Fields and Galois Theory: Fall 2004

41 Pages·2012·0.364 MB·English

by  Yu-Ru Liu, notes taken by Chris Almost

Fields and Galois Theory
Fall 2004
Professor Yu-Ru Liu
CHRIS ALMOST
Contents
1 Introduction 3
1.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 BriefReviewofRingTheory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Fieldextensions 4
2.1 DegreeofaFieldExtention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2 AlgebraicandTranscendentalNumbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.3 SimpleExtensions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.4 AlgebraicExtensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3 SplittingFields 7
3.1 Existenceofsplittingfields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3.2 Uniquenessofthesplittingfield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
4 SeparablePolynomials 9
4.1 PrimeFields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
4.2 FormalDerivativeandRepeatedRoots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
4.3 SeparablePolynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4.4 PerfectFields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
5 AutomorphismGroups 12
5.1 AutomorphismGroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
5.2 AutomorphismGroupsofPolynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
5.3 FixedFields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
6 GaloisExtensions 13
6.1 SeparableExtensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
6.2 Normalextensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
6.3 Conjugates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
6.4 GaloisExtensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
6.5 Artin’sTheorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1
2 CONTENTS
7 TheGaloisCorrespondence 19
7.1 TheFundementalTheorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
7.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
7.3 BriefReviewofGroupTheory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
7.4 ThePrimitiveElementTheorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
8 RulerandCompassConstructions 24
8.1 ConstructiblePoints. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
8.2 ConstructibleNumbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
8.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
9 CyclotomicExtensions 27
9.1 CyclotomicPolynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
9.2 CyclotomicFields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
9.3 AbelianExtensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
9.4 Constructiblen-gons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
10 GaloisGroupsofPolynomials 30
10.1 Discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
10.2 CubicPolynomials. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
10.3 QuarticPolynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
11 SolvabilitybyRadicals 33
11.1 Cardano’sFormula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
11.2 Solvablegroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
11.3 CyclicExtensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
11.4 RadicalExtensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
11.5 SolvingpolynomialsbyRadicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
11.6 ProbabilisticGaloisTheory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
INTRODUCTION 3
1 Introduction
GaloisTheoryistheinterplaybetweenfieldsandgroups.
1.1 Motivation
Considerthefollowinghistoricalproblems.
• Constructanarbitraryregularn-gonusingonlyarulerandacompass. Weknowhowtoconstructatriangle
andsquare,butwhatabout5-gon,etc.?
• Squarethecircleusingonlyarulerandcompass(i.e. constructasquareofareaπ).
• Solve an arbirary polynomial using only algebraic means (i.e. plus, minus, times, divides, and nthroot).
Thequadraticformulagivesasolutionforquadraticequations. Cubicandquarticequationscanbesolved
similarily. e.g. if x3+px =qthen
(cid:200) (cid:200)
(cid:114) (cid:114)
q p3 q2 q p3 q2
x = 3 + + + 3 − +
2 27 4 2 27 4
• Forwhichquinticequationsdowehaveradicalsolutions? Ifweknowthereissuchasolution,whatdoes
thesolutionlooklike?
Howcanwesolvetheseproblems? Themainstepsinapplyingthetheorythatwedevelopeinthiscourseare
asfollows:
(cid:112)
1. Associatethesolutionofinterest,sayα= πorα=therootofsomequintic,withthefield(cid:81)(α).
2. Associate (cid:81)(α) with the group of isomorphisms of (cid:81)(α) that fix (cid:81), Aut(cid:81)((cid:81)(α)). If α is algebraic then
Aut(cid:81)((cid:81)(α))isfinite. IfαisconstructablethentheorderofAut(cid:81)((cid:81)(α))isincertainforms.
HardQuestion: Howmanyintermediatefieldsbetween(cid:81)and(cid:81)(α)? Thereisa1-1correspondencebetweenthe
intermediatefieldsandthesubgroupsofAut(cid:81)((cid:81)(α))(thisistheFundementalTheoremofGaloistheory.)
1.2 Brief Review of Ring Theory
Forthiscoursewewillbedealingwithcommutativeringswithidentity.
