fields and galois theory-j.s. milne

8/16/2019 Fields and Galois Theory-J.S. Milne

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Fields and Galois Theory

J.S. Milne

Version 4.51August 31, 2015


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These notes give a concise exposition of the theory of elds, including the Galois theoryof nite and innite extensions and the theory of transcendental extensions. The rst six

chapters form a standard course, and the nal three chapters are more advanced.

BibTeX information@misc{milneFT,

author={Milne, James S.},title={Fields and Galois Theory (v4.51)},year={2015},note={Available at www.jmilne.org/math/},pages={138}

}

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 innite Galois theory; added material to the section on transcendentalextensions; 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; xed problem with hyperlinks; 111 pages.v4.22 (March 30, 2011). Minor changes; changed T EXstyle; 126 pages.v4.30 (April 15, 2012). Minor xes; added sections on étale algebras; 124 pages.v4.40 (March 20, 2013). Minor xes and additions; 130 pages.v4.50 (March 18, 2014). Added chapter explaining Grothendieck’s approach to Galois

theory (Chapter 8) and made many minor improvements; numbering has changed; 138

pages.v4.51 (August 31, 2015). Minor corrections; 138 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 c 1996–2015 J.S. Milne.Single paper copies for noncommercial personal use may be made without explicit permissionfrom the copyright holder.


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Contents

Contents 3Notations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1 Basic Denitions and Results 7Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8The characteristic of a eld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Review of polynomial rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Factoring polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Extension elds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13The subring generated by a subset . . . . . . . . . . . . . . . . . . . . . . . . . 14The subeld generated by a subset . . . . . . . . . . . . . . . . . . . . . . . . . 15Construction of some extension elds . . . . . . . . . . . . . . . . . . . . . . . 15Stem elds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Algebraic and transcendental elements . . . . . . . . . . . . . . . . . . . . . . . 17Transcendental numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19Constructions with straight-edge and compass. . . . . . . . . . . . . . . . . . . . 20Algebraically closed elds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2 Splitting Fields; Multiple Roots 27Maps from simple extensions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Splitting elds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Multiple roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3 The Fundamental Theorem of Galois Theory 35Groups of automorphisms of elds . . . . . . . . . . . . . . . . . . . . . . . . . 35Separable, normal, and Galois extensions . . . . . . . . . . . . . . . . . . . . . 37The fundamental theorem of Galois theory . . . . . . . . . . . . . . . . . . . . . 39Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42Constructible numbers revisited . . . . . . . . . . . . . . . . . . . . . . . . . . 43The Galois group of a polynomial . . . . . . . . . . . . . . . . . . . . . . . . . 44Solvability of equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4 Computing Galois Groups 47When is Gf An ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

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When is Gf transitive? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48Polynomials of degree at most three . . . . . . . . . . . . . . . . . . . . . . . . 49Quartic polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49Examples of polynomials with S p as Galois group over Q . . . . . . . . . . . . . 52Finite elds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53Computing Galois groups over Q . . . . . . . . . . . . . . . . . . . . . . . . . . 54Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5 Applications of Galois Theory 59Primitive element theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59Fundamental Theorem of Algebra . . . . . . . . . . . . . . . . . . . . . . . . . 61Cyclotomic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62Dedekind’s theorem on the independence of characters . . . . . . . . . . . . . . 65The normal basis theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66Hilbert’s Theorem 90 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

Cyclic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71Kummer theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72Proof of Galois’s solvability theorem . . . . . . . . . . . . . . . . . . . . . . . . 74Symmetric polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75The general polynomial of degree n . . . . . . . . . . . . . . . . . . . . . . . . 77Norms and traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

6 Algebraic Closures 85Zorn’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85First proof of the existence of algebraic closures . . . . . . . . . . . . . . . . . . 86

Second proof of the existence of algebraic closures . . . . . . . . . . . . . . . . 86Third proof of the existence of algebraic closures . . . . . . . . . . . . . . . . . 87(Non)uniqueness of algebraic closures . . . . . . . . . . . . . . . . . . . . . . . 88Separable closures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

7 Innite Galois Extensions 91Topological groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91The Krull topology on the Galois group . . . . . . . . . . . . . . . . . . . . . . 92

The fundamental theorem of innite Galois theory . . . . . . . . . . . . . . . . . 95Galois groups as inverse limits . . . . . . . . . . . . . . . . . . . . . . . . . . . 98Nonopen subgroups of nite index . . . . . . . . . . . . . . . . . . . . . . . . . 99

8 The Galois theory of étale algebras 101Review of commutative algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . 101Étale algebras over a eld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102Classication of ´ etale algebras over a eld . . . . . . . . . . . . . . . . . . . . . 1 0 4Comparison with the theory of covering spaces. . . . . . . . . . . . . . . . . . . 107

9 Transcendental Extensions 109Algebraic independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109Transcendence bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110Lüroth’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114Separating transcendence bases . . . . . . . . . . . . . . . . . . . . . . . . . . . 116Transcendental Galois theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

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Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

A Review Exercises 119

B Two-hour Examination 125

C Solutions to the Exercises 127

Index 137

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Notations.

We use the standard (Bourbaki) notations:

N D f0;1;2;: : :g;Z Dring of integers,R Deld of real numbers,C Deld of complex numbers,

Fp DZ =p Z Deld with p elements, p a prime number.Given an equivalence relation, Œ denotes the equivalence class containing . The cardinalityof a set S is denoted by jS j(so jS jis the number of elements in S when S is nite). Let I and A be sets. A family of elements of A indexed by I , denoted by .a i / i2I , is a functioni 7! a i WI ! A. Throughout the notes, p is a prime number: p D2;3;5;7;11;::: .X

Y X is a subset of Y (not necessarily proper).

X def DY X is dened to be Y , or equals Y by denition.X Y X is isomorphic to Y .X ' Y X and Y are 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.

Jacobson, N., 1964, Lectures in Abstract Algebra, Volume III — Theory of Fields and GaloisTheory, van Nostrand.Also, the following of my notes (available at www.jmilne.org/math/).

GT Group Theory, v3.13, 2013.ANT Algebraic Number Theory, v3.05, 2013.CA A Primer of Commutative Algebra, v3.00, 2013.

A reference monnnn is to http://mathoverflow.net/questions/nnnn/PARI is an open source computer algebra system freely available here .

ACKNOWLEDGEMENTS

I thank the following for providing corrections and comments for earlier versions of thenotes: Mike Albert, Lior Bary-Soroker, Maren Baumann, Leendert Bleijenga, TommasoCenteleghe, Sergio Chouhy, Demetres Christodes, 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, Geir Arne Magnussen,John McKay, Georges E. Melki, Courtney Mewton, Shuichi Otsuka, Dmitri Panov, AlainPichereau, David G. Radcliffe, Roberto La Scala, Chad Schoen, Prem L Sharma, DrorSpeiser, Bhupendra Nath Tiwari, Mathieu Vienney, Martin Ward (and class), Xiande Yang,Wei Xu, and others.

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


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C HAPTER 1Basic Denitions and Results

Rings

A ring is a set R with two composition laws Cand such that(a) .R; C/ is a commutative group;(b) is associative, and there exists 1 an element 1R such that a 1R Da D1R a for all

a 2RI(c) the distributive law holds: for all a;b;c 2R ,.a Cb/ c Da c Cb ca .b Cc/ Da b Ca c .

We usually omit “ ” and write 1 for 1R when this causes no confusion. If 1R D0, thenR

D f0

g.

A subring S of a ring R is a subset that contains 1R and is closed under addition, passageto the negative, and multiplication. It inherits the structure of a ring from that on R .

A homomorphism of rings ˛ WR ! R 0 is a map such that˛.a Cb/ D˛.a/ C˛.b/; ˛.ab/ D˛.a/̨ .b/; ˛.1 R / D1R 0

for all a; b 2R . A ring R is said to be commutative if multiplication is commutative:ab Dba for all a; b 2R:

A commutative ring is said to be an integral domain if 1R ¤0 and the cancellation lawholds for multiplication:

ab Dac , a ¤ 0, implies b Dc:An ideal I in a commutative ring R is a subgroup of .R; C/ that is closed under multiplicationby elements of R :

r 2R , a 2I , implies ra 2I:The ideal generated by elements a1 ; : : : ; a n is denoted by .a 1 ; : : : ; a n / . For example, .a/ isthe principal ideal aR .

We assume that the reader has some familiarity with the elementary theory of rings.For example, in Z (more generally, any Euclidean domain) an ideal I is generated by any“smallest” nonzero element of I .

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

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8 1. BASIC D EFINITIONS AND R ESULTS

Fields

D EFINITION 1.1 A eld is a set F with two composition laws Cand such that(a) .F;

C/ is a commutative group;

(b) .F ; / , where F DF Xf0g, is a commutative group;(c) the distributive law holds.

Thus, a eld is a nonzero commutative ring such that every nonzero element has an inverse.In particular, it is an integral domain. A eld contains at least two distinct elements, 0 and 1.The smallest, and one of the most important, elds is F 2 DZ =2Z D f0; 1g.

A subeld S of a eld F is a subring that is closed under passage to the inverse. Itinherits the structure of a eld from that on F .

L EMMA 1 .2 A nonzero commutative ring R is a eld if and only if it has no ideals other than .0/ and R .

P ROOF . Suppose R is a eld, and let I be a nonzero ideal in R . If a is a nonzero elementof I , then 1 Da 1 a 2I , and so I DR . Conversely, suppose R is a commutative ring withno proper nonzero ideals. If a ¤ 0, then .a/ DR , and so there exists a b in R such thatab D1. E XAMPLE 1.3 The following are elds: Q , R , C , Fp DZ =p Z (p prime) :

A homomorphism of elds ˛ WF ! F 0 is simply a homomorphism of rings. Such ahomomorphism is always injective, because its kernel is a proper ideal (it doesn’t contain 1),

which must therefore be zero.

