Field and Galois Theory by Milne

8/21/2019 Field and Galois Theory by Milne

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Fields and Galois Theory

J.S. Milne

Version 4.40

March 20, 2013


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These notes give a concise exposition of the theory of fields, including the Galois theory

of finite and infinite extensions and the theory of transcendental extensions. The first six

sections form a standard course. Chapters 7 and 8 are more advanced, and are required foralgebraic number theory and algebraic geometry repspectively.

BibTeX information

@misc{milneFT,

author={Milne, James S.},

title={Fields and Galois Theory (v4.40)},

year={2013},

note={Available at www.jmilne.org/math/},pages={130}

}

v2.01 (August 21, 1996). First version on the web.

v2.02 (May 27, 1998). Fixed about 40 minor errors; 57 pages.

v3.00 (April 3, 2002). Revised notes; minor additions to text; added 82 exercises with

solutions, an examination, and an index; 100 pages.

v3.01 (August 31, 2003). Fixed many minor errors; no change to numbering; 99 pages.v4.00 (February 19, 2005). Minor corrections and improvements; added proofs to the

section on infinite Galois theory; added material to the section on transcendental

extensions; 107 pages.

v4.10 (January 22, 2008). Minor corrections and improvements; added proofs for Kummer

theory; 111 pages.

v4.20 (February 11, 2008). Replaced Maple with PARI; 111 pages.

v4.21 (September 28, 2008). Minor corrections; fixed problem with hyperlinks; 111 pages.

v4.22 (March 30, 2011). Minor changes; changed TEXstyle; 126 pages.

v4.30 (April 15, 2012). Minor fixes; added sections on etale algebras; 124 pages.

v4.40 (March 20, 2013). Minor fixes and additions; 130 pages.

Available at www.jmilne.org/math/

Please send comments and corrections to me at the address on my web page.

The photograph is of Sabre Peak, Moraine Creek, New Zealand.

Copyright c1996, 1998, 2002, 2003, 2005, 2008, 2011 J.S. Milne.Single paper copies for noncommercial personal use may be made without explicit permission

from the copyright holder.


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Contents

Contents 3

Notations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1 Basic Definitions and Results 7Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

The characteristic of a field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Review of polynomial rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Factoring polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Extension fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Construction of some extension fields . . . . . . . . . . . . . . . . . . . . . . . 14

Stem fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

The subring generated by a subset . . . . . . . . . . . . . . . . . . . . . . . . . 15

The subfield generated by a subset . . . . . . . . . . . . . . . . . . . . . . . . . 16

Algebraic and transcendental elements . . . . . . . . . . . . . . . . . . . . . . . 16Transcendental numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Constructions with straight-edge and compass.. . . . . . . . . . . . . . . . . . . 20

Algebraically closed fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2 Splitting Fields; Multiple Roots 25

Maps from simple extensions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Splitting fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Multiple roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3 The Fundamental Theorem of Galois Theory 33

Groups of automorphisms of fields . . . . . . . . . . . . . . . . . . . . . . . . . 33

Separable, normal, and Galois extensions . . . . . . . . . . . . . . . . . . . . . 35

The fundamental theorem of Galois theory . . . . . . . . . . . . . . . . . . . . . 37

Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Constructible numbers revisited . . . . . . . . . . . . . . . . . . . . . . . . . . 41

The Galois group of a polynomial . . . . . . . . . . . . . . . . . . . . . . . . . 42

Solvability of equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4 Computing Galois Groups 45

When isGf An? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3


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When isGf transitive? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Polynomials of degree at most three . . . . . . . . . . . . . . . . . . . . . . . . 47

Quartic polynomials. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Examples of polynomials withSp as Galois group over Q . . . . . . . . . . . . . 49

Finite fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51Computing Galois groups over Q . . . . . . . . . . . . . . . . . . . . . . . . . . 52

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

5 Applications of Galois Theory 57

Primitive element theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Fundamental Theorem of Algebra . . . . . . . . . . . . . . . . . . . . . . . . . 59

Cyclotomic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

Dedekinds theorem on the independence of characters . . . . . . . . . . . . . . 63

The normal basis theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

Hilberts Theorem 90 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Cyclic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69Kummer theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

Proof of Galoiss solvability theorem . . . . . . . . . . . . . . . . . . . . . . . . 71

Symmetric polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

The general polynomial of degreen . . . . . . . . . . . . . . . . . . . . . . . . 75

Norms and traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

Etale algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

6 Algebraic Closures 85

Zorns lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

First proof of the existence of algebraic closures . . . . . . . . . . . . . . . . . . 86Second proof of the existence of algebraic closures . . . . . . . . . . . . . . . . 86

Third proof of the existence of algebraic closures . . . . . . . . . . . . . . . . . 87

(Non)uniqueness of algebraic closures . . . . . . . . . . . . . . . . . . . . . . . 88

Separable closures. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

7 Infinite Galois Extensions 91

Topological groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

The Krull topology on the Galois group . . . . . . . . . . . . . . . . . . . . . . 92

The fundamental theorem of infinite Galois theory . . . . . . . . . . . . . . . . . 94

Galois groups as inverse limits . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

Nonopen subgroups of finite index . . . . . . . . . . . . . . . . . . . . . . . . . 99Etale algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

8 Transcendental Extensions 101

Algebraic independence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

Transcendence bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

Luroths theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

Separating transcendence bases . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

Transcendental Galois theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

A Review Exercises 111

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B Two-hour Examination 117

C Solutions to the Exercises 119

Index 129

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Notations.

We use the standard (Bourbaki) notations:

ND f0;1;2;:::g;ZD ring of integers,RD field of real numbers,CD field of complex numbers,Fp D Z=pZD field withp elements,p a prime number.

Given an equivalence relation, denotes the equivalence class containing . The cardinalityof a setSis denoted by jSj (so jSj is the number of elements inSwhenSis finite). LetIandA be sets. A family of elements ofA indexed by I, denoted by.ai /i2I, is a functioni7! ai WI! A. Throughout the notes,p is a prime number: p D 2;3;5;7;11;:::.

X

Y Xis a subset ofY (not necessarily proper).

X defD Y Xis defined to beY, or equalsY by definition.

X Y Xis isomorphic toY.X' Y XandYare canonically isomorphic (or there is a given or unique isomorphism).

PREREQUISITES

Group theory (for example, GT), basic linear algebra, and some elementary theory of rings.

References.

Dummit, D., and Foote, R.M., 1991, Abstract Algebra, Prentice Hall.Jacobson, N., 1964, Lectures in Abstract Algebra, Volume III Theory of Fields and Galois

Theory, van Nostrand.

Also, the following of my notes (available at www.jmilne.org/math/).

GT Group Theory, v3.13, 2013.

ANT Algebraic Number Theory, v3.04, 2012.

CA A Primer of Commutative Algebra, v2.23, 2012.

A reference monnnn is to http://mathoverflow.net/questions/nnnn/

PARIis an open source computer algebra system freely available here.

ACKNOWLEDGEMENTS

I thank the following for providing corrections and comments for earlier versions of the

notes: Mike Albert, Lior Bary-Soroker, Maren Baumann, Leendert Bleijenga, Tommaso

Centeleghe, Sergio Chouhy, Demetres Christofides, Antoine Chambert-Loir, Dustin Clausen,

Keith Conrad, Hardy Falk, Jens Hansen, Albrecht Hess, Philip Horowitz, Trevor Jarvis,

Henry Kim, Martin Klazar, Jasper Loy Jiabao, Dmitry Lyubshin, John McKay, Georges

E. Melki, Courtney Mewton, Shuichi Otsuka, Dmitri Panov, Alain Pichereau, David G.

Radcliffe, Roberto La Scala, Chad Schoen, Prem L Sharma, Dror Speiser, Bhupendra Nath

Tiwari, Mathieu Vienney, Martin Ward (and class), Xiande Yang, W Xu, and others.

6
http://pari.math.u-bordeaux.fr/http://pari.math.u-bordeaux.fr/


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CHAPTER 1

Basic Definitions and Results

Rings

Aringis a setR with two composition laws C and such that(a) .R; C/is a commutative group;(b) is associative, and there exists1 an element1R such thata 1R D a D 1R a for all

a 2 RI(c) the distributive law holds: for alla;b;c 2 R,

.a Cb/ c D a c C b ca .b C c/ D a b C a c.

We usually omit and write 1 for 1R when this causes no confusion. If1RD 0, thenR

D f0g

.

Asubring Sof a ring Ris a subset that contains 1Rand is closed under addition, passage

to the negative, and multiplication. It inherits the structure of a ring from that onR.

Ahomomorphism of ringsWR ! R0is a map with the properties.a Cb/ D .a/ C .b/; .ab/ D .a/.b/; .1R/ D 1R0 ; alla; b 2 R:

A ringR is said to becommutativeif multiplication is commutative:

ab D ba for alla; b 2 R:A commutative ring is said to be an integral domain if1R 0 and the cancellation lawholds for multiplication:

ab D ac,a 0, impliesb D c:An idealI in a commutative ring R is a subgroup of.R; C/ that is closed under multiplicationby elements ofR:

r2 R,a 2 I, impliesr a 2 I:The ideal generated by elementsa1; : : : ; an is denoted by.a1; : : : ; an/. For example,.a/ is

the principal idealaR.

We assume that the reader has some familiarity with the elementary theory of rings.

For example, in Z (more generally, any Euclidean domain) an idealIis generated by any

smallest nonzero element ofI.

1We follow Bourbaki in requiring that rings have a 1, which entails that we require homomorphisms to

preserve it.

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8 1. BASIC D EFINITIONS AND R ESULTS

Fields

DEFINITION 1. 1 Afieldis a setFwith two composition laws C and such that(a) .F;

C/is a commutative group;

(b) .F; /, whereF D FXf0g, is a commutative group;(c) the distributive law holds.

Thus, a field is a nonzero commutative ring such that every nonzero element has an inverse.

In particular, it is an integral domain. A field contains at least two distinct elements, 0 and 1.

The smallest, and one of the most important, fields is F2 D Z=2ZD f0; 1g.A subfieldSof a field F is a subring that is closed under passage to the inverse. It

inherits the structure of a field from that onF.

LEMMA 1 .2 A nonzero commutative ringR is a field if and only if it has no ideals other

than.0/ andR.