1.1Example. Let R be a ring. We denote by R[x] the polynomial ring over R in indeterminant x. The degree
of a polynomial is the exponent on the leading term. By convention, deg0=−∞. If a polynomial has leading
coefficient1thenitiscalled“monic”.
AringRiscalledadomainifithasnozerodivisors. Anelementu∈Riscalledaunitifitisinvertible. Afield
isacommutativeringinwhicheachnon-zeroelementisaunitand0(cid:54)=1.
1.2Example. IfF isafield,thenF[x]isadomain(itissufficientthatF beadomain)andfor f,g∈F[x], deg(fg)=
deg(f)+deg(g). Thisdegreefunctionactuallymakes F[x]intoaEuclideandomain.
Therational(function)fieldoverafield F isdenoted F(x)andconsistsofallquotientsofpolynomials(with
non-zerodenominator)from F[x]. Itisthesmallestfieldthatcontains F[x].
Anideal I ofaringRisa(notnecessarilyunital)subringofRthatisabsorbingwithrespecttomultiplication
byelementsofR. WecannowconstructR/I,thequotientringmodulo I.
I issaidtobemaximalif I (cid:54)=Randforanyideal J wehave I ⊆J ⊆R⇒I =J∨J =R. I issaidtobeprime
if I (cid:54)=Rand ab∈ I ⇒a∈ I∨b∈ I. Noticethateverymaximalidealisprime, andinPIDseveryprimeidealis
maximal. Fieldshaveonlytrivialideals.
4 FIELDS ANDGALOIS
1.3Theorem. Let I beaproperidealofR. Then
1. R/I isafieldifandonlyif I ismaximal
2. R/I isadomainifandonlyif I isprime
1.4Theorem. (FirstIsomorphismTheorem)Ifϕ:R→S isaringhomomorphismandkerϕ=I thenthereisan
isomorphism
α:R/I →Imϕ:r+I (cid:55)→ϕ(r)
2 Field extensions
2.1Definition. If E is a field containing another field F then E is said to be a field extension of F, denoted by
E/F
2.1 Degree of a Field Extention
If E/F isafieldextensionthenwecanview E asavectorspaceover F.
• Additionisgiventoagreewiththefieldaddition
• Scalarmultiplicationisgiventoagreewiththefieldmultiplication
2.2Definition. The dimension of E viewed as a vector space over F is called the degree of E over F and is
denoted[E:F]. Ifthisquantityhappenstobefinite,then E/F issaidtobeafiniteextension,otherwiseitisan
infiniteextension.
∼
2.3Example. 1. (cid:67)=(cid:82)⊕i(cid:82),so[(cid:67):(cid:82)]=2
2. [(cid:82):(cid:81)]=∞
3. LetF beafield. Therationalfieldisaninfiniteextension. Aninfinitelinearlyindependentsetis{...,x−1,1,x,x2,...}
2.4Theorem. If E/K andK/F arefinitefieldextensions,then E/F isfiniteand
[E:F]=[E:K][K :F]
PROOF: Let {a1,...,am} be a basis for E over K and {b1,...,bn} be a basis for K over F. It suffices to prove
α:={a b |1≤i≤m, 1≤ j≤n}isabasisfor E over F. Everyelementof E isalinearcombinationofelements
i j
of α since each element of E is a linear combination of elements of {a ,...,a }, and each of the a’s (being
1 m i
elements of K) can be written as a linear combination of elements from {b ,...,b }. α is linearly independent
1 n
over F, for otherwise if (cid:80)m (cid:80)n c b a = 0, then {a ,...,a } a basis implies that (cid:80)n c b = 0 for all i.
i=1 j=1 i,j j i 1 m j=1 i,j j
Since{b ,...,b }isalsoabasis,wegetthatc =0foralli and j. (cid:131)
1 n i,j
2.5Definition. Let E/F be a field extension. If K is a subfield of E that contains F then we say that K is an
intermediatefieldof E/F.
2.6Corollary. If E/F isafiniteextensionandK isanintermediatefieldthen[E:K]and[K :F]aredivisorsof
[E:F].
FIELD EXTENSIONS 5
2.2 Algebraic and Transcendental Numbers
2.7Definition. Let E/F beafieldextensionandα∈E. Wesaythatαisalgebraicover F ifthereis f(x)∈F[x]
suchthat f (cid:54)=0and f(α)=0. Otherwiseαissaidtobetranscendentalover F.