The characteristic of a eld

One checks easily that the map

Z ! F; n 7! 1F C1F C C1F .n copies /;is a homomorphism of rings. For example,

.1 F C C1F

„ ƒ‚ …m

/ C.1 F C C1F

„ ƒ‚ …n

/ D1F C C1F

„ ƒ‚ …mCn

because of associativity. Therefore its kernel is an ideal in Z .C AS E 1: The kernel of the map is .0/ , so that

n 1F D0 H) n D0 (in Z ).Nonzero integers map to invertible elements of F under n 7! n 1F WZ ! F , and so this mapextends to a homomorphism

mn 7! .m 1F /.n 1F /

1WQ ,! F:

Thus, in this case, F contains a copy of Q , and we say that it has characteristic zero .


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Review of polynomial rings 9

CAS E 2: The kernel of the map is ¤ .0/ , so that n 1F D0 for some n ¤ 0. The smallestpositive such n will be a prime p (otherwise there will be two nonzero elements in F whoseproduct is zero), and p generates the kernel. Thus, the map n 7! n 1F WZ ! F denes anisomorphism from Z =p Z onto the subring

fm 1F jm 2Zgof F . In this case, F contains a copy of Fp , and we say that it has characteristic p .

The elds F 2 ; F 3 ; F 5 ; : : : ; Q are called the prime elds. Every eld contains a copy of exactly one of them.

R EMARK 1.4 The binomial theorem

.a Cb/ m Da m Cm1 a

m 1 b Cm2 a

m 2 b2 C Cbm

holds in every commutative ring. If p is prime, then p divides pn

r for all r with 1 r

p n 1. Therefore, when F has characteristic p ,.a Cb/ p

n

Da pn

Cbpn

all n 1;

and so the map a 7! a p WF ! F is a homomorphism. It is called the Frobenius endomorphism of F . When F is nite, the Frobenius endomorphism is an automorphism.

Review of polynomial rings

Let F be a eld.

1.5 The ring F ŒX of polynomials in the symbol (or “indeterminate” or “variable”) X with coefcients in F is an F -vector space with basis 1, X , . . . , X n , ... , and with themultiplication dened by

Xi a i X i Xj bj X j DXk Xi Cj Dk a i bj X k :For any ring R containing F as a subring and element r of R , there is a unique homomor-phism ˛ WFŒX ! R such that ˛.X/ Dr and ˛.a/ Da for all a 2F .1.6 Division algorithm : given f .X/ and g.X/ 2FŒX with g ¤ 0, there exist q.X/ ,

r.X/ 2FŒX with r D0 or deg . r /< deg .g/ such thatf Dgq Cr I

moreover, q.X/ and r.X/ are uniquely determined. Thus F ŒX is a Euclidean domain withdeg as norm, and so is a unique factorization domain.

1.7 Let f 2FŒX and a 2F . Thenf D.X a/q Cc

with q 2FŒX and c 2F . Therefore, if a is a root of f (that is, f .a/ D0), then X adivides f . From unique factorization, it now follows that f has at most deg .f / roots (seealso Exercise 1-3 ).


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10 1. B ASIC D EFINITIONS AND R ESULTS

1.8 Euclid’s algorithm : Let f and g 2FŒX have gcd d.X/ . Euclid’s algorithm constructspolynomials a.X/ and b.X/ such that

a.X/ f.X/

Cb.X/ g.X/

Dd.X/; deg .a/ < deg .g/; deg .b/ < deg .f /:

Recall how it goes. We may assume that deg .f / deg .g/ since the argument is the same inthe opposite case. Using the division algorithm, we construct a sequence of quotients andremainders

f Dq0 g Cr0g Dq1 r0 Cr1

r0 Dq2 r1 Cr2

rn 2 Dqn rn 1 Crnrn 1 DqnC1 rn

with r n the last nonzero remainder. Then, rn divides rn 1 , hence r n 2 ,.. . , hence g , andhence f . Moreover,

rn Drn 2 qn rn 1 Drn 2 qn .r n 3 qn 1 rn 2 / D Daf Cbgand so every common divisor of f and g divides rn : we have shown rn Dgcd .f;g/ .

Let af Cbg Dd . If deg .a/ deg .g/ , write a Dgq Cr with deg . r /< deg .g/ ; thenrf C.b Cqf /g Dd;

and b Cqf automatically has degree < deg .f / .PARI knows Euclidean division: typing divrem(13,5) in PARI returns Œ2;3, meaningthat 13 D2 5 C3, and gcd(m,n) returns the greatest common divisor of m and n.1.9 Let I be a nonzero ideal in F ŒX, and let f be a nonzero polynomial of least degree in

I ; then I D.f / (because FŒX is a Euclidean domain). When we choose f to be monic, i.e.,to have leading coefcient one, it is uniquely determined by I . Thus, there is a one-to-onecorrespondence between the nonzero ideals of F ŒX and the monic polynomials in F ŒX.The prime ideals correspond to the irreducible monic polynomials.

1.10 Since F ŒX is an integral domain, we can form its eld of fractions F.X/ . Itselements are quotients f =g , f and g polynomials, g ¤ 0:

Factoring polynomials

The following results help in deciding whether a polynomial is reducible, and in nding itsfactors.

P ROPOSITION 1.11 Suppose r 2Q is a root of a polynomial a m X m Ca m 1 X m 1 C Ca 0 ; a i 2Z ;

and let r Dc=d , c; d 2Z , gcd .c;d/ D1. Then cja 0 and d ja m :


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Factoring polynomials 11

P ROOF . It is clear from the equation

a m cm Ca m 1 cm 1 d C Ca 0 d m D0

that d ja m cm

, and therefore, d ja m : Similarly, cja 0 . E XAMPLE 1.12 The polynomial f .X/ DX 3 3X 1 is irreducible in Q ŒX because itsonly possible roots are ˙ 1, and f .1/ ¤ 0 ¤ f . 1/ .P ROPOSITION 1.13 (G AUSS ’S L EMMA ) Let f .X/ 2Z ŒX. If f .X/ factors nontrivially in Q ŒX, then it factors nontrivially in Z ŒX.

P ROOF . Let f Dgh in Q ŒX with g; h …Q . For suitable integers m and n , g1def

Dmg andh1

def

Dnh have coefcients in Z , and so we have a factorizationmnf Dg1 h1 in Z ŒX.

If a prime p divides mn , then, looking modulo p , we obtain an equation

0 Dg1 h1 in Fp ŒX.Since Fp ŒX is an integral domain, this implies that p divides all the coefcients of at leastone of the polynomials g1 ;h1 , say g1 , so that g1 Dpg 2 for some g2 2Z ŒX. Thus, we havea factorization

.mn=p/f Dg2 h1 in Z ŒX.Continuing in this fashion, we eventually remove all the prime factors of mn , and so obtain

a nontrivial factorization of f in Z ŒX.

P ROPOSITION 1.14 If f 2Z ŒX is monic, then every monic factor of f in Q ŒX lies inZ ŒX.

P ROOF . Let g be a monic factor of f in Q ŒX, so that f Dgh with h 2Q ŒX also monic.Let m; n be the positive integers with the fewest prime factors such that mg; nh 2Z ŒX. Asin the proof of Gauss’s Lemma, if a prime p divides mn , then it divides all the coefcientsof at least one of the polynomials mg;nh , say mg , in which case it divides m because g ismonic. Now mp g 2Z ŒX, which contradicts the denition of m. A SIDE 1.15 We sketch an alternative proof of Proposition 1.14 . A complex number ˛ is said to bean algebraic integer if it is a root of a monic polynomial in ZŒX. Proposition 1.11 shows that everyalgebraic integer in Q lies in Z . The algebraic integers form a subring of C — for an elementaryproof of this, using nothing but the symmetric polynomials theorem ( 5.35 ), see Chapter 1 of my noteson algebraic geometry. Now let ˛ 1 ; : : : ; ˛ m be the roots of f in C . By denition, they are algebraicintegers, and the coefcients of any monic factor of f are polynomials in (certain of) the ˛ i , andtherefore are algebraic integers. If they lie in Q , then they lie in Z .

P ROPOSITION 1.16 (E ISENSTEIN ’S CRITERION ) Let

f Da m X m Ca m 1 X m 1 C Ca 0 ; a i 2ZIsuppose that there is a prime p such that:


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˘ p does not divide a m ,˘ p divides am 1 ;:::;a 0 ,˘ p 2 does not divide a 0 .

Then f is irreducible in Q ŒX.

P ROOF . If f .X/ factors in Q ŒX, then it factors in Z ŒX, say,

a m X m Ca m 1 X m 1 C Ca 0 D.b r X r C Cb0 /.c s X s C Cc0 /with bi ; ci 2Z and r;s < m . Since p , but not p 2 , divides a0 Db0 c0 , p must divide exactlyone of b0 , c0 , say, b0 . Now from the equation

a 1 Db0 c1 Cb1 c0 ;we see that p

jb1 ; and from the equation

a 2 Db0 c2 Cb1 c1 Cb2 c0 ;that p jb2 . By continuing in this way, we nd that p divides b0 ;b1 ; : : : ; br , which contradictsthe condition that p does not divide am .

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

R EMARK 1.17 There is an algorithm for factoring a polynomial in QŒX. To see this,consider f 2Q ŒX. Multiply f .X/ by a rational number so that it is monic, and thenreplace it by D deg .f / f . X D / , with D equal to a common denominator for the coefcients

of f , to obtain a monic polynomial with integer coefcients. Thus we need consider onlypolynomialsf .X/ DX m Ca 1 X m 1 C Ca m ; a i 2Z :

From the fundamental theorem of algebra (see 5.6 below), we know that f splitscompletely in C ŒX:

f .X/ Dm

YiD1 .X ˛ i /; ˛ i 2C :From the equation

0 Df .˛ i / D˛ mi Ca 1 ˛ m 1i C Ca m ,it follows that

j˛ i

jis less than some bound depending only on the degree and coefcients of

f ; in fact,

j˛ i j maxf1;mB g, B Dmax ja i j.Now if g.X/ is a monic factor of f .X/ , then its roots in C are certain of the ˛ i , and itscoefcients are symmetric polynomials in its roots (see p .75). Therefore, the absolute valuesof the coefcients of g.X/ are bounded in terms of the degree and coefcients of f . Sincethey are also integers (by 1.14 ), we see that there are only nitely many possibilities forg.X/ . Thus, to nd the factors of f .X/ we (better PARI) have to do only a nite amount of checking .2

2Of course, there are faster methods than this. The Berlekamp–Zassenhaus algorithm factors the polynomialover certain suitable nite elds Fp , lifts the factorizations to rings Z =p m Z for some m, and then searches forfactorizations in Z ŒX with the correct form modulo p m .


