Galois Theory - James Lingard

1 Galois TheoryDr Wilson1 Michaelmas Term 20001 LATEXed by James Lingard please send all comments and corrections notes are based on a course of lectures given by Dr Wilson during Michaelmas Term 2000for Part IIB of the Cambridge University Mathematics general the notes follow Dr Wilson s lectures very closely, although there are certain particular, the organisation of Chapter 1 is somewhat different to how this part of the coursewas lectured, and I have also consistently avoided the use of a lower-casekto refer to a field in these notes fields are always denoted by upper-case roman notes have not been checked by Dr Wilson and should not be regarded as official notesfor the course.

2 In particular, the responsibility for any errors is mine please email me any comments or LingardOctober 2001 Contents1 Revision from Groups, Rings and Field extensions .. Classification of simple algebraic extensions .. Tests for irreducibility .. The degree of an extension .. Splitting fields .. 52 Separable polynomials and formal differentiation .. Separable extensions .. The Primitive Element Theorem .. Trace and norm .. 113 Algebraic Definitions .. Existence and uniqueness of algebraic closures .. 124 Normal Extensions and Galois Normal extensions.

3 Normal closures .. Fixed fields and Galois extensions .. The Galois correspondence .. Galois groups of polynomials .. 215 Galois Theory of Finite Finite fields .. Galois groups of finite extensions of finite fields .. 246 Cyclotomic Extensions277 Kummer Theory and Solving by Introduction .. Cubics .. Quartics .. Insolubility of the general quintic by radicals .. 3411 Revision from Groups, Rings and Field extensionsSupposeKandLare fields. Recall that a non-zero ring homomorphism :K Lis necessarilyinjective (since ker Kand so ker ={0}) and satisfies (a/b) = (a)/ (b). Therefore is ahomomorphism of extensionofKis given by a fieldLand a non-zero homomorphism :K a will also be called fact, we often identifyKwith its image (K) L, since :K (K) is an isomor-phism, and denote the extension byL/KorK {Ki}i Iis any collection of subfields of a fieldL, then i IKiis also a subfield exercise from the a field extensionL/Kand an arbitrary subsetS L, the subfield ofLgeneratedbyKandSisK(S) = {subfieldsM L|M K, M S}.

4 The lemma above implies that it is a subfield it is the smallest subfield { 1, .. , n}we writeK( 1, .. , n) forK(S).DefinitionA field extensionL/Kisfinitely generatedif for somenthere exist 1, .. , n LsuchthatL=K( 1, .. , n). IfL=K( ) for some L, the extension a field extensionL/K, an element LisalgebraicoverKif there exists a non-zeropolynomialf K[X] such thatf( ) = 0 inL. Otherwise, is algebraic, the monic polynomialf=Xn+an 1Xn 1+ +a1X+a0of smallest degree such thatf( ) = 0 is called theminimal polynomialoff. Clearly suchanfis unique and field extensionL/Kisalgebraicif every Lis algebraic overK. It ispure transcen-dentalif every L\Kis transcendental Classification of simple algebraic extensionsGiven a fieldKand an irreducible polynomialf K[X], recall that the quotient ringK[X]/(f)is a field.

5 Therefore we have a simple algebraic field extensionK K( ) =K[X]/(f), denoting the image ofXunder the quotient , for any simple algebraic field extensionK K( ) letfbe the minimal polynomial of overK. We then have a commutative diagramK//!!DDDDDDDDK[X] K( )inducing an isomorphism of fieldsK[X]/(f) =K( ). Thus up to field isomorphisms, any simplealgebraic extension ofKis of the formK K[X]/(f) for some irreduciblef K[X].Therefore, classifying simple algebraic extensions ofK(up to isomorphism) is equivalent toclassifying irreducible monic polynomials inK[X]. Tests for irreducibilityLetRbe a UFD andKits field of fractions, ,K= (Gauss Lemma)A polynomialf R[X]is irreducible inR[X]iff it is irreducible inK[X].