Inparticular,forα∈(cid:67)andαalgebraic(transcendental)over(cid:81)(cid:112),we(cid:112)saythatαisanalgebraic(transcendental)
number. For example, all rational numbers are algebraic, as are 2, 3 2+i, etc. The real numbers e (Hermite
1873)andπ(Lindemann1882)aretranscendentalnumbers.
2.8Theorem. (Liouville 1884) Let α ∈ (cid:82)\(cid:81) be a root of a polynomial f(x) ∈ (cid:81)[x] of degree n. Then there
existsaconstantc>0suchthatforanyrationalnumber p withq>0
q
(cid:12) (cid:12)
(cid:12)α− p(cid:12)> c
(cid:12) (cid:12)
(cid:12) q(cid:12) qn
PROOF: Withoutlossofgenerality, wecanassume|α− p|<1andthat f(x)∈(cid:90)[x]and f isirreducible. Then
q
f(α)=0and f(p)(cid:54)=0. BytheMeanValuetheorem,|f(p)|=|f(α)− f(p)|≤M|α− p|,where M =sup|f(cid:48)(x)|
q q q q
for|x−α|<1. Sinceαisirrational,deg(f)≥2andM (cid:54)=0. Furthermore,|f(p)|≥1/qn,andthus|α−p|≥ 1 1,
q q M qn
sotakec= 1. (cid:131)
M
Remark. Liouville’sTheoremsaysthatalgebraicnumbersare“harder”toapproximatebyrationalnumbersthan
transcendentalnumbers. Thue(1909)andSiegel(1921)improvedtheabovetheorembyreplacingnwith n+1
(cid:112) 2
and 2 n, respectively. In 1955, Roth improved the above theorem to |α− p| > c(cid:48) . This won him the Fields
q q2+ε
medalin1958.
2.9Example. z=(cid:80) 1 istrancendental.
n≥1 10n!
Supposethatz isalgebraicandisarootofapolynomialofdegreen. Thenthereisaconstantc>0suchthat
foranyrationalnumber p withq>0
q
(cid:12) (cid:12)
(cid:12)(cid:12)z− p(cid:12)(cid:12)> c
(cid:12) q(cid:12) qn
Consider(cid:80)s 1 = p , q=10s! Wehave
n=1 10n! 10s!
(cid:12) (cid:12) ∞
c <(cid:12)(cid:12)z− p(cid:12)(cid:12)= (cid:88) 1 < 1
qn (cid:12) q(cid:12) 10n! 10(s+1)!−1
n=s+1
Itfollowsthat
10n·s!
0<c< −→0
10(s+1)!−1
ass→∞. Thisimpliesthatc=0,acontradiction.
2.3 Simple Extensions
Let E/F beafieldextensionandα∈E. Let F[α]denotethesmallestsubringof E containing F andαand F(α)
denotethesmallestsufieldof E containing F andα.
2.10Definition. If E=F(α)thenwesaythat E isasimpleextensionof F.
[E:F]canbeeither∞orfinitedependingonwhetherαistranscendentaloralgebraicover F.
6 FIELDS ANDGALOIS
2.11Definition. IfRandR(cid:48) aretworingscontainingafield F,thenaringhomomorphismψ:R→R(cid:48) suchthat
ψ(c)=c ∀c∈F issaidtobean F-homomorphism.
∼
2.12Theorem. Let E/F be a field extension and α ∈ E. If α is transcendental over F then F[α] = F[x] and
∼ ∼
F(α)=F(x). Inparticular, F[α](cid:54)=F(α).
PROOF: The F-homomorphismα(cid:55)→x isclearlythedesiredisomorphismineachcase. (cid:131)
2.13Theorem. Let E/F be a field extension and α ∈ E. If α is algebraic over F then there is a unique monic
irreduciblepolynomial p(x)∈F[x]suchthatthereisan F-isomorphism
ψ:F[x]/〈p(x)〉→F[α]
withψ(x)=α. Fromthisweconcludethat F[α]=F(α).