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Extension elds 13

Therefore, we need not concern ourselves with the problem of factoring polynomi-als in the rings QŒX or Fp ŒX since PARI knows how to do it. For example, typingcontent(6*X^2+18*X-24) in PARI returns 6, and factor(6*X^2+18*X-24) returnsX 1 and X

C4, showing that

6X 2 C18X 24 D6.X 1/.X C4/in Q ŒX. Typing factormod(X^2+3*X+3,7) returns X C4 and X C6, showing that

X 2 C3X C3 D.X C4/.X C6/in F 7 ŒX.

R EMARK 1.18 One other observation is useful. Let f 2Z ŒX. If the leading coefcient of f is not divisible by a prime p , then a nontrivial factorization f Dgh in Z ŒX will give anontrivial factorization xf D xg xh in Fp ŒX. Thus, if f .X/ is irreducible in Fp ŒX for someprime p not dividing its leading coefcient, then it is irreducible in Z ŒX. This test is veryuseful, but it is not always effective: for example, X 4 10X 2 C1 is irreducible in Z ŒX butit is reducible 3 modulo every prime p .

Extension elds

A eld E containing a eld F is called an extension eld of F (or simply an extension of F , and we speak of an extension E=F ). Such an E can be regarded as an F -vector space,and we write ŒEWF for the dimension, possibly innite, of E as an F -vector space. Thisdimension is called the degree of E over F . We say that E is nite over F when it has nite

degree over F:When E and E 0 are extension elds of F , an F - homomorphism E ! E 0 is a homo-

morphism ' WE ! E 0 such that '.c/ Dc for all c 2F .E XAMPLE 1.1 9 (a) The eld of complex numbers C has degree 2 over R (basis f1; i g/:

(b) The eld of real numbers R has innite degree over Q : the eld Q is countable,and so every nite-dimensional Q -vector space is also countable, but a famous argument of Cantor shows that R is not countable.

(c) The eld of Gaussian numbers

Q .i / def D fa Cbi 2C ja; b 2Qg3Here 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 Exercise 1-3). If 2 is a square in Fp , then

X 4 10X 2 C1 D.X 2 2p 2X 1/.X 2 C2p 2X 1/:If 3 is a square in Fp , then

X 4 10X 2 C1 D.X 2 2p 3X C1/.X 2 C2p 3X C1/:If neither 2 nor 3 are squares, 6 will be a square in Fp , and

X 4 10X 2 C1 D.X 2 .5 C2p 6//.X 2 .5 2p 6//:The general study of such polynomials requires nonelementary methods. See, for example, the paperBrandl, R., Amer. Math. Monthly, 93 (1986), pp286–288, which proves that every nonprime integer n 1occurs as the degree of a polynomial in Z ŒX that is irreducible over Z but reducible modulo all primes :


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has degree 2 over Q (basis f1; i g).(d) The eld F.X/ has innite degree over F ; in fact, even its subspace F ŒX has

innite dimension over F (basis 1;X;X 2 ; : : :).

P ROPOSITION 1.20 ( MULTIPLICATIVITY OF DE GREES ) Consider elds L E F . ThenL=F is of nite degree if and only if L=E and E=F are both of nite degree, in which case

ŒLWF DŒLWE ŒEWF :P ROOF . If L is of nite degree over F , then it is certainly of nite degree over E . Moreover,E , being a subspace of a nite dimensional F -vector space, is also nite dimensional.

Thus, assume that L=E and E=F are of nite degree, and let .e i / 1 i m be a basis for Eas an F -vector space and let .l j /1 j n be a basis for L as an E -vector space. To completethe proof, it sufces to show that .e i lj / 1 i m;1 j n is a basis for L over F , because thenL will be nite over F of the predicted degree.

First, .e i lj / i;j spans L . Let 2L . Then, because .l j /j spans L as an E -vector space, DPj ˛ j lj ; some ˛ j 2E;and because .e i / i spans E as an F -vector space,

˛ j DPi a ij ei ; some a ij 2F :On putting these together, we nd that DPi;j a ij ei lj :Second, .e i lj / i;j is linearly independent. A linear relation

Pa ij ei lj

D0, a ij

2F ,

can be rewritten Pj .P

i a ij ei /l j D0. The linear independence of the lj ’s now shows thatPi a ij ei D0 for each j , and the linear independence of the ei ’s shows that each a ij D0.The subring generated by a subset

An intersection of subrings of a ring is again a ring. Let F be a subeld of a eld E , and letS be a subset of E . The intersection of all the subrings of E containing F and S is evidentlythe smallest subring of E containing F and S . We call it the subring of E generated by F and S (or generated over F by S ), and we denote it F ŒS. When S Df˛ 1 ;:::;˛ n g, we writeF Œ 1̨ ;:::;˛ n for F ŒS. For example, C DR Œp 1 .L EMMA 1.21 The ring F ŒS consists of the elements of E that can be expressed as nite sums of the form

Xa i1 in ˛ i11 ˛ inn ; ai1 in 2F; ˛ i 2S: (*)P ROOF . Let R be the set of all such elements. Evidently, R is a subring containing F and S and contained in every other such subring. Therefore R equals F ŒS.

E XAMPLE 1.22 The ring Q Œ , D3:14159::: , consists of the complex numbers that canbe expressed as a nite sum

a 0 Ca 1 Ca 2 2 C Ca n n ; a i 2Q :The ring Q Œi consists of the complex numbers of the form a Cbi , a; b 2Q .


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The subeld generated by a subset 15

Note that the expression of an element in the form (*) will not be unique in general. Thisis so already in R Œi.

L EMMA 1.23 Let R be an integral domain containing a subeld F (as a subring). If R is

nite dimensional when regarded as an F -vector space, then it is a eld.

P ROOF . Let ˛ be a nonzero element of R — we have to show that ˛ has an inverse in R .The map x 7! ˛x WR ! R is an injective linear map of nite dimensional F -vector spaces,and is therefore surjective. In particular, there is an element ˇ 2R such that ˛ˇ D1.

Note that the lemma applies to subrings (containing F ) of an extension eld E of F of nite degree.

The subeld generated by a subset

An intersection of subelds of a eld is again a eld. Let F be a subeld of a eld E , and letS be a subset of E . The intersection of all the subelds of E containing F and S is evidentlythe smallest subeld of E containing F and S . We call it the subeld of E generated byF and S (or generated over F by S ), and we denote it F.S/ . It is the eld of fractions of FŒS in E , since this is a subeld of E containing F and S and contained in every othersuch eld. When S D f˛ 1 ;:::;˛ n g, we write F .˛ 1 ;:::;˛ n / for F.S/ . Thus, F Œ 1̨ ; : : : ; ˛ nconsists of all elements of E that can be expressed as polynomials in the ˛ i with coefcientsin F , and F .˛ 1 ; : : : ; ˛ n / consists of all elements of E that can be expressed as the quotientof two such polynomials.

Lemma 1.23 shows that FŒS is already a eld if it is nite dimensional over F , in which

case F.S/ DFŒS.E XAMPLE 1.24 The eld Q . / , D3:14::: , consists of the complex numbers that can beexpressed as a quotient

g. /=h. /; g.X/;h.X/ 2Q ŒX ; h.X/¤ 0:The ring Q Œi is already a eld.

An extension E of F is said to be simple if E DF.˛/ some ˛ 2E . For example, Q . /and Q Œi are simple extensions of Q :

Let F and F 0 be subelds of a eld E . The intersection of the subelds of E containingF and F 0 is evidently the smallest subeld of E containing both F and F 0. We call it the composite of F and F 0 in E , and we denote it F F 0. It can also be described as the subeldof E generated over F by F 0, or the subeld generated over F 0 by F :

F .F 0/ DF F 0DF 0.F / .

Construction of some extension elds

Let f .X/ 2FŒX be a monic polynomial of degree m, and let .f / be the ideal gener-ated by f . Consider the quotient ring F ŒX =.f .X// , and write x for the image of X inFŒX =.f.X//, i.e., x is the coset X

C.f.X// . Then:


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(a) The mapP.X/ 7! P.x/ WFŒX ! FŒx

is a surjective homomorphism in which f .X/ maps to 0. Therefore, f .x/ D0.(b) From the division algorithm, we know that each element g of F ŒX =.f / is rep-resented by a unique polynomial r of degree < m . Hence each element of F Œx can be

expressed uniquely as a sum

a 0 Ca 1 x C Ca m 1 x m 1 ; a i 2F: (*)(c) To add two elements, expressed in the form (*), simply add the corresponding

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

use the relation f .x/ D0 to express the monomials of degree m in x in terms of lowerdegree monomials.

(e) Now assume f .X/ is irreducible. Then every nonzero ˛

2FŒx has an inverse,

which can be found as follows. Use (b) to write ˛ Dg.x/ with g.X/ is a polynomial of degree m 1, and use Euclid’s algorithm in F ŒX to obtain polynomials a.X/ and b.X/such that

a.X/f.X/ Cb.X/g.X/ Dd.X/with d.X/ the gcd of f and g . In our case, d.X/ is 1 because f .X/ is irreducible anddeg g.X/ < deg f .X/ . When we replace X with x , the equality becomes

b.x/g.x/ D1:Hence b.x/ is the inverse of g.x/ .

From these observations, we conclude:

1.25 For a monic irreducible polynomial f .X/ of degree m in F ŒX,

FŒx def DFŒX =.f.X//is a eld of degree m over F . Moreover, computations in F Œx reduce to computations in F .

