Es generate M. Proof By Lemma 2.3, this follows from the fact that J(R) = m. 0

Remark If R is Noetherian, then Lemma 2.3 may follow from Proposition 1.9. In particular, if R is a Noetherian local ring with maximal ideal m, then nr=lmn = (0).

A very important class of local rings from number theory and (analytic and algebraic) geometry is the class of discrete valuation rings, that is introduced as follows.

Let K be a field. Define in ZU {m} the rule m +a = a,03 + 03 = 00. A discrete valuation on K is an onto function

w : K -+ ZU {m} Local Rings, DVRs, and Localization 61 such that for all z,y E K (a) v(0) = 00; (b) v(zCy) = 4%)+ 4Y); (c) 42 + Y) 2 min{v(x), V(Y)). By definition, it is clear that v(1~)= 0 = v(-l~).Hence v(x) = v(-x) for all 3: E K. Thus, if we set

then it is easy to verify that R is a subring of K with 1~ E R.

2.5. Definition The ring R defined above is called the discrete valuation ring (abbreviated DVR) associated to the discrete valuation v.

2.6. Theorem Let R be a DVR associated to a discrete valuation w on a field K. Then R has the following properties. (i) R is a local ring with the unique maximal ideal

(ii) R is a PID, hence Noetherian. In particular, SpecR = ((01, m}. Proof (i) By the definition of v, v(z-') = -v(x) for any 0 # z E K, and x E R is a unit if and only if v(x) = 0. This proves (i). (ii) Let t E K be any element such that v(t) = 1. If x E m, then v(z) = n > 1, or equivalently, v(x) = v(tn). Hence v(xt-") = 0 and st-" = u for some unit in R. Thus, x = ut", and every ideal I of R is of the form (t") for some m 2 1. Therefore, R is a PID, and it is now clear that SpecR = {{0}, m}. 0

2.7. Corollary A DVR R is a UFD. If R has maximal ideal m = (t),then t (up to a unit multiple) is the unique prime in R. (In the literature, t is sometimes called a uniformizing element of R.) 0

Example (v) (Compare with previous Example (iii).) Let R be a UFD and K its field of fractions. Let p be a prime in R. Then for 0 # t E K, 62 Commutative Algebra f = pm$ with a unique m E Z,and p /y a', p 1b'. Thus

v: K- ZU{caI

m,ifO#t=pm$, p/ya'andp,/'b', -a b 00, if = 0 defines a discrete valuation on K, called the p-adic valuation on K, where the associated valuation ring

has the unique maximal ideal m={F,Kl m>O,p,/'bi =(p). (vi) The only discrete valuations on Q are p-adic valuations. To see this, let v: Q 4 Z U { ca} be a nonzero discrete valuation on Q and R its valuation ring with the maximal ideal m. Note that Z c R because 1 E R. Then

mnZ= (p) for some prime number p E Z. Since w is onto, it follows from the proof of Theorem 2.6(i) that m nZ = (p) for some prime number p. Thus, if 0 # 2 E Q then f = pk ($) with k E Z, gcd(p, a') = 1, and gcd(p, b') = 1. Hence a', b' $! m, or in other words, a' and b' are units of R. It follows that v (f) = kv(p). Suppose .(A) = 1. Then .(A) = mv(p) for some m E Z implies v(p)= 1 because p E rn. This shows that v is the padic valuation. (vii) Let K[z]be the polynomial ring in z over a field K and K(z)its field of fractions. Consider all discrete valuations v on K(z)such that .(A) = 0 for X E K x. Then there is only one such v which is not a padic valuation. To see this, let w: K(z)4 ZU{ca} be a nonzero discrete valuation on K(z) and R its valuation ring with the maximal ideal m. Below we consider two cases. Case I z E R. In this case K[z]c R and

mn K[z]= (p(z)) for some irreducible p(z) E K[z]. Local Rings, DVRs, and Localization 63

Arguing as in Example (vi), we may conclude that v is a p(x)-adic valuation. Case I1 x $2 R. It follows from the proof of Theorem 2.6(i) that y = $ E m. Hence, K(y] c R. Note that K[y] E K[x] and y E K[y]n m. So K[y]n m = (y) (why?). Thus, if f = a,xn + . . . + alx + a0 with a, # 0, then

This shows that v(q(y)) = 0 and consequently v(f(x)) = -nv(y). Now, for any 0 # # E K(x),if degf(x) = n and degg(x) = m, then 'u (#) = (rn - n)v(y). As in Example (v) we may derive from y E m that v(y) = 1. So eventually we have

that is, v is actually defined by the degree of polynomials.

2.8. Proposition Let R be a UFD and K its field of fractions. Then

R= nv, where V, runs over all p-adic valuation rings in K. Proof By the definition of a padic valuation on K, R 2 nv,. If 0 # x E nv,, say, x = %,a, b E R, then v(x)2 0 for any prime p. Suppose

where pi's and qj's are primes and ai,& > 0. If some qj does not appear in the numerator of x, then vgj(x) < 0; if qj = pj but Pj > aj, then v(qj) < 0. It follows that m 5 n, pj = qj for 1 5 j 5 m (up to a necessary re-arrangement of prime divisors), and ,Oj I cyj for 1 5 j 5 m. This shows that x E R. Therefore R S nv, 2 R.

To see under what condition a local domain is a DVR, we need the following result that is a special case of Chapter 1 Corollary 7.8. 64 Commutative Algebra

2.9. Lemma Let R be a Noetherian domain and I an ideal of R. Then

00 nI- = (0). n=l 0

2.10. Theorem Let R be a local domain and m its maximal ideal. Suppose m = (t)for some t # 0. If nr==lmn= {0}, then the following statements hold: (i) Every 0 # a E R is of the form a = tnu for some n 2 0 and some unit u. (ii) Let K be the field of fractions of R. Define .(a) = n for a = tnu as in part (i), and K- u {..I

Then v is a discrete valuation on K with R its associated valuation ring. (iii) Every nonzero ideal I of R is of the form I = (t") for some n 2 0.

Proof (i) By the assumption, if 0 # a E R is not a unit then a E me- me+' for some e 2 1, and it follows that a = tnu for some n 2 1 and some unit. (ii) By part (i), v is well-defined and surjective. Moreover, for s E K, v(s) 2 0 if and only if s E R. To verify that v is a discrete valuation on K, it is sufficient to note that for a,b E R with b # 0, if v(u) 2 v(b) then 9= f + 1 E R and v(a + b) 2 v(b); if v(a) > v(b) then 9 = + 1 E R is a unit and v(a + b) = v(b). (iii) By part (i), let n be the smallest integer such that t" E I where u is some unit. Then it is easy to see that I = (t").

A full structural characterization of a DVR is given in Chapter 3 The- orem 4.5.

Appendix. General valuation rings Let R be the associated discrete valuation ring of a discrete valuation w on some field K. If z E K, z @ R, then z # 0 and v(xU1) > 0, i.e., z-' E R. Local Rings, D VRs, and Localization 65

In a general valuation theory (cf. [Jac], [Coh]), all (not necessarily discrete) valuation rings are defined in this way,

2.11. Definition Let K be a field and R a subring of K. If for every x E KX,either x or x-' E R, then R is called a valuation ring of K.

If R is a valuation ring of some field K, then it is easy to see that K = Q(R),that is, K is necessarily the field of fractions of R.

To see why the above definition is really more general than a discrete valuation ring, we need to introduce general valuation functions. By an ordered abelian group (G,+; 3)we mean an abelian group (G, +) equipped with a total ordering 3 which is compatible with the binary operation +: a 3 b and c 4 d implies a + c 4 b + d, For example, (Z,+; I), (a,+; I)and (R, +; 5)are ordered abelian groups. Let K be a field and (G, +; 5) an ordered abelian group. Define in G U {a)the rule co + a, 03 + a = co, for all a E G, and 00 + 00 = co. If there is an onto function v : K - G U {co}

2.12. Proposition Let v be a valuation on the field K. Then

is a valuation ring in K. Moreover, R is a local ring with maximal ideal m = {z E R I w(~)+ 0). Proof Exercise. 0

Let R be the valuation ring associated to a valuation w on the field K. 66 Commutative Algebra

For x, y E KX,suppose u(x) = g, u(y) = h. Then

4 E R if and only if v (4) = u(y) - V(X)= h - g k 0

if and only if g 5 h. This, indeed, induces a total order on the abelian group K '. Motivated by the above idea, let R be a subring of the field K. If we write the binary operation of the abelian group G = KX/U(R)additively, that is, 3: + ij = 3:. ij, for x, y E KX, then a partial ordering 5 on G may be defined as

3: 57 if and only if -Y E R, x,y E KX. X 2.13. Proposition Let R c K and (Gl+;5) be as above. Then the following hold: (i) The correspondence

u: K- GU{4

00, if x = O; x H u(.) = 3, ifx#O, defines a valuation function with R as its associated valuation ring if and only if 5 is a total ordering on G. (ii) If (H,+; a) is some ordered abelian group and z1: K 3 H U {cm} is a valuation with the associated valuation ring R, then

'p: G?H as ordered abelian groups, i.e., g 5 h implies q(g) 9 cp(h) for g, h E G. Proof Exercise. 0

2.14. Theorem Let R be a valuation ring. Then R is a DVR if and only if R is Noetherian. Proof One direction is known by Theorem 2.6. Suppose R is Noetherian. Then the maximal ideal m of R is finitely generated, say m = (al,a2, ..., a,) with all ai # 0. Thus, we may assume E R, and so m = (u2,a3, ..., an). Repeating this procedure will finally yield m = (ui), a principal ideal. Hence R is a DVR by Theorem 2.10. Local Rings, D VRs, and Localization 67

Exercises 1. Let K[z)be the polynomial ring in z over a field K. Show that R = K[z]/(zn)is a local ring for any n E Z+. What is the maximal ideal of R? 2. Let R be a ring and M a maximal ideal of R. Show that R is a local ring with the unique maximal ideal M if and only if 1+ a is a unit in R for every a E M. 3. Show that Nakayama’s lemma holds for any ideal I C r(R). 4. Consider the local ring V, given in previous Example (v). Show that K = V, + $Vp+ $Vp + ... , and that K = pK. But K # (0). This illustrates that Nakayama’s lemma doesn’t work if the module considered is not finitely generated. (See also Chapter 1 (section 7, exercise 3).) 5. Let R be a local ring and m its maximal ideal. If m = (t)is principal-, show that either R is a DVR or tm = 0 for some m > 1. (Hint: Similar to the proof of Theorem 2.10.) 6. Complete the proof of Propositions 2.12 and 2.13.

3. The Ring of Fractions and Localization

Recall from (section 2, Examples (ii)-(iii)) that in the local ring

qp,={+/ PD}, all b E S1 = Z - (p),and in the local ring

all g(z) E S’z = R(z] - (z - a). 0bserve that

0 every s E Sl is invertible in Z(,), and every s’ E Sz is invertible in R[z],. We may say that Z(,) is the ring of fractions of Z with denominators in 5’1, and that R[x:], is the ring of fractions of R[x] with denominators in Sa. Also observe that 68 Commutative Algebra

the set of denominators S1,respectively 5’2, is closed under multiplication in Z, respectively in R[rc]. Considering any domain R and its field of fractions K, if S is a multiplicative set of R and 0 6 S (for instance, S = (1, r, r2, ...}, 0 # r E R), then a direct verification shows that Rs = {: I a E R, s E S} c K is a subring of K containing R. We may call Rs the ring of fractions with denominators in S. Our aim in this section is to demonstrate that the above described idea of establishing the ring of fractions may be carried out in a more general setting including the case where the ring R may have divisors of zero. Let R be a ring and S a multiplicative set in R. Define on R x S the relation

(al,s1) N (a2, s2) if and only if (*I there is s E S such that (sla2 - s2al)s = 0.

Then one easily sees that - is reflexive, symmetric. If (allsl) - (a2, s2) and (a2, ~2)- (a3, %), then

and thus, 0 = S~U~S’S~S- S~U~SS~S’ = (Sla3 - S~UI)SS~S’,i.e., (allsl) N (as, s3). This shows that - is also transitive, and hence - is an equivalence relation on R x S. Write Rs for the quotient set R x S/ N, and write for the equivalence class of (a,s) E R x S. Define the addition and multiplication on Rs as follows:

a1 a2 S2al fs1a2 --.-=-a1 a2 a1a2 -+-= 1 s1 s2 SlS2 s1 32 SlS2 Then a direct verification shows that Rs is an associative commutative ring with zero 0 = 4 and identity 1 = f.

3.1 Definition The ring Rs constructed above is called the ring of fractions with denominators in S. Local Rings, DVRs, and Localization 69

Remark Recall that if R is a domain and S = R - {0}, then the field K of fractions of R is given by R x S/ - where N is defined on R x S by (**I (UI,SI) - (UZ,SZ) if and only if 51u2 - SZUI = 0.

The reason why we did not define N in the foregoing (*) exactly as in (**) above will be clear if one checks the transitivity of - defined in (**). (Also see Example (i) below.)

3.2. Proposition Let Rs be the ring of fractions with denominators in S. (i) If 0 E S, then Rs is the zero ring. (ii) The correspondence

XR : R --+ Rs

X 2- - 1 defines a ring homomorphism (called the canonical homomorphism) such that (a) XR(S) = f is invertible in Rs for every s E S; and (b) KerXR = {u E R I as = 0 for some s E S}. Hence XR is injective if and only if S does not contain nontrivial divisor of zero (note that this is compatible with the case where R is a domain). (iii) XR has the universal property: If f: R + R' is any ring homomorphism such that f(1R) = 1~'and f(s)is invertible in R' for every s E S, then there exists a unique ring homomorphism f: Rs -+ R' with 7 (5) = % that yields the commutative diagram RhRs - f] Jr foXR=f R' Proof Exercise. 0

Example (i) Let K[z,y] be the polynomial ring in x and y over a field K. Consider the ring R = K[z,y]/(xy) and S = {1,Z,Z2,...}. Then (a) the previous (**) itself cannot define an equivalence relation on R x S; and (b) since ?@ = 0 and T is invertible in Rs, it follows that Rs K[t,t-'] c K(t),where K[t]is the polynomial ring in t over K. 70 Commutative Algebra

Note that R has nontrivial divisors of zero, but Rs is a domain.

(ii) Let R be a ring and f E R. Consider S = (1, f,f2, ...}. Then

where R[x]is the polynomial ring in x over R. (This trick plays a very important role in algebraic geometry, for example, in proving the Nullstel- lensatz and in obtaining the open afine covering of an algebraic set, etc.) Proof Note that 7 is invertible in R[x]/(xf- 1). By the universal property of Rs, the diagram R 3 Rs

is commutative, where a is the natural ring homomorphism R --f R[x]/(sf- 1). On the other hand, under the ring homomorphism P: R[z]-+ Rs with P(Crixi)= crib, P(xf - 1) = 0. By the first isomorphism theorem (or see section 0), there is a ring homomorphism p: R[x]/(xf- 1) 4 Rs that yields the commutative diagram

Tr R[Xl Rfzl - (xf - 1) 01 JP J. RS Now one checks that h and are inverses to each other.

The next proposition will be generalized in Chapter 3 Theorem 4.1.

3.3. Proposition Let R be a UFD, S a multiplicative set of R and 0 $ S. Then Rs is a UFD.

Proof View R as a subring via the canonical mapping XR: R 4 Rs. We show that factorization in Rs is completely determined by that in R. First note that if p is a prime in R then p[s,for some s E S, if and only if p is a unit in Rs. Secondly, note that if p is a prime of R but not a unit in Rs Local Rings, DVRs, and Localization 71 then p is a prime in Rs. (Indeed, using the factorization in R it is easy to see that every prime in Rs is of the form pu for some prime p of R and some unit in Rs). Finally, note that if 5 E Rs with r # 0 then

(21 (22 - = Pl P2 Pgm . u, S where the pi are primes of R but not units in Rs while u is a unit in Rs. It follows that Rs is a UFD by Chapter 1 Theorem 2.9. 0

Let R be a ring, Rs the ring of fractions of R with denominators in a multiplicative set S, and XR: R + Rs the canonical ring homomorphism. We now consider the relation between ideals of R and ideals of Rs. If I is an ideal of R, then a ~e = x~(I)R~= { - I a E I, s E S} S is an ideal of Rs and is called the extension of I in Rs. If J is an ideal of Rs, then the preimage

is an ideal of R and is called the contraction of J in R.

3.4. Proposition With notation as above, the following hold: (i) For any ideal J c Rs, J"" = J. (ii) For any ideal I c R,

rs E I for some s E S

(iii) If P is a prime ideal and P n S = 8, then Pe is a prime ideal of Rs.

Proof (i) If f E J then T E J", and hence 5 E J"". This shows that J JCe.The inclusion JCe J is clear. (ii) If r E I"" then 7 = $ E Rs for some r' E I and s E S. Thus, S'ST = S'T' E I for some s' E S. This proves the inclusion C. The inclusion 2 is again clear. (iii) This follows from (ii) immediately. 0

3.5. Corollary (i) Let I be an ideal of R. Then I"" = I if and only if

(0) rs E I implies r E I for r E R and s E S. 72 Commutative Algebra

(ii) There is a one-to-one and onto correspondence defined by extension and contraction of ideals:

{ideals of R satisfying (0) in part (i)} - {ideals of Rs} . It follows that if R is Noetherian then so is Rs. In particular, if R is a principal ideal ring then so is Rs. (iii) iec= R if and only if ie= Rs if and only if I n S # 0. (iv) If P E SpecR such that Pn S = 0, then (*) holds for P, and Pec= P. It follows that there is a one-to-one and onto correspondence defined by extension and contraction of ideals: {P E SpecR I Pn5’ = 0 1 +-+ SpecRs. Proof Exercise.

Let R be a ring and P E SpecR. Then Sp = R - P is a multiplicative set of R. Write Rp for the ring of fractions of R with denominators in Sp.

3.6. Proposition With notation as above, the following hold: (i) Rp is a local ring with the unique maximal ideal m = PRp = Pe. (ii) There is a one-to-one and onto correspondence defined by extension and contraction of ideals:

{ Q E SpecR 1 Q C P} H SpecRp.

Proof Note that 5 E Rp is a nonunit if and only if T E P. So (i) holds. (ii) follows from Corollary 3.5(iv).

3.7. Definition With notation as above, Rp is called the localization of Rat P.

The above definition comes from the “local study” of points in algebraic geometry (see (section 2, Example (ii)) and Chapter 5 section 3). For instance, if R = K[zl,..., xn]and K is algebraically closed, then the localization RM of R at a maximal ideal M is the ring of all rational functions in K(z1,..., 2,) that are defined on some (open) neighborhood of the point determined by M (Chapter 5 Proposition 3.6). Furthermore, the next theorem says that the set of “global rational functions” that are defined at every point is nothing but the ring R (Chapter 5 Theorem 3.4). Local Rings, DVRs, and Localization 73

3.8. Theorem Let R be a domain. Then

R = nRM = nRp, where M runs over all maximal ideals of R, and P runs over all prime ideals of R.

Proof The inclusion R C nRp is clear. Let x E nRp,and set Iz = {a E R I ax E R}. Then it is easy to see that I, is an ideal of R and x E R if and only if 1 E I,. If 1 # Iz, then there is a prime ideal P with P 2 I, by Proposition 1.5. But then x # Rp, for if x = f E Rp, b @ P, then bx E R and b E I, C P. This shows that x E R and hence R = nRp.

Exercises 1. For a,al, a2 E R, s, s' E S, verify in Rs

s'a - a - a1 fa2 = 5 + s. s's s' S ss 2. Let R be a domain and S a multiplicative set of R, where 0 # S. Show that R and Rs have the same field of fractions. 3. Complete the proof of Proposition 3.2. 4. Let S be a multiplicative set in Z, 0 # S. Find all primes in Zs. 5. Give a detailed proof of Corollary 3.5. 6. Let R = K[x,y]/(xy)be as in previous Example (i). (a) For every c E K, show that P = (Z - c, y) is a maximal ideal of R. (b) For every P in part (a) with c # 0, show that Rp is a DVR. What happens if c = O? (Hint: 7j = ?(Z - c)?j.) 7. Let R = K[z,y]/(y2 - x3),where K[x,y] is the polynomial in x,y over a field K. Show that Rp is not a DVR, where P = (Z,jj). (Hint: See Chapter 3 (section 3, Example (iii)).) 8. Let R be a PID. Show that if P E SpecR then Rp is a DVR.

4. The Module of Fractions

Let R be a ring and Rs the ring of fractions of R with denominators coming from a multiplicative set S, where 0 @ S. In this section we consider the relation between modules over R and modules over Rs. 74 Commutative Algebra

Let % be an Rs-module. Then also has an R-module structure via the canonical mapping XR: R 4 Rs: r r .z= XR(r)z = -2, r E R, z E 1 x.

In view of Chapter 1 section 7, there is a ring homomorphism p: R + EndzM, where p(r) = pr with pT(y) = r . y for y E %. Moreover, we observe that

0 for every s E S, ps is invertible in Endzx, that is, every ps is an isomorphism. Conversely, let M be an R-module. Then there is a ring homomorphism 'p: R + EndzM, where 'p(r)= pT with p,(m) = rm for r E R and m E M. Suppose that the property mentioned in the above (0) holds for M,i.e., ps is invertible for all s E S. Then, by the universal property of Rs,there is a unique ring homomorphism v: Rs 4 EndzM that yields the commutative diagram R 3 Rs Js FoXR=P

End&'

Consequently, M is equipped with an Rs-module structure. The above remark leads to the following construction.

Let M be an R-module. Define the equivalence relation N on M x S (check it!):

(ml,s1) - (m2,s2) if and only if s(slm2 - s2ml) = 0 for some s E S, and set MS = M x S/ N. Write for the equivalence class of (m,s)E M x S, and then define the R-module structure by

ml m2 s2ml+slm2 , r.-=-.m rm F+G= SlS2 S S A direct verification shows that MS is an R-module and every s E S defines an isomorphism ps in EndzMs. It follows from the foregoing remark that Ms is an Rs-module on which the Rs-action is given by -.-=-r m rm s1 '92 s1s2 Local Rings, D VRs, and Localization 75

4.1. Definition The Rs-module MS constructed above is called the module of fractions with denominators in S. If S = R - P for some P E SpecR, then Ms is called the localization of M at P and is denoted by Mp.