PROOF: Let ψ : F[x] → F(α) be the unique F-homomorphism with ψ(x) = α. Thus, Imψ = F[α] and let
∼
I =kerψ. Since α is algebraic, I (cid:54)=0. We have F[x]/I =Imψ, a subring of a field, so it is a (principal ideal)
domain. Therefore I is a prime ideal, so it must be generated by some irreducible polynomial p(x). We may
∼
assumethat p(x)ismonicwithoutlossofgenerality. Itfollowsthat F[x]/〈p(x)〉=F[α]isafield. F(α)isalsoa
field,andsinceitisthesmallestfieldthatcontains F[α],wemusthave F[α]=F(α). (cid:131)
2.14Definition. Themonicirreducibleinthelasttheoremiscalledtheminimalpolynomialofαover F.
2.15Theorem. Let E/F beafieldextensionandα∈E.
1. αistranscendentalover F ifandonlyif[F(α):F]=∞
2. αisalgebraicover F ifandonlyif[F(α):F]<∞
If p(x)istheminimalpolynomialofαover F thenwehave[F(α):F]=degp and{1,α,...,αdegp−1}isabasis
of F(α)/F.
2.16Example. Letpbeaprimeandζ betheprimitivepthrootofunity. Itisarootofthecyclotomicpolynomial
p
Φ (x). Fromtheassignment,thispolynomialisirreducibleover(cid:81)anditismonic,soitistheminimalpolynomial
p
ofζ . Thus[(cid:81)(ζ ):(cid:81)]=p−1. (cid:81)(ζ )iscalledthe pth cyclotomicextensionof(cid:81).
p p p
2.4 Algebraic Extensions
2.17Theorem. Let E/F beafieldextension. If[E:F]<∞thereexists{α ,...,α }⊆E suchthat F (cid:36)F(α )(cid:36)
1 n 1
F(α ,α )(cid:36)···F(α ...,α )=E
1 2 1 n
PROOF: Byinductionon[E:F]. If[E:F]=1, E=F andwearedone. Supposethat[E:F]>1. Thenthereis
α ∈E\F suchthat[E :F]=[E :F(α )][F(α ):F]. Since[F(α ):F]>1,wegetthat[E :F(α )]<[E :F].
1 1 1 1 1
Applyingtheinductionhypothesisto[E:F(α )],thereis{α ,...,α }⊆EsuchthatF(α )=F (cid:36)F (α )(cid:36)···(cid:36)
1 2 n 1 1 1 2
F (α ...,α )=E. Itfollowsthat E=F(α )(α ...,α )=F(α ...,α ). (cid:131)
1 2 n 1 2 n 1 n
2.18Definition. Afieldextension E/F isalgebraicifeveryα∈E isalgebraicover F. Otherwisetheextensionis
transcendental.
2.19Theorem. Let E/F beafieldextension. If[E:F]<∞then E/F isalgebraic.
SPLITTINGFIELDS 7
PROOF: Suppose that [E : F] = n. For α ∈ E the elements {1,α,...,αn} are not linearly independent over F.
Thusthereexistc ∈F,notallzero,suchthat
i
n
(cid:88)
cαi =0
i
i=0
Henceαisarootofthepolynomial(cid:80)n c xi ∈F[x]. (cid:131)
i=0 i
2.20Theorem. Let E/F beafieldextension. Definethesetofalgebraicelementstobe
L:={α∈E|[F(α):F]<∞}
Then L isanintermediatefield.
PROOF: If a,b ∈ L, then [F(a): F]<∞ and [F(b): F]<∞. Consider the field F(a,b). By assignment 1, we
have[F(a,b):F(a)]≤[F(b):F]. Itfollowsthat
[F(a,b):F]=[F(a,b):F(a)][F(a):F]≤[F(b):F][F(a):F]<∞
Thus F(a,b)/F isalgebraic,soa±b,ab,anda/b(b(cid:54)=0)areallin L,so L isafield. (cid:131)
2.21Definition. Let E/F beafieldextension. Theset
F ={α∈E|[F(α):F]<∞}
iscalledthealgebraicclosureof F in E.
2.22Example. Let(cid:81)bethealgebraicclosureof(cid:81)over(cid:67). Then[(cid:81):(cid:81)]=∞(Seeassignment2). Inparticular,
theconverseofTheorem2.19isfalse.
2.23Definition. Afield F issaidtobealgebraicallyclosedifforanyalgebraicextension E/F,then E=F.