E XAMPLE 1.26 Let f .X/ DX 2 C1 2R ŒX. Then R Œx has:elements: a Cbx , a; b 2RIaddition: .a Cbx/ C.a 0Cb0x/ D.a Ca0/ C.b Cb0/x Imultiplication: .a Cbx/.a 0Cb0x/ D.aa 0 bb 0/ C.ab 0Ca0b/x:

We usually write i for x and C for R Œx :

E XAMPLE 1.27 Let f .X/ DX 3 3X 1 2Q ŒX. We observed in ( 1.12 ) that this isirreducible over Q , and so Q Œx is a eld. It has basis f1;x;x 2gas a Q -vector space. Let

ˇ Dx 4 C2x 3 C3 2Q Œx :Then using that x 3 3x 1 D0, we nd that ˇ D3x 2 C7x C5. Because X 3 3X 1 isirreducible,

gcd .X 3 3X 1;3X 2 C7X C5/ D1:In fact, Euclid’s algorithm gives

.X 3 3X 1/ 737 X C 29111 C.3X 2 C7X C5/ 7111 X 2 26111 X C 28111 D1:


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Stem elds 17

Hence.3x 2 C7x C5/ 7111 x 2 26111 x C 28111 D1;

and we have found the inverse of ˇ:

We can also do this in PARI: beta=Mod(X^4+2*X^3+3,X^3-3*X-1) reveals that ˇ D3x 2 C7x C5 in Q Œx, and beta^(-1) reveals that ˇ 1 D 7111 x 2 26111 x C 28111 .

Stem elds

Let f be a monic irreducible polynomial in F ŒX. A pair .E;˛/ consisting of an extensionE of F and an ˛ 2E is called a stem eld 4 for f if E DFŒ˛ and f .˛/ D0. For example,the pair .E;˛/ with E DF ŒX =.f /DFŒx and ˛ Dx is a stem eld for f . Let .E;˛/ bea stem eld, and consider the surjective homomorphism of F -algebras

g.X/ 7! g.˛/ WFŒX ! E .Its kernel is generated by a nonzero monic polynomial, which divides f , and so must equalit. Therefore the homomorphism denes an F -isomorphism

x 7! ˛ WFŒx ! E; F Œx def

DF ŒX =.f /.In other words, the stem eld .E;˛/ of f is F -isomorphic to the standard stem eld.F ŒX =.f /; x/ . In particular, each element of a stem eld .E;˛/ for f has a uniqueexpression

a 0 Ca 1 ˛ C Ca m 1 ˛ m 1 ; a i 2F; m Ddeg .f / ,i.e., 1;˛; : : : ;˛ m 1 is an F -basis for F Œ˛, and arithmetic in F Œ˛ can be performed usingthe same rules as in F Œx. If .E;˛ 0/ is a second stem eld for f , then there is a uniqueF -isomorphism E ! E 0 sending ˛ to ˛ 0. We sometimes write “stem eld F Œ˛” instead of “stem eld .FŒ˛ ;˛/”.

Algebraic and transcendental elements

For a eld F and an element ˛ of an extension eld E , we have a homomorphism

f .X/ 7! f.˛/ WFŒX ! E:There are two possibilities.

C AS E 1: The kernel of the map is .0/ , so that, for f

2FŒX,

f.˛/ D0 H) f D0 (in F ŒX).In this case, we say that ˛ transcendental over F . The homomorphism F ŒX ! FŒ˛ is anisomorphism, and it extends to an isomorphism F.X/ ! F.˛/ .

CAS E 2: The kernel is ¤ .0/ , so that g.˛/ D0 for some nonzero g 2FŒX. In this case,we say that ˛ is algebraic over F . The polynomials g such that g.˛/ D0 form a nonzeroideal in F ŒX, which is generated by the monic polynomial f of least degree such f .˛/ D0.We call f the minimum polynomia l 5 of ˛ over F . It is irreducible, because otherwise there

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

5Some authors write “minimal polynomial” but the polynomial in question is “minimum” (it is the uniqueminimal element of the set of monic polynomials f with f .˛/ D0).


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would be two nonzero elements of E whose product is zero. The minimum polynomial ischaracterized as an element of F ŒX by each of the following sets of conditions:

f is monic; f .˛/ D0 and divides every other polynomial g in F ŒX with g.˛/ D0.f is the monic polynomial of least degree such that f .˛/

D0

If is monic, irreducible, and f .˛/ D0.Note that g.X/ 7! g.˛/ denes an isomorphism F ŒX =.f /! FŒ˛ . Since the rst is aeld, so also is the second:

F.˛/ DFŒ˛ :Thus, F Œ˛ is a stem eld for f .

E XAMPLE 1.28 Let ˛ 2C be such that ˛ 3 3˛ 1 D0. Then X 3 3X 1 is monic,irreducible, and has ˛ as a root, and so it is the minimum polynomial of ˛ over Q . The set

f1;˛;˛ 2gis a basis for Q Œ˛ over Q . The calculations in Example 1.27 show that if ˇ is theelement ˛ 4 C2˛ 3 C3 of Q Œ˛, then ˇ D3˛ 2 C7˛ C5, and

ˇ 1 D 7111 ˛ 2 26111 ˛ C 28111 :R EMARK 1.29 PARI knows how to compute in Q Œ˛. For example, factor(X^4+4) re-turns the factorization

X 4 C4 D.X 2 2X C2/.X 2 C2X C2/in Q ŒX. Now type nf=nfinit(a^2+2*a+2) to dene a number eld “nf” generated overQ by a root a of X 2 C2X C1. Then nffactor(nf,x^4+4) returns the factorization

X 4 C4 D.X a 2/.X a/.X Ca//.X Ca C2/;in Q Œa.

A eld extension E=F is said to be algebraic , and E is said to be algebraic over F , if all elements of E are algebraic over F ; otherwise it is said to be transcendental (or E issaid to be transcendental over F ). Thus, E=F is transcendental if at least one element of E is transcendental over F .

P ROPOSITION 1.30 A eld extension E=F is nite if and only if E is algebraic and nitely generated (as a eld) over F .

PROOF

. H): To say that ˛ is transcendental over F amounts to saying that its powers1;˛;˛ 2 ; : : : are linearly independent over F . Therefore, if E is nite over F , then it isalgebraic over F . It remains to show that E is nitely generated over F . If E DF , then itis generated by the empty set. Otherwise, there exists an ˛ 1 2E XF . If E ¤ F Œ 1̨ , thereexists an ˛ 2 2E XF Œ 1̨ , and so on. Since

ŒFŒ˛1 WF < Œ F Œ ˛1 ; ˛ 2 WF < < ŒEWFthis process terminates.

(H: Let E DF .˛ 1 ;:::;˛ n / with ˛ 1 ; ˛ 2 ; : : : ˛ n algebraic over F . The extension F .˛ 1 /=F is nite because ˛ 1 is algebraic over F , and the extension F .˛ 1 ; ˛ 2 /=F.˛ 1 / is nite because˛ 2 is algebraic over F and hence over F .˛ 1 / . Thus, by ( 1.20 ), F .˛ 1 ; ˛ 2 / is nite over F .Now repeat the argument.


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Transcendental numbers 19

C OROLLARY 1.31 (a) If E is algebraic over F , then every subring R of E containing F isa eld.

(b) If in L E F , L is algebraic over E and E is algebraic over F , then L is algebraic over F:

P ROOF . (a) We observed above (p. 18), that if ˛ is algebraic over F , then F Œ˛ is a eld. If ˛ 2R , then F Œ˛ R , and so ˛ has an inverse in R .

(b) Every ˛ 2L is a root of a monic polynomial f DX m Ca m 1 X m 1 C Ca 0 2EŒX. Now each of the extensions F Œa0 ; : : : ; a m 1 ; ˛ F Œa0 ; : : : ; a m 1 F is nite(1.20 ), and so F Œa0 ; : : : ; a m 1 ; ˛ is nite (hence algebraic) over F .

Transcendental numbers

A complex number is said to be algebraic or transcendental according as it is algebraic or

transcendental over Q . First some history:1844: Liouville showed that certain numbers, now called Liouville numbers, are tran-scendental.

1873: Hermite showed that e is transcendental.1874: Cantor showed that the set of algebraic numbers is countable, but that R is not

countable. Thus most numbers are transcendental (but it is usually very difcult to provethat any particular number is transcendental). 6

1882: Lindemann showed that is transcendental.1934: Gel’fond and Schneider independently showed that ˛ ˇ is transcendental if ˛ and

ˇ are algebraic, ˛ ¤ 0; 1, and ˇ …Q . (This was the seventh of Hilbert’s famous problems.)2013: Euler’s constant

D limn!1 n

Xk D1 1=k log n!has not yet been proven to be transcendental or even irrational (see Lagarias, Jeffrey C.,Euler’s constant: Euler’s work and modern developments. Bull. Amer. Math. Soc. 50(2013), no. 4, 527–628; arXiv:1303:1856).

2013: The numbers e C and e are surely transcendental, but again they have noteven been proved to be irrational!

P ROPOSITION 1.32 The set of algebraic numbers is countable.

P ROOF . Dene the height h.r/ of a rational number to be max .jmj;jnj/ , where r Dm=nis the expression of r in its lowest terms. There are only nitely many rational numberswith height less than a xed number N . Let A.N/ denote the set of algebraic numberswhose minimum equation over Q has degree N and has coefcients of height < N . ThenA.N/ is nite for each N . Choose a bijection from some segment Œ0;n.1/ of N onto A.10/ ;extend it to a bijection from a segment Œ0;n.2/ onto A.100/ , and so on.

6In 1873 Cantor proved the rational numbers countable. . . . He also showed that the algebraic numbers. . .were countable. However his attempts to decide whether the real numbers were countable proved harder. Hehad proved that the real numbers were not countable by December 1873 and published this in a paper in 1874(MacTutor ).

http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Cantor.htmlhttp://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Cantor.html


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A typical Liouville number is P1nD0 110 nŠ — in its decimal expansion there are increas-ingly long strings of zeros. Since its decimal expansion is not periodic, the number is notrational. We prove that the analogue of this number in base 2 is transcendental.

T HEOREM 1.33 The number ˛ DP 12 nŠ is transcendental.

P ROOF . 7Suppose not, and let

f .X/ DX d Ca 1 X d 1 C Ca d ; a i 2Q ;be the minimum polynomial of ˛ over Q . Thus ŒQ Œ˛WQ Dd . Choose a nonzero integer Dsuch that D f.X/ 2Z ŒX.

Let ˙ N DPN nD0

12 nŠ , so that ˙ N ! ˛ as N ! 1 , and let xN Df .˙ N / . As ˛ is not

rational, f .X/ , being irreducible of degree > 1 , has no rational root. Since ˙ N ¤ ˛ , it can’tbe a root of f .X/ , and so xN ¤0. Evidently, xN 2Q ; in fact .2 N Š/ d Dx N 2Z , and so

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

C ŒX, say,

f .X/ Dd

YiD1 .X ˛ i /; ˛ i 2C ; ˛ 1 D˛;and so

jxN j Dd

YiD1

j˙ N ˛ i j j˙ N ˛ 1j.˙ N CM / d 1 ; where M Dmaxi¤

1 f1; j˛ i jg.