PROOF. SupposeR is a field, and letIbe a nonzero ideal in R. Ifa is a nonzero elementofI, then1 D a1a 2 I, and soID R. Conversely, supposeR is a commutative ring withno proper nonzero ideals. Ifa 0, then .a/ D R, and so there exists a b in R such thatab D 1.

EXAMPLE 1.3 The following are fields: Q, R, C, Fp D Z=pZ (pprime):

A homomorphism of fields WF! F0 is simply a homomorphism of rings. Such ahomomorphism is always injective, because its kernel is a proper ideal (it doesnt contain 1),

which must therefore be zero.

The characteristic of a field

One checks easily that the map

Z! F; n 7! 1FC1FC C 1F .ncopies/;is a homomorphism of rings, and so its kernel is an ideal in Z.

CAS E 1: The kernel of the map is.0/, so that

n 1FD 0H) n D 0 (in Z).Nonzero integers map to invertible elements ofFundern 7! n 1FWZ! F, and so this mapextends to a homomorphism

mn7! .m 1F/.n 1F/1WQ ,! F:

Thus, in this case,Fcontains a copy ofQ, and we say that it has characteristic zero.

CAS E2: The kernel of the map is .0/, so thatn 1FD 0for somen 0. The smallestpositive suchn will be a primep (otherwise there will be two nonzero elements inF whose

product is zero), andp generates the kernel. Thus, the mapn 7! n 1FWZ! Fdefines anisomorphism from Z=pZ onto the subring

fm 1Fj m 2 ZgofF. In this case,Fcontains a copy ofFp , and we say that it has characteristicp.

The fields F2;F3;F5; : : : ;Q are called the prime fields. Every field contains a copy of

exactly one of them.


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Review of polynomial rings 9

REMARK 1.4 The binomial theorem

.a C b/m D am C m1

am1b C m

2

am2b2 C Cbm

holds in every commutative ring. Ifp is prime, thenp divides pnr for allr with1 rpn 1. Therefore, whenFhas characteristicp,.a C b/pn D apn Cbpn alln 1;

and so the mapa 7! apWF! Fis a homomorphism. It is called the Frobenius endomor-phismofF. WhenF is finite, the Frobenius endomorphism is an automorphism.

Review of polynomial rings

For more on the following, see Dummit and Foote 1991, Chapter 9. LetFbe a field.

1. 5 The ring F X of polynomials in the symbol (or indeterminate or variable) X

with coefficients in F is an F-vector space with basis 1, X, . . . , Xn, ... , and with the

multiplication defined byXi

ai XiX

jbjX

j

DX

k

XiCjDk ai bj

Xk:

For any ringR containingF as a subring and elementr ofR, there is a unique homomor-

phismWF X ! Rsuch that.X/ D r and.a/ D afor alla 2 F.

1. 6 Division algorithm: given f .X/ and g.X/ 2 F X with g 0, there exist q.X/,r.X/ 2 F XwithrD 0or deg. r / < deg.g/ such that

fD gq C r Imoreover, q.X/and r.X/are uniquely determined. Thus F Xis a Euclidean domain with

deg as norm, and so is a unique factorization domain.

1. 7 From the division algorithm, it follows that an elementa ofFis a root off (that is,

f.a/ D 0) if and only ifX adividesf. From unique factorization, it now follows that fhas at most deg.f / roots (see also Exercise1-3).

1. 8 Euclids algorithm: Let f and g 2 F X have gcd d.X/. Euclids algorithm constructspolynomials a.X/and b.X/such that

a.X/ f.X/ C b.X/ g.X/ D d.X/; deg.a/ < deg.g/; deg.b/ < deg.f /:Recall how it goes. We may assume thatdeg.f / deg.g/ since the argument is the same inthe opposite case. Using the division algorithm, we construct a sequence of quotients and

remainders

fD q0g C r0g D q1r0 C r1

r0 D q2r1 C r2

rn2 D qnrn1 C rnrn1 D qnC1rn


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with rn the last nonzero remainder. Then,rn dividesrn1, hencern2,.. . , henceg , andhencef. Moreover,

rn

Drn2

qnrn1

Drn2

qn.rn3

qn1rn2/

D Daf

Cbg

and so every common divisor off andg dividesrn: we have shownrn D gcd.f;g/.LetafC bg D d. If deg.a/ deg.g/, writea D gq C r with deg. r / < deg.g/; then

rfC .b C qf /g D d;

andb C qfautomatically has degree< deg.f /.PARI knows Euclidean division: typingdivrem(13,5)in PARI returns2;3, meaning

that13 D 2 5 C3, andgcd(m,n)returns the greatest common divisor ofm and n.

1. 9 LetIbe a nonzero ideal inF X, and letfbe a nonzero polynomial of least degree in

I; then ID .f / (because F X is a Euclidean domain). When we choose fto be monic, i.e.,to have leading coefficient one, it is uniquely determined by I. Thus, there is a one-to-onecorrespondence between the nonzero ideals ofF Xand the monic polynomials in F X.

The prime ideals correspond to the irreducible monic polynomials.

1.10 Since F X is an integral domain, we can form its field of fractions F.X/. Its

elements are quotientsf =g,f andg polynomials,g 0:

Factoring polynomials

The following results help in deciding whether a polynomial is reducible, and in finding its

factors.

PROPOSITION1.11 Supposer2Q is a root of a polynomial

amXm C am1Xm1 C Ca0; ai2 Z;

and letrD c=d,c; d2 Z,gcd.c;d/ D 1. Thencja0anddjam:PROOF. It is clear from the equation

amcm C am1cm1dC Ca0dm D 0

thatdjamc

m, and therefore,djam:Similarly,c

ja0.

EXAMPLE 1.12 The polynomialf .X/ D X3 3X 1is irreducible in QX because itsonly possible roots are 1, andf .1/ 0 f .1/.

PROPOSITION1.13 (GAUSSS L EMMA) Letf .X/ 2 ZX . Iff .X/ factors nontriviallyin QX , then it factors nontrivially in ZX .

PROOF. LetfD gh in QX withg; h Q. For suitable integersm and n,g1 defD mgandh1

defD nh have coefficients in Z, and so we have a factorization

mnf

Dg1

h1in ZX .


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Factoring polynomials 11

If a primep dividesmn, then, looking modulop, we obtain an equation

0 D g1 h1in FpX .Since FpX is an integral domain, this implies that p divides all the coefficients of at least

one of the polynomialsg1; h1, sayg1, so thatg1 D pg2for someg2 2 ZX . Thus, we havea factorization

.mn=p/fD g2 h1in ZX .Continuing in this fashion, we can remove all the prime factors ofm n, and so obtain a

nontrivial factorization off in ZX .

PROPOSITION1.14 Iff2 ZX is monic, then every monic factor off in QX lies inZX .

PROOF. Letg be a monic factor off in QX , so thatfD gh withh 2QX also monic.Letm; nbe the positive integers with the fewest prime factors such that mg; nh 2 ZX . Asin the proof of Gausss Lemma, if a primep dividesmn, then it divides all the coefficientsof at least one of the polynomialsmg;n h, saymg, in which case it dividesm becauseg is

monic. Now mp

g 2 ZX , which contradicts the definition ofm.

REMARK 1.15 We sketch an alternative proof of Proposition1.14.A complex number is

said to be analgebraic integerif it is a root of a monic polynomial in ZX . Proposition1.11

shows that every algebraic integer in Q lies in Z. The algebraic integers form a subring of

C for an elementary proof of this, using nothing but the symmetric polynomials theorem

(5.35), see Chapter 1 of my notes on algebraic geometry. Now let1; : : : ; mbe the roots of

f in C. By definition, they are algebraic integers, and the coefficients of any monic factor of

fare polynomials in (certain of) thei , and therefore are algebraic integers. If they lie in Q,

then they lie in Z.

PROPOSITION1.16 (EISENSTEINS CRITERION) Let

fD amXm C am1Xm1 C Ca0; ai2 ZIsuppose that there is a primep such that:

pdoes not divideam, pdividesam1;:::;a0, p2 does not dividea0.

Thenf is irreducible in QX .

PROOF. Iff .X/ factors in QX , then it factors in ZX , say,

amXm Cam1Xm1 C Ca0 D .br Xr C Cb0/.cs Xs C Cc0/

withbi ; ci2 Z andr;s < m. Sincep, but notp2, dividesa0 D b0c0,p must divide exactlyone ofb0,c0, say,b0. Now from the equation

a1 D b0c1 C b1c0;we see thatpjb1;and from the equation

a2 D b0c2 Cb1c1 Cb2c0;thatp

jb2. By continuing in this way, we find that p dividesb0; b1; : : : ; br , which contradicts

the condition thatp does not divideam.


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The last three propositions hold with Z replaced by any unique factorization domain.

REMARK 1.17 There is an algorithm for factoring a polynomial in QX . To see this,

consider f2 QX . Multiplyf .X/ by a rational number so that it is monic, and thenreplace it by D

deg.f /

f .XD /, withD equal to a common denominator for the coefficients

off, to obtain a monic polynomial with integer coefficients. Thus we need consider only

polynomials

f.X/ D Xm Ca1Xm1 C Cam; ai2 Z:From the fundamental theorem of algebra (see 5.6 below), we know that f splits

completely in CX :

f.X/ DmY

iD1.X i /; i2 C:

From the equation

0

Df .i /

Dmi

Ca1

m1i

C Cam,

it follows that ji j is less than some bound depending only on the degree and coefficients off; in fact,

ji j maxf1;mBg,BD max jai j.Now ifg.X/ is a monic factor off.X/, then its roots in C are certain of the i , and its

coefficients are symmetric polynomials in its roots. Therefore, the absolute values of the

coefficients ofg.X/are bounded in terms of the degree and coefficients off. Since they are

also integers (by1.14), we see that there are only finitely many possibilities for g.X/. Thus,

to find the factors off .X/we (better PARI) have to do only a finite amount of checking.2

Therefore, we need not concern ourselves with the problem of factoring polynomi-

als in the rings QX or FpX since PARI knows how to do it. For example, typing

content(6*X^2+18*X-24) in PARI returns 6, and factor(6*X^2+18*X-24) returnsX 1and XC4, showing that

6X2 C18X 24 D 6.X 1/.XC 4/

in QX . Typing factormod(X^2+3*X+3,7)returnsXC4and XC 6, showing that

X2 C3XC3 D .XC 4/.XC6/

in F7X .