4.2. Proposition Let M be an R-module and MS the Rs-module of fractions with denominators in the multiplicative set S. (i) If 0 E S, then Ms is the zero module. (ii) The correspondence

p~:M+Ms

m mw- 1 defines an R-module homomorphism (called the canonical homomorphism) with KerpM = {mE M I sm = 0 for some s E S}. (iii) p~ has the universal property: If u is an Rs-module and f: M + is an R-module homomorphism, then there is a unique Rs-module homomorphism 7: MS -+ =with 7 (:) = f . f (m)that yields the commutative diagram M 3 Ms I

Proof Exercise.

Let $: M + N be an R-module homomorphism with K = Ker$ and W = Im$. Then K is a submodule of M and W is a submodule of N. It is easy to check that llts: Ms - Ns

is an Rs-module homomorphism.

4.3. Theorem With notation as above, the following hold: (i) Ker$s = Ks, Im$s = Ws. 76 Commutative Algebra

(ii) If + is injective then so is +s; if + is onto then so is +s. (iii) (M/K)s E Ms/Ks.

Proof (i)-(ii) It is clear that Ks Ker$s. If 4s (7)= = 0, then s’$(m) = 0 for some s’ E S. Since $ is an R-module homomorphism, 0 = s’+(m) = +(s’m),that is, s’m E Ker$ = K. It follows that 7 = 2 E Ks. This shows that Ks = Ker$s. Similarly, one checks that Im$s = Ws. (iii) This is just a consequence of applying parts (i)-(ii) to the natural onto R-module homomorphism M -+ MIK. 0

Let R be a ring and I an ideal of R. In view of R3Rs and Theorem-- 4.3, one may verify that IS = IRs, and moreover, that the rule :. 5 = 5 defines a ring multiplication on the Rs-module (R/I)s.

4.4. Corollary (i) Let I be an ideal of R. Then there is a ring isomorphism &/Is E (R/I)s. (ii) If P E SpecR, then K = Rp/Pp is isomorphic to the field of fractions of the domain RIP. Proof This follows from Theorem 4.3 and the above remark.

Exercises

1. Verify that the relation N on M x S is an equivalence relation. 2. Verify that the Rs-module structure on Ms is well-defined. 3. For m,ml, m2 E M, s, s‘ E S, verify in MS

s’m m ml+m2 ml m2 - - +-. s’s s S S S 4. Complete the proof of Proposition 4.2. 5. Consider the prime ideal P = (p) in Z, where p is a prime number. Let m E Zf, m = p~’p~*...pgm the factorization of m into primes, and M = Z/(m). (a) If p Xm, show that Mp = (0). (b) If p = pi for some i, show that Mp E Z/(pq’). (Hint: Write m = pgim,. Then ml E Z - P. Moreover, if k E Z, then k = pqiq + r, 0 5 r < pqi.) 6. Let S be a multiplicative set of the ring R, and let M be a finitely generated R-module. Show that Ms is a finitely generated Rs-module. (Hint: Apply exercise 3 above to a direct verification, or use Proposition 4.3 and Chapter 1 Proposition 7.3.) Local Rings, DVRs, and Localization 77

7. Let S be a multiplicative set of the ring R, and let I be an ideal of R.

Viewing I as an R-module, show that Is = Ie via Is L) Rs, where Ie is the extension of I in Rs in the sense of section 3. Furthermore, if J is another ideal of R, verify that the following hold: (a) (I + J)s = IS + Js. (b) (IJ)s= ISJS. (c) (InJ)s = Is n Js. 8. State and verify a module-version of properties (a) and (c) above for submodules. This page intentionally left blank Chapter 3 Integral Extensions and Norrnalizat ion

1. Integral Extensions

In commutative algebra a natural generalization of field extension is the ring extension. Note that field extensions may be studied in terms of vector space. Here the structure of an R-algebra is convenient for the study of ring extensions. Let R and B be rings. B is said to be an R-algebra if there is a nonzero ring homomorphism 'p: R --f B. Recalling the convention 'p(1~)= lg,it follows that B forms an R-module with the module structure:

(*> ab = 'p(a)b, a E R, b E B.

If R is a field, then R E 'p(R) c B and B becomes an R-vector space.

An R-subalgebra of B is a subring B' B with lg~= lg, which is also an R-module with respect to the action defined by the above (*). If B is an R-algebra, then it is clear that any ideal of B is an R-module.

Let B1 and B2 be R-algebras. An R-algebra homomorphism $: B1 4 B2 is simultaneously a ring homomorphism and an R-module homomorphism.

Let B be an R-algebra defined by the ring homomorphism 'p: R 4 B. If cp is a ring monomorphism, then R is naturally viewed as an R-subalgebra of B. In this case we say that B is an extension ring or extension algebra of R. Bearing this in mind, if no confusion arises, we simply refer R C B to a ring extension.

The polynomial ring R[zl,...,z,] in variables 21, ..., z, over a ring R is

79 80 Commutative Algebra called a polynomial R-algebra.

1.1. Definition Let R 5 B be a ring extension. (i) If B = R[S]for some nonempty subset S c B, where R[S]is the subring of B generated by S over R (see Chapter 1 section 0 for the definition), we say that B is an R-algebra generated by S and call S a set of generators of B or a generating set of B. If S = {bl, ...,b,} is a finite set, then B is called a finitely generated R-algebra and is denoted by B = R[bl,..., b,]. In the latter case, we also say that B is a finitely generated extension ring (algebra) of R. (ii) If B is a finitely generated R-module, then we say that B is a module- finite extension ring (algebra) of R or B is module-finite over R.

From definition it is clear that any module-finite extension ring (algebra) is a finitely generated extension ring (algebra).

Example (i) Let R be a ring and B = R[x1,..., x,] the polynomial ring in variables xl,..., x, over R. Then B is a finitely generated extension ring (algebra) of R but B is not module-finite over R. (ii) Let d E Z be square-free (i.e., n2 ,/'d for any n E Z). Then Z c Z[&] = Z + Z& is a module-finite extension. (iii) Using the fact that Z is a UFD and there are infinitely many primes in Z, we conclude that Q is not a finitely generated extension ring of Z. (See also Chapter 1 (section 7, exercise 3).)

1.2. Definition (Compare with an algebraic element in a field extension.) Let B be an R-algebra defined by the ring homomorphism cp: R -+ B. Write R' = cp(R).An element b E B is said to be integral over R if there is a monic polynomial f (x)E R'[x],say f (x)= xn+X,-1xn-'+. . ~+Xlx+Xo, Xi E R', such that

If every b E B is integral over R, we say that B is integral over R. If R C B and B is integral over R, then we call B an integral extension ring (algebra) of R and refer R C B to an integral extension.

1.3. Theorem Let B be an R-algebra defined by the ring homomorphism Integral Extensions and Normalization 81

‘p: R --+ B. Write R’ = p(R). For b E B, the following are equivalent: (i) b is integral over R. (ii) The R-subalgebra R’[b] C B is a finitely generated R’-module (or equivalently, a finitely generated R-module). (iii) There exists an R-subalgebra B’ C B with the property that R’[b] 5 B’ and B’ is a finitely generated Rl-module (or equivalently, a finitely generated R-module). Proof (i) + (ii) b is integral over R implies

bn + X,-1bn-’ + . . . + Xlb + Xo = 0, Xi E R’.

Inductively we derive that

n- 1 bnii E R’bj, i E N. j=O

It follows that R’[b] = CYZ,’R‘bj. (ii) + (iii) Take B’ = R’[b]. (iii) + (i) Let R’[b] C B’ C B, where B’ is the R-subalgebra as described. Suppose B‘ = Cy=l R’Ji, Ji E B’. Then bJi E B‘, say

In a matrix form we have

Write M for the above n x n matrix. If we multiply the above equality by the adjoint matrix of M,then

det(M)Ji = 0, i = 1, ..., n.

This shows that det(M)B’ = 0. But lg = lgl E B‘ (by our convention for subrings). Hence det(M) = 0, and consequently b is a zero of some monic polynomial over R’. 82 Cornmutative Algebra

1.4. Corollary Let B be an R-algebra. If B is a finitely generated R- module, then B is integral over A. (One may compare this with a finite dimensional field extension.) 0

Example (iv) Let d E Z be square-free. By previous Example (ii), Z c Z[d]is a module-finite extension and hence an integral extension.

(v) Let K be a field and B = K[z] the polynomial ring in one variable z over K. Then for n 2 1, K[z"] K[z]is an integral extension. Indeed, one may check that B = K[zn]+K[zn]z++[s"]s2+...+K[zn]zn-1,that is, B is module-finite over K[zn].

(vi) Let R be a UFD and K its field of fractions. Then any y E K - R is not integral over R. To see this, let y = f # R, where gcd(a, b) = d E U(R).If yn + X,-ly"-l + . . . + Xly + XO = 0, Xi E R, then, multiplying by b", we have

an = -b (Xn-1un-' + Xn-2ban-' +.. . + X1bn-'u + b"-lXo) E R.

This shows that if a is a unit in R, then b is a unit in R and hence y E R. If a is not a unit in R, then b must be a unit in R and y = ab-I E R; otherwise a and b would have a nonunit common divisor.

(vii) Let K[z,y] be the polynomial ring in z, y over a field K. Consider the ring R = K[z,y]/(y - f(z)),where f is a monic polynomial with degf 2 1. Then, Z is integral over K[V]and R is module-finite over K[y].

(viii) Let K[z,y] be the polynomial ring in 2, y over a field K. Then, though - zy - 1 = 0 in the ring R = K[z,y]/(yz- l),T is not integral over K[V]. Indeed, by Chapter 2 (section 3, Example (ii)), with S = {l,y,y2,...},

while K[y][l/y] is not a finitely generated K[y]-module by Chapter 1 (section 7, exercise 3).

1.5. Proposition Let B be an R-algebra defined by the ring homomorphism p: R + B. Write R' = cp(R). The following hold: (i) The subset - R = { b E B 1 b is integral over R Integral Extensions and Normalization 83 is an R-subalgebra of B. - (ii) If b E B is integral over R, then b E 3. Hence R = R.

Proof (i) Let bl, b2 E R. We show that bl f b2, blb2 E 2. Consider the ring extension

R'[bi] C_ R'(bi][h]= R'[h1b2].

By Theorem 1.3 (ii),

n R'[bl] = CR'V~,~j E R'[bl]. j=1

Thus,

and it follows from Theorem 1.3(iii) that bl i b2, blbz E R. (ii) Let b E B be such that

b" + X,-lb"-l + . *. + Xlb + A* = 0, xi E R.

Then, inductively we derive that

n-1 bn+Z E cR'[Xo, XI, .", Xn-l]bj, 2 E N. j=O

Hence R'[b]C CyIi R'[Xo, XI,..., X,_l]bj. Note that the Xi are integral over R. A similar argument as for part (i) shows that R'[Xo,XI, ..., A,-,] is contained in an R-subalgebra of B which is also a finitely generated R'-module. By Theorem 1.3(iii), b E R. 13

1.6. Definition The subalgebra obtained in Proposition 1.5 is called the integral closure of R in B. If R C B and R = R C B, then we say that R is integrally closed in B. (Compare this with the algebraic closure in a field extension defined in Chapter 1 (section 3, exercise 3).) 84 Commutative Algebra

Combined with the forthcoming Noether normalization of section 2, the next proposition will play an important role in algebraic geometry (see the proof of Chapter 5 (Theorems 5.7 and 6.3)).

1.7. Proposition Let R C B be a module-finite extension and M E m- SpecR. Then MB # B and there exists a maximal ideal Q E m-SpecB such that Q n R = M. (A generalization of this result is given in exercise 8 below.) Proof Suppose B = C:==,R&, & E B. If B = MB = c:=,M&, then = El=,rij&, rij E MI i = 1, ... ,s. Arguing as in the proof ((iii) (i)) of Theorem 1.3, there would be 1 6 M. Hence MB # B. Let Q be a maximal ideal of B that contains MB (Chapter 2 Proposition 1.5). Note that R n Q # R (otherwise 1 E Q). Thus, M c R n Q implies M = R n Q. 0

Finally, let us see how integral extension is closely related to field extensions (see the proof of later Theorem 2.5, Chapter 5 (Theorems 1.5 and 5.3) for applications).

1.8. Theorem Let R C B be an integral extension, where R and B are domains. Then R is a field if and only if B is a field. Proof Suppose R is a field. If 0 # b E B then

b" + Xn-lb" +.. . + Xlb + Xo = 0, Xi E R, 2 1. Assume that n is the smallest degree. Then XO # 0 and

b-l = -x-l0 (b-1 + X,-,b"-2 + . . . + X2b + A,) E B.

Conversely, if B is a field and 0 # a E R, then a-1 E B and hence a-1 is integral over R. It follows that there is a relation of the form

(u-')~+ Xn-1(~-')"-' + * * + X~U-' + Xo = 0, Xi E R.

Consequently, a-' = -Xn-l - Xn-2a - . . . - Aoan-l, i.e., a-1 E R.

Exercises 1. Let R = Z[@ 1 pi's are distinct prime numbers, i 2 11. Show that R is neither module-finite nor finitely generated as an algebra over 25. (See also previous Example (iii) and Chapter 1 (section 7, exercise 3).) Integral Extensions and Normalization 85

2. By previous Example (v), K[s2]c K[s]is an integral extension. For any f E K[z],find a monic polynomial g(t) E k[s2][t]such that g(f) = 0. 3. Let A C B G C be ring extensions. Suppose that C is module-finite over B and B is module-finite over A. Show that C is module-finite over A. 4. Let R C B be a ring extension. Show that if bl, ..., b, E B are integral over R then R[bl,..., b,] is module-finite over R. Hence, B is module- finite over R if and only if B is finitely generated as an R-algebra and integral over R. 5. Let A 5 B & C be ring extensions. Suppose that C is integral over B and B is integral over A. Show that C is integral over A. 6. Let R c B be an integral extension. (a) For u E R, show that u E U(R) if and only if u E U(B). (b) Let H be a proper ideal of B and h = H n R. Show that (up to a ring monomorphism) R/h c B/H is an integral extension. 7. Let R C B be an integral extension, and let S be a multiplicative set of R, 0 @ S. Then clearly S is also a multiplicative set in B. (a) Show that Rs C Bs, and that Bs is integral over Rs. (Hint: Use Chapter 2 Proposition 4.3 and Chapter 2 (section 3, exercise 2).) (b) Show that if B is module finite over R then Bs is module finite over Rs. (Hint: Use Chapter 2 (section 4, exercise 6).) 8. (Lying over theorem) Let R G B be an integral extension. Show that if P E SpecR then there is some Q E SpecB such that Q n R = P. (Hint: First, by arguing as in the proof of Proposition 1.7, show that the assertion is true for every M E m-SpecR. Then, for each P E SpecR, use exercise 7 above, Chapter 2 (Corollary 3.5 and Proposition 4.3) in order to pass to the case Rp E Bp.)

2. Noether Normalization

Let R = K[al,..., a,] be a finitely generated K-algebra over the field K. Then by Chapter 1 section 0, R K[zl,..., zn]/I,where I is an ideal of the polynomial K-algebra K[zl,..., sn]. Hence R is Noetherian by Chapter 1 (section 1, exercise 2). In this section we show how to build R by a module-finite extension over some polynomial K-algebra.

By the definition of a polynomial ring we know that C Xa(j)sy31 . . . xnaJn 86 Commutative Algebra

= 0 in K[q,... ,zn] if and only if all X,(j) = 0. This property has the following generalization to an arbitrary K-algebra.

2.1. Definition Let R be a K-algebra over the field K, and r1, ..., rm E R. If there is some polynomial g E K[q,... ,z,] such that g(r1, ..., rm) = 0, then we say that r1, ..., r, are algebraically dependent over the field K; otherwise, r1, ..., r, are said to be algebraically independent over the field K.

Clearly, if K L is a field extension and 6 E L is a transcendental element over K (Chapter 1 section 3), then 6 is algebraically independent over K and K[8]is isomorphic to the polynomial algebra K[z]. More generally, if R is a K-algebra over the field K and r1, ..., r, E R are algebraically independent over K, then the subalgebra K[rl,..., r,] of R is isomorphic to the polynomial K-algebra K[zl, ..., z,].

To reach the main result of this section, the following technical prelim- inary is required.

2.2. Lemma Let a(j) = (ajl,..., q,),~(k) = (ffkl, ..., ffk,) E Nn. Sup- pose xy=lffj8 = cy=lffki = m, but lexicographically (see Chapter 1 section 4) ~(k)+lez a(j),that is, there is some s 5 n such that

ffkl = ffjl,..., ffk,-l = ffjs-,, while ffk, < CYjs.

Then cy=lmn-'aki = B < A = Cy=lmn-i aja.

Proof Without loss of generality we may assume ffkl < aj1. Then

/ n\n Integral Extensions and Normalization 87

i=3 i=3

i=l

= A.

F'urthermore, let f E Kjxl, ..., x,] be a polynomial of degf = m 2 1, where K is a field. Then by Chapter 1 (section 7, exercise 8), f may be written uniquely as

where each Fi is a homogeneous polynomial of degree nil i.e.,

with a(j)= (ajl,..., aj,) E Nn, E K.

Let us call F, the leading homogeneous part of f for convenience.

2.3. Proposition Let R = K[al,..., a,] be a finitely generated K- algebra over the field K. Suppose that there is a nonzero polynomial f E K[x~,..., x,] such that f(a1, ,.., a,) = 0. Then there exist bl, ..., bn-1 E R such that a, is integral over K[bl,..., bn-l] and R = K[bl,..., bn-1][~,].

Proof Write f = F, + F,- 1 + . . . + FO as a sum of homogeneous polynomials, where the leading homogeneous part off has the form

Since dlexis a total ordering on N",we may also assume that 88 Commutative Algebra

Now set

and define

in which the highest degree term in x, is given by

Frn (z; +xg1,...,x;-1 +xK*-',~n) that is,

Setting bk = ak - agk,k = 1, ..., n - 1, we see from (3) above that

H(b1,..., bn-l, a,) = f(a1,..., a,) = 0.

Note that 0 # Xa(l) E K. We conclude that a, is a zero of the monic polynomial X;tllH(bl, ..., b,-l,t) E K[bl,..., b,-l][t]. Finally, it IS clear that R = K[bl,..., b,-l][a,], and thus, the proof is completed. 0

2.4. Theorem (Noether normalization) Let R = K[al,..., a,] be a finitely generated K-algebra over a field K. Then there exist elements z1,..., zd E R, 0 5 d 5 n, such that (i) 21, ..., zd are algebraically independent over K; (ii) R is module-finite over the polynomial K-algebra B = K[zl,..., zd]. Proof We prove by induction on the number n of generators of R. For n = 1, the assertion follows from (section 1, exercise 4). So, For n 2 1, if al, ..., a, are algebraically independent over K, then the assertion is trivially true. Suppose that al, ..., a, are algebraically dependent over K, and that f E K[zl,..., x,] is such that f(a1,..., a,) = 0. Then, by Integral Extensions and Normalization 89

Proposition 2.3, there exist bl, ..., b,-l E R such that a, is integral over B = K[bl, ..., bn-l] and R = B[a,]. Applying the inductive hypothesis to B, we may find 21, ...,zd E B, that are algebraically independent over K, such that B is module-finite over A = K[zl, ..., zd]. Since an is integral over B, it follows that B[a,] is module-finite over B. If we look at each step of the tower A c B c B[a,] = R, then by (section 1, exercise 3), R is module finite over A, which completes the proof. 0

Theorems 1.8 and 2.4 enable us to derive the following wonderful result that is crucial in dealing with the Nullstellensatz in algebraic geometry (Chapter 5 section 1).

2.5. Theorem (Zariski) Let K L be a field extension. If L = K[al,...,a,] is a finitely generated K-algebra, then L is module-finite over K, i.e., [L: K] < 00, and hence L is algebraic over K.

Proof By the Noether normalization, there are 21, ..., Zd E L, which are algebraically independent over K, such that

Thus, K[zl,..., zd] 2 L is an integral extension. But since L is a field, it follows from Theorem 1.8 that K[z1, ..., zd] is a field. This shows that d = 0, and consequently L is module-finite over K.

Exercises 1. Given distinct a(l),a(2),..., a(m) E Nn with a(j)= (ajllaj2, ..., ajJ1 show that there are nonnegative integers p1, ..., pn-ll 1 = p, such that

(Hint: Use induction on n. If n = 1, it is trivially true. Suppose for a'(l),..., a'(m) E N"-l with a'(j) = (ai2,...,aim) we can choose p2, . ..,pn-1, pn = 1 SO that the conclusion is true. Now the choice

completes the induction process.) 90 commutative Algebra

2. Let R = K[z,y]/(y2 - x3), where K[z,y] is the polynomial ring in z, y over a field K. Show that Z is algebraically independent over K, y is integral over K[4,and R = k[Z,jj]is module-finite over K[Z]. 3. Let K[z,y,z]be the polynomial ring in z,y,z over a field K, R = K[z,y, z]/(xz,y2z). Write u = T + 2, v = ij. Show that u and v are algebraically independent over K and that R is module-finite over the subring K[u,v]. 4. Let K[z,y] be the polynomial ring in z, y over a field K. By (section 1, Example (viii)), in the ring R = K[z,y]/(yz - l),Z is not integral over K[jj],and similarly is not integral over K[z].Use Noether normalization to find some z E R that is algebraically independent over K, such that R is module-finite over K[z].

3. Normal Domains and Normalization

In this section we introduce the notion of a normal domain, and establish the module-finite property of the normalization (integral closure) of a finitely generated domain.

3.1. Definition Let R be a domain, K its field of fractions, and E its integral closure in K (Definition 1.6). (i) If R is integrally closed in K, i.e., R = R c K, then we say that R is a normal domain. (ii) If R = F[al,..., a,] is a finitely generated F-algebra over some field F, then is called the normalization of R (this name will be qualified by the proof of Corollary 3.3 below and by the behavior of R in Chapter 5 section 5).

Example (i) Let R be a domain and K its field of fractions. If R is the integral closure of- R in K, then, since the field of fractions of is also K (why?) and a = E c K, by Definition 3.1, R is a normal domain. By (section 1, Example (vi)), any UFD is normal. In particular, any DVR is normal.