But

j˙ N ˛ 1j D1

XnDN C11

2nŠ 1

2.N C1/Š 1

XnD01

2n !D 22.N C1/Š :Hence

jxN j 2

2.N C1/Š.˙ N CM / d 1

and

j.2 N Š/ d Dx N j 22d N ŠD2.N C1/Š

.˙ N CM / d 1

which tends to 0 as N ! 1 because 2d N Š

2 .N C 1/Š D 2d

2 N C 1N Š

! 0. This contradicts (*).

Constructions with straight-edge and compass.

The Greeks understood integers and the rational numbers. They were surprised to ndthat the length of the diagonal of a square of side 1, namely, p 2, is not rational. Theythus realized that they needed to extend their number system. They then hoped that the“constructible” numbers would sufce. Suppose we are given a length, which we call 1, astraight-edge, and a compass (device for drawing circles). A real number (better a length) is constructible if it can be constructed by forming successive intersections of

7This proof, which I learnt from David Masser, also works for P 1a nŠ for every integer a 2.


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Constructions with straight-edge and compass. 21

˘ 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 compassconstructions? We’ll see that the answer to all three is negative.Let F be a subeld of R . For a positive a 2F , p a denotes the positive square root of a

in R . The F - plane is F F R R . We make the following denitions:An F -line is a line in R R through two points in the F -plane. These are thelines given by equations

ax Cby Cc D0; a; b; c 2F:An F - circle is a circle in R R with centre an F -point and radius an elementof F . These are the circles given by equations

.x a/ 2 C.y b/ 2 Dc2 ; a; b; c 2F:L EMMA 1.34 Let L ¤ L 0 be F -lines, and let C ¤C 0 be F -circles.

(a) L \ L 0D ;or consists of a single F -point.(b) L \ C D ;or consists of one or two points in the F Œp e -plane, some e 2F , e > 0 .(c) C \ C 0D ;or consists of one or two points in the F Œp e -plane, some e 2F , e > 0 .

P ROOF . The points in the intersection are found by solving the simultaneous equations, andhence by solving (at worst) a quadratic equation with coefcients in F .

L EMMA 1.35 (a) If c and d are constructible, then so also are c Cd , c , cd , and cd

.d ¤0/ .(b) If c > 0 is constructible, then so also is p c .

S KETCH 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 c 1 . For (b), draw a circle of radius 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 p c . (For more details, see Artin, M., 1991, Algebra,Prentice Hall, Chapter 13, Section 4.)

T HEOREM 1.36 (a) The set of constructible numbers is a eld.(b) A number ˛ is constructible if and only if it is contained in a subeld of R of the form

Q Œp a 1 ; : : : ;p a r ; ai 2Q Œp a 1 ; : : : ;p a i 1 ; ai > 0 .P ROOF . (a) Immediate from (a) of Lemma 1.35 .

(b) It follows from Lemma 1.34 that every constructible number is contained in sucha eld Q Œp a 1 ; : : : ;p a r . Conversely, if all the elements of QŒp a 1 ; : : : ;p a i 1 are con-structible, then p a i is constructible (by 1.35b), and so all the elements of Q Œp a 1 ; : : : ;p a iare constructible (by (a)). Applying this for i

D0;1;::: , we nd that all the elements of

Q Œp a 1 ; : : : ;p a r are constructible.


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C OROLLARY 1.37 If ˛ is constructible, then ˛ is algebraic over Q , and ŒQ Œ˛WQ is a power of 2.

P ROOF . According to Proposition 1.20 , ŒQ Œ˛

WQ divides

ŒQ Œp a 1 Œp a r WQand ŒQ Œp a 1 ; : : : ;p a r WQ is a power of 2. C OROLLARY 1.38 It is impossible to duplicate the cube by straight-edge and compassconstructions.

P ROOF . The problem is to construct a cube with volume 2. This requires constructing thereal root of the polynomial X 3 2. But this polynomial is irreducible (by Eisenstein’scriterion 1.16 for example), and so ŒQ Œ 3p 2 WQ D3. C OROLLARY 1.39 In general, it is impossible to trisect an angle by straight-edge and compass constructions.

P ROOF . Knowing an angle is equivalent to knowing the cosine of the angle. Therefore, totrisect 3˛ , we have to construct a solution to

cos 3˛ D4 cos 3 ˛ 3 cos ˛:For example, take 3˛ D60 degrees. As cos 60ı D 12 , to construct ˛ , we have to solve8x 3 6x 1 D0, which is irreducible (apply 1.11 ). C OROLLARY 1.40 It is impossible to square the circle by straight-edge and compass con-structions.

P ROOF . A square with the same area as a circle of radius r has side p r . Since istranscendenta l8 , so also is p .

We next consider another problem that goes back to the ancient Greeks: list the n suchthat the regular n -sided polygon can be constructed. Here we consider the question for aprime p (see 5.12 for the general case). Note that X p 1 is not irreducible; in fact

X p 1

D.X 1/.X p 1

CX p 2

C C1/:

L EMMA 1.41 If p is prime, then X p 1 C C1 is irreducible; hence Q Œe2 i=p has degree p 1 over Q :

P ROOF . Let f .X/ D.X p 1/=.X 1/ DX p 1 C C1; thenf .X C1/ D

.X C1/ p 1X DX

p 1 C Ca 2 X 2 Ca 1 X Cp;with a i D

piC1 . Now p ja i for i D1;:::;p 2, and so f .X C1/ is irreducible by Eisenstein’scriterion 1.16 . This implies that f .X/ is irreducible.

8Proofs 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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Algebraically closed elds 23

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

cos 2p D.e2 i

p C.e2 i

p / 1 /=2:

ButQ Œe

2 ip Q Œcos 2p Q ;

and the degree of Q Œe2 i

p over Q Œcos 2p is 2 — the equation

˛ 2 2 cos 2p ˛ C1 D0; ˛ De2 i

p ;

shows that it is 2, and it is not 1 because Q Œe2 i

p is not contained in R . Hence

ŒQ Œcos 2p WQ D p 1

2 :

Thus, if the regular p -gon is constructible, then .p 1/=2 D2k for some k (later (5.12 ),we shall see a converse), which implies that p D2k C1 C1. But 2r C1 can be a prime onlyif r is a power of 2, because otherwise r has an odd factor t and for t odd,

Y t C1 D.Y C1/.Y t 1 Y t 2 C C1/Iwhence

2st C1 D.2 s C1/..2 s / t 1 .2 s / t 2 C C1/ .Thus primes for which the regular p -gon is constructible are exactly those of the formp D 22

k

C1 for some k. Such p are called Fermat primes (because he conjecturedthat all numbers of the form 22

k

C1 are prime). For k D0;1;2;3;4 , we have 22k

C1 D 3;5;17;257;65537 , which are indeed all prime, but Euler showed that 232 C1 D.641/.6700417/ , and we don’t know whether there are any more Fermat primes. Thus,we do not know the list of primes p for which the regular p -gon is constructible.

Gauss showed that 9

cos217 D

116 C

116

p 17C 116q 34 2p 17C18r 17 C3p 17 q 34 2p 17 2q 34 C2p 17

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

Algebraically closed eldsWe say that a polynomial splits in F ŒX (or, more loosely, in F ) if it is a product of polynomials of degree 1 in F ŒX.

P ROPOSITION 1.42 For a eld ˝ , the following statements are equivalent:

(a) Every nonconstant polynomial in ˝ŒX splits in ˝ŒX .(b) Every nonconstant polynomial in ˝ŒX has at least one root in ˝ .(c) The irreducible polynomials in ˝ŒX are those of degree 1.

9Or perhaps that

cos 217 D

1

16 C 1

16p 17

C 1

16

p 34 2p 17

C1

8

q 17

C3p 17 2

p 34 2p 17

p 170 26p 17

— both expressions are correct.


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(d) Every eld of nite degree over ˝ equals ˝ .

P ROOF . The implications (a) H) (b) H) (c) H) (a) are obvious.(c)

H)(d). Let E be a nite extension of ˝ , and let ˛

2E . The minimum polynomial of ˛

has degree 1, and so ˛ 2˝ .(d) H) (c). Let f be an irreducible polynomial in ˝ŒX . Then ˝ŒX =.f / is an extensioneld of ˝ of degree deg .f / (see 1.30 ), and so deg .f / D1. D EFINITION 1.43 (a) A eld ˝ is said to be algebraically closed if it satises the equivalentstatements of Proposition 1.42 .

(b) A eld ˝ is said to be an algebraic closure of a subeld F when it is algebraicallyclosed and algebraic over F .

For example, the fundamental theorem of algebra (see 5.6 below) says that C is alge-

braically closed. It is an algebraic closure of R .P ROPOSITION 1.44 If ˝ is algebraic over F and every polynomial f 2FŒX splits in˝ŒX , then ˝ is algebraically closed (hence an algebraic closure of F ).

P ROOF . Let f be a nonconstant polynomial in ˝ŒX . We have to show that f has a root in˝ . We know (see 1.25 ) that f has a root ˛ in some nite extension ˝ 0 of ˝ . Set

f Da n X n C Ca 0 , a i 2˝;and consider the elds

F F Œa0 ; : : : ; a n F Œa0 ; : : : ; a n ;˛ :Each extension is algebraic and nitely generated, and hence nite (by 1.30 ). Therefore ˛lies in a nite extension of F , and so is algebraic over F — it is a root of a polynomial gwith coefcients in F . By assumption, g splits in ˝ŒX , and so the roots of g in ˝ 0 all lie in˝ . In particular, ˛ 2˝: P ROPOSITION 1.45 Let ˝ F ; then

f˛ 2˝ j˛ algebraic over F gis a eld.

P ROOF . If ˛ and ˇ are algebraic over F , then F Œ˛;ˇ is a eld (by 1.31 ) of nite degreeover F (by 1.30 ). Thus, every element of F Œ˛;ˇ is algebraic over F , including ˛ ˙ ˇ , ˛=ˇ ,˛ˇ .