REMARK 1.18 One other observation is useful. Letf2 ZX . If the leading coefficient off is not divisible by a primep, then a nontrivial factorization fD gh in ZX will give anontrivial factorization xfD xgxhin FpX . Thus, iff.X/is irreducible in FpX for someprimep not dividing its leading coefficient, then it is irreducible in ZX . This test is very

useful, but it is not always effective: for example,X4 10X2 C 1is irreducible in ZX butit is reducible3 modulo every primep .

2Of course, there are faster methods than this. The BerlekampZassenhaus algorithm factors the polynomial

over certain suitable finite fields Fp , lifts the factorizations to rings Z=pmZ for somem, and then searches for

factorizations in ZX with the correct form modulopm.3Here is a proof using only that the product of two nonsquares in Fp is a square, which follows from the

fact that Fp is cyclic (see Exercise1-3). If2 is a square in Fp , then

X4 10X2 C 1 D .X2 2p2X1/.X2 C 2p2X 1/:


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Extension fields 13

Extension fields

A fieldE containing a fieldFis called anextension fieldofF(or simply anextensionof

F). Such anE can be regarded as anF-vector space, and we writeE

WF for the dimension,

possibly infinite, ofE as anF-vector space. This dimension is called thedegreeofE overF. We say thatE isfiniteoverFwhen it has finite degree over F:

EXAMPLE 1.1 9 (a) The field of complex numbersC has degree2 over R (basis f1; ig/:(b) The field of real numbers R has infinite degree over Q: the field Q is countable,

and so every finite-dimensional Q-vector space is also countable, but a famous argument of

Cantor shows that R is not countable.

(c) The field ofGaussian numbers

Q.i /defD fa C bi2C j a; b 2Qg

has degree2 over Q (basis f1; ig).(d) The field F.X/ has infinite degree over F; in fact, even its subspace F X has

infinite dimension overF (basis1;X;X2; : : :).

PROPOSITION1.20 (MULTIPLICATIVITY OF DE GREES) Let L E F (all fields andsubfields). ThenL=F is of finite degree if and only ifL=E andE =Fare both of finite

degree, in which case

LWF D LWEEWF :PROOF. IfLis of finite degree over F, then it is certainly of finite degree overE . Moreover,

E , being a subspace of a finite dimensionalF-vector space, is also finite dimensional.

Thus, assume thatL=E and E=Fare of finite degree, and let.ei /1imbe a basis forEas anF-vector space and let.lj/1

j

n be a basis for L as anE-vector space. To complete

the proof, it suffices to show that .ei lj/1im;1jn is a basis forL overF, because thenLwill be finite overFof the predicted degree.

First,.ei lj/i;j spansL. Let2 L. Then, because.lj/j spansL as anE -vector space,DPjjlj; somej2 E;

and because.ei /i spansE as anF-vector space,

jDP

i aijei ; someaij2 F :On putting these together, we find that

DPi;jaijei lj:Second, .ei lj/i;j is linearly independent. A linear relation

Paijei ljD 0, aij2 F,

can be rewrittenP

j.P

iaijei /ljD 0. The linear independence of theljs now shows thatPiaijeiD 0for eachj , and the linear independence of the ei s shows that eachaijD 0.

If3 is a square in Fp , then

X4 10X2 C 1 D .X2 2p

3XC 1/.X2 C 2p

3XC 1/:If neither2 nor 3 are squares,6 will be a square in Fp , and

X4 10X2 C 1 D .X2 .5 C 2p

6//.X2 .5 2p

6//:

The general study of such polynomials requires nonelementary methods. See, for example, the paper

Brandl, R., Amer. Math. Monthly, 93 (1986), pp286288, which proves that every nonprime integern

1

occurs as the degree of a polynomial in ZX that is irreducible over Z but reducible modulo all primes:


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Construction of some extension fields

Let f.X/ 2 F X be a monic polynomial of degree m, and let .f / be the ideal gener-ated byf . Consider the quotient ringF X=.f .X//, and writex for the image ofX in

FX=.f.X//, i.e.,x is the cosetXC .f.X//. Then:(a) The map

P.X/ 7! P.x/WF X ! Fxis a surjective homomorphism in whichf.X/maps to0. Therefore,f .x/ D 0.

(b) From the division algorithm, we know that each element g ofF X=.f / is rep-

resented by a unique polynomial r of degree < m. Hence each element ofF x can be

expressed uniquely as a sum

a0 Ca1x C Cam1xm1; ai2 F: (*)

(c) To add two elements, expressed in the form (*), simply add the corresponding

coefficients.(d) To multiply two elements expressed in the form (*), multiply in the usual way, and

use the relationf .x/ D 0to express the monomials of degree min x in terms of lowerdegree monomials.

(e) Now assumef.X/is irreducible. To find the inverse of an element 2 Fx, use (b)to write D g.x/withg.X/is a polynomial of degree m 1, and use Euclids algorithminF Xto obtain polynomialsa.X/and b.X/such that

a.X/f.X/C b.X/g.X/ D d.X/

with d.X/ the gcd off and g . In our case,d.X/ is 1 because f .X/ is irreducible and

deg g.X/ < degf.X/. When we replaceXwithx , the equality becomes

b.x/g.x/ D 1:

Henceb.x/is the inverse ofg.x/.

From these observations, we can conclude:

1.21 For a monic irreducible polynomialf .X/ of degreem inF X,

Fx D FX=.f.X//

is a field of degreem overF. Moreover, computations inF xreduce to computations inF.

EXAMPLE 1.22 Letf.X/ D X2 C1 2 RX . Then Rx has:elements: a Cbx ,a; b 2 RIaddition: .a C bx/ C .a0 Cb0x/ D .a C a0/ C .b C b0/xImultiplication: .a C bx/.a0 C b0x/ D .aa0 bb 0/ C .ab0 Ca0b/x:

We usually writei forx and C for Rx:

EXAMPLE 1.23 Let f.X/ D X3 3X 1 2 QX . We observed in (1.12) that this isirreducible over Q, and so Qx is a field. It has basis f1;x;x2g as a Q-vector space. Let

D x4 C 2x3 C 3 2Qx:


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Stem fields 15

Then using thatx3 3x 1 D 0, we find that D 3x2 C7x C5. BecauseX3 3X 1isirreducible,

gcd.X3 3X 1;3X2 C 7XC 5/ D 1:

In fact, Euclids algorithm gives

.X3 3X 1/737

XC 29111

C .3X2 C 7XC 5/ 7111

X2 26111

XC 28111

D 1:Hence

.3x2 C 7x C5/ 7111

x2 26111

x C 28111

D 1;and we have found the inverse of:

We can also do this in PARI: beta=Mod(X^4+2*X^3+3,X^3-3*X-1)reveals that D3x2 C7x C 5in Qx, andbeta^(-1)reveals that1 D 7

111x2 26

111x C 28

111.

Stem fields

Letfbe a monic irreducible polynomial inF X. We say thatF is astem field4 forf if

f./ D 0. Then $ xWF ' Fx defD F X=.f /:

Therefore, stem fields always exist, and each element of a stem fieldF forfhas a unique

expression

a0 Ca1 C Cam1m1; ai2 F; m D deg.f /,i.e.,1;;::: ;m1 is a basis forF overF. Arithmetic inF can be performed usingthe same rules as in F x. IfF 0 is a second stem field for f, then there is a uniqueF-isomorphismF ! F 0sending to 0.

The subring generated by a subset

An intersection of subrings of a ring is again a ring. LetFbe a subfield of a field E , and let

Sbe a subset ofE . The intersection of all the subrings ofE containingF andSis evidently

the smallest subring ofE containingF andS. We call it the subring ofE generated byF

andS(orgenerated overF byS), and we denote itF S. WhenSD f1;:::;ng, we writeF 1;:::;nfor F S. For example, CD R

p1.LEMMA 1.24 The ringF Sconsists of the elements ofE that can be expressed as finite

sums of the form

Xai1ini11 inn ; ai1in2 F; i2 S: (*)PROOF. LetR be the set of all such elements. Evidently,R is a subring containing F andS

and contained in every other such subring. Therefore R equalsF S.

EXAMPLE 1.25 The ring Q,D 3:14159:::, consists of the complex numbers that canbe expressed as a finite sum

a0 C a1 C a22 C Cann; ai2Q:The ring Qi consists of the complex numbers of the forma C bi ,a; b 2Q.

4Following A. Albert, Modern Higher Algebra, 1937, who calls the splitting field of a polynomial its root

field. More formally, a stem field forf

is a pair.E;/

consisting of a fieldE

containingF

and a generator

forE overFsuch thatf./ D 0.


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Note that the expression of an element in the form (*) willnotbe unique in general. This

is so already in Ri .

LEMMA 1.26 LetR be an integral domain containing a subfieldF(as a subring). IfR is

finite dimensional when regarded as anF-vector space, then it is a field.PROOF. Let be a nonzero element ofR we have to show that has an inverse inR.

The mapx 7! x WR ! Ris an injective linear map of finite dimensionalF-vector spaces,and is therefore surjective. In particular, there is an element 2 Rsuch that D 1.

Note that the lemma applies to subrings (containingF) of an extension field E ofF of

finite degree.

The subfield generated by a subset

An intersection of subfields of a field is again a field. Let Fbe a subfield of a fieldE , and letSbe a subset ofE . The intersection of all the subfields ofE containingF andSis evidently

the smallest subfield ofE containingF andS. We call it the subfield ofE generated by

F andS (orgenerated overF byS), and we denote itF.S/. It is the field of fractions of

F Sin E , since this is a subfield ofE containingF andSand contained in every other

such field. When SD f1;:::;ng, we writeF .1;:::;n/ for F.S/. Thus, F 1; : : : ; nconsists of all elements ofE that can be expressed as polynomials in thei with coefficients

inF, andF .1; : : : ; n/consists of all elements ofE that can be expressed as the quotient

of two such polynomials.

Lemma1.26shows that F Sis already a field if it is finite dimensional over F, in which

caseF.S/ D F S.EXAMPLE 1.27 The field Q./,D 3:14:::, consists of the complex numbers that can beexpressed as a quotient

g./=h./; g.X/;h.X/ 2QX; h.X/ 0:

The ring Qi is already a field.