(ii) Let K = Q(s)be a number field and dK its ring of algebraic integers. Then dKis normal and Noetherian but not necessarily a UFD (see Chapter 4 for details). Integral Extensions and Normalization 91

(iii) Let K[x,y] be the polynomial ring in variables x,y over a field K, and let R = K[rc,y]/(y2 - x3) (see section 2, exercise 2). Then jj2 = I3 in R. Thus, any element of R has the form

= ~Animly1+3m1-y + c XnzmzP+3mz. Let K[t]be the polynomial ring in variable t over K. Then the ring homomorphism K[x,y] -+ K[t2,t3] with x H t2 and y H t3 induces a ring homomorphism (see Chapter 1 section 0):

R -+ K[t2,t3] p : - x H t2 - y H t3

By the remark made above, it is not difficult to see that p is an isomorphism. Note that $ = t in the field of fractions K(t)of K[t].Identifying R with the subring K[t2,t3]of K[t],R and K[t]have the same field of fractions K(t),i.e., we have the relation

R c K[t]c K(t)

Clearly, t is integral over R but is not contained in R. Hence, R is not normal. But since K[t]is a UFD, the normalization of R is K[t]by Example (i) above. A geometric explanation of this example will be given in Chapter 5 section 5.

(iv) Let K[x,y,z] be the polynomial ring in x,y,z over a field K, R = K[x,y, z]/(zy2 - x2). If K[u,u] is the polynomial ring in u,v over K, then the ring homomorphism K[x,y, 21 + K[uv,u, u2] with x H UV, y H u,and z H v2 induces a ring homomorphism

$ : R K[uu,u,v~] - - XH 2/21 - YH U - ZH V2

A similar argumentation as in Example (iii) above shows that $ is an isomorphism, and the normalization of R is K[u,u] # K[uv,u, u2]. 92 Commutative Algebra

By (section 1, exercise 4), the normalization R of R obtained in Example (iii), respectively in Example (iv) above, is module-finite over R (see also other examples provided by exercises of the current section). Below we will see that this is by no means accidental.

3.2. Theorem Let R be a normal domain and K its field of fractions. If K c L = K(t9) is an n-dimensional separable field extension, then the integral closure of R in L, denoted ?i, is contained in a finitely generated R-submodule of L. If R is Noetherian then i? is module-finite over R. Proof Choose elements a1, ...,an E R to form a K-basis for L (why is this possible?). Then C:=l Rai C x. Consider the trace function TL/K on L as defined in Chapter 1 section 5. By Chapter 1 (Theorems 5.5 and 5.7), TL/K(uv) is a nondegenerate bilinear form and there exists a dual K-basis {PI, ...,On} for L such that

We claim that x & C:=, RPi. To see this, let cl,...,cn be all distinct K-linear monomorphisms from L to a splitting field of pe(x) (the minimal polynomial of 6 over K). Then for u E R C L, we have ai(u) E XI i = 1, ..., n. But Cy=lei(u) = TL/K(u)E K, it follows that TL/K(u) E R because R is normal. Now, for y E R, y = CZ1Xipi with Xi E K, we have TL/K(yaj) E R and by the above (l),

This shows that R C CZl RPi, as claimed. The last assertion follows from Chapter 1 Theorem 7.6. 0

3.3. Corollary Let R = F[al,...,a,] be a finitely generated F-algebra over a field F. If R is a domain and K is its field of fractions, then the normalization of R in K is module-finite over R and hence a Noetherian ring. Proof By the Noether normalization theorem, there are algebraically in- Integral Extensions and Normalization 93 dependent zl, ..., Zd E R over F such that n F C: F [ZI,.. . , Zd] 2 R = C F [ZI,.. . , zd]

Assume that K is a separable extension of F(z~,..., zd) that is the field of fractions of the polynomial ring F[zl,..., zd]. Then since R is also the integral closure of F[zl,..., zd] in K, it follows from Theorem 3.2 that R is module-finite over R (note that R is Noetherian). 0

Applications of Theorem 3.2 and Corollary 3.3 are given in Chapter 4 and Chapter 5.

Remark In the proof of Corollary 3.3 we argued by assuming the separability of K over F(z1,..., zd). If the ground field F is of characteristic 0, the separability of K is naturally guaranteed. In general case, the feasibility of this assumption is proved in detail by Zariski and Samuel in Theorem 9 (p. 267) of Commutative Algebra, Vol. I (New York: Springer-Verlag, 1958). We do not quote Zariski-Samuel’s detailed argumentation here because it involves more deeper field theory that is not included in Chapter 1. So the reader may accept Corollary 3.3 and use it without any doubt.

Exercises 1. Let R = K[z,y]/(y - f(z)),where K[z,y] is the polynomial ring in z, y over a field K and degf(x) 2 1. Is R a normal domain? 2. Let R = K[z,y]/(y2 - x2 - x3) where K[z,y] is the polynomial ring in z and y over a field K. Prove that the normalization of R is isomorphic to the polynomial ring K[t]. (See also Chapter 5 (section 7, Example (iv>1.1 3. Let R = K[z,y,z]/(y - x2,z- z3),where K[z,y, z] is the polynomial ring in z, y, z over a field K. Show that R is a normal domain. (Hint: R 2 K[t] with z H t, y H t2, and z H t3.) 94 Commutative Algebra

4. Normal Domains and DVRs

This section is devoted to a local study of normal domains in terms of DVRS. First, a typical example of “global concern versus local solutions”.

4.1. Theorem Let R be a domain and K its field of fractions. The following statements are equivalent. (i) R is normal. (ii) Rs is normal for any multiplicative set S of R. (iii) Rp is normal for all P E SpecR. (iv) RM is normal for all M E m-SpecR. Proof (i) 3 (ii) let x E K be integral over Rs. Then

Ti zn + Xn-1xn-l +.. . + Xlx + XO = 0, Xi = -, ri E R, si E S. Si

Multiplying the relation by (SOS~. . . s,-l),, we see that (~0~1. . . S,_~)X is integral over R. By the fact that R is normal, (soq . . . s,-~)x = y E R, i.e.,

2= €Rs. SOSl * . S,-1

(ii) j (iii) + (iv) is obvious. (iv) + (i) let z E K be integral over R. Then

Zn + X,-lXn--l + . . . + XlX + A0 = 0

holds over both R and RM for each M E m-SpecR (note that R 2 RM). Since RM is normal, x E nRM = R by Chapter 2 Theorem 3.8. 0

Comparing with Chapter 2 Proposition 2.8, our goal is to show that the equality R = nRp holds for a normal and Noetherian domain R, where P runs over all minimal nonzero primes of R and every Rp is a DVR, though a normal and Noetherian domain is not necessarily a UFD (section 3, Example (ii)). To this end, we need the following preparation. Let R be any ring, M an R-module, and x E M. Then the set of annihilators of x in R, denoted

AnnRX = {a E R 1 ax = 0} Integral Extensions and Normalization 95 is an ideal of R. Indeed, AnnRx coincides with the kernel of the R-module homomorphism

px: R-M a H ax

Hence R/Kerpx = R/AnnRx is isomorphic to an R-submodule of M. Note that if 3: # 0 then AnnRx # R, for 1~ E R and all modules considered in this book are unitary. If AnnRx = P E SpecR, we say that P is an associated prime of M. Set

P E SpecR P an associated prime of M

Example (i) For any P E SpecR, AssR(R/P) = {P}.

(ii) Let R be a PID, and let M = R/(a), where a = pypy for two primes pl # pa. Then ii = pT-'py E M has AnnRii = (PI), and similarly V = pn,p2 m-1 E M has AnnRij = (p~).Moreover, if w E M with AnnRw = (T) E SpecR, then, it follows from the first isomorphism theorem that rla. Hence AssRM = {(Pl), (P2)).

4.2. Proposition Let R be a ring and M an R-module. The following hold: (i) Any maximal element (with respect to the inclusion relation) in the set

fl= {AnnRx 1 OfxE M} is prime and hence belongs to ASSRM. (ii) If R is Noetherian and M # {0}, then ASSRM# 0. Proof (i) Let x E M be chosen such that AnnRx is maximal in a. If a, b E R, ab E AnnRx, then abx = 0. It follows that if bx = 0 then b E AnnRx; if bx # 0, then since bx E M and AnnRx C AnnR(bx), we have AnnR(bx) = AnnRx by the choice of AnnRx. Therefore, abx = 0 yields a E AnnRx. (ii) If R is Noetherian and M # {0}, then

R = {AnnRx I 0 # x E M} # 0 and has a maximal element. Hence ASSRM# 8 by (i). 96 Commutative Algebra

The next proposition provides the key result in the local study of a normal Noetherian domain.

4.3. Proposition Let R be a local domain with the unique maximal ideal m. Assume that R is normal and Noetherian. If m E Ass~(R/(a))for some 0 # a E R, then R is a DVR, and hence m is a minimal nonzero prime ideal. Proof Let u E R - (a) such that mu c (a). Then m: C R. If m: 5 m, then since m is a finitely generated R-module, a similar demonstration as in the proof of Theorem 1.3 ((iii) * (i)) shows that is integral over R (note that now R is a domain) and hence in R. Thus, u = ar E (a) for some T E R. This contradicts the choice of u. Therefore, we must have m: = R because rn: is an ideal of R. Let 1 = v: for some v E m. Then, for any y E m, y: = E R implies y E (v). This shows that m = (v). It follows from Chapter 2 Theorem 2.10 that R is a DVR and m is a minimal nonzero prime ideal. 0

4.4. Corollary Let R be a normal and Noetherian domain and P E Ass~(R/(a))for some 0 # a E R. Then P is a minimal nonzero prime ideal of R.

Proof Consider the localization (R/(a))p of R/(a) at R - P. Then

by Chapter 2 Corollary 4.4. Now P E Ass~(R/(a))implies PRp E ASSR~(Rp/aRp) by later Exercise 4, and the assertion follows from Propo- sition 4.3 and Chapter 2 Proposition 3.6. 0

Comparing with Chapter 2 (Theorems 2.6 and 2.10), we now have the following characterization of a DVR.

4.5. Theorem Let R be a domain. The following statements are equivalent. (i) R is a DVR. (ii) R is a normal and Noetherian local domain with SpecR = {{O},m} where m is the maximal ideal of R. (Therefore, this theorem may also be viewed as the converse of Chapter 2 Theorem 2.6.) Integral Extensions and Normalization 97

Proof (i) + (ii) This follows from (section 3, Example (i)) and Chapter 2 Theorem 2.6. (ii) + (i) By Chapter 2 Theorem 2.10, we need only to prove that (ii) implies m is a principal ideal. By Nakayama’s lemma, m # m2. Let 0 # u E m - m2. If (u) is a prime ideal then m = (a) as desired. If (u) is not a prime ideal, then (u) # m and there are y,z E R, y,z # (u) but yz E (u). This shows that AnnRF # {0}, where F is the image of z in M = R/(u). By Proposition 4.2, ASSRM# ((0)) and hence m E ASSRM. It follows from Proposition 4.3 (or its proof) that m is principal and R is a DVR. 0

4.6. Corollary Let R be normal and Noetherian. If P is a minimal nonzero prime ideal of R (i.e., P is minimal in SpecR with respect to the inclusion relation), then Rp is a DVR. Proof Rp is normal by Theorem 4.1, and Noetherian by Chapter 2 Corol- lary 3.5. Since P is minimal in SpecR, it follows from Chapter 2 Proposition 3.6 that SpecRp = (0, PRp}. Hence Rp is a DVR by the theorem above. 0

We are ready to prove the previously promised result.

4.7. Theorem Let R be a normal and Noetherian domain, K its field of fractions. Then

where P runs over all minimal nonzero prime ideals of R. In particular, R is an intersection of DVRs. Proof nRpR is clear. To get the inclusion nRp C R, a trick used in the proof of Chapter 2 Theorem 3.8 is employed again. Let x = E K, and consider the ideal of R given by

I, = {a E R I ax E R} .

Then Iz = {u E R I ub E (c)} = Ann& where 5 is the class of b in R/(c). If i; = 0, then b E (c) and hence x E R. Suppose # 0. Then

O#EER/(c)}#0. 98 Commutative Algebra

Since R is Noetherian, I, is contained in a maximal element P of 0. By Proposition 4.2(i), P E Ass~(R/(c)).It follows from Corollary 4.4 that P is a minimal nonzero prime. However, note that I, C P implies z e Rp. This proves the following: z @ R implies 3: 6Rp for some minimal nonzero prime P.

Exercises 1. Let M be an R-module and N an R-submodule of M. Show that ASSRM C Ass~NuAssR(M/N). Can you find an example to show that g holds? 2. Show that for any n 2 1, there is a Zmodule M such that AsszM contains exactly n prime ideals. (Hint: Consider m = pyl . . .p,"" with distinct pimes pi.) 3. Use Proposition 4.2(i) to show that if R is Noetherian and M an R- module, then u AnnRz= u p. O#XEM PEASS~M 4. Let M be a nonzero R-module and P E ASSRM. If S is a multiplicative set of R with P n S = 0, show that PRs E ASSR~Ms. Chapter 4 The Ring & in K = Q(8)

This chapter introduces the normal and Noetherian domain AK that has its location among numbers as indicated by the following Hasse diagram (of the obvious partial order):

K A

\"r=Kndz

99 100 Commutative Algebra where

A = { p E c I g(p) = o for some monic g(z) E ~[z]},

K c C a subfield with [K : Q] < 00.

Consequently, 6 is the algebraic closure of Q in C (Chapter 1 section 3, Exercise 3), A is the integral closure of Z in C (Chapter 3 Definition 1.6), and AK is the integral closure of Z in K (Chapter 3 Definition 1.6). Very soon in section 1 below we will see that AK is the integral closure of Z[8], where K = Q(8). For any /3 E A, since A is integrally closed in C,we have fi E A (Chapter 3 Proposition 1.5(ii)). It follows that

holds for all a E A. Note that not every a E A is a unit (for instance 6). We conclude that factorization into irreducible elements in A is impossible, and consequently, A is not Noetherian (Chapter 1 Proposition 2.5). Thus, A is usually not involved in practice. Why AK? First and foremost, AK is a normal and Noetherian domain (section 1, Theorem 1.2 below). So factorization into irreducible elements in AK is always feasible (Chapter 1 Proposition 2.5), and in many cases AK is a UFD (section 3). Perhaps a motive example to this topic should be the following oldest example.

Example Find all integer solution (z, y, z) of the equation

(1) x2 + y2 = z2 where gcd(z, y, z) = 1. Solution In order to solve this problem, one way is to factorize the left of equation (1) in the ring of Gaussian integers Z[i]= {a + bi I a, b E Z} with i=m

2 (2) (. + yi)(z - yi) = 2 . The Ring AK in K = Q(8) 101

Note that Z[i]is a UFD (see section 3). The assumption (x,y, z) = 1 entails that x and y cannot be both even or odd. So, z must be odd. Thus, x + yi and x - yi must be coprime, and a comparison of the prime factorizations on both sides of equation (2) yields z + yi = %a2,where u is a unit of Z[i] and Q = c + di which is not a unit. Since V(Z[i]) = {fl, f i} (see section 3), it follows that the desired solutions (2,y, z) are given by z = f(c2-d2), y = f2cd, z = f(c2 + d2). In Chapter 5 (section 7, Example (iii)), another way to find the integer solutions of equation (1) will be given in terms of algebraic geometry.

In section 2 we will see that actually Z[i]= Q(i)n A.

A generalization of last example is the famous Fermat’s last theorem which states:

0 For n > 2, the equation xn + yn = zn does not have nonzero integer solutions (z, y, z). Using the above example, one may show that the theorem is true for n = 4 and hence (automatically) also for n = 4k. It is therefore sufficient to consider the case where n is an odd prime p, for if no solutions exist when n = p then no solutions exist when n = pe. Now, for p, an odd prime,

yp = zp - xp = (% - z)(z - WX)(% - w”) . . . (2 - wp-l x) in Z[w],where w is a primitive pth root of unity, i.e., w = e2?ri/P. Historically in the literature (cf. Borevich and Shafarevich: Number Theory, Academic Press, 1966, pp. 378-381), it was shown that if Z[w] is a UFD, then the theorem is true for p. But unfortunately, Z[w]is not always a UFD (see section 3). Actually, we have Z[w]= Q(w) n A, where w = e2?ri/P,p an odd prime (see section 2). Though Fermat’s last theorem was finally proved by Wiles in a very deep algebraic-geometric way in 1994, it was the ring AK that started algebraic number theory and led to the ideal structure theory in both commutative and noncommutative algebra. 102 Commutative Algebra

1. AK is Normal and Free of ZRank [K : Q]

Adopting the classical definition in number theory, if a field extension Q c K,where K c C, is finite dimensional, then K is called a number field. By Chapter 1 Theorem 3.11,

K = Q(Q),Q E K is algebraic over Q. Using the notation as in the introduction of this chapter, let dK = An K = {Ip E K f(P)= 0 for a monk f(z)E Z[z]}. Elements in dK are called algebraic integers, and so, dKis called the ring of algebraic integers of K. Note that since the generator a: of K is algebraic over Q, there is some 0 # c E Z such that CQ E dK (later Exercise 1). It turns out that we may write

Thus,

Z 2 2[6]2 dK C K = Q(8).

1.1. Lemma With notation- as above, dK is equal to the integral closure of Z[8] in K, that is, dK= 2[8].Hence AK is a normal domain. Proof This follows from Chapter 3 (section 1, exercise 5). 0

1.2. Theorem (i) AK is a finitely generated abelian group (or Z-module). Therefore, dK is normal and Noetherian. (ii) dK is a free abelian group of Zrank [K : Q]. Proof (i) Since Z is a normal Noetherian domain (a UFD is normal), Z C Q C Q(6),and dK is the integral closure of Z in K, the assertion follows from Chapter 3 Theorem 3.2 and Chapter 1 Theorem 7.6. (ii) By Chapter 1 (section 6, exercise 4), dKis a free abelian group of finite rank. Note that any Zbasis of dK is also a Q-basis for K (exercise 4). Hence the Z-rank of dKmust be equal to [K : Q]. 0

In the literature, a Zbasis of dK is also called an integral basis of K, due to the fact that a Zbasis of dKis necessarily a Q-basis for K (exercise 4). The Ring AK in K = Q(0) 103

Let K = Q(19) be as before, where 6 E Arc, [K : Q] = n. From now on we let p(z) be the minimal polynomial of 19 over Q.

1.3. Proposition For a E K, let pa(.) be the minimal polynomial of a over Q. Then a E dK if and only ifp,(x) E Z[z]. In particular, p(z)E Z[Z]. Proof The (‘if” part is clear, and the “only if” part follows from Chapter 1 Theorem 2.17 (note that a E dK if f(a)= 0 for some monk f(z)E Z[z]).

Remark From Proposition 1.3 we derive immediately that Q n A = Z. This recaptures the fact that Z is integrally closed in Q.

Concerning integral bases of K, we first note an easy fact in view of Proposition 1.3.

Observation Let a E dK,then pa(.) E Z[x]. Since pa(.) is monic, a division algorithm by pa(.) in Z[z] plus the ring homomorphism Z[z] -+ Z[a] with z H a yields Z[a]= xyr: Zai, where n = degp,(z). Consequently, Z[a]is a free abelian group of rank n, and {1,a,a2,...,an-1} is a Zbasis of Z[a].So, if Z[a]= dK1then we have found a Zbasis for AK (in section 2 we will see that many number fields have this property).

However, finding an integral basis for an arbitrary number field K is by no means an easy job in algebraic number theory. As to this topic, a very useful invariant of K is introduced and discussed below. Consider the n distinct zeros of p(z) in Q: (Chapter 1 Theorem 3.6), say

191 = 6,292, ..., 6,.

By Chapter 1 Proposition 5.1, there are exactly n distinct Q-linear ring monomorphisms

a2 : K = Q(19)- c 19 H 192 i = 1, ..., n.

1.4. Definition Let 01, ...,a,be as above. If {a~,...,a,}is a Q-basis for 104 Commutative Algebra

K = Q(6),then

A[al, ..., a,] = [det (ai(aj))]' is called the discriminant of the given basis.

1.5. Lemma Let {al,..., a,} and {PI,..., ,&} be two Q-bases of K = Q(6). Then

where Pk = CT=, cjkaj, k = 1, .., n, cjk E Q. Proof Exercise.

1.6. Proposition With notation as above, the following hold: (i) 0 # A[al, ..., a,] E Q for any Q-basis {al, ..., a,} of K = Q(6). (ii) If all gi(6),i = 1, ..., n, are real numbers, then A[al, ..., a,] > 0 for any Q-basis {a~,..., a,} of K. Proof (i) Consider the Q-basis { 1,6,d2, ..., P-'}and note that oi(6) = 6i, i = 1, ..., n. Then

Since applying any permutation 7~ of (61, ..., 6,) to ni.,,(6i - 8j) is the same as applying 7~ to det(ai(@)), it interchanges rows of det(oi(6j)) and hence the sign of det(oi(6j)). But then det(ai(z9j))2 is symmetric on 61, ..., 6,. It follows from Chapter 1 Corollary 7.2 that A[1,6,d2,..., Sn-'] E Q. Since all 6i's are distinct, A[l,6,d2, ..., P-']# 0 and the conclusion (i) follows from Lemma 1.5. (ii) This follows from part (i) and Lemma 1.5. The Ring Ax in K = Q(19) 105

1.7. Corollary If (01, ...,a,} is a Q-basis of K = Q(29)in which all cq's are algebraic integers, then 0 # A[al, ...,an] E d n Q = Z. Proof This follows from Proposition 1.6(i) and the fact that all ai(aj)'s are algebraic integers.

As argued in [ST], there is a direct proof of Theorem 1.2(ii) by using only discriminant.

1.8. Another proof of Theorem 1.2(ii) Since { 1,8,f12, ..., P-'}is a Q- basis of K consisting of algebraic integers and 0 # A[l,29, ..., 29"-'] E Z by Corollary 1.7, we may select a Q-basis W = {[I, ..., <,} of algebraic integers for which /A[&,...,&]I is the least. We claim that {el, ...,en} is a Zbasis for dK. To see this, note that since {el, ...,en} is also Zlinearly independent, we need only to show that, for w E dK, if w = al

we have

Let P be the above n x n matrix. Then det(P) = a1 - a = T > 0. This shows that {ql, ..., 7,) is also a 0-basis of K consisting of algebraic integers. Now

contradicts the choice of ((1, ..., en}. Therefore a1 E Z.Similarly, all ai E Z, as desired. 0 106 Commutative Algebra

It is not difficult to verify that any two integral bases of K have the same discriminant (exercise 4). Thus, the discriminant of any integral basis of K is called the discriminant of K.