The eld constructed in the proposition is called the algebraic closure of F in ˝ .

C OROLLARY 1.46 Let ˝ be an algebraically closed eld. For any subeld F of ˝ , the algebraic closure of F in ˝ is an algebraic closure of F:

P ROOF . From its denition, we see that it is algebraic over F and every polynomial in F ŒXsplits in it. Now Proposition 1.44 shows that it is an algebraic closure of F .


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Exercises 25

Thus, when we admit the fundamental theorem of algebra ( 5.6 ), every subeld of C hasan algebraic closure (in fact, a canonical algebraic closure). Later (Chapter 6) we shall prove(using the axiom of choice) that every eld has an algebraic closure.

A SIDE 1.47 Although various classes of eld, for example, number elds and function elds, hadbeen studied earlier, the rst systematic account of the theory of abstract elds was given by Steinitzin 1910 (Algebraische Theorie der K örper, J. Reine Angew. Math., 137:167–309). Here he introducedthe notion of a prime eld, distinguished between separable and inseparable extensions, and showedthat every eld can be obtained as an algebraic extension of a purely transcendental extension. Healso proved that every eld has an algebraic closure, unique up to isomorphism. His work inuencedlater algebraists (Noether, van der Waerden, Artin, . . . ) and his article has been described by Bourbakias “. . . a fundamental work which may be considered as the origin of today’s concept of algebra”.See: Roquette, Peter, In memoriam Ernst Steinitz (1871–1928). J. Reine Angew. Math. 648 (2010),1–11.

Exercises

1-1 Let E DQ Œ˛, where ˛ 3 ˛ 2 C˛ C2 D0. Express .˛ 2 C˛ C1/.˛ 2 ˛/ and .˛ 1/ 1in the form a˛ 2 Cb˛ Cc with a;b;c 2Q .1-2 Determine ŒQ .p 2;p 3/WQ .1-3 Let F be a eld, and let f .X/ 2FŒX.

(a) For every a 2F , show that there is a polynomial q.X/ 2FŒX such thatf .X/ Dq.X/.X a/ Cf .a/:

(b) Deduce that f .a/ D0 if and only if .X a/ jf .X/ .(c) Deduce that f .X/ can have at most deg f roots.(d) Let G be a nite abelian group. If G has at most m elements of order dividing m for

each divisor m of .G W1/ , show that G is cyclic.(e) Deduce that a nite subgroup of F , F a eld, is cyclic.

1-4 Show that with straight-edge, compass, and angle-trisector, it is possible to construct aregular 7-gon.

1-5 Let f .X/ be an irreducible polynomial over F of degree n, and let E be a eld

extension of F with ŒEWF Dm . If gcd .m;n/ D1, show that f is irreducible over E .


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27/138

C HAPTER 2Splitting Fields; Multiple Roots

Maps from simple extensions.

Let E and E 0 be elds containing F . Recall that an F -homomorphism is a homomorphism

' WE ! E 0such that '.a/ Da for all a 2F . Thus an F -homorphism ' maps a polynomial

Xa i1 im ˛ i11 ˛ imm ; ai1 im 2F;to

Xa i1 im '.˛ 1 / i1 '.˛ m / im :An F -isomorphism is a bijective F -homomorphism.An F -homomorphism E ! E 0 of elds is, in particular, an injective F -linear map

of F -vector spaces, and so, if E and E 0 have the same nite degree over F , then everyF -homomorphism is an F -isomorphism.

P ROPOSITION 2 .1 Let F .˛/ be a simple eld extension of a eld F , and let ˝ be a second eld containing F .

(a) Let ˛ be transcendental over F . For every F -homomorphism ' WF.˛/ ! ˝ , '.˛/ istranscendental over F , and the map ' 7! '.˛/ denes a one-to-one correspondence

fF -homomorphisms ' WF.˛/ ! ˝ g $ felements of ˝ transcendental over F g:(b) Let ˛ be algebraic over F with minimum polynomial f .X/ . For every F -homomorphism

' WFŒ˛ ! ˝ , '.˛/ is a root of f .X/ in ˝ , and the map ' 7! '.˛/ denes a one-to-one correspondence

fF -homomorphisms ' WFŒ˛ ! ˝ g $ froots of f in ˝ g:In particular, the number of such maps is the number of distinct roots of f in ˝ .

P ROOF . (a) To say that ˛ is transcendental over F means that F Œ˛ is isomorphic to thepolynomial ring in the symbol ˛ with coefcients in F . For every 2˝ , there is a uniqueF -homomorphism ' WFŒ˛ ! ˝ sending ˛ to (see 1.5). This extends to the eld of fractions F .˛/ of F Œ˛ if and only if all nonzero elements of F Œ˛ are sent to nonzeroelements of ˝ , which is so if and only if is transcendental.

27


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28 2. S PLITTING F IELDS ; M ULTIPLE ROOTS

(b) Let f .X/ DPa i X i , and consider an F -homomorphism ' WFŒ˛ ! ˝ . On applying' to the equation Pa i ˛ i D0, we obtain the equation Pa i '.˛/ i D0, which shows that'.˛/ is a root of f .X/ in ˝ . Conversely, if 2˝ is a root of f .X/ , then the mapFŒX

!˝ , g.X/

7!g. / , factors through FŒX =.f.X//. When composed with the inverse

of the isomorphism X Cf .X/ 7! ˛ WFŒX =.f.X// ! FŒ˛ , this becomes a homomorphismFŒ˛ ! ˝ sending ˛ to . We shall need a slight generalization of this result.

P ROPOSITION 2 .2 Let F .˛/ be a simple eld extension of a eld F , and let ' 0WF ! ˝ be a homomorphism of F into a second eld ˝ .

(a) If ˛ is transcendental over F , then the map ' 7! '.˛/ denes a one-to-one correspon-dence

fextensions '

WF.˛/

!˝ of ' 0

g $ felements of ˝ transcendental over ' 0 .F /

g:

(b) If ˛ is algebraic over F , with minimum polynomial f .X/ , then the map ' 7! '.˛/denes a one-to-one correspondence

fextensions ' WFŒ˛ ! ˝ of ' 0g $ froots of ' 0 f in ˝ g:In particular, the number of such maps is the number of distinct roots of ' 0 f in ˝ .

By ' 0 f we mean the polynomial obtained by applying ' 0 to the coefcients of f :if f D

Pa i X i then ' 0 f D

P'.a i /X i . By an extension of ' 0 to F.˛/ we mean a

homomorphism '

WF.˛/

!˝ such that '

jF

D' 0 .

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

Splitting elds

Let f be a polynomial with coefcients in F . A eld E containing F is said to split f if f splits in EŒX: f .X/ DQ

miD1 .X ˛ i / with ˛ i 2E . If, in addition, E is generated by theroots of f ,

E DF Œ 1̨ ; : : : ; ˛ m ;then it is called a splitting or root eld for f . Note that Qf i .X / m i (m i 1) and Qf i .X /have the same splitting elds. Also, that if f has deg .f / 1 roots in E , then it splits inEŒX.E XAMPLE 2 .3 (a) Let f .X/ DaX 2 CbX Cc 2Q ŒX, and let ˛ Dp b2 4ac . The sub-eld Q Œ˛ of C is a splitting eld for f .

(b) Let f .X/ DX 3 CaX 2 CbX Cc 2Q ŒX be irreducible, and let ˛ 1 ; ˛ 2 ; ˛ 3 be itsroots in C . Since the nonreal roots of f occur in conjugate pairs, either 1 or 3 of the ˛ i arereal. Then Q Œ1̨ ; ˛ 2 ; ˛ 3 DQ Œ1̨ ; ˛ 2 is a splitting eld for f .X/ . Note that ŒQ Œ1̨ WQ D3and that ŒQ Œ1̨ ; ˛ 2 WQ Œ1̨ D1 or 2, and so ŒQ Œ1̨ ; ˛ 2 WQ D3 or 6. We’ll see later ( 4.2) thatthe degree is 3 if and only if the discriminant of f .X/ is a square in Q . For example, thediscriminant of X 3 CbX Cc is 4b 3 27c 2 , and so the splitting eld of X 3 C10X C1has degree 6 over Q .


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Splitting elds 29

P ROPOSITION 2 .4 Every polynomial f 2FŒX has a splitting eld Ef , and ŒEf WF . deg f /Š . factorial deg f /:

P ROOF . Let F 1 DF Œ 1̨ be a stem eld for some monic irreducible factor of f in F ŒX.Then f .˛ 1 / D0, and we let F 2 DF 1 Œ2̨ be a stem eld for some monic irreducible factorof f .X/=.X ˛ 1 / in F 1 ŒX. Continuing in this fashion, we arrive at a splitting eld Ef .

Let n Ddeg f . Then ŒF 1WF Ddeg g1 n , ŒF 2WF 1 n 1;:::, and so ŒEf WE nŠ.R EMARK 2 .5 Let F be a eld. For a given integer n , there may or may not exist polynomialsof degree n in F ŒX whose splitting eld has degree nŠ — this depends on F . For example,there do not for n > 1 if F DC (see 5.6 ), nor for n > 2 if F DFp (see 4.22 ) or F DR .However, later ( 4.33 ) we shall see how to write down innitely many polynomials of degreen in Q ŒX whose splitting elds have degree nŠ.

E XAMPLE 2 .6 (a) Let f .X/ D.X p 1/=.X 1/ 2Q ŒX, p prime. If is one root of f ,then the remaining roots are 2 ; 3 ; : : : ; p 1 , and so the splitting eld of f is Q Œ .(b) Suppose F is of characteristic p , and let f DX p X a 2FŒX. If ˛ is one root

of f , then the remaining roots are ˛ C1;:::;˛ Cp 1, and so any eld generated over F by˛ is a splitting eld for f (and F Œ˛ ' F ŒX =.f / if f is irreducible).

(c) If ˛ is one root of X n a , then the remaining roots are all of the form ˛ , where n D1. Therefore, if F contains all the nth roots of 1 (by which we mean that X n 1 splits

in F ŒX), then F Œ˛ is a splitting eld for X n a . Note that if p is the characteristic of F ,then X p 1 D.X 1/ p , and so F automatically contains all the p th roots of 1.P ROPOSITION 2 .7 Let f

2FŒX. Let E be a eld generated over F by roots of f , and let

˝ be a eld containing F in which f splits.