An extensionE ofFis said to be simpleifE D F./some 2 E. For example, Q./and Qi are simple extensions ofQ:

LetF andF0 be subfields of a fieldE . The intersection of the subfields ofE containingF andF0 is evidently the smallest subfield ofE containing bothF andF0. We call it thecompositeofF andF0in E , and we denote itF F0. It can also be described as the subfieldofE generated overF byF0, or the subfield generated overF0 by F:

F .F0/ D FF0 D F0.F /.

Algebraic and transcendental elements

For a fieldFand an element of an extension fieldE , we have a homomorphism

f.X/ 7! f./WF X ! E:

There are two possibilities.


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Algebraic and transcendental elements 17

CAS E 1: The kernel of the map is.0/, so that, forf2 F X,

f./ D 0H) fD 0(inF X).

In this case, we say that transcendental overF. The homomorphismF X ! Fis anisomorphism, and it extends to an isomorphismF.X/ ! F./.

CAS E2: The kernel is .0/, so thatg./ D 0for some nonzerog 2 F X. In this case,we say that is algebraic overF. The polynomialsg such thatg./ D 0form a nonzeroideal inF X, which is generated by the monic polynomialfof least degree such f ./ D 0.We callf theminimum polynomialof overF. It is irreducible, because otherwise there

would be two nonzero elements ofE whose product is zero. The minimum polynomial is

characterized as an element ofF Xby each of the following sets of conditions:

f is monic;f ./ D 0and divides every other polynomialg in F Xwithg./ D 0.f is the monic polynomial of least degree such thatf ./ D 0If is monic, irreducible, andf ./ D 0.

Note thatg.X/ 7! g./ defines an isomorphism F X=.f / ! F. Since the first is afield, so also is the second:

F./ D F:Thus,F is a stem field forf.

EXAMPLE 1.28 Let 2 C be such that 3 3 1 D 0. ThenX3 3X 1 is monic,irreducible, and has as a root, and so it is the minimum polynomial of over Q. The set

f1;;2g is a basis for Q over Q. The calculations in Example1.23show that if is theelement4 C23 C3ofQ, then D 32 C 7 C 5, and

1 D 7111

2 26111

C 28111

:

REMARK 1.29 PARI knows how to compute in Q. For example, factor(X^4+4)re-

turns the factorization

X4 C 4 D .X2 2XC2/.X2 C 2XC 2/

in QX . Now type nf=nfinit(a^2+2*a+2)to define a number field nf generated over

Q by a roota ofX2 C 2XC 1. Then nffactor(nf,x^4+4)returns the factorization

X4 C 4 D .X a 2/.X a/.XC a//.XC a C 2/;

in Qa.

A field extensionE=Fis said to be algebraic, andE is said to bealgebraic overF, if

all elements ofE are algebraic overF; otherwise it is said to be transcendental(or E is

said to betranscendental overF). Thus,E=Fis transcendental if at least one element of

E is transcendental overF.

PROPOSITION1.30 A field extensionE=Fis finite if and only ifE is algebraic and finitely

generated (as a field) overF.

PROOF.H): To say that is transcendental over Famounts to saying that its powers1;;2; : : : are linearly independent over F. Therefore, ifE is finite over F, then it is

algebraic overF. It remains to show thatE is finitely generated overF. IfED

F, then it


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is generated by the empty set. Otherwise, there exists an1 2 E X F. IfE F 1, thereexists an2 2 E X F 1, and so on. Since

F1

WF < F 1; 2

WF 1, has no rationalroot. SinceN , it cant be a root off.X/, and soxN 0. Evidently,xN2 Q; in fact.2N /dDxN2 Z, and so

j.2N /dDxNj 1. (*)From the fundamental theorem of algebra (see 5.6 below), we know that f splits in

CX , say,

f.X/ DdY

iD1.X i /; i2 C; 1 D ;

and so

jxNj DdY

iD1jN i j jN 1j.N C M /d1; whereMD max

i1f1; ji jg.

But

jN 1j D1X

nDNC1

1

2n 1

2.NC1/

1XnD0

1

2n

!D 2

2.NC1/:

Hence

jxNj 2

2.NC1/ .N CM /d1

and

j.2N /dDxNj 2 2dN D2.NC1/

.N CM /d1

which tends to0 as N! 1 because 2d N2.NC1/

D

2d

2NC1

N ! 0. This contradicts (*).

6More precisely, choose a bijection from some segment0;n.1/ofN ontoA.10/; extend it to a bijection

from a segment0;n.2/ontoA.100/, and so on.7This proof, which I learnt from David Masser, also works for P 1an for every integera 2.8In fact is not rational because its expansion to base 2 is not periodic.


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Constructions with straight-edge and compass.

The Greeks understood integers and the rational numbers. They were surprised to find

that the length of the diagonal of a square of side 1, namely,p

2, is not rational. They

thus realized that they needed to extend their number system. They then hoped that theconstructible numbers would suffice. Suppose we are given a length, which we call 1, a

straight-edge, and a compass (device for drawing circles). A real number (better a length) is

constructibleif it can be constructed by forming successive intersections of

lines drawn through two points already constructed, and circles with centre a point already constructed and radius a constructed length.

This led them to three famous questions that they were unable to answer: is it possible

to duplicate the cube, trisect an angle, or square the circle by straight-edge and compass

constructions? Well see that the answer to all three is negative.

LetFbe a subfield ofR. For a positivea

2F,

padenotes the positive square root ofa

in R. TheF-planeis FF RR. We make the following definitions:AnF-lineis a line in RR through two points in theF-plane. These are thelines given by equations

ax C by C c D 0; a; b; c 2 F:

AnF-circleis a circle in RR with centre anF-point and radius an elementofF. These are the circles given by equations

.x a/2 C .y b/2 D c2; a; b; c 2 F:

LEMMA 1.34 LetL L0 beF-lines, and letC C0 beF-circles.(a) L \ L0 D ; or consists of a singleF-point.(b) L \ CD ; or consists of one or two points in theF pe-plane, somee 2 F,e > 0.(c) C\ C0 D ; or consists of one or two points in theF pe-plane, somee 2 F,e > 0.

PROOF. The points in the intersection are found by solving the simultaneous equations, and

hence by solving (at worst) a quadratic equation with coefficients in F.

LEMMA 1.35 (a) Ifc andd are constructible, then so also arec C d,c, cd, and cd

.d 0/.(b) Ifc > 0is constructible, then so also is

pc.

SKETCH OF PROOF. First show that it is possible to construct a line perpendicular to a givenline through a given point, and then a line parallel to a given line through a given point.

Hence it is possible to construct a triangle similar to a given one on a side with given length.

By an astute choice of the triangles, one constructs cd and c1. For (b), draw a circle ofradius cC12 and centre.

cC12

; 0/, and draw a vertical line through the point A D .1;0/to meetthe circle at P. The length AP is

pc. (For more details, see Artin, M., 1991, Algebra,

Prentice Hall, Chapter 13, Section 4.)

THEOREM 1.36 (a) The set of constructible numbers is a field.

(b) A number is constructible if and only if it is contained in a subfield ofR of the form

Qpa1; : : : ;par ; ai2Qpa1; : : : ;pai1; ai > 0.


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Constructions with straight-edge and compass. 21

PROOF. (a) Immediate from (a) of Lemma1.35.

(b) It follows from Lemma1.34that every constructible number is contained in such

a field Qp

a1; : : : ;p

ar . Conversely, if all the elements ofQp

a1; : : : ;p

ai1 are con-structible, then

pai is constructible (by1.35b), and so all the elements ofQ

pa1; : : : ;

pai

are constructible (by (a)). Applying this for iD 0;1;:::, we find that all the elements ofQ

pa1; : : : ;

par are constructible.

COROLLARY 1.37 Ifis constructible, then is algebraic overQ, andQWQ is a powerof2.

PROOF. According to Proposition1.20,QWQdivides

Qp

a1 par WQ

andQp

a1; : : : ;p

ar WQis a power of2.

COROLLARY 1.38 It is impossible to duplicate the cube by straight-edge and compass

constructions.

PROOF. The problem is to construct a cube with volume2. This requires constructing the

real root of the polynomial X3 2. But this polynomial is irreducible (by Eisensteinscriterion1.16for example), and soQ

3p

2WQ D 3.

COROLLARY 1.39 In general, it is impossible to trisect an angle by straight-edge and

compass constructions.

PROOF. Knowing an angle is equivalent to knowing the cosine of the angle. Therefore, to

trisect3, we have to construct a solution to

cos 3 D 4 cos3 3 cos :

For example, take 3D 60 degrees. As cos 60D 12

, to construct , we have to solve

8x3 6x 1 D 0, which is irreducible (apply1.11).

COROLLARY 1.40 It is impossible to square the circle by straight-edge and compass con-

structions.

PROOF. A square with the same area as a circle of radius r has sidep

r . Since is

transcendental9,so also isp

.

We now consider another famous old problem, that of constructing a regular polygon.

Note thatXm 1is not irreducible; in fact

Xm 1 D .X 1/.Xm1 C Xm2 C C1/:

LEMMA 1.41 Ifp is prime thenXp1 C C1is irreducible; henceQe2i=phas degreep 1overQ:

9Proofs of this can be found in many books on number theory, for example, in 11.14 of

Hardy, G. H., and Wright, E. M., An Introduction to the Theory of Numbers, Fourth Edition, Oxford, 1960.


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PROOF. Letf .X/ D .Xp 1/=.X 1/ D Xp1 C C1; then

f .XC 1/ D .XC 1/p 1

X D Xp1 C Ca2X2 Ca1XCp;

with ai D piC1. Now pjai for i D 1;:::;p 2, and so f .XC1/ is irreducible by Eisensteinscriterion1.16.This implies thatf.X/is irreducible.

In order to construct a regularp-gon,p an odd prime, we need to construct

cos 2pD .e 2ip C .e 2ip /1/=2:

But

Qe2ip Qcos 2

p Q;

and the degree ofQe2ip over Qcos2

p

is 2 the equation

2 2 cos2p C1 D 0; D e 2ip ;

shows that it is 2, and it is not1 because Qe 2ip is not contained in R. Hence

Qcos 2p

WQ D p 12

:

Thus, if the regularp-gon is constructible, then.p 1/=2 D 2k for somek (later (5.12),we shall see a converse), which impliesp D 2kC1 C1. But2r C1can be a prime only ifris a power of2, because otherwiser has an odd factort and fort odd,

Yt C 1 D .YC 1/.Yt1 Yt2 C C1/I

whence

2st C 1 D .2s C1/..2s /t1 .2s /t2 C C1/.Thus if the regularp-gon is constructible, then p D 22k C 1for somek . Fermat conjecturedthat all numbers of the form 22

k C1are prime, and claimed to show that this is true for k 5 for this reason primes of this form are calledFermat primes. For 0 k 4, the numbersp D 3;5;17;257;65537, are prime but Euler showed that232 C 1 D .641/.6700417/, andwe dont know of any more Fermat primes.