Bonus Question: Let F be a field with characteristic p, and assume that F ⊆ E, where E is algebraically
closed. Istheresuchafield E/F suchthat[E:F]<∞?
3 Splitting Fields
3.1Definition. Forafield F,weconsiderthepolynomialring F[x]. For f(x)∈F[x]andafieldextension E/F,
we say that f(x)splits over E if it is a product oflinear factors in E[x]. In other words, E contains all roots of
f(x). Iffurthermorethereisnopropersubfieldof E that f(x)splitsover, thenwesaythat E isasplittingfield
of f(x)in E.
3.1 Existence of splitting fields
3.2Theorem. Let p(x)∈F[x]beirreducible. Thequotientring F[x]/〈p(x)〉isafieldcontaining F andaroot
of p(x).
PROOF: Since p(x) is irreducible, the ideal I = 〈p(x)〉 is maximal. Hence E := F[x]/I is a field. Consider the
map
ψ:F →E:a(cid:55)→a+I
Thismapisinjectivesincekerψisanidealofthefield F (andhencetrivial). Byidentifying F withψ(F), F isa
subfieldof E. Moreover,letα=x+I ∈E.
8 FIELDS ANDGALOIS
Claim. αisarootof p(x)
Writep(x)=a +a x+···+a xn∈F[x],sop(x)=(a +I)+(a +I)x+···+(a +I)xn∈E[x]. Thuswehave
0 1 n 0 1 n
p(α)=(a +I)+(a +I)(x+I)+···+(a +I)(x+I)n=p(x)+I =0
0 1 n
in E. Thusα=x+I ∈E isarootof p(x). (cid:131)
3.3Theorem. (Kronecker)Let f(x)∈F[x]. Thereexistsafield E/F suchthat f(x)splitsover E
PROOF: By induction on degf. If degf = 1, then E = F. If degf > 1 then write f(x) = p(x)g(x) where
p(x) is irreducible. By the previous theorem there is a field K/F containing a root α of p(x). Hence f(x) =
(x−α)h(x)g(x)∈K[x], for some h(x)∈K[x]. Since deg(hg)<degf, by induction there is a field E/K over
which ghisaproductoflinearfactors. Itfollowsthat f(x)splitsover E/F. (cid:131)
3.4Theorem. Every f(x)∈F[x]hasasplittingfieldthatisafiniteextensionof F.
PROOF: For f(x)∈F[x],thereexistsafield E/F suchthat f(x)splitsover E. Say a1,...,an aretheroots. Con-
siderthealgebraicextension F(a ,...,a ). Thisextensionisfinite,and f(x)splitsover F(a ,...,a ). Moreover,
1 n 1 n
f(x)doesnotsplitoveranypropersubfieldof F(a ,...,a ),sinceanysuchsubfieldwillomitatleastoneofthe
1 n
a’s. Therefore F(a ,...,a )isasplittingfieldof f(x)in E. (cid:131)
i 1 n
3.2 Uniqueness of the splitting field
3.5Lemma. Letϕ:R→R bearinghomomorphism. ThenthereisauniqueringhomomorphismΦ:R[x]→
1
R [y]suchthatΦ| =ϕ andΦ(x)= y. WesaythatΦextendsthemapϕ.
1 R
PROOF: Trivial. (cid:131)
3.6Theorem. Letϕ:F →F beanisomorphismoffields,and f(x)∈F[x]. LetΦ:F[x]→F [x]betheunique
1 1
ringisomorphismwhichextendsϕ andmaps x to x. Let f (x)=Φ(f(x))and E/F and E /F besplittingfields
1 1 1
of f and f ,respectively. Thenthereexistsanisomorphismψ:E→E whichextendsϕ.
1 1
PROOF: By induction on [E : F]. If [E : F] = 1, f is a product of linear factors in F[x]. Thus E = F and
E = F . Take ψ = ϕ and we are done. If [E : F] > 1 then let p(x) be an irreducible factor of f(x) with
1 1
degp ≥2. Write p (x)=Φ(p(x)). Let α∈ E and α ∈ E be roots of p and p , respectively. Then we have an
1 ∼ 1 1 ∼ 1
F-isomorphism F(α)=F[x]/〈p(x)〉andan F -isomorphism F (α )=F [x]/〈p (x)〉. Considertheisomorphism
1 1 1 1 1
Φ. Since p (x)=Φ(p (x))theremustexistafieldisomorphism
1 1
∼
Φ :F[x]/〈p(x)〉→F [x]/〈p (x)〉=F (α )
1 1 1 1 1
whichextendsϕ. Itfollowsthatthereexistsafieldisomorphismϕ :F(α)→F (α )whichextendsϕandsends
1 1 1
αtoα .