1.9. Proposition Let (a1,...,a,} be a Q-basis of K = Q(d) with all ai E dK. If A[al, ...,a,] is square-free then (a1, ...,a,}is an integral basis of K. Proof Let {PI, .. . , ,On} be an integral basis and

n ai = ccij,Oj, i = 1, ..., n, cij E Z. j=1

By lemma 1.5, A[al, ..., a,] = (det(cij))2A[,0,,...,,On]. Since A[al, ..., a,] is square-free, it follows that det(cij) = fl. This shows that the matrix (cij) is unimodular and hence {al,...,a,} is a Zbasis for dK by Chapter 1 Lemma 6.5. 0

Example (i) Let K = a(&), where d E Z is square-free and 41(d - 1). Consider a = 5 + &. Then, a has minimal polynomial pol(.) = x2 - z + in Z[z] (Chapter 1 section 3, exercise 6). Hence a E dK,but a q! Z[&] (why?). So dK # Z[&]. Note that (1, a} forms a Q-basis of K and the Q-linear ring monomorphisms K 4 C are defined by al(&) = & and a2(&) = -& respectively. We have

It follows from Proposition 1.9 that (1, a} is an integral basis of K, for d is square-free by the assumption. In section 2, we will establish integral bases for quadratic number fields and cyclotomic fields.

We end this section by giving two easy but important facts concerning the trace function TK/Qand the norm function NK/Qon K (Chapter 1 section 5) determined by the n distinct Q-linear ring monomorphisms aj: K = Q(d) 4 C. With no confusion we write T and N in place of TK/a 107 and NKIQ respectively. By definition, for a E K,

1.10. Lemma Let K = Q(8) and AK be as before. With notation as above, the following hold for a E AK: (i) oj(a)E A = {p E C I ,B is an algebraic integer}, j = 1, ..., n. (ii) T(a),N(a) E AnQ = Z. Proof Exercise.

Exercises 1. Show that if a E C is an algebraic element over Q,then there is some 0 # c E Z such that ca is an algebraic integer. 2. Show that a Zbasis of AK is also a Q-basis for the number field K = a(@. 3. Complete the proof of Lemma 1.5. 4. Show that any two integral bases of the number field K = Q(8) have the same discriminant. (Hint: Use Lemma 1.5 and Chapter 1 Lemma 6.5.) 5. Show that if two number fields K1 and K2 are isomorphic, then they have the same discriminant. 6. Find an example to show that the converse of Proposition 1.9 is not true in general. 7. Let K = Q(8)be a number field where 8 has the minimal polynomial p(z) of degree n. Verify

(Hint: See the proof of Proposition 1.6 and calculate the right-hand side of the equation.) 8. Complete the proof of Lemma 1.10 and answer whether oj(a)E AK or not. (Hint: Consider K = Q(8)where fi is the real root of x3 - 2.) 108 Commutative Algebra

This section aims to establish the integral bases for quadratic number fields and cyclotomic number fields.

(I) dK in K = a(&). Let K = Q(6)be a number field with 6 E dK and [K : Q]= 2. Then K is called a quadratic number field. Since 6 E dK, by Proposition 1.3 its minimal polynomial over Q is of the form

x2 + bx + c, b, c E Z.

Thus, 6 = -bfF.Factorizing b2 - 4c in Z, we may write b2 - 4c = 12d where d is square-free. It follows that 6 = w.Hence, we conclude that K = a(&),where d E Z is square-free.

2.1. Theorem Let d E Z be square-free, K = Q(&). The following hold: (i) If 4 1 (d - 1) then dK = Z[&]. Hence (1, &} is an integral basis for K. (ii) If41(d-l), thendK =Z [i + ;&I. Hence {l,i+$&}isanintegral basis for K.

Proof Let Q E dK. Then Q = r + s&, r, s E Q.Write

(a) If b = 0, then Q E dK implies Q E Z C z[&]. (b) If a = 0, b # 0, since d is square-free, it follows from Proposition 1.3 that Q E dK implies Q E Z[&]. (c) If a # 0, b # 0, we may assume gcd(a,b, c) = 1. The final conclusion will follow from a careful analysis of the coefficients of the minimal polynomial of a: 2a a2 - b2d pa(.) = x - -x+ C c2 . By Proposition 1.3, 2a a'- b2d Q E AK if and only if -, -E C C2 z. The Ring AK in K = Q(0) 109

a2-bzd But cz E Z implies gcd(a,c) = 1 because d is square-free and gcd(a, b, C) = 1. Thus 2a - E Z implies c = 1 or c = 2. C

If c = 1, then a + b& = a E Z[fi]. If c = 2, then a and b are odd as gcd(a, c) = 1, az$d E Z,and gcd(a, b, c) = 1. This implies a E Z , 4((a2- l), and 41(b2 - 1). Writing

- - - - a2 b2d - (a2 1) (b2 l)d - (d - 1) -- 7 C2 4

then E Z implies 41(d - 1). In conclusion, if 4 /y (d - I), then c = 1 and AK = Z[&]; if 41(d - l),then ,B = + has minimal polynomial z2 - IC + = pp(z) E Z[z] and hence ,B E AK. But clearly ,B $! Z[&]. Therefore, c = 2 0

(11) dK in K = Q(w). For n 2 1, consider u,= {Iw€C w"-1=0 I . By Chapter 1 Proposition 3.4, Un is a cyclic subgroup of Cx. U, is called the group of the nth roots of unity. If w E Un is a generator of Un, i.e., Un = (w),then w is called a primitive nth root of unity. For instance, w = e2rri/n= cos % + i sin $. From a first course of group theory one knows that there are cp(n) = nu:=, (1 - $) primitive roots of unity, where n = p;"' . . .pFr is the factorization of n in Z and cp(n) is the Euler's cp-function. If w E U,, the number field K = Q(w) is called a cyclotomic field. That is, the roots of unity are the vertices of a regular n-gon located in the unit circle {a E C I la1 = 1) of the complex a-plane with one vertex of the n-gon at a = 1. In view of the example given in the introductory part of this chapter, we restrict our attention to n = p, a prime number. If p = 2, then U2 = { 1, -1) and Q(w) = Q. So we assume from now on that 110 Commutative Algebm

p is an odd prime number.

2.2. Lemma Let w E Up,w # 1. Then the minimal polynomial p(x) of w is

zp--l + xp--2 + . . . + z + 1.

Proof Write f(z) = xp-' + xp-' + . .. + x + 1. Note that wp - 1 = 0. We have p(z)](xP- 1). Since xp - 1 = (x - l)f(z) and w # 1, it follows that p(x)lf(x). But f(z) is irreducible by Chapter 1 Corollary 2.22. Hence P(X) = f(X).

Let w U, be a primitive pth root of unity. Then Up = (w) = {1,W,W2, ..., wp-l}, and by Lemma 2.1, in C

(1) p(x) = (x - w) (x - w2) . . . (x - wp-l )

Moreover, for K = Q(w), [K : Q]= p - 1, {l,w,..., wPp2}is a Q-basis of K, and all Q-linear ring monomorphisms K -+ C are given by

~j:K+C j = 1,2, ...,p - 1. w t-+ WJ

Our aim is to show dK = Z[w]by making use of the trace function TK/Q and the norm function NK/Qon K (Chapter 1 section 5) determined by oj described in (2) above. As in Lemma 1.10, we write T and N in place of TK/Qand NK/Qrespectively.

2.3. Lemma With notation as above, the following hold: (i) For (Y E dK,T(cY), N(a) E Z. (ii) For i = 1, ...,p - 1, T(w2)= -1, N(w2) = 1. (iii) For a E Q,T(a) = (p- l)a, N(a)= up-'. (iv) For 0 # s E Z, s = ph + T, 0 5 T 5 p - 1,

p- 1, if T = 0, T(wS)= -1, if T # 0.

N(w") = 1. The Ring AK in K = Q(e) 111

(v) For a0 + alw + . . . + up-2wP-2 = a E K,

v- 2 2)- 2

(vi) N(1 - w)= p = T(1- w)= ~(1- u2) = . . . = ~(1-wp-l).

Proof (i) By Lemma 1.10, if a E AK then T(cx),N(cx) E An Q = Z. (ii) By the foregoing (1)-(2),T(w2) = p(w)-1 = -1, and N(w2) = p(0) = 1. (iii) Since all uj's are @linear, this follows from the definition of T and N. (iv) This follows from (ii) and (iii). (v) Since T is a linear function by Chapter 1 Proposition 5.4, this follows from (ii)-(iii). (vi) Note that by previous (1)-(2), N(l - w) = n;l;(l - wj) = p(1) = p. That T(1 - w)= T(1 - w2) = . . . = T(1- wp-') = p follows from (iii)-(iv). 0

2.4. Proposition For K = Q(w), where w is a primitive pth root of unity, the following hold: (i) Let (1 - w) be the ideal of A generated by 1 - w, where A = {p E C 1 /? is an algebraic integer}. Then (1 - w)n Z = pZ. (ii) For cx E AK,T((1 - w)a) E pZ. Proof (i) By Lemma 2.3(vi),

(*) p = ~(i- W) = (1 - w)(i - w2)... (1 - UP-') E (1 - W)n Z, and hence pZ C (1 - w) n Z. Note that pZ is a maximal ideal of Z. If (1 - w)n Z # pZ, then (1 - w)n Z = Z. This implies that 1 - w is a unit of A. Consequently, all 1 - wj, j = 1, ..., p - 1, and also p, are units of A by the formula (*) above. Thus, p-l E An Q = Z, that is impossible. This shows that (1 - w) n Z = pZ. (ii) Let (1-w) be asin part (i) and let ao+ulw+...+a,-2~P-~ = cx E AK. Then by Lemma 1.10,

Oj((1 - w)a) = (1 - W+Jj(cx) E (1 - w) forj = 1, ...,p- 1. Thus, aj((l-w)a) = T((1-w)tr) E (l-w)nZ = pZ by (i) and Lemma 2.3(i). 0

Remark In Proposition 2.4 the reason for using the ring A instead of AK is that we do not know if uj(cx) E AK (section 1, exercise 8). 112 Commutative Algebra

2.5. Theorem For K = Q(w), where w is a primitive pth root of unity, we have dK = Z[w]. Hence {l,w, is an integral basis of K.

Proof Let a0 + alw +... + = LY E dK where ai E 0. Then

(1 - w)a = ao(1 - w) + a1(w - w2) + . . . + aP-2(wP-2 - wP--l), and by Lemma 2.3 and Proposition 2.4(ii),

T((1 - w)a) = aoT( 1 - w) = aop E pZ.

Hence a0 E Z. Note that w-’ = wp-’ E AK. It follows that

(a - a0)w-1 = a1 + a2w + . . . + ap--2wp-3.

Proceeding as with ao, it turns out that a1 E Z.After repeating a similar argumentation p - 1 times successively, we conclude that each ai E Z. This shows that LY E Z[w].

Exercises 1. Let p be an odd prime number and K = Q(w), where w is a primitive pth root of unity. Use the integral basis {l,w, ...,~p-~}of K to show that K has the discriminant

(Hint: Use section 1, exercise 7.) 2. Let p be a prime number, T 2 1, and w a primitive prth root of unity. Show that the minimal polynomial of w over Q is

Furthermore, use a similar argumentation of this section to show that for K = Q(w) the equality dK = Z[w] also holds. Can this be generalized to an arbitrary positive integer n = p;lp? . . .pis (a product of primes)?

3. Factorization of Elements in dK

This section specifies examples of JtK in which some are UFDs and some are not UFDs. The Ring dK in K = Q(0) 113

Let K = Q(d) be a number field, where 6 E AK, [K : Q] = n. Since dK is a Noetherian domain (Theorem 1.2), factorization into irreducible elements is feasible in dK (Chapter 1 Proposition 2.5). In order to realize the factorization of elements in dK, it is necessary to know how to recognize units and irreducible elements in dK. As before we write U(dK) for the multiplicative group of units in dK; and as before, for a E dK, we write T(a)for the trace of a and N(a)for the norm of a. By Lemma 1.10, T(a),N(a) E Z.

3.1. Lemma For a,p E dK, the following hold: (i) a E U(dK) if and only if N(a)= fl. (ii) If a and ,B are associates to each other (see Chapter 1 Definition 2.1), then N(a)= fN(P). (iii) If "(a)(is a prime number in Z, then a is an irreducible element in dK. (The converse of this statement is not true as shown in later Example (ii).)

Proof (i) If a, a-l E dK, then aa-' = 1 implies N(aa-l) = N(a)N(a-l) = 1, and hence N(a) = fl. Conversely, if N(a) = al(a)a2(a)...an(a)= fl, where all..., an are all n distinct Q-linear ring monomorphisms K + C, then, assuming a1(a) = a without loss of generality, we have fl = aaz(a)...an(a). By Proposition l.lO(i), a-1 = *a2(a)...an(a)E dnK = AK, that is, a E U(dK). (ii) This follows from (i). (iii) Suppose N(a) = p is a prime number in 27,. If a = ,BE in dK, then P = IN(a)I = IN(P)IIN(J)J. Thus, plN(P) or PIN([). If plN(P), then N([)= fl and [ E U(dK) by (i). Similarly, if plN([) then ,L? E U(dK). This shows that a has no proper divisors in dK. 0

The proposition given below finds U(dK) in a quadratic field K of complex numbers.

3.2. Proposition Let d E Z, d < 0 and square-free, K = Q(&). Write i=J--r. (i) U(dK) = {fl, *i} if d = -1. (ii) U(&) = {fl, f w, f w2}if d = -3, where w = eZnil3. (iii) U(dK) = {fl}if d < -3.

Proof If a E U(dK) then N(a)= fl by Lemma 3.1(i). Write a = r+s& 114 Commutative Algebra in K. Since d < 0, it follows from Chapter 1 (section 5, exercise 3) that

(0) il = N(a)= r2 - s2d implies N(a)= +l.

If 4 /y (d - l),then a = a + b& with a,b E Z by Theorem 2.1(i), and by (0) above,

(1) N(a)= a2 - b2d = 1.

If 41(d - l),then

2a+b b 2 +5&, U,bEZ,

and by (0) above,

(2) N(a)= 1 implies (2a + b)2 - b2d = 4.

If d = -1, then (i) follows from (1) above. If d = -3, then by (2) above, (2a + b)2 + 3b2 = 4 and this yields

b = 0 implies a = fl;

2a + 1 = $1 implies a = 0 2a + 1 = -1 implies a = -1 b = *Iimplies (2a + bl2 = 1 implies 2a - 1 = +l implies a = 1 2a - 1 = -1 implies a = 0.

This proves (ii). If d < -3 and 4 /I (d - l), then by (1) above, b = 0 and a = fl. If d < -3 and 4[(d- l),then by (2) above, b = 0 and a = *l. This proves (iii) .

Remark If d 2 2 and d is square-free, then K = Q(&) is a subfield of R and U(dK) is no longer necessarily finite. For instance, in K = Q(fi), l+fi=a~U(A~)becauseN(a)=-l(indeed,(l+fi)(-l+fi)=l); and since an > 1 for n 2 1, it follows that {*an},>l- C U(dK), i.e., U(dK) is infinite. For square-free d 2 2, K = a(&),a general fact is that the positive units of dKform a multiplicative group isomorphic to Z (cf. [Sam] section 4.6, Proposition 1). The Ring AK in K = Q(8) 115

For a general structure theory on U(dK), where K = Q(8)is an arbitrary number field, we refer the reader to the famous Dirichlet units theorem (cf. [Sam]).

The following proposition enables us to see how the norm function on a number field K can help to construct a Euclidean function on dK.

3.3. Theorem Let K = Q(8)be a number field, d: = dK - (0). For CY E K, let N(Q)be the norm of a. Then the correspondence #: d$- N

Q "(Q)I

defines a Euclidean function on dKif and only if for any q E K there exists E E dK such that IN(? - [)I < 1. Proof If the function has the stated property, we show that 4 satisfies the conditions (i)-(ii) of Chapter 1 Definition 1.11. By Lemma 1.10, N(Q)E Z for Q E dK. If QIQ' in dK then a' = ~p for some p E dK. Thus, IN(Q')l= $(a') = #(QM(P) = IN(Q)l. IN(P)l and hence #(Q) I#(Q'). This shows that Chapter 1 Definition 2.11(i) holds. To see that Chapter 1 Definition 2.11(ii) holds as well, for alpE d:, set q = 2. Then there is E dK such that IN(q - <)I < 1, that is,

Put r = CY - p<. Then Q = ,BE + r where #(r)< #(P), as desired. Conversely, suppose that # defines a Euclidean function. Let 17 E K. Then there is some c E Z such that cq E dK (section 1, Exercise 1). Now for Q = c~,= c, we have a = pp + r, where p, r E dKwith the property that r = 0 or 4(r) < #(P). If r = 0, then a = Pp, i.e., cq = cp and 77 = p E dK. In this case, taking (' = p fulfils the job. If r # 0, then c # 0 implies

#(?-I = IN(r)l = "(Q -@)I = IN(crl- WL)I < IN(c)I = #(PI. After multiplying both sides of the formula above by IN( :)I, we have /N(q- p)I < 1. This shows that the choice < = p is a desired object. 0

Example Let r E Q be a rational number and [r]the largest integer smaller 116 Commutative Algebra than or equal to T. Then [TI 5 T < [TI + 1. Thus, 1 1 either IT - [TI( 5 -, or (T - [TI - 11 5 -. 2 2 With the help of this observation, the following two examples are easily presented.

(i) Let K = a(&) with d = 2,3. Then dK= Z[&]. For any T+S& = q E K and a+b& = < E dK, )N(v-<))= ((r-a)2-d(s-b)2(. So it is possible to choose a+b& = < E AK such that IN(v-<)( = I(~-a)'-d(s-b)'1 < 1. This shows that Z[d]is Euclidean and hence a UFD. (ii) Let K = Q(z) where i = n.Then AK = Z[i].For any r+si = 7 E K, we may find a, b E Z,such that (T - u)~,(s - b)2 5 i. Thus, for < = a + bi we have "(7 -()I = I(T - a)2 + (s - b)2( < 1. This shows that Z[i]is Euclidean and hence a UFD. Example (iii) Let K = a(-). Then dK = Zi-1, and by Proposition 3.2, U(dK) = {il}. Using Proposition 3.1, one checks that 2, 3, 1 f fl are irreducible elements but not primes in AK, because 6=2.3= (l+q.(l-q.

This shows that AK is not a UFD. Moreover, N(l+-) = 6. This shows that the converse of Proposition 3.l(iii) is not true.

Without proof we mention the following result from algebraic number theory (for instance, cf. [ST]).

3.4. Theorem Let d E Z be square-free, K = Q(4).If d < 0, then dK is Euclidean if and only if d = -1, - 2, - 3, - 7, - 11. And for each of these listed numbers, the Euclidean function is the one defined in Theorem 3.3. 0

Finally, the reader is referred to [Edw] for a detailed argumentation about the fact that for the cyclotomic field K = Q(w), where w = e2?ri/23, AK is not a UFD. The Ring AK in K = Q(29) 117

Exercises 1. Let K = Q(i), i = m. Which of 2, 3, 5 and 7 are reducible in dK? Which of 1 + i, 3 - 7i, and -4 + 5i are irreducible in dK. 2. Let K = ~(fi). (a) Show that a = 5 + 3fi is irreducible in dK, and that /3 = 7 + 4fi is a unit in dK. (b) Use part (a) to show that 71 + 41fi and 5 + 3fi are associates to each other. (c) Is 2 a prime in dK? (Hint: Consider 6 = 2 * 3 = (3 + G)(3 - fi), 2 = (5 + 3&)(-5 + 3fi).) 3. Use the factorizations 18 = 2.3.3 = (1 + J=is). (1 - m),

10 = 2.5 = (5 + fl).(5 - a) to show that dK is not a UFD for K = Q(m),K = Q(m)and K = ~(a),respectively. 4. Show that dK is Euclidean where K = Q(&). (Hint: Consider the function 4: dc + N with

and show that for anyrfsfi = 77 E K,there is a+b($++fi) = < E dK such that 4(q - 6) I i.)

4. From AK to Dedekind Domains

Let K = Q(8)be a number field, [K : Q] = n, and dKits ring of algebraic integers. We have seen in section 3 that dK is not necessarily a UFD. In this section we clarify how far is dK from being a UFD, by studying the structure of ideals in dK. More precisely, we show that every nonzero ideal I of dK can be expressed as a product of finitely many prime (indeed maximal) ideals uniquely up to the order of factors, that 1 is generated by at most two elements, and that dK is a UFD if and only if it is a PID. We start with a useful fact. 118 commutative Algebra

4.1. Lemma Let I be an ideal of dK. If 0 # a E I,then 0 # N(a)E InZ, where N(a)stands for the norm of a.

Proof Let 01,..., an be the n distinct Q-linear ring monomorphisms K + C, where al(a)= a. By Lemma 1.10, a1(a)a2(a)...an(a)= N(a) E Z,and since a # 0, a2(a)...an(a)= CU-~N(~) E K n d = dK. It follows that N(a)E I n Z. 0

4.2. Theorem (i) Every nonzero prime ideal of AK is a maximal ideal (hence nonzero and minimal). (ii) If P is a nonzero prime ideal of AK, then the localization (dK)pof AK at P is a DVR.

Proof (i) Let P be a nonzero prime ideal of dK and 0 # cy E P. By Lemma 4.1, 0 # N(a) E P n Z. Let N(a)= q and (q) the ideal of dK generated by q. Then (q) C PI and we have the natural onto ring homomorphism dK/(q) + dK/P. If {el, ...,en} is an integral basis of K (or by Theorem 1.2, a Z-basis of dK), then it is easy to see that every element of AK/(q) is of the form

that is, dK/(q) has at most qn elements. Consequently, AK/Pis a finite domain and hence a field (Chapter 1 Proposition 0.1). It follows from Chapter 2 Proposition 1.2(ii) that P is maximal. (ii) This follows from part (i) and Chapter 3 Corollary 4.6. 0

4.3. Proposition Let I be a nonzero ideal of dK. Then there exist prime ideals PI,..., P, such that PI . ..P, I. Proof If I is a prime ideal, it is done. Suppose that I is not a prime ideal. If the assertion was not true for I, then the set

R = (ideals of dK not containing any product of prime ideals} # 8.

Since dK is Noetherian, 52 has a maximal member with respect to c,say M. Thus M is not a prime ideal, and there are ideals J1,J2 of AK with JlJ2 g M but J1 M,52 M.Put Wi= J1 + MI W2 = Jz + M.Then M c Wl, M c W2,and by the choice of M,Wl, W2 # R. Thus, both W1 and W2 contain some product of prime ideals. But Wl W2 M and this is a contradiction. Therefore R = 8 and the assertion is proved. 0 The Ring AK in K = Q(@) 119

Our aim is to show that every nonzero ideal I is a product of finitely many prime ideals. In doing so, we need to study special dK-submodules of K. Note that since dK is a subring of K, K forms an dK-module in a natural way (Chapter 1 section 7).