Gauss showed that10

cos2

17D 1

16C 1

16

p17C 1

16

q34 2

p17C 1

8

r17 C 3

p17

q34 2

p17 2

q34 C2

p17

when he was 18 years old. This success encouraged him to become a mathematician.

10Or perhaps that

cos 217D 116 C 116p

17 C 116p

34 2p17 C 18q

17 C 3p17 2p

34 2p17 p

170 26p17 both expressions are correct.


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Algebraically closed fields 23

Algebraically closed fields

We say that a polynomial splits in F X (or, more loosely, in F) if it is a product of

polynomials of degree1 in F X.

PROPOSITION1.42 For a field , the following statements are equivalent:

(a) Every nonconstant polynomial in Xsplits inX.

(b) Every nonconstant polynomial in Xhas at least one root in .

(c) The irreducible polynomials inXare those of degree1.

(d) Every field of finite degree over equals .

PROOF. The implications (a) H) (b) H) (c) H) (a) are obvious.(c) H) (d). LetE be a finite extension of . The minimum polynomial of any element ofE has degree1, and so 2 .(d) H) (c). Letfbe an irreducible polynomial inX. ThenX =.f /is an extensionfield of of degree deg.f / (see1.30), and so deg.f /

D1.

DEFINITION1.43 (a) A field is said to bealgebraically closedif it satisfies the equivalent

statements of Proposition1.42.

(b) A field is said to be analgebraic closureof a subfield Fwhen it is algebraically

closed and algebraic overF.

For example, the fundamental theorem of algebra (see5.6 below) says that C is alge-

braically closed. It is an algebraic closure ofR.

PROPOSITION1.44 If is algebraic overF and every polynomialf2 F X splits in

X, then is algebraically closed (hence an algebraic closure ofF).PROOF. Letfbe a nonconstant polynomial in X. We have to show that fhas a root in

. We know (see1.21) thatfhas a root in some finite extension 0of . Set

fD anXn C Ca0,ai2 ;

and consider the fields

F F a0; : : : ; an F a0; : : : ; an;:

Each extension is algebraic and finitely generated, and hence finite (by1.30). Therefore

lies in a finite extension ofF, and so is algebraic over F it is a root of a polynomial g

with coefficients inF. By assumption,g splits inX, and so the roots ofg in 0all lie in . In particular, 2 :

PROPOSITION1.45 Let F; then

f 2 j algebraic overFg

is a field.

PROOF. If and are algebraic overF, thenF ;is a field (by1.31) of finite degree

overF (by1.30). Thus, every element ofF ;is algebraic overF, including ,= ,.


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The field constructed in the proposition is called the algebraic closure ofF in .

COROLLARY 1.46 Let be an algebraically closed field. For any subfieldF of , the

algebraic closure ofF in is an algebraic closure ofF:

PROOF. From its definition, we see that it is algebraic overFand every polynomial inF Xsplits in it. Now Proposition1.44shows that it is an algebraic closure ofF.

Thus, when we admit the fundamental theorem of algebra (5.6), every subfield ofC has

an algebraic closure (in fact, a canonical algebraic closure). Later (Chapter 6) we shall prove

(using the axiom of choice) that every field has an algebraic closure.

Exercises

Exercises marked with an asterisk were required to be handed in.

1-1 (*) Let ED Q, where 3 2 C C 2 D 0. Express . 2 C C 1/.2 / and. 1/1 in the forma2 C b C c witha;b;c 2Q.

1-2 (*) DetermineQ.p

2;p

3/WQ.

1-3 (*) LetFbe a field, and let f .X/ 2 F X.(a) For everya 2 F, show that there is a polynomialq.X/ 2 F Xsuch that

f.X/ D q.X/.X a/ Cf .a/:

(b) Deduce thatf .a/ D 0if and only if.X a/jf.X/.(c) Deduce thatf .X/can have at most deg f roots.

(d) LetG be a finite abelian group. IfG has at mostm elements of order dividing m for

each divisorm of.GW1/, show thatG is cyclic.(e) Deduce that a finite subgroup ofF,Fa field, is cyclic.

1-4 (*) Show that with straight-edge, compass, and angle-trisector, it is possible to con-

struct a regular7-gon.


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CHAPTER 2

Splitting Fields; Multiple Roots

Maps from simple extensions.

LetE andE 0 be fields containingF. AnF-homomorphismis a homomorphism

'WE ! E0

such that'.a/ D afor alla 2 F. Thus anF-homorphism' maps a polynomialXai1im

i11 imm; ai1im2 F;

to

Xai1im'.1/

i1 '.m/im:

AnF-isomorphismis a bijectiveF-homomorphism. Note that ifE andE 0have the samefinite degree overF, then everyF-homomorphism is anF-isomorphism.

PROPOSITION2 .1 LetF ./be a simple field extension of a fieldF, and let be a second

field containingF.

(a) Let be transcendental overF. For everyF-homomorphism'WF./ ! ,'./istranscendental overF, and the map' 7! './defines a one-to-one correspondence

fF-homomorphisms ' WF./ ! g $ felements of transcendental overFg:

(b) Letbe algebraic overF with minimum polynomialf.X/. For everyF-homomorphism

'W

F!

,'./is a root off .X/in , and the map'7!

'./defines a one-to-

one correspondence

fF-homomorphisms' WF ! g $ froots off ing:

In particular, the number of such maps is the number of distinct roots off in .

PROOF. (a) To say that is transcendental over Fmeans thatF is isomorphic to the

polynomial ring in the symbol with coefficients inF. For every2 , there is a uniqueF-homomorphism 'WF ! sending to (see 1.5). This extends to the field offractions F ./ ofF if and only if all nonzero elements ofF are sent to nonzero

elements of , which is so if and only if is transcendental.

(b) Let f.X/DPai Xi , and consider an F-homomorphism 'WF ! . On applying' to the equation Pai i D 0, we obtain the equation Pai './i D 0, which shows that

25


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26 2. SPLITTING F IELDS; MULTIPLE ROOTS

'./ is a root of f.X/ in . Conversely, if 2 is a root of f.X/, then the mapF X ! , g.X/ 7! g./, factors through FX=.f.X//. When composed with the inverseof the isomorphism XCf.X/ 7! WFX=.f.X// ! F, this becomes a homomorphismF

! sending to .

We shall need a slight generalization of this result.

PROPOSITION2 .2 LetF ./be a simple field extension of a fieldF, and let'0WF! bea homomorphism ofFinto a second field .

(a) If is transcendental overF, then the map' 7! './defines a one-to-one correspon-dence

fextensions'WF./ ! of'0g $ felements of transcendental over'0.F /g:

(b) If is algebraic overF, with minimum polynomialf.X/, then the map'

7!'./

defines a one-to-one correspondence

fextensions'WF ! of'0g $ froots of'0f ing:

In particular, the number of such maps is the number of distinct roots of'0f in .

By '0f we mean the polynomial obtained by applying '0 to the coefficients off:

if fD Pai Xi then '0fD P'.ai /Xi . By an extension of '0 to F./ we mean ahomomorphism' WF./ ! such that'jFD '0.

The proof of the proposition is essentially the same as that of the preceding proposition.

Splitting fields

Letfbe a polynomial with coefficients in F. A fieldE containingFis said tosplitf iff

splits inEX:f.X/ DQmiD1.X i /withi2 E. If, in addition, E is generated by theroots off,

E D F 1; : : : ; m;then it is called a splittingor root fieldforf. Note that

Qfi .X /

mi (mi 1) andQ

fi .X /

have the same splitting fields. Also, that iff has deg.f / 1 roots in E , then it splits inEX.

EXAMPLE 2 .3 (a) Letf .X/

DaX2

CbX

Cc

2QX , and let

D

pb2

4ac. The sub-

field Q ofC is a splitting field forf.(b) Letf.X/ D X3 C aX2 C bXC c2QX be irreducible, and let 1; 2; 3 be its

roots in C. Since the nonreal roots offoccur in conjugate pairs, either 1 or 3 of the i are

real. Then Q1; 2; 3 DQ1; 2is a splitting field for f.X/. Note thatQ1WQ D 3and thatQ1; 2WQ1 D 1or2, and soQ1; 2WQ D 3or6. Well see later (4.2) thatthe degree is3 if and only if the discriminant off.X/is a square in Q. For example, the

discriminant ofX3 C bXC c is 4b 3 27c2, and so the splitting field ofX3 C 10XC 1has degree6 over Q.

PROPOSITION2 .4 Every polynomialf2 F Xhas a splitting fieldEf, and

Ef WF .deg f / .factorialdegf /:


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Splitting fields 27

PROOF. LetF1 D F 1be a stem field for some monic irreducible factor off inF X.Thenf .1/ D 0, and we letF2 D F12be a stem field for some monic irreducible factoroff .X/=.X 1/in F1X . Continuing in this fashion, we arrive at a splitting fieldEf.

Letn

Ddegf. ThenF1

WF

Ddegg1

n,F2

WF1

n

1;:::, and soEf

WE

n.

REMARK 2 .5 For a given integern, there may or may not exist polynomials of degree n

inF X whose splitting field has degree n this depends onF. For example, there do

not forn > 1 ifFD C (see5.6), nor for n > 2 ifFD Fp (see4.21) orFD R. However,later(4.32) we shall see how to write down infinitely many polynomials of degree nin QX

whose splitting fields have degreen.

EXAMPLE 2 .6 (a) Letf .X/ D .Xp 1/=.X 1/ 2QX ,p prime. Ifis one root off,then the remaining roots are2; 3; : : : ; p1, and so the splitting field off is Q.