1
ϕ (cid:47)(cid:47)
F(cid:127)(cid:95) ∼= F1(cid:127)(cid:95)
(cid:15)(cid:15) (cid:15)(cid:15)
F(α(cid:127)(cid:95)) ϕ1 (cid:47)(cid:47) F1(α(cid:127)(cid:95)1)
(cid:15)(cid:15) (cid:15)(cid:15)
ψ (cid:47)(cid:47)
E E
1
Byinduction,since[E:F(α)]<[E:F],thereexistsψ:E→E whichextendsϕ ,andthusextendsϕ. (cid:131)
1 1
SEPARABLEPOLYNOMIALS 9
3.7Corollary. Anytwosplittingfieldsofanon-zeropolynomial f(x)∈F[x]over F are F-isomorphic.
3.8Corollary. (E.H.Moore)Anytwofinitefieldsoforder pn forsomeprime pareisomorphic.
PROOF: Anyfinitefield F oforder pn isasplittingfieldof xpn−x over(cid:70)p (cid:131)
3.9Theorem. Let F be a field and f(x)∈ F[x] have degree n≥1. Let E/F be a splitting field of f(x). Then
[E:F]dividesn!.
PROOF: By induction on degf. If degf = 1 then [E : F] = 1 and it’s trivial. Suppose degf > 1. If f is
∼
irreducibleandα∈ E isarootof f,thenthereexistsasimpleextension F(α)/F suchthat F(α)= F[x]/〈f(x)〉
and[F(α): F]=degf =n. Write f(x)=(x−α)g(x)∈ F(α)[x]anddegg =n−1. Byinduction, [E : F(α)]
isadivisorof(n−1)!. Itfollowsthat[E :F]=[E :F(α)][F(α):F]divides n!. If f(x)isnotirreducible,write
f =g·h,wheredegg=manddegh=k. LetK beasplittingfieldof g over F. Byinduction,[K :F]dividesm!.
Also,[E:K]dividesk!(E isasplittingfieldofhoverK). Thus[E:F]dividesm!k!,whichisafactorofn!. (cid:131)
4 Separable Polynomials
4.1 Prime Fields
4.1Definition. Theprimefieldofafield F istheintersectionofallofthesubfieldsof F.
4.2Theorem. If F isafield,thenitsprimefieldisisomorphicto(cid:81)orto(cid:70) forsomeprime p.
p
PROOF: Considertheringmap
χ :(cid:90)→F :n(cid:55)→1+1+···+1
(cid:124) (cid:123)(cid:122) (cid:125)
ntimes
Let I =kerχ. Then (cid:90)/I is a domain (since it is isomorphic to the image of χ((cid:90)), a subring of F). Hence I is a
primeidealof(cid:90),andsoeitheris〈0〉or〈p〉forsomeprime p. If I =〈0〉then(cid:90)⊆F. Itfollowsthatallsubfields
of F containFrac(F)=(cid:81),andsotheprimefieldof F is(cid:81). If I =〈p〉thenbythefirstisomorphismtheorem,
∼ ∼
(cid:70) =(cid:90)/〈p〉=Imχ ⊆F
p
andsotheprimefieldof F is(cid:70) . (cid:131)
p
4.3Definition. Given a field F, if the prime field is isomorphic to (cid:81) then we say that F has characteristic 0,
denotedchF =0. Ontheotherhand,iftheprimefieldisisomorphicto(cid:70) thenwesaychF = p. Noticethatif
p
chF =pthen(a+b)p=ap+bp.
4.2 Formal Derivative and Repeated Roots
4.4Definition. If F isafield,themonomials{1,x,x2,...}forman F-basisfor F[x]. Definethelinearoperator
D : F[x] → F[x] by D1 = 0 and Dxn = nxn−1. D is called the formal derivative, and is also denoted with a
prime.