4.4. Definition An &-submodule J of K is called a fractional ideal of AK if there exists some 0 # c E AK such that CJ G AK.

4.5. Lemma (i) Every ideal I C dK is a fractional ideal of dK. (ii) An dK-submodule J is a fractional ideal of AK if and only if there is an ideal I C dK and some 0 # c E dK such that c-' I = J. Proof Exercise. 0

Example (i) If dK is a PID, then every fractional ideal J of d,y is of the form J = $dK, where $ E K. To see this, note that if c E dK is such that CJ C AK, then CJ is an ideal of dK, and hence CJ = ddK for some d E AK, that is, J = $AK.

Put F(AK)= { J c K I J a fractional ideal of dK and define the multiplication on ~(AK)by setting

JiJ2 = {~ZYI z E 51, Y E J2>, J1, J2 E F(dK). The reader will be asked in later exercise 4 to check that this operation is well-defined and it makes F(dK) into a commutative associative semigroup with the identity element dK. We now proceed to show that F(dK) is indeed a group.

For each ideal I of dK, set

Then it is clear that dK C I-'.

4.6. Proposition With notation as above, the following statements hold: (i) If I is an ideal of dKthen I-' E F(dK). Moreover, IT-' is an ideal of dK. 120 Commutative Algebra

(ii) If M is a maximal ideal of dK, then dK # M-'. Consequently, dK # 1-l for any proper ideal I of dK. (iii) Let I be an ideal of dK, I # (0). If SI G I for some subset S K, then S C dK. (iv) If M is a maximal ideal of dK, then MM-l = dK.

Proof (i) If 0 # c E I then cT-' G dK. So I-' E F(dK). That 11-' is an ideal of dK may be verified directly. (ii) Let 0 # a E M and (a)the ideal of dK generated by a. By Proposition 4.3, PI... P, C (a) C M for some prime ideals PI,..., P,. Since M is a prime ideal, M = Pi for some i by Theorem 4.2. Assume Pi = PI without loss of generality. Choosing T to be smallest, we have P2 .. . P, 9 (a). Taking b E PZ . . . P,, b @ (a),we see that

bM = bP1 C (a)= adK implies a-lb E M-', b @ (a)= adK implies a-lb @ dK.

Thus, if I is a proper ideal of dK, then I C M for some maximal ideal M. It follows that M-l G I-l and I-' # dK. (iii) Since dK is Noetherian, assume I = czldKwi, vi E I. Let s E s. Then SI 5 I yields

This means that the system of linear equations

(a11 - S)Vl + a12212 + . . . + a1m21, = 0 a21211 + (a22 - s)v2 + . . . + a2m21, = 0 1;am1211 + am2212 + . . . + (amrn- s)21, = 0 has a nonzero solution (211, ..., vm). Write A for the coefficient matrix of the system. Working over the field K, A is singular and hence det(A) = 0. This shows that s is a zero of a monic polynomial in dK[2]. Thus, s E dK by Lemma 1.1. (iv) By (i), MM-' is an ideal of dK. Since dK C M-l, M C MM-l C dK. The maximality of M entails M = MM-l or MM-l = dK. If M = MM-l, then (iii) implies M-' C_ dK, contradicting (ii). It follows that MM-l = dK. 0

4.7. Theorem With notation as before, the following hold: The Ring AK in K = Q(8) 121

(i) If I is a nonzero ideal of dK, t,hen 11-' = dK. (ii) Every nonzero fractional ideal J E F(dK) has an inverse J-l E F(dK), i.e., JJ-l = dK. Therefore, F(AK)is a commutative multiplicative group.

Proof (i) Suppose that I1-l # dK, and that I is maximal among all such ideals (with respect to C). Let M be a maximal ideal of AK containing I. Then

dK c M-' c 1-l implies I c IM-' 5 II-' c dK.

By Proposition 4.6(ii-iii), I # IMP'. Note that IM-' is an ideal of dK. By the choice of I,

(1M-l) (IM-')-' = dK implies M-' (IMP1)-' 2 I-' and this shows that dK = (1M-l) (IM-')-' 2 IT-' 2 dK, that is, II-' = dK, a contradiction. It follows that we must have 1I-l = dK. (ii) This follows from part (i) and Lemma 4.5(ii). 0

Example (ii) Let K = a(-). Then dK = Z[fl]. The ideal I = (2,l - G)of dK has inverse

We are ready to reach the unique factorization theorem of ideals in dK.

4.8. Theorem Every nonzero ideal of dK can be written as a product of finitely many prime ideals, uniquely up to the order of factors.

Proof If there is some nonzero ideal I of dK which cannot be written as a product of finitely many prime ideals, we may assume I is maximal among all such nonzero ideals (with respect to C). Then I is not a prime and it is properly contained in some maximal ideal M. Thus, M-' 5 I-', I C IMP' dK, and by Proposition 4.6(ii-iii), I # 1M-l. Since 1M-l is an ideal of AK, by the choice of I we have IM-' = PI . . . P, for some prime ideals PI,..., P,. But this yields I = MPl... P,, a contradiction. 122 Commutative Algebra

Therefore, every nonzero ideal of dKcan be written as a product of finitely many prime ideals. Now if I = pl. . . Pr = Q1 . . . Qs = J for prime ideals Pi, ..., Pr, &I, ..., QS, then I C PI implies Qi PI for some Qi.So PI = Qi by their maximality. Assuming Qi = &I and multiplying I and J by PF', we have P2.. . pr = QZ. . . Qs and the proof of uniqueness may be done by an inductive demonstration. 0

Example (iii) In dK = Z[G]C K = a(-), we know that 6 = 2 x 3 = (1 + -)(1 - fl).But we have a unique factorization of (6) into prime ideals:

Moreover, in dK = Z[m]C K = Q)(m),we know that 18 = 2 x 3 x 3 = (1 + m)(1 - m).But we have a unique factorization of (18) into prime ideals:

The reader is asked to check that each ideal used in the above factorizations is prime.

Finally, with the help of Theorem 4.8 we are ready to see how far is dK from being a UFD.

4.9. Theorem dK is a UFD if and only if it is a PID. Proof The "if" part is known. Now suppose dK is a UFD. In view of Theorem 4.8 it is sufficient to show that every prime ideal of dK is principal. If P is a nonzero prime ideal of dK, then by Lemma 4.1 we may pick up some integer 0 # m E P and consider its unique factorization into irreducible elements in dK:

m=YlY2"'Ys.

Then some ~i E P or (ri)G P. But each yi is a prime because dK is a UFD (Chapter 1 Theorem 2.9). It follows that (yi) is maximal and hence (Ti) = P. 0

The next theorem will show that dK is only one pace away from being a PID (just as indicated in the diagram given in the preface of this book!). The Ring AK in K = Q(8) 123

We demonstrate this result through several technical preliminaries.

4.10. Lemma (i) If I and J are ideals of dK such that I + J = dK, then InJ=IJ. (ii) Let PI,..., P, be distinct prime ideals of dK, then PI . . . P, = nL='=,Pi. Proof Exercise. 0

4.11. Proposition (i) Let I be a nonzero ideal of dK and P a prime ideal (hence maximal) of JtK. Then for any integer s >_ 1 there exists b E I such that bI-' + Ps = dK. (ii) Let I, J be nonzero ideals of dK, and J = P:' . . . PTST the unique prime factorization of J in dK, where T 2 1, si 2 1, i = 1, ..., T. If there is some 0 # b E n;='(I - IPi), then bl-' + J = dK. (iii) Let I and J = Pfl. . . P:T be as in part (ii), then there is some 0 # b E nZ=.=,(I- IPi).

Proof (i) Note that IP c I but IP # I (otherwise P = I-l(IP) = I-'I= dK). Let 0 # b E I - IP. Then (1) bI-' 9 P.

If bl-' = dK, then bI-' + P" = dK. If bI-' # dK, then by the above (1) we have,

(2) bI-l+ P = dK.

It follows that

bI-lP + P2 = P, and by substitution in (2), bl-' + P2 = dK.

After repeating the same process for s - 2 times we arrive at

bI-'+ Ps = dK.

(ii) By the assumption, bI-'Pi 9 Pi, i = 1, ..., T. If b1-l = dK, then bI-' + J = dK. If b1-l # dK, then a similar argumentation as in the proof of part (i) shows that

b1-l + P," = dK, i = 1, ..., T. It follows that 124 Commutative Algebra

After substitution step by step, we arrive at

bI-l + J = bI-' + P:' 1.. P,? = dK.

(iii) For each i = 1, ..., r, let Ii = IP1 . . . P,PT-'. Then

(3) Ii = IP1...Pi-lPi+l...Pr g IPj, j #i.

Since Ii # IiPi (otherwise Pi = dK),there is bi E Ii - IiPi, i = 1, ..., r. Set

b = bl + . . . + b,.

Then b E I and we claim that b @ IPi, i = 1, ..., r. To see this, suppose that b E IPj for some j. Then by the construction of b and (3) above, bj E IPj would hold. But we know that bj@IjPj=(IPi...P,P~~l)Pj=IPl...P,.

By Lemma 4.10(ii), this yields r (4) bjI-l Pl...P,.= nPi. i= 1

Moreover, bj E Ij implies

(5) bjI-' C I-lIj = PI... Pj-lPj+l ...P,. Thus, by (5) above, bj E IPj, or equivalently, bjI-l g Pj would yield r bjI-l c pj n p1... pj-lpj+l... P, = npi, i=l contradicting the foregoing (4). Therefore, we must have b E n;='(I- IPi).

4.12. Theorem Let I be a nonzero ideal of dK,0 # a E I. Then there is some b E I such that I = adK + bdK. Proof For 0 # a E I, set J = aI-'. Then by Proposition 4.ll(ii-iii), there is b E I such that bI-l + aI-' = dK, and it follows that I = bdk + adK.

To measure the extent to which ideals in dK are not principal, another topic in algebraic number theory is to study the class-group of K,that is the The Ring AK in K = Q(8) 125 quotient group of 3(d~)by the (normal) subgroup of principal fractional ideals.

From dK to Dedekind domains Historically in the literature, if a normal Noetherian domain R also has the property that every nonzero prime ideal is maximal, then R is called a Dedekind domain after the mathematicians, such as Kummer, Dedekind in the 19th century and Noether in the 1930~~who developed the ring theoretic approach to the study of dK. With no modification, all results about dK given after Theorem 4.2 in this section hold for a Dedekind domain R, in particular, 0 every nonzero ideal of R has a unique factorization into primes; every ideal of R can be generated by two elements; and 0 R is a UFD if and only if it is a PID. Moreover, there is also the class-group theory associated to the ideals of a Dedekind domain that has an intimate connection with the divisor class-group of an algebraic curve in algebraic geometry.

Exercises

1. Let a E dKbe an irreducible element but not a prime. If (a)= PI + . . P, is the unique factorization of (a)into prime ideals in dK, show that none of Pi is principal. 2. Show that (3,l + G)is a prime ideal in dK = z[G],and that (1 + a)is a prime ideal in dK = Z[m]. 3. Complete the proof of Lemma 4.5. (Hint: To reach (ii), note that if CJ c dK for some 0 # c E dK then CJ is an ideal of dK.) 4. Show that the operation on 3(d~)is well-defined and it makes 3(d~) into a commutative associative semigroup with the identity element dK. 5. Complete the proof of Lemma 4.10. This page intentionally left blank Chapter 5 Algebraic Geometry

This chapter demonstrates how Noetherian rings, field extensions, local rings, DVRs, normalization and localization stem naturally from algebraic geometry.

Throughout the text K[zl,..., 2,] denotes the polynomial ring in z1, ..., z, over a field K. If n = 2 we use K[z,y] instead of K[21,22], and similarly we use K[z,y, z] in place of K[Q,22,231. Moreover, let

A" = A"4 - P = (a1,..., a,) ,...,a, E K 1 stand for the n-dimensional ufine space (or afine n-space) over K. An element P E A" is called a point, and if P = (al,..., a,) then ui is called the ith coordinate of P. It is clear that in the case where K = R, A' = R is the real affine line, A2 = R2 is the real affine plane, and A3 = R3 is the real affine space.

1. Finite Field Extension and Nullstellensatz

Let f E K[zl,..., 2,]. Then f defines a function

4f: A"=AE + K P = (Ul, ..., a,) H f(P)= f(a,..., a,)

If f = c E K is a constant, then $f(P) = c defines a constant function. If f,g E K[zl,..., z,], then, as usual functions, (4f+#g)(P) = 4p(P)+4,(P) and (4f49HP) = 4f(P)49(P).

127 128 Commutative Algebra

1.1. Definition Let T be a subset of K[zl,..., z,], and V(T) = { P E A" I dP(p)= f(p)= 0, for all f E T 1, that is, V(T) is the set of common zeros of all polynomials in T, or equivalently the set of solutions of the system of polynomial equations

We call V(T) an afine algebraic set (or just an algebraic set) of the affine n-space A" = A;. If T = {fl, ..., fs} is a finite subset, we also write

V(T) = V(f1, ...7 fs). If f E K[xl,..., z,] is not a constant, the algebraic set V(f) is called the hypersurface defined by f;if deg(f) = 1, V(f) is called a hyperplane in A". A hypersurface in A2 is called a plane curve; and a hyperplane in A2 is called a line.

Example (i) A" and 0 are algebraic sets (indeed A" = V(O), 0 = V(1) = V(c) for any c E KX). (ii) Conics in IR2 defined by quadratic polynomials in R[z,y] are plane curves; while C = V(y2 - x3) c A: is known as the cuspidal curve and C = V(y2 - z3 - z2)c A: is known as the nodal curve. Moreover, V = V(y - x2,z - x3) c A; is known as the twisted cubic in IR3.

1.2. Proposition (i) Let T be a nonempty subset of K[xl,..., zn] and I = (T) the ideal generated by T in K[zl,..., z,]. Then V(T) = V(I) = V(f1,..., fs) for finitely many fl, ..., fs E T. (ii) If TI C T2 are subsets of K[zl,..., z,], then V(T1) 2 V(T2). (iii) If Y1,Y2 are algebraic sets, then so is Y1 U Y2. Consequently, the union of any finitely many algebraic sets is an algebraic set. (iv) If {Y,}iG~is a family of algebraic sets, then nicJyZ is an algebraic set.

Proof The existence of fl, ..., fs follows from the fact that K[Q,..., zn]is Noetherian (Chapter 1 Theorem 1.3). All other assertions may be directly verified by means of part (i) and Definition 1.1. I3

Proposition 1.2(i) establishes the initial connection between ideals and algebraic sets that makes the flexibility of finding easier defining equations of a given system of polynomial equations. Algebraic Geometry 129

Example (iii) Let V = V(fl, f2) and I = (fl, f2), where fl = z2 - zy - y2 + z2, f2 = z2 - y2 + z2 - z E R[z,y,z]. Then since I = (fl,f2) = (fi, fi + cfi) for any nonzero c E R, it follows that, after setting c = -1, V = V(f1, z - zy), i.e., every point P E V is of the form P = (z,y,zy). (iv) Since for ai E k, V(z1 - al, ..., z, - a,) = {P = (al, ..., a,)} c A", it follows from Proposition l.a(iii) that any finite subset of A" is an algebraic set.

For ideals in K[zl,..., z,], the operations sum, product, and intersection correspond to operations of algebraic sets.

1.3. Proposition Let I and J be ideals in K[zl,..., x,]. The following hold: (i) V(I + J)= V(I) n V(J). (ii) V(I . J) = V(I) U V(J). (iii) V(I n J) = V(I)U V(J). Proof Exercise.

Now, if we start with a subset Y E A", then Y may be associated to a set of polynomials

It is easily seen that I(Y) is an ideal of K[zl,..., z,]. I(Y) is called the ideal of Y in K[zl,..., x,]. As we will see from now on, it is this ideal that makes the essential connection between algebraic structure theory and the geometry of algebraic sets.

1.4. Proposition (i) If J is an ideal of K[zl, ..., z,], then J C_ I(V(J)), and V(I(V(J)))= V(J). (ii) For any subset Y g A",V(I(Y)) = B is the smallest affine algebraic set containing Y. (iii) For any two subsets Y1, Y2 of A",we have I(Y1 U Y2) = I(Y1) n I(Y2). (iv) Let V and W be affine algebraic sets in A". Then: (a) V C W if and only if I(V) 2 I(W). (b) V = W if and only if I(V) = I(W). Proof Exercise. 0 130 Commutative Algebra

Example (v) If K is infinite then I(A") = (0) (see later exercise 2). If P = (a1,a2,..., a,) E A", then I({P})= (z1-al,z2-a2, ..., z,-a,) which is a maximal ideal of K[zl,..., z,] (Chapter 2 section 1, Example (iii)).

(vi) For V = V(y - x2) c A;, I(V) = (y - z2). This may be verified by noticing that any monomial z*yP in R[z,y] can be written as Ci(y - z2)hi + za+2P where hi E R[z,y], and any point P E V is of the form P = (z,z2), z E R.

(vii) Note that (z2,y2)is properly contained in I(V(z2,y2))= (z,y). So in general J is not necessarily equal to I(V(J)).

So far we have built two mappings {ideals} 3 { affine algebraic sets}

JH V(J) and {affine algebraic sets} A {ideals} V - I(V) Note that by Proposition 1.4 the second mapping is injective. The first mapping, however, is not necessarily injective: diflerent ideals can define the same algebraic sets. For example, the ideals (z) and (x2) are different in K[z], where K is an arbitrary field, but V(x) = V(z2) = (0) in Ak. Also consider the ideals of R[x]:

J1 = (1) = R[z], 52 = (1 + z2), 53 = (1 + z2 + z4).

We see that Ji # Jk if i # k. But V(1l) = V(l2) = V(13)= 0. Similarly, (z3- 1) # (zy2 - y2 + z - 1) in R[z, y], but both ideals correspond to the same algebraic set ((1,y) I y E EX}. The wonderful thing is that if the ground field K is algebraically closed then the Hilbert's Nullstellensatz solves the problem entirely.

1.5. Theorem (weak Nullstellensatz) If the field K is algebraically closed, then there is a natural one-to-one and onto correspondence A" - m-SpecK[xI, ..., zn] (a1,...,a,) H (21 - a1, "',Z, - a,) Algebraic Geometry 131

Proof For each point P = (al,..., a,) E An, that (51 -a], ..., z, - a,) is a maximal ideal of K[zl, ..., z,] follows from Chapter 2 (section 1, Example (iii)). Conversely, let M E m-SpecK[zI, ..., z,] and write K[&,...,%] = K[zl,..., z,]/M,where each Tiis the image of zi in K[zl,..., z,]/M. Then the natural K-algebra homomorphisms - K Lf K[q,..., z,] IK[Tl, ..., z,] yield a finitely generated field extension K c K[T1,..., Cn] (here K and x(K)are identified). By Chapter 3 Theorem 2.5, K[31,..., ?En] is algebraic over K. But K is algebraically closed by the assumption. Hence K = K[Tl,..., En].Writing Ti = bi E K, i = 1,..., n, then x is just given by

fb1, ...,4H f(h, *..I bn) and it follows that Kerx = M = (z1 - bl, ..., z, - b,) by Chapter 2 (section 1, Example (iii)). That the correspondence is one-to-one and onto is easily seen now.

1.6. Theorem (Nullstellensatz) Let K be an algebraically closed field, and let J be an ideal of K[q,..., z,]. (i) If J # K[q,..., z,], then V(J)# 0. (ii) I(V(J))= a,where fl = {f E K[zl,..., x,] 1 f" E J for some m 2 l} is the radical of J (Chapter 2 section 1, exercise 2). Proof (i) By the assumption and Theorem 1.5,

J & M = (XI- all ..., Z, - a,) for some P = (al,..., a,) E An. Thus, P E V(M) V(J). (ii) By the definition of a,the inclusion fi I(V(J))is clear. To reach the inclusion 2 I(V(J)),let J = (fl, ..., fs), and 0 # f E I(V(J)). Then S = (1, f,f2, ..., f", ...} is a multiplicative set of K[xl,..., zn].Con- sider the ring of fractions K[zl,..., z,]~.By Chapter 2 (section 3, Example

(4)I

Note that K[zl,..., zn]s# (0). So (yf - 1) # K[zi, ..., zn][y]. But we claim 132 Commutative Algebra for, Theorem 1.5 asserts that there is no maximal ideal of K[zl,..., zn,y] containing J'. Thus,

S (1) 1 = cPi(51,...,z,, Y)fi +q(zl,.-,z,,Y)(l- Yf) i= 1 for some polynomials pi,q E K[q,..., x,,y], NOW, going back to K[x1,..., x,]~via y H $ (see Chapter 2 (section 3, Example (ii))), the relation (1) above implies that

After clearing all denominators of Eq. (2) by a suitable power fm, we obtain

S (3) i=7 for some polynomials gi E K[zl,..., xn],as desired. 0

If an ideal J c K[xl,..., z,] has the property that J = a,then J is called a radical ideal. It is clear that I(V(J)) is radical, and every prime ideal is radical. But not every ideal is radical, as easily illustrated by the ideal (z2,y2) c R[x,y] (also see later exercise 9). Working with radical ideals, the Nullstellensatz establishes the following perfect correspondences.

1.7. Theorem If K is algebraically closed, then the mappings {affine algebraic sets} -I {radical ideals) and {radical ideals} -V {affine algebraic sets} are inclusion-reversing bijections which are inverses to each other. I7

Theorem 1.6 answered when an algebraic set is nonempty, provided the ground field is algebraically closed. The next proposition answers how to algebraically recognize the finiteness of an algebraic set V(J). Algebraic Geomety 133

1.8. Theorem Let J be an ideal of K[zl,..., x,], V = V(J), and R = K[Xl, ...,xn]/J. (i) If dimKR < 00, then V is finite. (ii) If K is algebraically closed and V is finite, then dimKR < 00. Proof (i) Let dimKR = rn and Zi the image of xi in R, i = 1, ..., n. Then AmZy + X,_I?E~-~ + . . . + X1Zi + XO = 0, for some Xi E K. This implies the one-variable polynomial C,"=,xi = f(q)E J. Hence, f (xi)vanishes at every point of V. Since f (xi) can have only finitely many zeros in K, it follows that the points of V have only finitely distinct ith coordinates, i = 1, ..., n. This shows that V is finite. (ii) Note that K is algebraically closed. If V = 0, then 1 E J by Nullste- lensatz, and hence dimKR = 0. Suppose V # 0, and let aij be the distinct ith coordinates of all s points in V,i = 1, ..., n, j = 1, ..., s. Define the one-variable polynomials

S = n(xi- aij), i = 1,..., n. j=1

Then fi vanishes at every point of V. By Nullstellensatz, fyi E J for some mi 2 1. Noticing that fi is a monic polynomial in one variable, a formal division on elements of K[q,..., xi-1, xi+l, ..., xCn][xi] by fi, i = 1, ...,n, shows that dimKR 5 00.

