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1 galois theory : THE PROOFS, THE WHOLE PROOFS, ANDNOTHING BUT THE PROOFSMARK DICKINSONC ontents1. Notation and conventions12. Field extensions13. Algebraic extensions44. Splitting fields65. Normality76. Separability77. galois extensions88. Linear independence of characters109. Fixed fields1310. The Fundamental Theorem14I ve adopted a slightly different method of proof from the textbook for many ofthe galois theory results. For your reference, here s a summary of the main resultsand their proofs, without any of that pesky history and motivation or distractingexamples to get in the way.

2 Just the proofs1. Almost all of the hard work liesin three main theorems: Corollary (a splitting field of a separable polynomialis galois ), Theorem (linear independence of characters), and Theorem (thedegree ofKoverKHis bounded by the order ofH). and conventionsFor groupsHandG, we writeH < Gto mean thatHis a (not necessarilyproper) subgroup ofG. Similarly, for setsSandT, we writeS Tto indicatethatSis a (not necessarily proper) subset extensionsDefinition extensionK/Fis a triple (F,K,i) consisting of fieldsFandKtogether with a field homomorphismi:F K.

3 When there is no dangerof confusionFwill be identified with its imagei(F) underi, and so regarded as asubfield field extensionK/FisfiniteifKis finite-dimensional as avector space overF. Thedegree[K:F] of a finite extensionK/Fis the dimensionofKas a vector space the occasional definition or two. Not to mention the theorems, lemmas and so DICKINSONP roposition (Tower Law).Suppose thatK/EandE/Fare field finite if and only ifK/EandE/Fare finite, and in this case[K:F] = [K:E][E:F]. finite then any basis forKas a vector space overFspansKas a vector space overE, hence contains a basis forKoverE, soK/Eis , any basis forE/Fgives a subset ofKlinearly independent overF, whichcan be completed to a basis ofK/F.

4 HenceE/Fis also , suppose thatK/EandE/Fare finite, and letSbe a basis forKas a vector space overEandTa basis forEas a vector space overF. We ll showthat the [K:E][E:F] elementstsforsinSandtinTare distinct, and that theyprovide a basis forKas vector space thatvis an element ofK. SinceSspansKoverE,v= s S ssfor some elements sinE. SinceTspansEoverF, for everysinSthere is arelation s= t T s,ttfor some elements s,tofF. Substituting givesv= s S t T s,tts,hence the elementstsspanKas a vector space suppose that there are elements s,tofFsuch that s S t T s,tts= s= t T s,ttfor eachsinS, we can rewrite this as s S ss= each sis inEand the elements ofSare linearly independent overE, itfollows that s= 0 for eachs, hence t T s,tt= 0for eachsinS.

5 Since the elements ofTare linearly independent overF, it followsthat s,t= 0 for allsinSandtinT. Hence the elementstsare distinct and linearlyindependendent overF. SoTS={ts|s S,t T}gives a basis forKoverF, andit follows that [K:F] =|TS|=|T||S|= [K:E][E:F] as required. Lemma thatK/Fis a field extension and thatEis a subset ofKthat containsFand is closed under addition and multiplication. IfEis finite-dimensional as a vector space overFthenEis a closed under addition and multiplication, and containsF, itfollows that it is a vector space overF.

6 For each nonzero inE, multiplicationby gives an injectiveF-linear map fromEto itself. Since any injective linearmap from a finite-dimensional vector space to itself is automatically surjective, galois theory : THE PROOFS3multiplication by must be surjective. In particular, 1 is in the image, so 1 = for some inE. Hence has an inverse inE. Since this is true for arbitrarynonzero ,Eis a field. Definition thatK/Fis a field extension and thatSis a subset ofK. Then we writeF(S) for the intersection of all subfields ofKcontaining bothFandS, and call it the fieldgenerated bySoverF.

7 If is an element ofKthenwe writeF( ) for the fieldF({ }) generated byFand . An extensionK/Fis asimple extensionifK=F( ) for some thatKis a field, and thatE1andE2are subfieldsofK. ThecompositumorcompositeE1E2ofE1andE2is the intersection of allsubfields ofKcontaining bothE1andE2. So in the notation of Definition ,E1E2=E1(E2) =E2(E1).Corollary thatK/Fis an extension, and thatE1andE2are subfieldsofKcontainingF. IfE1/FandE2/Fare finite thenE1E2/Fis finite, and[E1E2:F] [E1:F][E2:F]. a basisSforE1overF, and consider the subsetEofKconsistingof linear combinations of the elements ofSwith coefficients inE2:E={ s S ss s E2}.

8 Since 1 is inE1andSspansE1overF, there are elements sofFsuch that1 = s S for anyxinE2,x= s S(x s)sis an element ofE. SinceE1is closed under multiplication, for everytanduinSthere are elements t,usofFsuch thattu= s S t, for elementsx= s S ssandy= s S ssofE,x+y= s S( s+ s)sis inEandxy= t S,u S t utu= t S,u S t u( s S t,uss)= s S( t S,u S t u t,us)sis an element ofE. SoEcontainsE2and is closed under addition and mul-tiplication. Furthermore,SspansEas a vector space overE2, soEis finite-dimensional overE2, of dimension at most|S|= [E1:F].

9 Hence by ,Eis a subfield ofK. SinceEcontains bothE1andE2, and is generatedby elements ofE1E2,E=E1E2. By the Tower Law,E1E2/Fis finite, and[E1E2:F] = [E1E2:E2][E2:F] [E1:F][E2:F] as required. 4 MARK extensionsThroughout this section,K/Fwill be a field extension and will denote anelement 0 i daixibe a polynomial inF[x]. Then we writef( )for the elementf( ) = 0 i dai follows directly from the definitions that forfandginF[x] and inK,(f g)( ) =f( ) g( ), (fg)( ) =f( )g( ) anda( ) =afor anyainF(whereais interpreted as the constant polynomialaon the left-hand side).

10 Remark the notation above resembles the usual one for application ofa functionfto an argument , it s important not to forget that a polynomial isnot, strictly speaking, a function fromFtoF. For example, whenF=Fpis thefinite field withpelements, the polynomialsf=xpandg=xinF[x] are distinct,even though as functions onFthey re identical:f( ) =g( ) for all a nonzero polynomial inF[x]. The element ofKis arootoffiff( ) = 0 the notation of the previous definition, is a root ofFif and only if there is a factorizationf= (x )ginK[x]for some nonzeropolynomialginK[x].