(b) SupposeF is of characteristicp, and letfD Xp X a 2 F X. If is one rootoff, then the remaining roots are

C1;:::;

Cp

1, and so any field generated over F by

is a splitting field forf (andF ' F X=.f /iff is irreducible).(c) If is one root ofXn a, then the remaining roots are all of the form , where

n D 1. Therefore, ifFcontains all thenth roots of1 (by which we mean thatXn 1splitsinF X), thenF is a splitting field forXn a. Note that ifp is the characteristic ofF,thenXp 1 D .X 1/p , and soFautomatically contains all the pth roots of1.

PROPOSITION2 .7 Letf2 F X. LetE be a field generated overFby roots off, and let be a field containingFin whichf splits.

(a) There exists anF-homomorphism' WE ! ; the number of such homomorphisms isat mostE

WF , and equalsE

WF iff has distinct roots in .

(b) IfE and are both splitting fields forf, then eachF-homomorphismE ! is anisomorphism. In particular, any two splitting fields forf areF-isomorphic.

PROOF. Byf splitting in , we mean that

f.X/ DYdeg.f /

iD1 .X i /; i2 ;

inX. Byfhaving distinct roots in , we mean thati j ifi j .We begin with an observation: letF,f, and be as in the statement of the proposition,

letL be a subfield of containingF, and letg be a factor off inLX; theng divides

f inXand so (by unique factorization in X),g is product of certain number of the

factorsX

i off inX; in particular, we see thatg splits in , and that its roots are

distinct if the roots off are distinct.

(a) By assumption, ED F 1;:::;m with thei (some of the) roots off .X/. Theminimum polynomial of1 is an irreducible polynomial f1 dividing f, anddeg.f1/ DF1WF . From the initial observation with L D F, we see thatf1 splits in , and thatits roots are distinct if the roots offare distinct. According to Proposition2.1,there exists

anF-homomorphism'1WF 1 ! , and the number of such homomorphisms is at mostF1WF , with equality holding whenf has distinct roots in .

The minimum polynomial of2overF 1is an irreducible factorf2off inF 1X.

On applying the initial observation with L D '1F 1and g D '1f2, we see that '1f2splitsin , and that its roots are distinct if the roots off are distinct. According to Proposition

2.2,each'1extends to a homomorphism'2W

F 1; 2!

, and the number of extensions

is at mostF1; 2WF 1, with equality holding whenf has distinct roots in :


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On combining these statements we conclude that there exists an F-homomorphism

'WF 1; 2 ! ;and that the number of such homomorphisms is at most F1; 2

WF , with equality holding

iff has distinct roots in:After repeating the argumentm times, we obtain (a).

(b) EveryF-homomorphismE ! is injective, and so, if there exists such a homomor-phisms,E WF WF . IfE and are both splitting fields forf, then (a) shows that thereexist homomorphismFE, and soE WF D WF . Therefore, everyF-homomorphismE ! is an isomorphism.

COROLLARY 2 .8 LetE andL be extension fields ofF, withE finite overF.

(a) The number ofF-homomorphisms E ! Lis at mostE WF .(b) There exists a finite extension=Land anF-homomorphismE ! :

PROOF. WriteED

F 1; : : : ; m, andf be the product of the minimum polynomials of

thei . Let be a splitting field for fregarded as an element ofLX. The proposition

shows that there is an F-homomorphismE ! , and the number of such homomorphismsis EWF . This proves (b), and since anF-homomorphism E ! Lcan be regarded as anF-homomorphismE ! , it also proves (a).

REMARK 2 .9 (a) LetE1; E2; : : : ; Embe finite extensions ofF, and letLbe an extension of

F. The corollary implies that there exists a finite extension /Lcontaining an isomorphic

copy of everyEi .

(b) Letf2 F X. IfE and E 0 are both splitting fields off, then we know there is anF-isomorphismE ! E0, but there will in general be no preferredsuch isomorphism. Errorand confusion can result if you simply identify the fields. Also, it makes no sense to speak

of the field F generated by a root off unlessfis irreducible (the fields generated

by the roots of two different factors are unrelated). Even when fis irreducible, it makes

no sense to speak of the field F ;generated by two roots; off (the extensions of

Fgenerated by the roots of two different factors off inF X may be very different).

Multiple roots

Letf; g 2 F X. Even whenf andg have no common factor in F X, one might expectthat they could acquire a common factor in X for some F. In fact, this doesnthappen greatest common divisors dont change when the field is extended.

PROPOSITION2.10 Letf andg be polynomials in F X, and let F. If r.X/ isthe gcd off andg computed in F X, then it is also the gcd off andg in X. In

particular, distinct monic irreducible polynomials in F Xdo not acquire a common root in

any extension field ofF:

PROOF. LetrF.X/and r .X/be the greatest common divisors off andg in F X and

X respectively. CertainlyrF.X/jr .X/in X, but Euclids algorithm (1.8) showsthat there are polynomials a and b in F Xsuch that

a.X/f.X/C b.X/g.X/ D rF.X/;and sor .X/dividesrF.X /in X.

For the second statement, note that the hypotheses imply that gcd.f;g/D

1(in F X),

and sof andg cant acquire a common factor in any extension field.


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Multiple roots 29

The proposition allows us to speak of the greatest common divisor off andg without

reference to a field.

Letf2 F X, and let

f.X/ D arY

iD1.X i /mi ; i distinct,mi 1,

rXiD1

miD deg.f /; (*)

be a splitting off in some extension field ofF. We say that i is a root off of

multiplicitymi . Ifmi > 1,i is said to be amultiple rootoff, and otherwise it is a simple

root.

The unordered sequence of integersm1; : : : ; mr in (*) is independent of the extension

field in whichf splits. Certainly, it is unchanged when is replaced with its subfield

F 1; : : : ; m, butF 1; : : : ; mis a splitting field for f, and any two splitting fields are

isomorphic (2.7b).

We say thatfhas a multiple rootwhen at least one of themi > 1, and we say that f

hasonly simple roots when allmiD 1.We wish to determine when a polynomial has a multiple root. Iffhas a multiple factor

inF X, sayfD Qfi .X/mi with somemi > 1, then obviously it will have a multiple root.IffD Qfi with thefi distinct monic irreducible polynomials, then Proposition2.10showsthatfhas a multiple root if and only if at least one of the fi has a multiple root. Thus, it

suffices to determine when an irreducible polynomial has a multiple root.

EXAMPLE 2.11 LetFbe of characteristicp 0, and assume thatFcontains an elementathat is not apth-power, for example,a D Tin the field Fp.T/:ThenXp ais irreducibleinF X, butXp a 1:4D .X /p in its splitting field. Thus an irreducible polynomial canhave multiple roots.

Define the derivativef0.X/of a polynomialf.X/ DPai Xi to be P i ai Xi1. Whenf has coefficients in R, this agrees with the definition in calculus. The usual rules for

differentiating sums and products still hold, but note that in characteristic p the derivative of

Xp is zero.

PROPOSITION2.12 For a nonconstant irreducible polynomialf inF X, the following

statements are equivalent:

(a) fhas a multiple root;

(b) gcd.f;f0/ 1;(c) Fhas characteristicp

0andfis a polynomial inXp;

(d) all the roots off are multiple.

PROOF. (a)H) (b). Let be a multiple root off, and writefD .X /mg.X/,m > 1,in some splitting field. Then

f0.X / D m.X /m1g.X/ C .X /mg0.X/:

Hencef0./ D 0, and so gcd.f;f0/ 1.(b)H) (c). Sincef is irreducible and deg.f0/ < deg.f /,

gcd.f;f0/ 1H) f0 D 0:

But, becausef is nonconstant,f0can be zero only if the characteristic is p

0andf is a

polynomial inXp .


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(c)H) (d). Supposef.X/ D g.Xp/, and letg.X/ DQi .X ai /mi in some splittingfield forf. Then

f.X/

Dg.Xp/

DYi .Xp

ai /

mi

DYi .X i /pmi

wherepi D ai . Hence every root off .X/has multiplicity at leastp.

(d)H) (a). Obvious.

DEFINITION2.13 A polynomial f2 F Xis said to be separable overF if none of itsirreducible factors has a multiple root (in a splitting field).1

The preceding discussion shows thatf2 F Xwill be separable unless(a) the characteristic ofF isp 0,and(b) at least one of the irreducible factors offis a polynomial inXp .

Note that, iff2 F Xis separable, then it remains separable over every field containingF (condition (b) of2.12continues to hold see2.10).

DEFINITION2.14 A fieldFis said to be perfectif all polynomials inF Xare separable

(equivalently, all irreducible polynomials inF Xare separable).

PROPOSITION2.15 A field of characteristic zero is always perfect, and a fieldFof charac-

teristicp 0is perfect if and only if every element ofF is apth power.PROOF. A field of characteristic zero is obviously perfect, and so we may suppose Fto be of

characteristic p 0. IfFcontains an element athat is not a pth power, then Xp a 2 F Xis not separable (see2.11). Conversely, if every element ofF is apth power, then everypolynomial inXp with coefficients inF is apth power inF X,X

ai Xp D .

Xbi X /

p if aiD bpi ,

and so is not irreducible.

EXAMPLE 2.16 (a) A finite field F is perfect, because the Frobenius endomorphism

a 7! apWF! Fis injective and therefore surjective (by counting).(b) A field that can be written as a union of perfect fields is perfect. Therefore, every field

algebraic over Fp is perfect.

(c) Every algebraically closed field is perfect.(d) IfF0has characteristicp 0, thenFD F0.X/is not perfect, becauseXis not apthpower.

1This is the standard definition, although some authors, for example, Dummit and Foote 1991, 13.5, give a

different definition.


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Exercises 31

Exercises

2-1 (*) LetFbe a field of characteristic 2.(a) LetE be quadratic extension ofF (i.e.,E

WF

D2); show that

S.E/ D fa 2 F j ais a square inEg

is a subgroup ofF containingF2.(b) LetE and E 0 be quadratic extensions ofF; show that there is an F-isomorphism

'WE ! E0 if and only ifS.E/ D S.E 0/.(c) Show that there is an infinite sequence of fields E1; E2; : : : with Ei a quadratic

extension ofQ such thatEi is not isomorphic toEj fori j .(d) Letp be an odd prime. Show that, up to isomorphism, there is exactly one field with

p2 elements.

2-2 (*) (a) LetFbe a field of characteristic p. Show that ifXp X a is reducible inF X, then it splits into distinct factors inF X.

(b) For every primep, show thatXp X 1is irreducible in QX .

2-3 (*) Construct a splitting field forX5 2over Q. What is its degree over Q?