(a) There exists an F -homomorphism ' WE ! ˝ ; the number of such homomorphisms isat most ŒEWF , and equals ŒEWF if f has distinct roots in ˝ .

(b) If E and ˝ are both splitting elds for f , then each F -homomorphism E ! ˝ is anisomorphism. In particular, any two splitting elds for f are F -isomorphic.

By f splitting in ˝ , we mean that

f .X/ DYdeg .f /

iD1.X ˛ i /; ˛ i 2˝;

in ˝ŒX . By f having distinct roots in ˝ , we mean that ˛ i ¤ ˛ j if i ¤ j .P ROOF . We begin with an observation: let F , f , and ˝ be as in the statement of theproposition, let L be a subeld of ˝ containing F , and let g be a factor of f in LŒX; then gdivides f in ˝ŒX and so (by unique factorization in ˝ŒX ), g is product of certain numberof the factors X ˛ i of f in ˝ŒX ; in particular, we see that g splits in ˝ , and that its rootsare distinct if the roots of f are distinct.

(a) By assumption, E DF Œ 1̨ ;:::;˛ m with the ˛ i (some of the) roots of f .X/ . Theminimum polynomial of ˛ 1 is an irreducible polynomial f 1 dividing f , and deg .f 1 / DŒFŒ˛1 WF . From the initial observation with L DF , we see that f 1 splits in ˝ , and thatits roots are distinct if the roots of f are distinct. According to Proposition 2.1, there existsan F -homomorphism ' 1

WF Œ 1̨

!˝ , and the number of such homomorphisms is at most

ŒFŒ˛1 WF , with equality holding when f has distinct roots in ˝ .


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The minimum polynomial of ˛ 2 over F Œ 1̨ is an irreducible factor f 2 of f in F Œ 1̨ ŒX.On applying the initial observation with L D' 1 F Œ 1̨ and g D' 1 f 2 , we see that ' 1 f 2 splitsin ˝ , and that its roots are distinct if the roots of f are distinct. According to Proposition2.2 , each ' 1 extends to a homomorphism ' 2

WF Œ 1̨ ; ˛ 2

!˝ , and the number of extensions

is at most ŒFŒ˛1 ; ˛ 2 WF Œ 1̨ , with equality holding when f has distinct roots in ˝ :On combining these statements we conclude that there exists an F -homomorphism' WF Œ 1̨ ; ˛ 2 ! ˝;

and that the number of such homomorphisms is at most ŒFŒ˛1 ; ˛ 2 WF , with equality holdingif f has distinct roots in ˝ :

After repeating the argument m times, we obtain (a).(b) Every F -homomorphism E ! ˝ is injective, and so, if there exists such a homomor-

phisms, ŒEWF Œ˝WF . If E and ˝ are both splitting elds for f , then (a) shows that thereexist homomorphism F E , and so ŒEWF DŒ˝WF . Therefore, every F -homomorphismE

!˝ is an isomorphism.

C OROLLARY 2 .8 Let E and L be extension elds of F , with E nite over F .

(a) The number of F -homomorphisms E ! L is at most ŒEWF .(b) There exists a nite extension ˝=L and an F -homomorphism E ! ˝:P ROOF . Write E DF Œ 1̨ ; : : : ; ˛ m , and f be the product of the minimum polynomials of the˛ i . Let ˝ be a splitting eld for f regarded as an element of LŒX. The proposition showsthat there exists an F -homomorphism E ! ˝ , and the number of such homomorphisms is

ŒEWF . This proves (b), and since an F -homomorphism E ! L can be regarded as anF -homomorphism E ! ˝ , it also proves (a). R EMARK 2 .9 (a) Let E 1 ;E 2 ; : : : ;E m be nite extensions of F , and let L be an extension of F . The corollary shows that there exists a nite extension L 1 =L containing an isomorphicimage of E 1 ; then that there exists a nite extension L 2 =L 1 containing an isomorphicimage of E 2 . On continuing in this fashion, we nd that there exists a nite extension ˝ / Lcontaining an isomorphic copy of every E i .

(b) Let f 2FŒX. If E and E 0 are both splitting elds of f , then we know there is anF -isomorphism E ! E 0, but there will in general be no preferred such isomorphism. Errorand confusion can result if the elds are simply identied. Also, it makes no sense to speak of “the eld F Œ˛ generated by a root of f ” unless f is irreducible (the elds generatedby the roots of two different factors are unrelated). Even when f is irreducible, it makesno sense to speak of “the eld F Œ˛;ˇ generated by two roots ˛; ˇ of f ” (the extensions of FŒ˛ generated by the roots of two different factors of f in F Œ˛ ŒX may be very different).

Multiple roots

Let f; g 2FŒX. Even when f and g have no common factor in F ŒX, one might expectthat they could acquire a common factor in ˝ŒX for some ˝ F . In fact, this doesn’thappen — greatest common divisors don’t change when the eld is extended.

P ROPOSITION 2.10 Let f and g be polynomials in F ŒX, and let ˝ be an extension of F .If r.X/ is the gcd of f and g computed in F ŒX, then it is also the gcd of f and g in ˝ŒX .In particular, distinct monic irreducible polynomials in F ŒX do not acquire a common root in any extension eld of F:


31/138

Multiple roots 31

P ROOF . Let rF .X/ and r ˝ .X/ be the greatest common divisors of f and g in F ŒX and˝ŒX respectively. Certainly rF .X/ jr ˝ .X/ in ˝ŒX , but Euclid’s algorithm ( 1.8 ) showsthat there are polynomials a and b in F ŒX such that

a.X/f.X/ Cb.X/g.X/ DrF .X/;and so r ˝ .X/ divides rF .X / in ˝ŒX .

For the second statement, note that the hypotheses imply that gcd .f;g/ D1 (in F ŒX),and so f and g can’t acquire a common factor in any extension eld.

The proposition allows us to speak of the greatest common divisor of f and g withoutreference to a eld.

Let f 2FŒX. Then f splits into linear factors

f .X/

Da

r

YiD1.X ˛ i / m i ; ˛ i distinct, mi

1,

r

XiD1m i

Ddeg .f /; (*)

in ˝ŒX for some extension eld ˝ of F (see 2.4). We say that ˛ i is a root of f of multiplicity mi in ˝ . If mi > 1 , ˛ i is said to be a multiple root of f , and otherwise it is a simple root .

The unordered sequence of integers m1 ; : : : ; m r in (*) is independent of the extensioneld ˝ chosen to split f . Certainly, it is unchanged when ˝ is replaced with its subeldF Œ 1̨ ; : : : ; ˛ m , but F Œ 1̨ ; : : : ; ˛ m is a splitting eld for f , and any two splitting elds areF -isomorphic ( 2.7 b). We say that f has a multiple root when at least one of the m i > 1 ,and we say that f has only simple roots when all mi D1.

We wish to determine when a polynomial has a multiple root. If f has a multiple factor

in F ŒX, say f DQf i .X/ m i with some mi > 1 , then obviously it will have a multiple root.

If f DQf i with the f i distinct monic irreducible polynomials, then Proposition 2.10 showsthat f has a multiple root if and only if at least one of the f i has a multiple root. Thus, itsufces to determine when an irreducible polynomial has a multiple root.

E XAMPLE 2.11 Let F be of characteristic p ¤ 0, and assume that F contains an element athat is not a p th-power, for example, a DT in the eld Fp .T/: Then X p a is irreduciblein F ŒX, but X p a

1:4

D .X ˛/ p in its splitting eld. Thus an irreducible polynomial canhave multiple roots.

Dene the derivative f 0.X/ of a polynomial f .X/DP

a i X i to be

Pi a i X i 1 . When

f has coefcients in R , this agrees with the denition in calculus. The usual rules fordifferentiating sums and products still hold, but note that in characteristic p the derivative of X p is zero.

P ROPOSITION 2.12 For a nonconstant irreducible polynomial f in F ŒX, the following statements are equivalent:

(a) f has a multiple root;(b) gcd .f;f 0/ ¤ 1;(c) F has characteristic p ¤ 0 and f is a polynomial in X p ;(d) all the roots of f are multiple.


32/138


P ROOF . (a) H) (b). Let ˛ be a multiple root of f , and write f D.X ˛/ m g.X/ , m > 1 ,in some splitting eld. Then

f 0.X /

Dm.X ˛/ m 1 g.X/

C.X ˛/ m g0.X/: (1)

Hence f 0.˛/ D0, and so gcd .f;f 0/ ¤ 1.(b) H) (c). Since f is irreducible and deg .f 0/ < deg .f / ,

gcd .f;f 0/ ¤ 1 H) f 0D0:But, because f is nonconstant, f 0 can be zero only if the characteristic is p ¤ 0 and f is apolynomial in X p .

(c) H) (d). Suppose f .X/ Dg.X p / , and let g.X/ DQi .X a i / m i in some splittingeld of f . Thenf .X/

Dg.X p /

DYi.X p a i / m i

DYi.X ˛ i / pm i

where ˛ pi Da i . Hence every root of f .X/ has multiplicity at least p .(d) H) (a). Obvious.

P ROPOSITION 2.13 For a nonzero polynomial f in F ŒX, the following statements are equivalent:

(a) gcd .f;f 0/ D1;(b) has only simple roots (in any eld splitting of f ).

P ROOF . Let ˝ be an extension of F such that f splits in ˝ŒX . A root ˛ of f in ˝ ismultiple if and only if it is also a root of f 0 (see ( 1)).

If gcd .f;f 0/ D1, then f and f 0 have no common factor X ˛ in ˝ŒX (see 2.10), andso they have no common root. Hence f has only simple roots.

If f has only simple roots, then d def

Dgcd .f;f 0/ must be constant, because otherwise itwould have a root in ˝ which would be a common root of f .X/ and f 0.X/ .

D EFINITION 2.14 A polynomial f 2FŒX is said to be separable if it is nonzero andsatises the equivalent conditions of ( 2.13 )1

According to this denition, a polynomial with multiple factors is not separable. The

preceding discussion shows that f 2FŒX without multiple factors will be separable unless(a) the characteristic of F is p ¤ 0, and (b) at least one of the irreducible factors of f is a polynomial in X p .