Exercises 1. Prove Propositions 1.3-1.4. 2. Let f be a nonconstant polynomial in K[xl,..., x,], where K is algebraically closed. Show that A" - V(f) is infinite if n 2 1, and V(f) is infinite if n 2 2. Conclude that the complement of any algebraic set is infinite. (Hint: Note that K is infinite by Chapter 1 (section 3, exercise 5). f defines the zero function on K" if and only if f = 0 in K[zl,...,xn].) 3. Show that if K is infinite then I(A") = (0). Is this still true if K is finite? 4. Consider the plane curve C = V(f) c A& where f is a polynomial of degree n in K[z,y]. If L is a line in A$ and L $ C, show that L n C is a finite set of no more than n points. (Hint: Suppose L = V (y - (ax + b)), and consider f(x,ax + b) E K[x].) 5. Let V = V(flg) c A:, where f = x2 - y + 1, g = y + x2 - 5. 134 Commutative Algebra

(a) Show that V = V(z2 - y + 1,z2 - 2). (b) Use part (a) to determine V. 6. This exercise is to demonstrate that factorization of polynomials may help to determine an algebraic set. (a) Show that if g E K[z~,..., zn] factors as g = 9192, where K is a field, then for any f,V(f,g) = v(f, 91) u v(ft 92). (b) Show that in A$, V(y - x2,zt - y3) = V(y - x2,xz- z4). (c) Use part (a) to determine the algebraic set in part (b). 7. For V = V(zy - 1) c A;, show that I(V) = (zy - 1). 8. For the twisted cubic V = V(y - z2,z - z3)c A:, show that I(V) = (y-x2, z-z3). (Hint: Any monomial ~~yflz?in R[z, y, z] can be written as ci(y- z2)hi+ C .(z - z3)gj + za+2fl+3Y, and every point P E V is of the form P = (z, z4 ,z3).) 9. Let J = (z,y3 + 1) c R[z,y]. Show that J # fi, and that dimRR[z, y]/ais finite. 10. Let I = (2, y) c K[z,y], where K is a field. Show that dimK(K[x,y]/I”) - “(“:l) for all n 2 1. 11. Let J be a proper ideal of K[z1, ..., z,]. If K is algebraically closed, show that fi = I(V(J)) = nM, where M runs over all maximal ideals containing J.

2. Irreducible V and the Prime I(V)

To study the points of an algebraic set in an “analytic-like” way, we need irreducible algebraic sets and use only “polynomial functions and rational functions” that respect the Zariski topology on the afine n-space An = AnK over a field K (this viewpoint will he clarified in the next section). Some basic notions and examples on topological space are given in the appendix of this section.

2.1. Definition The Zariski topology on A” is defined by taking the open subsets to be the complements of affine algebraic sets. For any subset Y c A”, the Zariski topology on Y is the induced topology.

Example (i) Let us consider the Zariski topology on the affine line A1 = K. Every ideal of K[z] is principal, so every algebraic set is the zero locus of a single polynomial. Thus the algebraic sets in A’ are just the finite subsets Algebraic Geometry 135

(including the empty set) and the whole space (corresponding to f = 0). Consequently the open sets are the empty set and the complements of finite subsets.

(ii) In general, the Zariski topology on A" (TI 2 1) is not the product topology of the Zariski topology on A'. For example, if we identify A' x A' with A2 and consider any curve C = V(y - f(z))in A', where f(z)is a polynomial in K[z]of degree 2 1, then from (i) above it is easy to see that the open subset U = A2- C in A2 cannot be a product of two open subsets of A'.

2.2. Definition A nonempty subset Y of a topological space X is irreducible if it cannot be expressed as the union Y = Yl U Y2 of two proper subsets, each of which is closed in Y(where Y has the induced topology). The empty subset 0 is not considered to be irreducible. If X is not irreducible then it is reducible.

Example (iii) If K is infinite, then A:, is irreducible, for its only proper closed subsets are finite sets. (iv) Similarly, by Proposition 1.3(ii) and (section 1, exercise 3), An is irreducible provided K is infinite.

(v) V = V(32+y-1, y2 -2) is reducible in A;, for V consists of two points. More reducible algebraic sets may be obtained via Proposition 1.3(ii).

The following proposition states two basic properties of irreducible subsets in a topological space. The first one plays a key role in algebraic geometry because of previous Example (iv).

2.3. Proposition (i) Any nonempty open subset of an irreducible space is dense and irreducible. (ii) Y is an irreducible subset of the topological space then its closure - If X, Y in X is also irreducible. Proof (i) Let U, V be nonempty open subsets of the irreducible space X. Then U n V = 0 leads to X = (X - U)U (X- V), a contradiction. Hence U is dense. If U = U1 U Uz for closed U1, U2 in U, then there are proper closed V1, V2 in X such that V1 n U = U1, V2 n U = U2. But then X = (VI U V2) U (X - U),both V, UV2 and X - U are proper closed subsets. 136 Commutative Algebra

Once again this is a contradiction. (ii) Exercise. 0

To determine the irreducibility of an algebraic set V c An in a purely algebraic way, let us write

K[V] = K[q,..., z,]/I(V).

K[V] is called the Coordinate ring of V (a remark on this name is given in the beginning of section 3).

2.4. Theorem Let V C An be an affine algebraic set. Then V is irreducible if and only if I(V) is a prime ideal if and only if its coordinate ring K[V] is a domain.

Proof First, assume that V is irreducible and let fg E I(V). Set V1 = V n V(f) and V2 = V n V(g). Then fg E I(V) implies that V = V1 U V2. Since V is irreducible, we have either V = V1 or V = Vz. Say the former holds, so that V = Vl = V n V(f). This implies that f vanishes on V, so that f E I(V). Thus, I(V) is prime. Next, assume that I(V) is prime and let V = V1 U V2. Suppose that V # V1. We claim that I(V) = I(V2). To prove this, note that I(V) c I(V2) since V2 c V. For the opposite inclusion, first note that I(V) is properly contained in I(V1) since V1 is properly contained in V. Thus, we can pick f E I(V1) - I(V). Now take any g E I(V2). Since V = V1 U Vz, it follows that fg vanishes on V, and, hence, fg E I(V). But I(V) is prime, so that f or g lies in I(V). We know that f $ I(V). So, g E I(V). This proves I(V) = I(V2), whence V = V2 by Proposition 1.4(iv)(b). Therefore V is an irreducible algebraic set. 0

In the “algebra-geometry” dictionary of section 1, algebraic sets in An are associated with ideals in K[zl,..., z,]. Note that the ring homomorphism K[zl,..., z,] 4 K[zl,..., z,]/I(V) = k[V] yields the following one- to-one and onto mappings between the given sets: (a) {ideals of K[zl,..., z,] containing I(V)} and {ideals of K[V]}. (b) {radical ideals of K[zl,..., x,] containing I(V)} and {radical ideals of K[VI}. (c) {prime ideals of K[x1, ..., z,] containing I(V)} and {prime ideals of K[VIl. (d) { maximal ideals of K[zl,..., zn] containing I(V)} and {maximal ideals Algebraic Geometry 137

of K[V]}. Consequently, the following corollary of Theorem 1.7 is obtained.

2.5. Corollary Let K be an algebraically closed field and let V c A" be an affine algebraic set. (i) The correspondences

affine algebraic subsets { wcv I are inclusion-reversing bijections and are inverses to each other, where IV denotes taking ideal I(W) with W as indicated above and VV denotes taking algebraic set V(J) with J as indicated above; we use IV, respectively Vv,just for the restrictions of I and V respectively. (ii) Under the correspondence given in part (i) , irreducible algebraic subsets, respectively points of V, correspond to prime ideals, respectively correspond to maximal ideals of K[V]. Proof Exercise. 0

Example (vi) If we plot the twisted cubic V = V(y - x2,z - x3) c A; in a limited space, then an intuitive feeling tells that V is irreducible.

To convince ourselves, however, Theorem 2.4 requires proving that I(V) 138 Commutative Algebra is a prime ideal of R[z,y,z], that is, if fg E I(V), we have to show that f E I(V) or g E I(V). For any P = (a, b, c) E V, note that setting X = a yields

a=X, b=X2, c=X3.

Conversely, any X E A' = R determines a point of V as defined above. Thus, we have f(X,X2,X3)g(X,X2,X3) = 0 for all X E A'. Consider the one-variable polynomials fl(t) = f(t,t2, t3), g1(t) = g(t,t2,t3) E Rjt], and let Vj, = {A E A' I fl(X) = f(X,X2,X3) = O}. If Vj, = A' then f vanishes on V and hence f E I(V). If Vj, # A', then g1 vanishes on the nonempty open subset U = A' - Vj, of A'. It follows from Proposition 2.3(i) that g1 vanishes on A', and consequently g vanishes on V. This shows that g E I(V), as desired. (Without using the density of U,the proof may also be completed by applying the infiniteness of U to a one-variable polynomial.)

Observe that the last example has used two important facts: (a) There is a one-to-one and onto correspondence

A1=K+ V

X H (XlX2,X3)

(b) A1 is irreducible (so that Proposition 2.3(i) can be used). The above observation shows the correct way to introduce polynomial mappings and rational mappings between algebraic sets, in order to compare the properties of two algebraic sets (such as irreducibility) with respect to the Zariski topology.

2.6. Definition Let V 5 A", W C A" be afine algebraic sets. (i) If there exist polynomials fl, ...,fn E K[zl,..., zm]that define a mapping

#: v- W

then # is called a polynomial mapping from V to W;if there is also a polynomial mapping $I: W 4 V such that the composite mapping $4 = lv, the identity mapping on V, and $$I = lw, then # is called a (polynomial) Algebraic Geometry 139 isomorphism with the inverse (this name is qualified by Proposition 2.7 below). If furthermore V = A" then # is called a polynomial parametrization of w. (ii) If there exist elements k,..., 5 E K(x1,..., x,), where K(x1,..., x,) is the field of fractions of the polynomial ring K[z~,..., x,], that define a mapping

n # : u = v - UV(g,)- W i= 1

then # is called a rational mapping from V to W;if furthermore V = A" then # is called a rational parametrization of W.

By definition, in Example (vi) above the correspondence X H (A, X2, X3) defines a polynomial parametrization of the twisted cubic V = V(y-x2,z- 23).

2.7. Proposition (i) Polynomial and rational mappings are continuous with respect to Zariski topology. (ii) Let V C Am, W A" be affine algebraic sets, and let # be a (polynomial) rational mapping from V to W. (a) Suppose 4 is onto. If X is a closed subset of W and its preimage d-'(X) in V is irreducible, then X is irreducible. (b) If V = Am and the image of # in W is dense, then W is irreducible. Proof (i) It is sufficient to prove the assertion for rational mappings. Sup- pose # is defined by

n 4 : u = v - UV(g2)+ W i= 1

where E = ,..., Snli K(z1,..., x,). If WI V(h1,..., h,) n W is a closed subset of W,where hl, ..., h, E K[yl,..., yn], then P E U and #(P>E WI 140 Commutative Algebra implies for i = 1, ..., s

Write hi (t, ..., k) = 5,i = 1, ..., S, where Fi, Gi E K[xl,..., x,]. Then it follows from equation (1) that Fi(P) = 0, G,(P) # 0, i = 1, ..., s, and hence

(2) P E V(Fl,..., F,) n U.

On the other hand, if Q E V(F1, ..., F,) n U,then Fi(Q) = 0 implies G,o-F'(Q) - 0, i = 1, ..., s, and hence equation (1) holds for Q, that is, $(Q) E V(hl,..., h,) n W = W1. This shows that #(V(Fl,...,F,) n U)= W1, and consequently # is continuous. (ii) The assertion of part (a) follows from part (i) directly. We prove (b) by showing that I(W) is a prime ideal. Let # be defined by 4: u- V

where U = A" - (uZlV(gi)), gl,..., ,L E K(yi, ...,ym), and #(U) is LL 9n dense in W by the assumption. Suppose fg E I(W) where f,g E K[zl,..., x,]. Then 0 = (fg)(Q) = f(Q)g(Q) for all Q E W. But this implies 0 = f ( k,..., 5)(P)g (2, ..., k) (P)for all P E U. Since f (k,..., k),g (k,..., k) E ~(yl,...,ym), we can write j(k,..., 5) = G>g(gll"'lgnfi fik) = G'l where F, F', G, G' E Kjyl, ..., ym] with G(P)# 0 and G'(P)# 0 for all P E U. Thus, if we put

v, = { P E u I f (k,...) 5) (P) = o}

vg = { P E u 1 g (k,...) 5) (P)= o} then V, = V(F) n U, V, = V(F') n U. Now, if V, = U,then f(#(U))= 0 and hence f(W) = 0 because +(U) is dense in W. So f E I(W). If V, # U, then since (fg)(Q) = 0 for all Q E W,V,, and consequently V(F')must contain the nonempty open subset U - V' = U - V(F). It follows from Algebraic Geometry 141

Proposition 2.3(i) that V, = U. Therefore] g(+(U)) = 0 and g(W)= 0 by the density of +(V)in W. Consequently, g E I(W) as desired. 0

Remark (i) By Definition 2.6(i) and Proposition 2.7, algebraic sets may be classified in terms of “polynomial isomorphism”. In a general theory of algebraic geometry] rational mappings provide models of abstract rational morphisms that yield classification of (quasi-)varieties in terms of “birational isomorphism”. (ii) Practically, polynomial parametrization and rational parametrization are very useful in 3D-plotting of curves and surfaces, geometric modelling, and CAD (computer aided design).

Since K[q,...] z,] is Noetherian, it follows from Proposition 1.4(iv) that

0 A” is a Noetherian space in the sense that every collection of algebraic sets in A” has a minimal member with respect to (check it!). This observation enables us to prove the following decomposition theorem for affine algebraic sets.

2.8. Theorem Let V be an algebraic set in A”. Then V can be expressed uniquely as a union of finitely many irreducible algebraic sets Vl1...] V,, that is, V = Vl U . . . U V,, and V, @ V, for all i # j. Proof Let

V is not a union of finitely many R = {algebraic sets V An C irreducible algebraic sets

We claim that R is empty. If not, let V be a minimal member of R. Since V E R, V is not irreducible. Thus, V = Vl U V, where V, are proper closed subsets of V, and V, @ R. Hence V, = V,1 U . . . U with V,j irreducible. But then V = Ui,j&jl a contradiction. Therefore 0 = 0. So any algebraic set V may be written as V = V1 U . . . U V,, where each Q is irreducible. To get the second condition] simply throw away any & such that V, c vj for i # j. To show uniqueness, let V = Wl U . . . U W, be another such decomposition. Then V, = Uj(Wj n V,), so V, c Wj(i)for some j(i). Similarly] Wj(i)c Vk for some k. But V, c Vk implies i = kl so V, = Wj(i).Likewise each Wj is equal to some 17 142 Commutative Algebra

The V, appearing in the theorem are called the irreducible components of V;V = Vl U . . . U V, is the minimal decomposition (or sometimes, the irredundant union) of V into irreducible components.

Example (vii) Consider the algebraic set V = V(zz,yz). Then V is a union of a line (the z-axis) and a plane (the zy-plane), both of which are irreducible. (viii) Let f E K[zl,..., zn]and f = fr'f;' . . . f,". the factorization off into irreducible polynomials. If K is algebraically closed, then, by Nullstellen- satz, it may be derived that V(f) = V(f1)L.J...UV(fr) is the decomposition of V(f) into irreducible components. Moreover, it is an exercise to check that I(V(f))= (flf2 . . . fr). (ix) Note that any P E V = V(zz - y2, z3 - z5) c A; is either the form (A3, A4, A5) or the form (-A3, -A4, -A5), where A E R. If we plot a part of V,then it seems that V is a union of two irreducible curves.

In later exercise 8 the reader will be asked to prove that it is the case indeed.

Theorem 2.8 can also be stated in a purely algebraic way using the one-to-one correspondence between radical ideals and algebraic sets. Algebraic Geometq 143

2.9. Theorem If K is algebraically closed, then every radical ideal I in K[xl, ..., z,] may be written uniquely as a finite intersection of prime ideals: I = PI n . . .nP, where Pi @ Pj for i # j. (Such a presentation is called the minimal decomposition or the irredundant intersection of a radical ideal.) Proof This follows immediately from Theorems 1.7 and 2.8. 0

Remark Theorem 2.9 is indeed the geometric model of a general primary decomposition theory in commutative algebra.

We complete this section by seeking irreducible hypersurfaces V(f) defined by f E K[zl, ..., z,]. Iff is irreducible then the ideal (f)is a prime ideal, and hence m = (f). (a) If K is algebraically closed, then I(V(f))= m = (f) and hence V(f) is irreducible. (b) If K is not algebraically closed, then exercise 2 of the current section shows that V(f) is not necessarily irreducible. (c) If K is not algebraically closed but f E K[x,y] and V(f) is infinite, then I(V(f))= (f) and consequently V(f) is irreducible by Theorem 2.4. The assertion (c) is argued in detail as follows.

2.10. Proposition Let f and g be polynomials in K[z,y] with no common divisors, and let V = V(f, g) be the algebraic set defined by f and g. Then V = V(f) n V(g) is a finite set of points. Proof f and g have no common divisors in K[x][y],so they also have no common divisors in K(z)[y] by Chapter 1 (section 2, exercise 7). Since K(5)[y] is a PID, pf + qg = 1 for some p,q E K(x)[y]. Let d E K[z]be such that dp = a, dq = b E K[z,y]. Then af + bg = d. If P = (u,v) E V, then d(P)= d(u) = 0. But d has only a finite number of zeros. This shows that only a finite number of x-coordinates appear among the points of V. Since the same reasoning applies to the y-coordinates, there can only be a finite number of points. 0

2.11. Corollary Iff is an irreducible polynomial in K[x,y], and if V(f) is infinite, then I(V(f))= (f)and V(f) is irreducible. Proof If g E I(V(f)) then V(f,g) is infinite. By the proposition, fig, i.e., 9 E (f). 0 144 Commutative Algebra

2.12. Corollary Suppose K is infinite. The irreducible algebraic subsets of A2 are: A2, points, and irreducible plane curves V(f), where f is an irreducible polynomial and V(f)is infinite. (Note that we do not view 0 as an irreducible algebraic set, as we indicated in Definition 2.2.)

Proof Let V be an irreducible algebraic set in A2. If V is finite or I(V) = (0), V is of the required type. Otherwise, I(V) contains an irreducible polynomial f because I(V) is now a prime ideal. Then we claim I(V) = (f); for if g E I(V), f A g, then V c V(f,g) is finite by Proposition 2.10, contradicting the infiniteness of V.

Appendix. Topological space Given a nonempty set X,a topology on X is a collection T of subsets of X satisfying (a) 0 E T,X E T, (b) the union of any family of sets in T belongs to T,and (c) the intersection of finitely many sets in T is also in T. The sets in T are called open sets of X, and their complements in X are called closed sets. A set X equipped with a topology T is then called a topological space, and the elements of X are usually called points. Given a topological space X, a neighborhood of a point x E X is any set that contains some open set containing x.

Example (i) Any set X can be made into a topological space by giving the discrete topology, in which every subset (including empty set) is open. (ii) The real line R is a topological space, where a set U is open if it is a union of (finitely or infinitely many) open intervals. (iii) The real Euclidean space R" is a topological space, where the open sets are defined as the union of open balls. (iv) Any metric space X is equipped with the topology in which open sets are unions of open balls. (v) Every manifold (if you know what it is) is a topological space.

Topological spaces appear in all branches of modern mathematics, in which it is necessary to study the convergence, connectedness and continu- ity. Algebraic Geometry 145

Let X and Y be topological spaces and f: X + Y a mapping. f is said to be a continuous function if the preimage of every open set in Y is an open set in X. Suppose that f is a one-to-one and onto mapping with inverse f-l. If both f and f-' are continuous, then f is called a homeomorphism, and in this case, X and Y are said to be homeomorphic. Homeomorphic topological spaces have essentially the same topological structure. Any subset Y of a topological space X has the induced topology, in which, every open set is of the form Y n U where U is an open set of X. Given topological spaces XI,X2, ..., X,, the Cartesian product XI x X2 x . . . x X, has the product topology given by the products of open sets. Finally, let X be a topological space, Y a subset of X. The closure of Y in X, denoted is the smallest closed subset in X containing Y,i.e., - Y, Y = nw where W runs over all closed subsets of X containing Y. A subset Z of X is said to be dense in X if Z n U # 0 for every nonempty open subset U c X,or equivalently, if z= X.

Exercises 1. Show that V = V(y - f (x)) c A; is irreducible. 2. (This exercise and the next help us to better understand the Nullstel- lensatz.) Show that f = y2 + x2(x - 1)2 E IR[x,y] is an irreducible polynomial, but that V(f) is reducible. 3. Show that if f E @[XI,..., x,] is irreducible, then V(f)is irreducible. Also show that if V = V(g) is an irreducible hypersurface in A& there is no irreducible algebraic set W such that V c W c A& W # V, W # A;. 4. Show that any linear subspace of K" = A;, where K is a field, is an irreducible algebraic set. (Hint: Use Proposition 2.7.) 5. Use exercise 3 above to show that V(y2-x(x-l)(x-X)) is an irreducible curve in A& where X E @. 6. Let$: AA -+ V =V(y2-x3-x2) bedefinedby$(c) = (c2-1,c(c2-l)). Show that q5 is one-to-one and onto, except that $(fl) = (0,O). Hence V is irreducible by Proposition 2.7 (see also section 6, Example (i)). 7. Find the irreducible components of V(y2 - xy - x2y + x3) in A;, and also in A;. 8. Let 1 = (ZZ - y2, z3 - x'). Show that V(1) is a union of two irreducible algebraic sets. 9. Let f E C[q,..., z,] and let f = fr' .. . f,". be the decomposition of f into irreducible factors. Show that V(f) = V(f1) U ... U V(fr) is 146 Commutative Algebra

the decomposition of V(f) into irreducible components and I(V(f)) = (flfi...fr). 10. Give a detailed proof of Proposition 2.3(ii) and Proposition 2.5, respectively. 11. Let V C A", W C A" be algebraic sets. If (p: V -+ W is a polynomial mapping defined by P H (fl(P), ..., fn(P)) where fi ,...,fn E K[zi,..., x,], show that (p*: K[W]-+ K[V]defined by (p*(?j) = g(f1, ...fn) is a K-linear ring homomorphism (i.e., a K-algebra homomorphism); if furthermore (p is an isomorphism then (p* is a ring isomorphism.