The formal derivative has all the usual algebraic properties of the differential operator from calculus, in
particular
1. D(f +g)=Df +Dg
2. D(fg)=(Df)g+f(Dg)
10 FIELDS ANDGALOIS
4.5Theorem. Let F befieldand f(x)∈F[x].
1. IfchF =0andDf =0then f(x)=c forsomec∈F
2. IfchF =pandDf =0then f(x)=g(xp)forsome g(x)∈F[x]
PROOF: Trivial. (cid:131)
4.6Definition. Let E/F be a field extension and f(x)∈ F[x]. We say that α∈ E is a repeated root of f(x) if
f(x)=(x−α)2g(x)forsome g(x)∈E[x].
4.7Lemma. If E[x],αisarepeatedrootof f(x)ifandonlyif x−αdividesboth f andDf.
PROOF: If f(x)=(x−α)2g(x)thenDf(x)=2(x−α)g(x)+(x−α)2Dg(x),so x−αisacommonfactorof f
andDf. Supposeconverselythat x−αdividesboth f andDf. Write f(x)=(x−α)h(x),forsomeh(x)∈E[x].
ThenDf(x)=h(x)+(x−α)Dh(x). Df(α)=0impliesthath(α)=0,andsowearedone. (cid:131)
4.8Theorem. Let f(x)∈F[x]. Then f hasnorepeatedrootsinanyextensionofF ifandonlyifgcd(f,Df)=1
in F[x]
Notice that the condition of repeated roots depends on the extension of F, while the gcd condition involves
only F.
PROOF: Let g = gcd(f,Df). Write g = sf +tDf for some polynomials s(x),t(x) ∈ F[x] (F[x] is a Euclidean
domain). Suppose f(x)hasarepeatedrootαinsomeextension E/F. Thenclearly x−αisacommonfactorof
f andDf,andso g (cid:54)=1. Supposenowthat g (cid:54)=1. Thenthereisanextension E/F suchthat E containsarootα
of g. Then x−αdividesboth f andDf,andsoαisarepeatedrootof f. (cid:131)
4.3 Separable Polynomials
4.9Definition. Let F beafieldand f(x)∈ F[x]notzero. If f(x)isirreducible,thenwesay f(x)isseparable
over F ifithasnorepeatedrootsin anyextensionof F. If f(x)isnotirreducible, thenwesayitisseparableif
allofit’sirreduciblefactorsareseparable.
4.10Example. Considerthepolynomial f(x)= xt−a∈F[x],with t ≥2. Ifa=0,then f isclearlyseparable,
astheonlyirreduciblefactorof f is x. Alinearpolynomialisalwaysseparable. Nowweassumethata(cid:54)=0. Note
thatDf(x)=txt−1.
1. IfchF =0thengcd(f,Df)=1,so f isseparable.
2. IfchF =pandgcd(p,t)=1thengcd(f,Df)=1,so f isseparable.
3. IfchF =pand t=pthenDf =0,sogcd(f,Df)(cid:54)=1. However,itisstillpossiblethatalloftheirreducible
factors p(x) have the property that gcd(p,Dp)=1. To decide, we need to find the irreducible factors of
f. Define Fp ={ap |a∈ F}, a subfield of F. If a∈ Fp then there is some b∈ F such that a= bp, and so
f(x) = (x−b)p, and f is separable. There is another case, although it only comes up if F is an infinite
field of characteristic p. If a (cid:54)∈ Fp then we claim that f(x)= xp−a is irreducible. Assume that we may
write xp−a=g(x)h(x),where g,h∈F[x]aremonic. Let E/F beaextensionsuchthat xp−ahasaroot
β ∈E. Thenβp=a,andsoβ (cid:54)∈F. Wehave
xp−a=xp−βp=(x−β)p
Thus g(x) = (x −β)r and h(x) = (x −β)s for some r+s = p. Write g(x) = xr +rβxr−1+···. Then
since rβ ∈ F, r =0in F. Thus r =kp forsome k. Thisshowsthateither r =0ors=0,andso xp−a is
irreducibleover F. Therefore xp−aisnotseparableinthiscase. Wesaythat f ispurelyinseparablesince
alloftherootsof f arethesame.