6 Theorem (Eisenstein s Criterion)Supposef=anXn+an 1Xn 1+ +a1X+a0 R[X]and there exists an irreduciblep Rsuch thatp-an,p|aifori=n 1, .. , irreducible inR[X]and hence irreducible inK[X].ProofsSee Groups, Rings and Fields . The degree of an extensionDefinitionIfL/Kis a field extension, thenLhas the structure of a vector space overK. Thedimension of the vector space is called thedegreeof the extension, written [L:K].We say thatLisfiniteoverKif [L:K] is a field extensionL/Kand an element L, is algebraic overKiffK( )/Kisfinite. When is algebraic,[K( ) :K]is the degree of the minimal polynomial of .Proof( )If [K( ) :K] =n, then 1.

7 , nare linearly dependent overK, so there exists apolynomialf K[X] withf( ) = 0, as claimed.( )If is algebraic overKwith minimal polynomialf, thenf( ) = n+an 1 n 1+ +a1 +a0= 0( ) K[X] withg( )6= 0. Sincefis irreducible we have hcf(f, g) = 1. Euclid salgorithm implies that there existx, y K[X] such thatxf+yg= 1 and soy( )g( ) = 1inL(sincef( ) = 0). Sog( ) 1 1, , 2, .. , the subspace ofLgenerated by powersof .NowK( ) consists of all elements of the formh( )/g( ) forh, g K[X] polynomials,g( )6= 0, and soK( ) is spanned as aK-vector space by 1, , 2, ..and hence fromrelation ( ) by 1, , .. , n ofnimplies that the spanning set 1.

8 , n 1is a basis and hence[K( ) :K] = (Tower Law)Given atowerof field extensionsK L M,[M:K] = [M:L][L:K].ProofLet (ui)i I, be a basis forMoverLand let (vj)j J, be a basis for be a basis shall show that (uivj)i I,j Jis a basis forMoverK, from which the result we show that theuivjspanMoverK. Now any vectorx Mmay be written as alinear combination of theui, that isx= i I iuifor some i L. But since thevjspanLoverKwe can write each ias a linearcombination of thevj, that is i= j J ijvj4for some ij K. But thenx= i Ij J ijuivjas we shall show that theuivjare linearly independent overK. Suppose that we have i Ij J ijuivj= 0for some ij L.

9 But then i I j J ijvj ui= 0and then since theuiare linearly independent overLwe must have j J ijvj= 0for eachj J. But then since thevjare linearly independent overKwe must have that ij= 0 for eachi I, j J, as finitely generated,L=K( 1, .. , n), with each ialgebraic overK, thenL/Kis a finite iis algebraic overK( 1, .. , i 1) and so by ( ) we have that for eachi,[K( 1, .. , i) :K( 1, .. , i 1)] is finite. Induction and the Tower Law give the Splitting fieldsRecall that ifL/Kis a field extension andf K[X] we say thatfsplits(completely) overLif it may be written as a product of linear factorsf=k(X 1) (X n),wherek Kand i called asplitting fieldforfifffails to split over any propersubfield ofL, that is, ifL=K( 1.)

10 , n).RemarkSplitting fields always ifgis any irreducible factor off, thenK[X]/(g) =K( ) is an extension ofKforwhichg( ) = 0, where denotes the image ofX. The remainder theorem implies thatg(and hencef) splits off a linear factor. Induction implies that there exists a splitting fieldLforf, with [L:K] n! (n= degf) by ( ).5 Splitting fields are unique up to isomorphisms :K K is an isomorphism of fields, with the polynomialf K[X]corresponding tog= (f) K [X]. Then any splitting fieldLoffoverKis isomorphicover to any splitting fieldL ofgoverK , and we have the commutative diagramL L x x K K ProofSincefsplits inL, so does any irreducible factorf1.