Conversely, if $I: K[W]-+ K[V]is a K-linear ring homomorphism such that $I(?ji)= Tii = 1, ..., n, show that q!Y: V 4 W defined by $I*(P)= (fl(P),..., fn(P)) is a polynomial mapping. Thus we may conclude that (p is an isomorphism if and only if & is an isomorphism.

3. Point P and the Local Ring OP,"

By Theorem 2.8, the study of an algebraic set is reduced to the study of its irreducible components. In this section we demonstrate how a point P of an irreducible algebraic set V determines a unique local ring Op," consisting of rational functions defined at P. Let V C An = A& be an irreducible algebraic set, I(V)its ideal, and K[V]= K[x~,...,~~]/I(V)its coordinate ring. For f,g E K[x~,...,x~], write 7,sfor their images in K[V].Then 7= 3 if and only if f - g E I(V). Thus, every 7E K[V]defines a function (pf: V+ K p - f(P) which is independent of the choice of representatives of 7.Moreover, let W be a closed subset of K = A' with respect to the Zariski topology on A', i.e., W = V(g) for some g E K[z]and W is finite. Then one checks that

*(V(g(f)) n V)= W. This shows that (pf is a continuous function with respect to the Zariski topology. (pf is called a polynomial function on V. Note that if P = (a', ..., a,) E V, then (pei(P)= ai for i = 1,..., n. And since K[V]is Algebraic Geometry 147 generated by all Zi’s over K, this qualifies the name “coordinate ring” KFI. Also note that K[V]is a domain by Theorem 2.4. We seek more continuous functions on V by looking at the field of fractions of K[V],denoted K(V).For 0 # $ E K(V),set - d E K[m,..., z,] 1 2.5 E K[V]}.

Then Jz- is an ideal of K[q,..., x,] and I(V)C Jf (check this!). 9 9

3.1. Lemma V (Jf J = {P E v 1 f = Proof By definition, this can be verified directly. 0

- The algebraic set V is called the pole set of f in V.

- = Example (i) Let V = V(y3 - xz) c A3. Then for f9 = E K(V), = {(z,y,z) E A3 I y = t = 0).

For $ E K(V),let

Note that if V # 8 then U-j # 8 (why?). Thus, defines a function - $ 9

- which is independent of the choice of representatives of f. Moreover, if W = V(h) for some EX& = h E K[x],and thus h(f/g) = fl/gm, then one checks that 148 Commutative Algebra i.e., 4~- is continuous on Uz. B

By Proposition 2.3(i), the open subset Uf- is dense in V. Therefore, from 9 a topological point of view, 4~- is “locally” defined “almost everywhere” on 9 V. It follows that K(V)is also called the field of rational functions of V. The key property of 4~- is that it is an L‘analytic-like”function in the sense 9 of the next proposition.

3.2. Proposition Let V and k(V) be as before. Two rational functions 42 and 45 of V are equal if and only if they agree on some open subset ugcv. a

- = Proof Suppose 41- and 4~ agree on U. Then U Uf- n Uh, and 4~- (P) 9 a 9d 9 &(P)- implies (gh - df)(P) = 0 for all P E U. Let s = gh - df and d GS : V -+ K be the polynomial function defined by s. Then U G W,where W is the preimage of (0) under GS. Note that (0) is closed in K = A’ and U is dense in V. It follows that W = V and s E I(V). Thus, in K[V], O=S=~~-~,andhence4~=4~. I7 9 d

The nice “analytic-like” property of a rational function discussed above makes the basis to define abstract regular functions on a quasi-variety in modern algebraic geometry in the schematic language. Now, we can associate to each P E V a subring of K(V)as follows. Notation is maintained as before.

3.3. Definition For P E V,we say that is defined at P if P E Uz. 4~- - 9 9

Remark In fact, this definition says that $2- is defined at P if there are -- some h,s E K[xl,... ,xn] such that s(P) # 0 and $ = f. For instance, consider P = (0,O) E V = V(y2 - x3) c A%,($)2 = in K(V).

Put Algebraic Geometry 149

Then it is clear that Ic c Ic[V]c 0p,v c Ic(V).

3.4. Theorem (i) Op," is a local ring with the maximal ideal

Moreover, Op,v is uniquely determined by P. (ii) (Compare with Chapter 2 Theorem 3.8.) If K is algebraically closed, then

KIVI= n~~,~, PEV i.e., the rational functions that are defined at every P E V are nothing but all polynomial functions on V.

Proof (i) That Op,v is -a subring of K(V)may be verified directly. Note that if &(P) # 0 then f is invertible in Op,v. Hence 0p,v is a local ring B with the described maximal ideal Mp,v. Indeed, this result can also be obtained by using the surjective ring homomorphism

ForP=(al, ..., a,) ~V,zi-ai~Mp,~,i=l,..., n. IfQ=(bl, ..., b,)EV and Q # P, then it is clear that Mp,v # MQ,V and hence Op,v # OQ,V. This proves- (i). (ii) If f E npEvOp,v, then V(J-,)- = 0, so 1 E J-gy b Nullstellensatz, i.e., -- 9 1 . $ = f E Ic[V],as desired. I7

3.5. Definition Op,v is called the local ring of V at P.

3.6. Proposition Op," = K[V]M,where the latter is the localization of K[V]at the maximal ideal

A4 = (f E K[V]1 *(P) = o}. Hence Op,v is Noetherian. 150 Commutative Algebra

Proof If P = (all... ,a,) E V,then

M = (21 - all ..., zn - a,)(mod I(V)) by Chapter 2 (section 1, Example (iii)). In view of Chapter 2 section 3, K[V]Mis a subring of K(V)and it is (by definition) nothing but Op,v. Thus, Op,v is Noetherian by Corollary 3.5(ii) of Chapter 2. 0

The last proposition qualifies Definition 3.7 of Chapter 2.

Exercises 1. Verify Lemma 3.1. 2. Let V = V(y2 - x3 - x2)c AZz be the nodal curve, and Z,?j the image of x, y in k[V];let t = E k(V). Find the pole sets of z and of z2. (Note that V is irreducible by (section 2, Exercise 6).) 3. Let Op,v be the local ring of an irreducible algebraic set V at a point P. Show that there is a natural one-to-one correspondence between the prime ideals of Op," and the irreducible algebraic subsets of V containing P. (Hint: If I is prime in Op,v, Ink[V] is prime in k[V], and I is generated by Ink[V]; use Theorem 3.4.) 4. Let V be an irreducible algebraic set in A", I = I(V),P E V, and let J be an ideal of K[Q,..., x,] which contains I. Let 5 be the image of J in k[V].Show that there is a natural homomorphism p from Op,An/JOp,An to op,v/JOp,v,and p is an isomorphism. In particular, Op,An/IOpAn is isomorphic to Op,v. Also show that if V = V(I)= {P},then

(Hint: Use Chapter 2 Corollary 4.4.) 5. Let V C A", W C A" be irreducible algebraic sets. If 4: U -+ W is a rational mapping defined by P H (al(P),..., a,(P)) where a1 = ..., a, = JL K(z1,..., 2,) and U = V - u?==,V(gi),show that &: b,91 Sn E K(W) 4 K(V)defined by +* (5) = F(al,...,a,)/ G(a1,...,a,) is a K-linear ring homomorphism. Conversely, if $I: K(W)-+ K(V)is a K-linear ring homomorphism such that $I(Zi)= ai = E K(z1,..., x"), i = 1, ..., n, then show that P H (al(P),..., a,(P)) defines a rational mapping $*: U -+ W where u = v - u;=lv(gi). Algebraic Geometry 151

4. Nonsingular Points and DVRs

This section is devoted to showing how to algebraically recognize singularities in algebraic curves.

Let A2 = A$. If f E K[z,y] is a nonconstant polynomial, then we write C = V(f) for the plane curve defined by f (and hence by cf for any 0 # c E K), and K[C]= K[z,y]/I(C) for its coordinate ring; in the case where C is irreducible we also write K(C) for its field of rational functions. If f has degree n 2 1, we say that the curve C is of degree n. To have a more convenient discussion, the first thing to be explored is that after an (affine) change of coordinates nothing about C = V(f) is “intrinsically” changed. (Although there is a result that applies polynomial isomorphisms to general affine algebraic sets, we do not have space to discuss it.) Let a,b E K. Define the (affine) change of coordinates (as in R2) 4: A2 - A2 (z1?4) (z+a,y+b) and put C’ = V(f(z+ a,y + b)).

4.1. Lemma With notation as above, 4(C’) = C. Proof Exercise.

4.2. Proposition (i) The foregoing q5 induces a K-linear ring isomorphism #*: K[C]-K[C’]. (ii) is irreducible, then q5 induces a K-linear ring isomorphism - If C -4*: K(C)-K(C’) that yields a ring isomorphism O~(Q),C-OQ,C/with $*(M+(Q),~)= MQ,P for every Q E c’. Proof (i) Consider the K-linear ring homomorphism 4* : KbIYl - K[5,9l dz,Y) 9(z + a,!/ + b) It is well-known that $* is an isomorphism (or use (section 2, exercise 11) as 4 is now an isomorphism). Let h E I = I(C). Then &(h) = h(z+a, y + b). For any Q = (c,d) E C’, 0 = h($(Q))= h(c + a,d + b) = &(h)(Q) by Lemma 4.1. This shows that 4,(I) I’ = I(C’). Conversely, if g E I’, then 152 Commutative Algebra since (u,v) E C implies (‘~1- a,v - b) E C’, we have g(u - a,v - b) = 0. It follows that g1 = g(x - a, y - b) E I and &(g1) = g E I’. Therefore, +,(I) = I’ and

(ii) If C is irreducible, by part (i) we have a K-linear ring isomorphism induced by 4* - 4* : K(C) - K(C’)

- 4J It is not hard to check that 04(Q),C*0Q,CJ with $*(kf4(Q),c)= MQ,CJfor - every Q E C‘. 0

For a curve C = V(f) c A’, in order to introduce the singularity in C, we first provide an algebraic way to define the tangent line of C at a point P. Let L be a line passing P = (a,b) E C, which is parametrized as

x=a+ct y=b+dt where t is a variable. Put g(t) = f(a + ct, b + dt). Then t = 0 is a root of g(t). Considering the formal MacLaurin series expansion of g(t) at 0, we have

If g(m)(0)# 0, and ~(~)(0)= 0 for f? < m, then 0 is a root of multiplicity m for g(t), and we say that L intersects C at P = (a,b) with multiplicity m. It is easy to show that this definition is independent of the particular parametrization of the line L.

4.3. Proposition Put Of = (the gradient vector off). (i) If Vf(P) # (O,O), then there is a unique line through P which intersects C with multiplicity 2 2. Algebraic Geometry 153

(ii) If Of(P) = (O,O), then evey line through P intersects C with multiplicity 2 2.

Proof Note that t = 0 is a root of multiplicity 2 2 if and only if g’(0) = 0 if and only if

E(P).c+ afdY-(P) .d= 0.

Hence, if V f (P)= (0, 0), then L always intersects C at P with multiplicity 3 2. This proves (ii). Now suppose Vf(P) # (0,O). Then the solution space of the equation (1) with unknowns c and d is 1-dimensional. Thus, there is (CO,do) # (0,O) with the property that (c,d) satisfies (1) if and only if (c,d) = X(co,do) for some X E k. It follows that the pairs (c,d) that make g’(0) = 0 all parametrize the same line. This shows that there is a unique line which intersects C at P with multiplicity 2 2, and hence (i) is proved. 0

4.4. Definition If Of(P) # (O,O), then the tangent line of C at P is defined to be the unique line through P intersecting C with multiplicity 2 2 at P. In this case, P is called a nonsingular point (or a simple point) of c. If Vf(P) = (O,O), then P is called a singular point of C, or we say that P defines a singularity in C. If every point P E C is nonsingular, then C is called a nonsingular curve (or a smooth curve).

Example (i) Any irreducible conic C c A$ is nonsingular. To see this, let P E C. After a change of coordinates we may assume P = (0,O). If g(P)= g(P)= 0, then f would be of the form ax2 + by2 + my, which can be factorized into a product of linear forms.

(ii) P = (0,O) is a singular point in the cuspidal curve C = V(y2- x3) 154 Commutative Algebra

and P = (0,O) is also a singular point in the nodal curve C = V(y2-z3--z2).

Let C = V(f) c A2 be a curve of degree n 2 1, P = (0,O) E C. Then f Algebraic Geometry 155 must be of the form

f=Fm+Fm+l+...+Fqwithm>l, Fj= XaoxaylP, a+o=j i.e. , every Fj is a homogeneous polynomial of degree j 2 m 2 1. In view of the previous discussion on singularity, this number m is usually called the multiplicity of C at P = (O,O), denoted mp(C) = m. Observation P = (0,O) is a simple point if and only if m = 1.

If C is irreducible, then K E Op,c/Mp,c by the proof of Theorem 3.4(i). So ME,c/Mp,:l may be regarded as a K-vector space for every n 2 0. This enables us to characterize mp (C) algebraically.

4.5. Theorem Let C = V(f) c A2 be an irreducible curve of degree n 2 1, P = (0,O) E C. Write 0 for the local ring of C at P and M for the maximal ideal of 0. If C has the multiplicity mp(C) = m as defined above, then

mp(C) = dimk (M"/M"+l) for large n.

Proof From the isomorphism ( (O/Mnsl) / (Mn/Mnfl)) (O/Mn) it follows that it is sufficient to prove that

dimK (O/Mn)= n + mp(C)+ s for some constant s and all n 2 m = mp(C). Since P = (O,O), we have Mn = mnO, where m = (2,y) c K[x,y]. Note that V(m") = {P} = V(mn,f). By section 3, exercise 5,

where C?p,A2 C K(z,y) is the local ring of A2 at P. Thus, the problem is reduced to the calculation of the dimension of K[x,y]/(mn, f). Since m = mp(C) implies fg E mn whenever g E mn--m,there is a natural ring homomorphism p: K[x,y]/mn + K[x,y]/(mn, f), and a K-linear mapping $: K[~,y]/m~-~--t K[x, y]/mn defined by $(g) = z. It is easy to verify 156 Commutative Algebra that cp o II, induces the isomorphism of K-vector spaces:

It follows from section 1, Exercise 14 that

as desired. 0

Remark (i) By the theorem, mp(C) depends only on the local ring of P = (0,0), that is, it is an intrinsic property. Considering any point Q = (a,b) E C, by Lemma 4.1 and Proposition 4.2, the multiplicity (for singularity) of C at Q, denoted ms(C), may be defined to be mco,o,(C’), where C’ = V(f(.+ a,Y + b)). (ii) It can be shown that the function x(n) = dimK (Op,c/M;,,) is a polynomial in n (for large n) which is called the Hilbert-Samuel polynomial of the local ring Op,c.

We are ready to reach the main result of this section.

4.6. Theorem Let C = V(f)c A2 be an irreducible curve of degree n 2 1, P = (a,b) E C. Then P is a simple point if and only if the local ring Op,c of C at P is a DVR in K(C),and in this case, the maximal ideal Mp,c of Op,c can be generated by where V(L= ax + by + c) is any line through P but not tangent to C at P. Proof First suppose that P is a simple point in C. Since L is not the tangent line of C at P, after an appropriate (affine) change of coordinates we may assume that P = (O,O), that y is the tangent line, and that L = 2 by Proposition 4.2. Then by Chapter 2 Theorem 2.10, it is sufficient to show that Mp,c is generated by the image of x in Op,c. Write z, for the images of x,y in k[C].Then P = (0,O) implies M~,c= @,g)c Op,c,whether P is simple or not. Moreover, since y is the tangent line to C at P = (0,O) (recall that the tangent line at P is defined by the equation g(P)x+ g(P)y= 0), we have

f = y + higher terms. Algebraic Geometry 157

Grouping together terms involving y, we may write

f = yg - z2h, where g = 1 + higher terms, and h E K[x] - Thus, j@ = 22h E k[C], so jj = x2hij-’ since g(P) # 0. It follows that MP,C= (T,jj) = (T),as desired. Conversely, if 0 = 0p,c is a DVR, then M = Mp,c = (t)for some prime t E M. Thus, dimK(Mn/MnS1) = 1 for all n 2 0, and hence rnp(C) = 1 by Theorem 4.5. This shows that P is simple.

Exercises 1. Complete the proof of Lemma 4.1. 2. Find all singular points in C = V(x2y2+ x2 + y2 + 2zy(x + y + 1))C A; and C = V(xy + z3+ y3) c A;. 3. Show that the elliptic curve y2 = (x- a1)(x - a2)(x- a3) in A;, where al,a2, a3 are distinct complex numbers, has no singularities. 4. Use the notation as in the proof of Theorem 4.5 to show that the function x(n) = dimK (Op,cjM;,,) is a polynomial in n (for large n).

5. Normalization of Algebraic Curves

In the light of Proposition 3.6, Theorem 4.6 and Chapter 3 Theorem 4.7, we explore in this section the relation between the singularity of an algebraic curve C and the normality of the coordinate ring of C. As a consequence, this leads to the “resolution” of singularities in algebraic curves by means of “normalization” .

Let K be a field and let C = V(f) c A2 = A& be an irreducible plane curue, where f E K[x,y] is a nonconstant polynomial. As before, K[C] = K[z,y]/I(C) denotes the coordinate ring of C and K(C) stands for the field of rational functions of C (i.e., the field of fractions of K[C]).Put

5.1. Theorem With notation as above, S(C) is a finite subset of C. Hence C - S(C), the set of nonsingular points in C, is a dense open subset of C. 158 Commutative Algebra

Proof We may assume that f is irreducible and S(C) # 8. Note that

If % # 0, then since deg( g)< deff and f is irreducible, by Proposition 2.10, V (2, f) and hence S(C) is a finite set. A similar argumentation holds provided # 0. Below we show that and % cannot both be identically zero. If chark = 0, then = % = 0 only when x and y do not appear in f. Hence f is a constant in K, a contradiction. If chark = p > 0, then 2 = = 0 only when x and y appear in f in the form of xffpand ypp. Hence f can be expressed as f = CuapxPayPp. Assuming that K is algebraically closed, the equation a,p = tP has a solution, say u,p = spap for some sap E K. Thus, putting g = Cs,pxayP we have gp = f , contradicting the irreducibility of f. I?

In view of the last theorem,- almost all points in C are simple. Next, we show that the normalization K[C]of K[C]in K(C) (i.e., the integral closure of K[C]in K(C) in the sense of Chapter 3 (Definitions 1.6 and 3.1)) defines a nonsingular algebraic curve C' (see later Definition 5.5) which projects onto C with respect to the Zariski topology. To this end, we first generalize Chapter 4 Theorem 4.2(i) (concerning Z c &) to a more general setting.

5.2. Proposition Let R C B be a module-finite ring extension, where R and B are domains. If every nonzero prime ideal of R is maximal, then every nonzero prime ideal of B is maximal.

Proof Suppose B = Cz,RJi where Ji E B. Let P be a nonzero prime ideal of B, 0 # b E P. Then b is integral over R, that is,

Assume that n is the smallest degree. Then 0 # bo E P n R. Thus, P n R is a nonzero prime ideal of R and hence maximal by the assumption. Put K = R/(P n R). Then B/P may be viewed as an extension ring of K via the embedding K = R/(Pn R) 4 B/P. By Chapter 3 (section 3, exercise 6(b)), B/P is integral (indeed module-finite) over K. It follows from Chapter 3 Theorem 1.8 that B/P is a field. Therefore, P is maximal. Algebraic Geometry 159

5.3. Corollary Let C = V(f) c A2 be an irreducible- plane curve, where f E K[z,y] is a nonconstant polynomial, and let K[C]be the normalization of K[C]in K(C). Then every- nonzero prime ideal of K[C],-respectively every nonzero prime ideal of K[C],is maximal. In particular, K[C]is a Dedekind domain (in the sense of the final remark given in Chapter 4 section 4). Proof By Corollary 2.12 we may assume that C is infinite. It follows from Proposition 1.8(i) and the Noether normalization (Chapter 3 Theorem 2.4) that K[C]is module-finite- over a polynomial subring K[t].Furthermore by Chapter 3 Corollary 3.3, K[C]is module-finite over K[C].Now Proposition 5.2 and Chapter 2 (section 1, exercise 2) yield the assertion. 0

5.4. Theorem Let C = V(f) c A2 be an irreducible- plane curve, where f E K[z,y] is a nonconstant polynomial, and let K[C]be the normalization of K[C]in K(C). If K is algebraically closed, then the following statements are equivalent. (i) C is nonsingular. (ii) The local- ring Op,c of C at every P E C is a DVR. (iii) K[C]= K[C],that is, K[C]is a Dedekind domain.

Proof By Theorem 4.6, the equivalence (i) @ (ii) holds even if K is not algebraically closed. The implication (ii) + (iii) follows from Corollary 2.5(ii), Proposition 3.6, and Chapter 2 Theorem 3.1. And the implication (iii) + (ii) follows from Corollary 2.5(ii), Proposition 3.6, Chapter 2 Proposition 3.6, Chapter 3 (Theorem 4.1 and Corollary 4.6).

Example (i) The fact that the cuspidal curve C -= V(y2 - z3)c A; has one singular point (0,O) is indicated by R[C] # R[C] where the latter is isomorphic to the polynomial ring R[t] (see Chapter 3 (section 3, Example (iii))). (ii) The fact that the nodal curve C = V(y2 - x3 - z2) c A; has one singular point (0,O) is indicated by later section 7, Example (i) which shows R[C] # Iw[cl R[t].

Now let V be an algebraic set in the affine n-space An = A;, n 2 1. By the Noetherian normalization theorem, the coordinate ring K[V]of V is module-finite over a polynomial subring K[tl,..., td], d 5 n. The last theorem leads to the following definition. 160 commutative Algebra

5.5. Definition Let V C A" be an algebraic set, n 2 1. If V is irreducible and its coordinate ring K[V]is module-finite over a polynomial subring K[t],then V is called an algebraic curve in A". Let V c A" be an algebraic curve, P E V. If the local ring Op," of V at P is a DVR then P is called a nonsingular point of V;otherwise, P is a singular point (or P defines a singularity in V).If every point of V is nonsingular then V is called a nonsingular curve.