Note that, if f 2FŒX is separable, then it remains separable over every eld ˝ containingF (condition (a) of 2.13 continues to hold — see 2.10 ).

D EFINITION 2.15 A eld F is said to be perfect if all irreducible polynomials in F ŒX areseparable.

1This is Bourbaki’s denition. Often (e.g., in the books of Jacobson and in earlier versions of these notes) apolynomial f is said to be separable if none of its irreducible factors has a multiple root.


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Exercises 33

P ROPOSITION 2.16 Every eld of characteristic zero is perfect. A eld F of characteristic p ¤ 0 is perfect if and only if every element of F is a p th power.P ROOF . A eld of characteristic zero is obviously perfect, and so we may suppose F tobe of characteristic p ¤ 0. If F contains an element a that is not a p th power, then thepolynomial X p a 2FŒX is not separable (see 2.11 ). Conversely, if every element of F isa p th power, then every polynomial in X p with coefcients in F is a p th power in F ŒX,

Xa i X p D.Xbi X/ p if a i Dbpi ,and so is not irreducible. E XAMPLE 2.17 (a) A nite eld F is perfect, because the Frobenius endomorphism

a 7! a p WF ! F is injective and therefore surjective (by counting).(b) A eld that can be written as a union of perfect elds is perfect. Therefore, every eldalgebraic over Fp is perfect.

(c) Every algebraically closed eld is perfect.(d) If F 0 has characteristic p ¤ 0, then F DF 0 .X/ is not perfect, because X is not a p th

power.

Exercises

2-1 Let F be a eld of characteristic ¤ 2.(a) Let E be quadratic extension of F (i.e., ŒEWF D2); show that

S.E/ D fa 2F ja is a square in E gis a subgroup of F containing F

2

.(b) Let E and E 0 be quadratic extensions of F ; show that there is an F -isomorphism' WE ! E 0 if and only if S.E/ DS.E 0/ .(c) Show that there is an innite sequence of elds E 1 ; E 2 ; : : : with E i a quadraticextension of Q such that E i is not isomorphic to E j for i ¤ j .(d) Let p be an odd prime. Show that, up to isomorphism, there is exactly one eld withp 2 elements.

2-2 (a) Let F be a eld of characteristic p . Show that if X p X a is reducible in F ŒX,then it splits into distinct factors in F ŒX.

(b) For every prime p , show that X p X 1 is irreducible in Q ŒX.

2-3 Construct a splitting eld for X 5 2 over Q . What is its degree over Q ?

2-4 Find a splitting eld of X pm

1 2Fp ŒX. What is its degree over Fp ?2-5 Let f 2FŒX, where F is a eld of characteristic 0. Let d.X/ Dgcd .f;f 0/ . Show

that g.X/ Df.X/d.X/ 1 has the same roots as f .X/ , and these are all simple roots of g.X/ .

2-6 Let f .X/ be an irreducible polynomial in F ŒX, where F has characteristic p . Showthat f .X/ can be written f .X/ Dg.X p

e/ where g.X/ is irreducible and separable. Deduce

that every root of f .X/ has the same multiplicity p e in any splitting eld.


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C HAPTER 3The Fundamental Theorem of Galois

Theory

In this chapter, we prove the fundamental theorem of Galois theory, which gives a one-to-onecorrespondence between the subelds of the splitting eld of a separable polynomial and thesubgroups of the Galois group of f .

Groups of automorphisms of elds

Consider elds E F . An F -isomorphism E ! E is called an F -automorphism of E .The F -automorphisms of E form a group, which we denote Aut .E=F/ .

E XAMPLE 3 .1 (a) There are two obvious automorphisms of C , namely, the identity mapand complex conjugation. We’ll see later (9.18 ) that by using the Axiom of Choice one canconstruct uncountably many more.

(b) Let E DC .X / . An automorphism of E sends X to another generator of E overC . It follows from ( 9.24 ) below that these are exactly the elements aX CbcX Cd ,

ad bc ¤0.Therefore Aut .E= C / consists of the maps f .X/ 7! f aX CbcX Cd , ad bc ¤ 0, and so

Aut .E= C / ' PGL 2 . C /;the group of invertible 2 2 matrices with complex coefcients modulo its centre. Analystswill note that this is the same as the automorphism group of the Riemann sphere. This is not acoincidence: the eld of meromorphic functions on the Riemann sphere P 1C is C .z/

'C .X / ,

and so there is certainly a map Aut . P 1C / ! Aut . C .z/= C / , which one can show to be anisomorphism.

(c) The group Aut . C .X 1 ;X 2 /=C / is quite complicated — there is a map

PGL 3 . C / DAut . P 2C / ,! Aut . C .X 1 ;X 2 /=C /;but this is very far from being surjective. When there are more X ’s, the group is not known.The group Aut . C .X 1 ; : : : ; X n /=C / is the group of birational automorphisms of P nC , and iscalled the Cremona group. Its study is part of algebraic geometry. See the Wikipedia .

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

35

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


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36 3. T HE F UNDAMENTAL T HEOREM OF G ALOIS T HEORY

P ROPOSITION 3 .2 If E is a splitting eld of a separable polynomial f 2 F ŒX, thenAut .E=F/ has order ŒEWF :P ROOF . As f is separable, it has deg f distinct roots in its splitting eld E . Now Proposition2.7 shows that there are ŒEWF distinct F -homomorphisms E ! E . Because E has nitedegree over F , they are automatically isomorphisms. E XAMPLE 3 .3 (a) Consider a simple extension E DFŒ˛ , and let f be a polynomial withcoefcients in F having ˛ as a root. If f has no root in E other than ˛ , then Aut .E=F/ D1:For example, if 3p 2 denotes the real cube root of 2, then Aut . Q Œ 3p 2 =Q / D1. Thus, in theproposition, it is essential that E be a splitting eld.

(b) Let F be a eld of characteristic p ¤ 0, and let a be an element of F that is not a p thpower. Then f DX p a has only one root in a splitting eld E , and so Aut .E=F/ D1.Thus, in the proposition, it is essential that E be a splitting eld of a separable polynomial.

When G is a group of automorphisms of a eld E , we set

E G DInv .G/ D f˛ 2E j ˛ D˛ , all 2Gg:It is a subeld of E , called the subeld of G -invariants of E or the xed eld of G .

In this section, we shall show that, when E is the splitting eld of a separable polynomialin F ŒX and G DAut .E=F/ , then the maps

M 7! Aut .E=M/; H 7! Inv .H /give a one-to-one correspondence between the set of intermediate elds M , F M E ,and the set of subgroups H of G .

T HEOREM 3.4 (E. A RTIN ) Let G be a nite group of automorphisms of a eld E , and let F DE G ; then ŒEWF .G W1/:P ROOF . Let G D f 1 D1; : : : ; m g, and let ˛ 1 ; : : : ; ˛ n be n > m elements of E . We shallshow that the ˛ i are linearly dependent over F . Consider the system of linear equations

1 .˛ 1 /X 1 C C 1 .˛ n /X n D0:::

m .˛ 1 /X 1 C C m .˛ n /X n D0with coefcients in E . There are m equations and n > m unknowns, and hence there arenontrivial solutions in E . We choose one .c 1 ; : : : ; cn / having the fewest possible nonzeroelements. After renumbering the ˛ i ’s, we may suppose that c1 ¤ 0, and then (after multiply-ing by a scalar) that c1 2F . With these normalizations, we’ll show that all ci 2F . Then therst equation

˛ 1 c1 C C˛ n cn D0(recall that 1 D1) will be a linear relation on the ˛ i .

If not all ci are in F , then k .c i / ¤ ci for some k and i , k ¤ 1 ¤ i . On applying k tothe equations

1 .˛ 1 /c 1 C C 1 .˛ n /c n D0:::

m .˛ 1 /c 1 C C m .˛ n /c n D0


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Separable, normal, and Galois extensions 37

and using that f k 1 ; : : : ; k m gis a permutation of f 1 ; : : : ; m g, we nd that.c 1 ; k .c 2 / ; : : : ; k .c i /;:::/

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

C OROLLARY 3 .5 For any nite group G of automorphisms of a eld E ,

G DAut .E=E G /:P ROOF . As G Aut .E=E G / , we have inequalities

ŒEWE G3.4

.G W1/ . Aut .E=E G /W1/2.8a

ŒEWE G :These must be equalities, and so G DAut .E=E G /:

Separable, normal, and Galois extensions

D EFINITION 3 .6 An algebraic extension E=F is said to be separable if the minimumpolynomial of every element of E is separable; otherwise, it is inseparable .

Thus, an algebraic extension E=F is separable if every irreducible polynomial in F ŒXhaving a root in E is separable, and it is inseparable if

˘ F is nonperfect, and in particular has characteristic p

¤0, and

˘ there is an element ˛ of E whose minimal polynomial is of the form g.X p / , g 2FŒX.For example, E DFp .T / is an inseparable extension of Fp .T p /:D EFINITION 3 .7 An algebraic extension E=F is normal if the minimum polynomial of every element of E splits in EŒX.

In other words, an algebraic extension E=F is normal if every irreducible polynomialf 2FŒX having a root in E splits in EŒX.

Let f be an irreducible polynomial of degree m in F ŒX. If f has a root in E , then

E=F separable H) roots of f distinctE=F normal H) f splits in E H)

f has m distinct roots in E:

Therefore, E=F is normal and separable if and only if, for each ˛ 2E , the minimumpolynomial of ˛ has ŒF Œ˛WF distinct roots in E .E XAMPLE 3 .8 (a) The eld Q Œ 3p 2 , where 3p 2 is the real cube root of 2, is separable butnot normal over Q (because X 3 2 doesn’t split in Q Œ˛).

(b) The eld Fp .T / is normal but not separable over Fp .T p / — the minimum polynomialof T is the inseparable polynomial X p T p .

D EFINITION 3 .9 Let F be a eld. A nite extension E of F is said to be Galois if F isthe xed eld of the group of F -automorphisms of E . This group is then called the Galois group of E over F , and it is denoted by Gal .E=F/ .


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38 3. T HE F UNDAMENTAL T HEOREM OF G ALOIS T HEORY

T HEOREM 3.10 For an extension E=F , the following statements are

fields and galois theory-j.s. milne

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