2-4 (*) Find a splitting field ofXpm 1 2 FpX . What is its degree over Fp?

2- 5 Letf2 F X, whereFis a field of characteristic 0. Letd.X/ D gcd.f;f0/. Showthat g.X/ D f.X/d.X/1 has the same roots as f.X/, and these are all simple roots ofg.X/.

2-6 (*) Let f.X/ be an irreducible polynomial in F X, whereFhas characteristic p .

Show thatf.X/can be written f .X/ D g.Xpe/whereg.X/is irreducible and separable.Deduce that every root off .X/ has the same multiplicitype in any splitting field.


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CHAPTER 3

The Fundamental Theorem of Galois

Theory

In this chapter, we prove the fundamental theorem of Galois theory, which gives a one-to-one

correspondence between the subfields of the splitting field of a separable polynomial and the

subgroups of the Galois group off.

Groups of automorphisms of fields

Consider fieldsE F. An F-isomorphismE! E is called an F-automorphismofE .TheF-automorphisms ofE form a group, which we denote Aut.E=F/.

EXAMPLE 3 .1 (a) There are two obvious automorphisms ofC, namely, the identity map

and complex conjugation. Well see later(8.18) that by using the Axiom of Choice one canconstruct uncountably many more.

(b) LetED C.X /. An automorphism ofE sendsX to another generator ofE overC. It follows from (8.24) below that these are exactly the elements aXCb

cXCd, ad bc 0.Therefore Aut.E=C/consists of the mapsf.X/ 7! f

aXCbcXCd

,ad bc 0, and so

Aut.E=C/ ' PGL2.C/;

the group of invertible2 2matrices with complex coefficients modulo its centre. Analystswill note that this is the same as the automorphism group of the Riemann sphere. This is not a

coincidence: the field of meromorphic functions on the Riemann sphere P1Cis C.z/

'C.X /,

and so there is certainly a map Aut.P1C/ ! Aut.C.z/=C/, which one can show to be anisomorphism.

(c) The group Aut.C.X1; X2/=C/is quite complicated there is a map

PGL3.C/ D Aut.P2C/ ,! Aut.C.X1; X2/=C/;

but this is very far from being surjective. When there are moreXs, the group is not known.

The groupAut.C.X1; : : : ; X n/=C/is the group of birational automorphisms ofPnC, and is

called theCremona group. Its study is part of algebraic geometry. See theWikipedia.

In this section, we shall be concerned with the groups Aut.E=F/when E is a finite

extension ofF.

33
http://en.wikipedia.org/wiki/Cremona_grouphttp://en.wikipedia.org/wiki/Cremona_grouphttp://en.wikipedia.org/wiki/Cremona_group


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34 3. THE F UNDAMENTAL T HEOREM OF G ALOIS T HEORY

PROPOSITION3 .2 If E is a splitting field of a separable polynomial f2 F X, thenAut.E=F/has orderE WF :PROOF. LetfD

Qf

mii , with thefi irreducible and distinct. The splitting field off is

the same as the splitting field ofQfi . Hence we may assumef is a product of distinctseparable irreducible polynomials, and so has deg fdistinct roots in E . Now Proposition

2.7shows that there areE WF distinctF-homomorphismsE ! E . BecauseE has finitedegree overF, they are automatically isomorphisms.

EXAMPLE 3 .3 (a) Consider a simple extensionE D F, and letfbe a polynomial withcoefficients inF having as a root. Iffhas no root inE other than, thenAut.E=F/ D 1:For example, if

3p

2denotes the real cube root of2, thenAut.Q 3p

2=Q/ D 1. Thus, in theproposition, it is essential thatE be a splittingfield.

(b) LetFbe a field of characteristicp 0, and leta be an element ofFthat is not apthpower. ThenfD Xp a has only one root in a splitting field E , and soAut.E=F/ D 1.Thus, in the proposition, it is essential thatE be a splitting field of a separablepolynomial.

WhenG is a group of automorphisms of a fieldE , we set

EG D Inv.G/ D f 2 E j D , all2 Gg:It is a subfield ofE , called the subfield ofG-invariants ofE or thefixed fieldofG .

In this section, we shall show that, when Eis the splitting field of a separable polynomial

inF XandG D Aut.E=F/, then the mapsM7! Aut.E=M /; H 7! Inv.H /

give a one-to-one correspondence between the set of intermediate fieldsM,F M E,and the set of subgroupsH ofG .THEOREM 3.4 (E. ARTIN) LetG be a finite group of automorphisms of a fieldE , and let

FD EG; thenE WF .GW1/:PROOF. LetG D f1 D 1; : : : ; mg, and let1; : : : ; n ben > m elements ofE . We shallshow that thei are linearly dependent overF. In the system of linear equations

1.1/X1 C C1.n/Xn D 0

m.1/X1 C Cm.n/Xn D 0there arem equations andn > munknowns, and hence there are nontrivial solutions inE

choose one.c1; : : : ; cn/having the fewest possible nonzero elements. After renumbering thei s, we may suppose thatc1 0, and then (after multiplying by a scalar) that c1 2 F. Withthese normalizations, well show that allci2 F. Then the first equation

1c1 C Cncn D 0(recall that1 D 1) will be a linear relation on the i .

If not allci are inF, thenk .ci / ci for somek and i ,k 1 i . On applyingk tothe equations

1.1/c1 C C1.n/cn D 0 (*)

m.1/c1 C Cm.n/cn D 0


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Separable, normal, and Galois extensions 35

and using that fk 1; : : : ; kmg is a permutation off1; : : : ; mg, we find that

.c1; k .c2/ ; : : : ; k .ci /;:::/

is also a solution to the system of equations (*). On subtracting it from the first, we obtain asolution.0;:: :;ci k.ci /;:::/, which is nonzero (look at the i th coordinate), but has morezeros than the first solution (look at the first coordinate) contradiction.

COROLLARY 3 .5 For any finite groupG of automorphisms of a fieldE ,G D Aut.E=EG /.PROOF. AsG Aut.E=EG /, we have inequalities

EWEG 3.4.G W1/ .Aut.E=EG /W1/ 2.8a EWEG:

These must be equalities, and soG D Aut.E=EG /:

Separable, normal, and Galois extensions

DEFINITION3 .6 An algebraic extension E=F is said to be separable if the minimum

polynomial of every element ofE is separable; otherwise, it is inseparable.

Thus, an algebraic extensionE=F is separable if every irreducible polynomial in F X

having a root inE is separable, and it is inseparable if

Fis nonperfect, and in particular has characteristic p 0,and there is an element ofEwhose minimal polynomial is of the form g.Xp/, g 2 F X.

For example,E

DFp.T /is an inseparable extension ofFp.T

p/:

DEFINITION3 .7 An algebraic extension E=F is normalif the minimum polynomial of

every element ofE splits inEX.

In other words, an algebraic extensionE=F is normal if every irreducible polynomial

f2 F Xhaving a root in E splits inE .Letfbe an irreducible polynomial of degree m inF X. Iffhas a root inE , then

E=F separable H) roots off distinctE=F normal H) fsplits inE

H) f hasm distinct roots inE:

Therefore, E=F is normal and separable if and only if, for each 2

E, the minimum

polynomial of has F WF distinct roots inE .EXAMPLE 3 .8 (a) The field Q

3p

2, where 3p

2is the real cube root of 2, is separable but

not normal over Q (X3 2doesnt split in Q).(b) The field Fp.T /is normal but not separable over Fp.T

p/ the minimum polyno-

mial ofT is the inseparable polynomialXp Tp.

DEFINITION3 .9 LetFbe a field. A finite extension E ofFis said to be GaloisifF is

the fixed field of the group ofF-automorphisms ofE . This group is then called the Galois

groupofE overF, and it is denoted by Gal.E=F/.

THEOREM 3.10 For an extensionE=F, the following statements are equivalent:


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(a) E is the splitting field of a separable polynomialf2 F X.(b) FD EG for some finite groupG of automorphisms ofE .(c) E is normal and separable, and of finite degree, overF.

(d) E is Galois overF.

PROOF. (a) H) (d). Let GD Aut.E=F/, and let F0D EG F. ThenE is also thesplitting field offregarded as a polynomial with coefficients in F0, andf is still separablewhen it is regarded in this way. Hence Proposition3.2shows that

EWF0 D Aut.E=F0/EWF D jAut.E=F/j :

Since Aut.E=F0/ (3.5)D G D Aut.E=F/, we conclude thatFD F0, and soFD EG .(d)H) (b). According to (2.8a) , Gal.E=F/is finite, and so this is obvious.(b)H) (c). By Proposition3.4,we know thatE WF .GW1/; in particular, it is finite.

Let 2 E and letfbe the minimum polynomial of; we have to prove thatf splits intodistinct factors inEX. Let f1 D ;:::;mg be the orbit of under the action ofG on E ,and let

g.X/ DY

.X i / D Xm Ca1Xm1 C Cam:Every2 G merely permutes the i . Since the ai are symmetric polynomials in the i ,we find that aiD ai for all i , and so g.X/ 2 F X. It is monic, and g./ D 0, and sof.X/jg.X/(see the definition of the minimum polynomial p.17). But alsog.X/jf.X/,because eachi is a root off.X/(ifiD , then applyingto the equation f ./ D 0givesf .i / D 0). We conclude thatf.X/ D g.X/, and sof .X/splits into distinct factorsinE .

(c)

H) (a). BecauseE has finite degree over F, it is generated over Fby a finite

number of elements, say, ED F 1;:::;m, i2 E, i algebraic over F. Letfi be theminimum polynomial ofi overF. BecauseE is normal overF, eachfi splits inE , and

soE is the splitting field offD Qfi :BecauseE is separable overF,f is separable. REMARK 3.11 (a) Let E be Galois over F with Galois group G, and let 2 E. Theelements1 D ,2;:::;mof the orbit of are called theconjugatesof. In the course ofthe proof of (b)H) (c) of the above theorem we showed that the minimum polynomial of is

Q.X i /:

(b) Note that ifFD EG for some finite group G, then, because E is the splittingfield of a separable polynomial, Proposition2.7shows thatGal.E=F/hasE WF elements.Combined with Artins theorem (3.4), this shows thatG

DGal.E=F/and.G

W1/

DE

WF].

COROLLARY 3.12 Every finite separable extensionE ofFis contained in a finite Galois

extension.