Example (iii) By later exercise 1-2, the curve C = V(y - f(z))c A; is nonsingular, where f(x) E K[5];and the twisted cubic curve V = V(y - x2,2 - x3) c A; is nonsingular.

5.6. Proposition Let V c A; be an algebraic curve. If K is algebraically closed, then there exists a plane curve C c A$ such that K(V)= K(C). (Note that this qualifies Definition 5.5.) Proof By the definition of an algebraic curve, K[V]= CEl K[t]& where K[t]is a polynomial subring of K[V]and E K[V].Then it is easy to see that K(V)= K(t)[&,...,Im] where each is algebraic over the field K(t). We may assume that all ti's are separable over K(t). It follows from Chapter 1 Theorem 3.12 that K(V) = K(t,6) for some 6 E K(V). Let K[zl,221 be the polynomial ring in 21, 22 over K and consider the onto ring homomorphism K[zl,Q] -+ K[t,61. Then K[t,61 2 K[z1,z2]/1 for some ideal I c K[zl,221. Note that K is algebraically closed and hence infinite (Chapter 1 section 3, exercise 5). By the Nullstellensatz and Corollary 2.12, K[tl,z2]/I is the coordinate ring of the plane curve C = V(I). Hence K(V)= K(C).

5.7. Theorem (normalization of a plane curve) Let C = V(f) C A2 be an irreducible- plane curve, where f E K[x,y] is a nonconstant polynomial, and let K[C]be the normalization of K[C]in K(C). If K is algebraically closed, then the following statements hold: (i) There exists a nonsingular algebraic curve V c A" for some m 2 1 such that - K[V]= K[C].

(ii) Let V be the algebraic curve obtained in part (i). There exists an onto polynomial mapping 4: V -+ C and an open subset U c V such that 4(U)= C - S(C),the open subset of all nonsingular points in C. Algebraic Geometry 161

Proof (i)- Assume that C is infinite. It follows from the proof of Corollary 5.3 that K[C]is module-finite over a polynomial subring K[t] K[C], - - - c say K[C] = xi=lK[t]&, & E K[C]. Hence K[C] = K[t,&,...,<,I K[zl,..., z,+1]/I where I is an ideal of the polynomial ring K[zl,..., z,+1]. Since K is algebraically closed, K[zl,..., z,+1]/1 may be viewed as the coordinate ring of V = V(I)c- A$' by the Nullstellensatz. V is nonsingular because the localization of K[C]at every nonzero prime ideal is a DVR by Corollary 5.3 and Chapter 3 Corollary 4.6. (ii) Let a be the composite ring homomorphism K[C] - K[Cl% K[V] and suppose a(:) = 3, a@) = E K[V],where g,h E K[z1,..., zs+l]. Thus, ~(0)= a(f)= f(g, h) = 0 E K[V],that is, f(g,h) E I. It foliows that 0 = f(g,h)(Q) = f(g(Q),h(Q)) for every Q E V,and that there is a polynomial mapping

f#J: v- C

Now, if P = (a,b) E C and (x-a,y-b) is the maximal ideal of P, then 3-a, - - h - b E (Z1 - cl, ..., z,+1 - c,+1) c K[V]by the weak Nullstellensatz and Chapter 3 Proposition 1.7, where (c1,..., c,+1) = Q E V. Thus, g(Q) = a, h(Q) = b and hence 4(Q)= (a,b) = P. This shows that $ is onto. Finally, since 4 is continuous with respect to the Zariski topology by Proposition 2.7, Theorem 5.1 concludes that there is some open subset U c V such that $(U) = C - S(C). 0

5.8. Definition The algebraic curve V obtained in Theorem 5.7 is called the normalization of C, denoted c.

It follows from Chapter 3 Proposition 1.7 and part (ii) in the above proof of Theorem 5.7 that the following property holds for the normalization of an irreducible plane curve C. This property will be a key to approach section 6 Theorem 6.3(ii),

5.9. Corollary With the assumption and notation as in Theorem 5.7, a maximal ideal of K[c]contains no more than one maximal ideal of K[C]. 0 162 Commutative Algebra

Remark (i) With the discussion on algebraic curves given in this section, the interested reader can move forward to the core of algebraic curve theory - Ftiemann-Roch Theorem that deals with the divisor theory and consequently provides the way to compute one of the most important invariants - the genus of an algebraic curve. (ii) For an irreducible algebraic set V C A", after introducing the "dimension" of V and translating it into an algebraic version about K[V]and K(V),singularities in V can also be defined both geometrically and algebraically, and the singularity of a point P E V may be characterized by the property of Op," as well. In this lecture, of course, we do not go that far. -(iii) For an irreducible algebraic set V A", if we look at the normalization K[V]of K[V]in K(V),its structure is immediately clear by the Noether normalization, Chapter 3 (Corollary 3.3 and Theorem- 4.7). The interested reader may start a study of the geometric impact of K[V]on V by means of exercise 3 below.

Exercises 1. Show that for the curve C = V(y - f(x)) c A;, R[C] E K[t],the polynomial ring in t over R. 2. For the twisted cubic V = V(y - x2,z - z3)c A;, show that R[V] % K[t],the polynomial ring in t over R. (Hint: In both exercises above apply section 2, exercise 11 to t H (t,f(t)), and t H (t,t2,t3>.) - 3. Let V be an irreducible algebraic set in A&, and let K[V]be the normalization of K[V].If K is algebraically closed,- show that there is an irreducible algebraic set W such that K[W]2 K[V]and W projects onto V with respect to the Zariski topology; moreover,- every maximal ideal of K[V]is contained- in some maximal ideal of K[V],and every maximal ideal of K[V]contains no more than one maximal ideal of WI.

6. Parametrize a Rational Curve via Normalization

In section 2 we have remarked the significance of polynomial and rational parametrization. This section focuses on plane curves such that all but finitely many points can be rationally parametrized. Notations are maintained as before. Algebraic Geometry 163

6.1. Definition Let C c A2 = A> be a plane curve, and let K(t)be the field of fractions of the polynomial ring K[t].C is called a rational cume if there is a rational parametrization (in the sense of Definition 2.6) C

such that C - +(U) is finite.

Remark In the literature, a rational curve is usually defined by assuming that C is irreducible. Though Definition 6.1 above does not assume the irreducibility of C, the next theorem tells that a rational curve defined in this way is necessarily irreducible.

6.2. Theorem Let C = V(f) c A2 be a rational plane curve with the rational parametrization 4 as in Definition 6.1. If C is infinite then the following statements hold: (i) @(U)is open and dense in C. (ii) C is irreducible. (iii) 4 induces a K-linear ring isomorphism K[C] K [$,21 and hence induces an injective K-linear ring homomorphism K(C)+ K(t). (iv) K(C)2 K(t).

Proof (i) The finiteness of C - 4(U) implies $(U) is open. Since every closed subset of A:, is finite and + is continuous with respect- to the Zariski topology (Proposition 2.7), we conclude that the closure +(U) of #(U) in C is equal to C,i.e., 4(U) is dense in C. (ii) By part (i), this follows from Proposition 2.7. (iii) Consider the ring homomorphism $J : K[Z,Yl - K(t)

which is clearly K-linear. We claim that KerG = I(C). To see this, suppose - 0. Then F (m = 0 for all c E U,and hence $J(F)= F ( 3,g) - hz(c)l sz(c)a) F($(U)) = 0. Thus, #(U) c V(F) n C. But #(U) is open and dense in C by part (i). Hence F(C) = 0, i.e., F E I(C). On the other hand, for 164 Commutative Algebra

G E I(C), G($(U))= 0 implies that G (-, %) = 0 for all c E U, or equivalently, the rational function r(t) = G (<,h Ll)92 E K(t) vanishes on U and hence on A& = K. This shows that 0 = r(t) = $(G) because K is infinite (we assumed that C is infinite). Therefore, Ker$ = I(C), and

K[C]%KN [k,E] c K(t)with &(3) = k,$@) = g. (iv) This follows from part (iii) and Luroth's theorem (Chapter 1 Theorem 3.16). 0

In view of section 4, if C c A' is a plane curve and P E C, then, after a change of coordinates we can always assume P = (0,O) E C. Thus, f is of the form

f=Fm+Fm+l+...+Fq, m21, Fj= 1 XaosayP, ff+o=j where mp = m is the multiplicity of C at P.

6.3. Proposition Let K be an algebraically closed field, and let C = V(f) c A; be a plane curve with P = (0,O) E C. Suppose f = Fm+l +Fm with m 2 1, as described above. If F, # As" with X E KXand f(0,y) is not the zero polynomial, then C is a rational curve, and hence irreducible. Proof Cutting the curve with the line y = tx of variable slope t, or equivalently, substituting y = tx into the equation f(s,y) = 0, we have

x"Fm(l,t) + x"flF(1,t) = 0.

Then for x # 0, (*I

Since K is algebraically closed and Fm(l,t) is not a constant by the assumption, it follows that P = (0,O) may be recaptured by (*) above. Moreover, for any constant c E K, f(c,y) has only finitely many zeros. Hence C is a rational curve. 0

Example (i) Clearly any curve of the form C = V(y - f(s)) c A; is rational, where 4: A; + C is defined by $(c) = (c, f(c)). Algebraic Geometry 165

(ii) Every curve C c A; defined by an irreducible quadratic polynomial f E C[x,y] is rational. (iii) Curves C1 = V(y2 - x3), C2 = ~(y2 - 23 - x2), c3 = v(y3 - %4 - x3), C4 = V(x3 + y3 - 3x9) are rational in A;.

Remark Not every irreducible plane curve is rational. For instance, after a change of coordinates, the polynomial y2 - x3-px - q with 4p3 + 27q2 # 0 in C[x,y] becomes y2 - x(z - l)(x - A). By section 2, exercise 5, the curve C = V(y2 - x(x - 1)(x- A)) is irreducible. But if charK # 2 and A # 0, 1, then C is not rational. There are two different ways to know this. One way may be explained as follows. If 4: U + C was a rational parametrization with d(c) = (El%) where U = Ah - (V(h2) u V(g2)), then by Theorem 6.2(iii) there would be an injective K-linear ring homomorphism &: K(C) -+ K(t)with $*(T) = 2, &(y) = 2, and thus there would be

Note that hl,h2,g1, g2 E K[t].One may further try to show that the above equality implies that 2, E K. Another way is to learn a theory on the genus of curves. The theory on genus asserts that an irreducible plane curve C is rational if and only if its genus g(C) is zero, where if C is defined by a polynomial of degree m, then

(m- l)(m - 2) - #(singularities of C properly counted). g(C) = 2

By Theorem 6.2, if C is a rational curve, then K(C)2 K(t). We now consider the converse by passing to the normalization of C. Part (ii) of the next theorem seems missing in the literature.

6.4. Theorem Let C = V(f) c A& be an irreducible plane curve, and suppose that C is infinite. (i) If there is an injective K-linear ring homomorphism K(C)-+ K(t),then there is a rational parametrization for C. (ii) (Huishi Li) If K is algebraically closed and K(C) 2 K(t),then there exists a rational parametrization q!~ as in Definition 6.1 such that either C - $(U)= 8 or C - $(U)contains only one point. Therefore C is rational. 166 Commutative Algebra

Proof (i) By the assumption we have the following composite ring homomorphism

cp: K[C] L) K(C) - K(t) with cp(:) = 2,p(Y) = E E K(t). Set U = K - V(h2) U V(g2). Then since 0 = cp(f) = f (2,E) , the latter defines a rational parametrization: $: u- C (W sdd) h2 (4’ Q2 (c) (ii) Let K[c]be the coordinate ring of the normalization c of C, and consider the composite ring homomorphism

cp : K[C] - K[C] Lf K(C) 2- K(t) Let $J: U 4 C be the parametrization as obtained in part (i). If P = (a,b) E C,and @-a,y-b) is the maximal ideal of P in K[C],then (?i-a,y--b) c M for some maximal ideal M of K[E]by Chapter 3 Proposition 1.7. Let R be the DVR in K(t) that corresponds to K[~]M,where the latter is the localization of K[c]at M, which, by Corollary 5.3 and the proof of Chapter 2 Theorem 2.10, is a DVR containing K and has its field of fractions K(C). By Chapter 2 (section 2, Example (vii)), the only DVRs in K(t) that contain K and have the field of fractions K(t)are those localizations K[t-l](t--l)and K[tl(,-~),X E K. So, if Q(K[~]M)= R # K[t-’](t--l)then a(K[z]~)= R = K[tl+~)for some X E K (weak Nullstellensatz). In the second case we have 2 - a, E - b E (t - A) c K[t](t-~).Hence 2,2 are defined at X and

By Corollary 5.9, at most one maximal ideal of K[C]is contained in some maximal ideal M of K[c]where K[~]]Mcorresponds to K[t-l](,-1).There- fore, there is at most one P = (a,b) E C which is not an image of the rational mapping $J defined in part (i). Algebraic Geometry 167

Exercises 1. Find a rational parametrization of the curve C = V(y2- x4 - x2)c A;. (Hint: Use the line y = tx to cut C first and then use the line t = u(x- 1) to cut the obtained hyperbola C’ = V(x2 - t2 - l).) 2. Show that the curve C = V(y2(a - x) - x3) c A; is rational, where a E R. 3. Show that the curve C = V((x2+ y2)2+ 3x2y - y3) c AS is rational.

7. Rational Curves and Diophantine Equations

In number theory one of the themes is to find the integer solutions of the Diophantine equation f(z1,...,x,) = 0 where f E Q[x~,...,x,],or in the language of algebraic geometry, is to find the points P = (all... ,a,) on the hypersurface V(f) c A: with all a%E Z. The subject of rational curves leads in a natural way to some connection with this aspect, though not every curve is rational. We close this course by examples that illustrate such a point of view.

Example (i) Consider the nodal curve C = V(y2 - x3 - x2)c A;. After cutting the curve with the line y = tx of variable slope t,we obtain t2x2= x3 + x2. It follows that if x # 0 then

x=t2-1 { y = t(t2 - 1)

But t = +1 also gives (0,O) in C. This shows that every point of C can be obtained by system (1) with t E R. By Definition 6.1 and Theorem 6.2, C is rational and irreducible. Moreover, putting h = t2 - 1, g = t(t2- l), -then R[C] 2 R[h,g] c R(t) by Theorem 6.2. Thus, it is clear that R[C] # R[C] R[t]. If P = (a,b) E C with a,b E Z, then we say that P is an integral point. By system (1) it is clear that C has infinitely many integral points if we let t E Z. Conversely, suppose (a,b) is an integral point. We may assume a # 0 (note that (0,O) can be obtained from system (1)). Then t = is a rational number but t2 = a+ 1 is an integer by system (1). Thus, t must be an integer, and hence all integral points on C are given by system (1) with t E Z. In other words, we obtain all integer solutions of the Diophantine equation y2 - x3 - x2 = 0 by letting t E Z in the equation system (1). 168 Commutative Algebra

(ii) Similarly as dealing with Example (i), the curve C = V(x6-z2y3-y‘) c A; may be parametrized by intersecting it with the line y = tx of variable slope t: x = t3 + t5 iy = t4 + t6 Every point of C is given by system (2) with t E R, C is irreducible and rational. Clearly, if t E Z then from system (2) we get integral points on the curve. We claim that these are all integral points on the curve. Suppose (x,y) is an integral point. Then z and y = tx are integers. Since x = t3(l +t2),the only way that x can be zero is for t to be zero. Thus it may be assumed that x # 0 and so t = f is rational. We write t in the form t = where a and b are relatively prime integers. Now x = t3 + t5 and so z = w. This must be an integer and so bl(a3b2 + a’). Therefore bla’. Since a and b are relatively prime, it follows that b = fl. Thus t must be an integer, as we claimed. Therefore, all integral points in C can be obtained by system (2) with t E Z. In other words, we obtain all integer solutions of the Diophantine equation x6 - x2y3- y5 = 0 by letting t E Z in the system (2). (iii) It is well known that the unit circle C = V(x2+ y2 - 1) C A: can be parametrized by using trigonometric functions:

x = cos(t), iy = sin(t). However, if we intersect the circle with the line y = tz - 1 of variable slope t (using y = tx will cut the circle in two points), we obtain the rational parametrization of C: x=- 2t 1 +t2’ (3) 1 -t2 y=- i 1 +t2’ Note that the system (3) does not describe the whole circle since y = &$ can never equal -1, that is, the point (0,-1) is not covered. However, by Definition 6.1 and Theorem 6.2 we know that C is irreducible and rational. We now use the system (3) to find the integer solutions (X, Y,2) of the Diophatine equation Z2 + Y2= Z2 with gcd(X, Y,2) = 1 (recall the ar- Algebraic Geometry 169 gumentation on this problem given in the introduction part of Chapter 4), that is the same as to find the rational solutions (x = $, y = 5) for the equation x2 + y2 = 1. From (3) it is clear that x and y are rational if and only if t is rational. So assume that t = f where u,v E Z and gcd(u, v) = 1, and for the moment assume that X, Y,and Z are positive. Thus, x v2-u2 Y 2uv x=--=- y=-=- 2 v2+u2’ 2 v2+u2 and since gcd(X,Y) = gcd(X,Z) = gcd(Y,Z) = 1, there is a positive integer m such that

mZ = u2 + u2, mX = v2 - u2, my = 2uv.

We claim that m = 1. To see this, note that m[(v2+ u2),ml(v2 - u2).It follows that m1(2v2), m1(2u2). But gcd(u,v) = 1. So m = 1 or m = 2. Since X and Y cannot both be odd or even (otherwise 41 (Z2- 2)), we may assume Y is even. Then, 21(uv) and one of u,‘u is even. Consequently, one of u2,v2 is even and the other is odd. Hence u2 + v2 is odd, contradicting 22 = u2 + u2. Therefore, m = 1. This shows that all desired integer solutions (X, Y,2) are given by

x = f(v2 - u2), Y = f2uv, 2 = f(v2 + 2).

Exercises 1. Indicate why in Example (iii) X and Y cannot both be odd or even. 2. Find the integer solutions of the equation y3 - x4 - x3 = 0. This page intentionally left blank References

[Coh] P. M. Cohn, Algebra I & 11, John Wiley and Sons Ltd., 1982. [Edw] H. M. Edwards, Fermat’s Last Theorem, Springer, 1997. [Full W. Fulton, Algebraic Curves, W. A. Benjamin, New York, 1969. [Jac] N. Jacobson, Basic Algebra, Vol. 11, San Francisco, 1974-1980. [LVO] H. Li and F. Van Oystaeyen, A Primer of Algebraic Geometry, Mar- cel Dekker, 2000. [Mar] D. A. Marcus, Number Fields, Springer-Verlag, 1977. [Mat] H. Matsumura, Commutative Algebra (second edition), The Ben- jaminlCummings Publishing Company, Inc., 1980. [Rei] M. Reid, Undergraduate Commutative Algebra, LMS, Student Texts 29, 1995. [Sam] P. Samuel, Algebraic Theory of Numbers, English translation (by A. J. Silberger), Hermann, 1971. [ST] I. N. Stewart and D. 0. Tall, Algebraic Number Theory (second edition), Chapman and Hall, 1987. [Van] B. L. Van der Waerden, Algebra I & 11, Springer-Verlag, 1985.

171 This page intentionally left blank Index

affine space, 127 direct sum of modules, 48 algebra, 79 discrete valuation, 60 algebra homomorphism, 79 discrete valuation ring, 61 algebraic closure, 28 discriminant, 104 algebraic curve, 160 divisible, 9 algebraic clement, 22 divisor, 9 algebraic field extension, 22 domain, 3 algebraic integer, 102 DVR, 61 algebraic set, 128 algebraically closed, 25 extension field, 18 algebraically independent, 86 extension of an ideal, 71 ascending chain condition, 6 associate, 9 field, 3 associated prime, 95 field extension, 18 field of fractions, 4 basis, 40, 48 field of rational functions, 148 bilinear form, 37 finitely generated extension ring, 80 finitely generated ideal, 2 characteristic, 3 finitely generated module, 47 closed set, 144 finitely generated subring, 2 closure of a subset, 145 fractional ideal, 119 continuous function, 145 free abelian group, 40 contraction of an ideal, 71 free module, 48 coordinate ring, 136 cuspidal curve, 128 group of units, 9 cyclotomic field, 109 hypersurface, 128 Dedekind domain, 125 degree of a curve, 151 ideal of an algebraic set, 129 degree of a field extension, 22 induced topology, 145 degree of a polynomial, 4 inseparable field extension, 24 dense subset, 145 inseparable polynomial, 22

173 174 Commutative Algebra integral basis, 102 pole set, 147 integral closure, 83 polynomial function, 146 integral element, 80 polynomial mapping, 138 integral extension, 80 polynomial parametrization, 139 integrally closed, 83 prime, 11 irreducible, 9 prime ideal, 53 irreducible algebraic set, 135 prime spectrum, 53 irreducible component, 142 prime subfield, 3 primitive element, 25 Jacobson radical, 57 primitive polynomial, 14 principal ideal ring, 7 local ring, 58 product topology, 145 local ring at P, 149 proper factorization, 9 localization at P, 72 quadratic number field, 108 maximal condition, 6 quotient module, 47 maximal ideal, 53 maximal spectrum, 53 R-basis, 48 minimal polynomial, 23 radical ideal, 132 module, 45 rational curve, 163 module homomorphism, 47 rational mapping, 139 module of fractions, 75 rational parametrization, 139 module-finite, 80 reducible, 9 monic polynomial, 5 ring extension, 79 multiplicative set, 53 ring of algebraic integers, 102 multiplicity at P, 155 ring of fractions, 68 neighborhood of a point, 144 separable field extension, 24 nilradical, 56 separable polynomial, 22 nodal curve, 128 simple extension field, 18 Noetherian module, 49 simple point, 153 Noetherian ring, 6 singular point, 153 Noetherian space, 141 splitting field, 19 nonsingular curve, 153 square-free, 80 nonsingular point, 153, 160 subalgebra, 79 norm, 35 submodule, 46 normal domain, 90 symmetric polynomial, 29 normalization, 90 normalization of a curve, 161 tangent line, 153 number field, 102 topological space, 144 topology, 144 open set, 144 total polynomial, 34 trace, 35 p-adic valuation, 62 transcendental element, 22 PID, 7 transcendental field extension, 22 plane curve, 128 trivial divisor, 9 References 175

twisted cubic, 128

UFD, 11 unimodular, 41 unit, 3 valuation ring, 65

Z-basis, 40 Zariski topology, 134