PROOF. LetE D F 1;:::;m. Letfi be the minimum polynomial ofi overF, and takeE 0to be the splitting field of

Qfi overF.

COROLLARY 3.13 LetE M F; ifE is Galois overF, then it is Galois overM:PROOF. We knowE is the splitting field of some separablef2 F X; it is also the splittingfield offregarded as an element ofM X:


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The fundamental theorem of Galois theory 37

REMARK 3.14 When we drop the assumption thatE is separable overF, we can still say

something. An element of an algebraic extension ofFis said to beseparableoverF if

its minimum polynomial overFis separable. The proof of Corollary3.12shows that every

finite extension generated by separable elements is separable. Therefore, the elements of a

finite extensionE ofFthat are separable overFform a subfieldEsep ofE that is separableover F; write EWF sep D EsepWF (separable degreeofEover F /. If is an algebraicallyclosed field containingF, then everyF-homomorphism Esep ! extends uniquely toE ,and so the number ofF-homomorphismsE ! isE WF sep. WhenE M F (finiteextensions),

EWF sep D EWM sepMWF sep:In particular,

E is separable overF E is separable overM andMis separable over F:

See Jacobson 1964, I 10, for more details.

DEFINITION3.15 A finite extensionE Fis called acyclic,abelian,..., solvableexten-sion if it is Galois with cyclic, abelian, ..., solvable Galois group.

The fundamental theorem of Galois theory

THEOREM 3.16 (FUNDAMENTAL THEOREM OFG ALOIS THEORY) LetE be a Galois ex-

tension ofF, and letG D Gal.E=F/. The maps H7! EH andM7! Gal.E=M/ are inversebijections between the set of subgroups ofG and the set of intermediate fields betweenE

andF:

fsubgroups ofGg $ fintermediate fieldsF M Eg:Moreover,

(a) the correspondence is inclusion-reversing: H1 H2 EH1 EH2I(b) indexes equal degrees: .H1WH2/ D EH2WEH1;(c) H 1 $ M, i.e.,E H1 D .EH/;Gal.E=M/ D Gal.E=M/1:(d) His normal inG EH is normal (hence Galois) overF, in which case

Gal.EH=F / ' G=H:PROOF. For the first statement, we have to show thatH7! EH andM7! Gal.E=M/areinverse maps.

LetHbe a subgroup ofG . Then, as we observed in(3.11b), Gal.E=EH/ D H:LetMbe an intermediate field. ThenE is Galois overM by (3.13), which means that

EGal.E=M/ D M.(a) We have the obvious implications:

H1 H2H) EH1 EH2 H) Gal.E=E H1/ Gal.E=EH2/:

But Gal.E=EHi / D Hi .(b) As we observed in(3.11b), for every subgroupHofG,E WEH D .Gal.E=EH/W1/.

This proves (b) in the caseH2 D 1, and the general case follows, using that

.H1W1/ D .H1WH2/.H2W1/ and EWEH1

D EWEH2

E

H2

WEH1

:


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(c) For2 G and 2 E, D 1./ D . Therefore,Gal.E=M/ DGal.E=M/1 , and soGal.E=M/1 $ M:

(d) LetHbe a normal subgroup ofG . BecauseH 1 D Hfor all2 G, we musthaveE H

DEH for all

2G, i.e., the action ofG on E stabilizesE H. We therefore have

a homomorphism7! jEHWG ! Aut.EH=F /

whose kernel isH. As.EH/G=H D F, we see thatE H is Galois overF(by Theorem3.10)and thatG=H' Gal.EH=F /(by3.11b).

Conversely, assume thatMis normal overF, and writeMD F 1;:::;m. For2 G, i is a root of the minimum polynomial ofi overF, and so lies in M. HenceMD M,and this implies thatH 1 D H(by (c)).

REMARK 3.17 The theorem shows that there is an order reversing bijection between the

intermediate fields ofE=Fand the subgroups ofG . Using this we can read off more results.

(a) Let M1; M2; : : : ; M r be intermediate fields, and let Hi be the subgroup correspondingto Mi (i.e., HiD Gal.E=Mi /). Then (by definition) M1M2 Mr is the smallest fieldcontaining allMi ; hence it must correspond to the largest subgroup contained in all Hi ,

which isT

Hi . Therefore

Gal.E=M1 Mr / D H1 \ ::: \ Hr :

(b) LetHbe a subgroup ofG and letMD EH. The largest normal subgroup containedinH isNDT2G H1 (see GT4.10), and soE N, which is the composite of the fieldsM, is the smallest normal extension ofFcontainingM. It is called the normal, or Galois,

closure ofM inE .

PROPOSITION3.18 LetE andLbe field extensions ofFcontained in some common field.

IfE=F is Galois, thenEL=LandE=E \ Lare Galois, and the map

7! jEWGal.EL=L/ ! Gal.E=E \ L/

is an isomorphism.

PROOF. BecauseE is Galois overF, it is the splitting field of a separable polynomial

f2 F X. ThenEL is the splitting field off overL, andE is the splittingfield off over E \ L. Hence EL=L and E=E \ L are Galois. EveryautomorphismofEL fixing the elements ofL maps roots off to roots

off, and soE D E. There is therefore a homomorphism7! jEWGal.EL=L/ ! Gal.E=E \ L/.

If2 Gal.EL=L/fixes the elements ofE , then it fixes the elements ofEL, and hence is1. Thus,7! jE is injective. If 2 E is fixed by all2 Gal.EL=L/, then 2 L\ E . By the fundamental theorem,

EL

E L

E \ L

F

D

D

this implies that the image of7! jE is Gal.E=E \ L/.

COROLLARY 3.19 Suppose, in the proposition, thatL is finite overF. Then

ELWF DE

WF L

WF

E \ LWF .
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Examples 39

PROOF. According to Proposition1.20,

ELWF D ELWLLWF ;

butELWL 3:18D EWE \ L 1:20D EWF

E \ LWF .

PROPOSITION3.20 LetE1 andE2 be field extensions ofFcontained in some common

field. IfE1andE2are Galois overF, thenE1E2andE1 \ E2are Galois overF, and

7! .jE1; jE2/WGal.E1E2=F / ! Gal.E1=F / Gal.E2=F /

is an isomorphism ofGal.E1E2=F /onto the subgroup

HD f.1; 2/ j 1jE1 \ E2 D 2jE1 \ E2gofGal.E1=F / Gal.E2=F /.

PROOF: Leta 2 E1 \ E2, and letf be its minimum polynomial overF. Thenf hasdeg fdistinct roots inE1and deg fdistinct roots inE2. Sincef

can have at mostdeg f roots inE1E2, it follows that it has deg f

distinct roots inE1 \ E2. This shows thatE1 \ E2is normal andseparable over F, and hence Galois(3.10). As E1 and E2 are

Galois over F, they are splitting fields of separable polynomials

f1; f2 2 F X. NowE1E2is a splitting field for f1f2, and henceit also is Galois over F. The map 7! .jE1; jE2/ is clearlyan injective homomorphism, and its image is contained inH. We

prove that the image is the whole ofHby counting.

E1E2

E1 E2

E1 \ E2

F

From the fundamental theorem,

Gal.E2=F/= Gal.E2=E1 \ E2/ ' Gal.E1 \ E2=F /,

and so, for each1 2 Gal.E1=F /,1jE1 \ E2 has exactlyE2WE1 \ E2extensions to anelement of Gal.E2=F /. Therefore,

.H

W1/

DE1

WFE2

WE1

\E2

DE1WF E2WF

E1 \ E2WF ;

which equalsE1E2WF by (3.19):

Examples

EXAMPLE 3.21 We analyse the extension Q=Q, whereis a primitive7th root of1, say

D e2i=7.


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Note that Q is the splitting field of the polyno-

mialX7 1, and thathas minimum polynomial

X6 C X5 C X4 C X3 CX2 CXC 1

(see 1.41). Therefore, Q is Galois of degree

6 over Q. For any 2 Gal.Q=Q/, D i ,somei ,1 i 6, and the map 7! i defines anisomorphismGal.Q=Q/ ! .Z=7Z/. Letbethe element ofGal.Q=Q/ such that D 3.Then generatesGal.Q=Q/because the class

of3 in.Z=7Z/generates it (the powers of3mod7 are 3;2;6;4;5;1). We investigate the subfields

ofQ corresponding to the subgroups h3i andh2i.

Q

QCx Qp7

Q

h3i h2i

hi=h3i hi=h2i

Note that

3

D 6

D x(complex conjugate of/. The subfield ofQ corresponding toh3i is QCx, andCxD 2 cos27

. Since h3i is a normal subgroup ofhi, QCx isGalois over Q, with Galois group hi=h3i: The conjugates of1 defD Cxare 3 D 3 C3,2 D 2 C 2. Direct calculation shows that

1 C2 C 3 DX6

iD1 i D 1;

12 C13 C23 D 2;123 D .C 6/.2 C 5/.3 C 4/

D .C 3 C 4 C 6/.3 C 4/

D.4

C6

C1C

2

C5

C1C

C

3/

D 1:

Hence the minimum polynomial1 ofCxis

g.X/ D X3 C X2 2X 1:

The minimum polynomial of cos 27D 1

2 is therefore

g.2X/

8 D X3 CX2=2 X=2 1=8:

The subfield ofQ corresponding to h2

i is generated by D C2

C4

. Let 0 D .Then. 0/2 D 7. Hence the field fixed by h2i is Qp7:

EXAMPLE 3. 22 We compute the Galois group of a splitting fieldE ofX5 2 2QX .1More directly, on setting XD Cxin

.X3 3X /C .X2 2/ C XC 1one obtains1 C C 2 C C 6 D 0.


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Constructible numbers revisited 41

Recall from Exercise2-3 thatEDQ; whereis a primitive5th root of1, and is a root ofX5 2. For example, we could takeE to be the splitting field ofX5 2in C, with D e2i=5 and equal to the real5th root of2. We have the picture at right, and

Q WQ D 4; Q WQ D 5:

Because4 and5 are relatively prime,

Q; WQ D 20:

Q;

Q Q

Q

N H

G=N

HenceG D Gal.Q;=Q/has order20, and the subgroupsN andHfixing Q and Qhave orders 5 and4 respectively. Because Q is normal over Q (it is the splitting field

ofX5 1),Nis

Field and Galois Theory by Milne

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