Graduate algebra: numbers, equations, symmetries [Lecture notes]

96 Pages·2013·0.595 MB·English

by  Jenia Tevelev

GRADUATE ALGEBRA: NUMBERS, EQUATIONS, SYMMETRIES
JENIATEVELEV
CONTENTS
§1. AlgebraicExtensions 3
§1.1. Fieldextensions 3
§1.2. Multiplicativityofdegree 5
§1.3. Algebraicextensions 5
§1.4. Adjoiningroots 6
§1.5. Splittingfields 6
§1.6. Algebraicclosure 8
§1.7. Finitefields 10
§1.8. Compositefield 10
§1.9. Exercises 11
§2. GaloisTheory 12
§2.1. Separableextensions 12
§2.2. Normalextensions 14
§2.3. MainTheoremofGaloisTheory 14
§2.4. Fieldsofinvariants 17
§2.5. Exercises 18
§3. ApplicationsofGaloisTheory 18
§3.1. Intermediatesubfieldsinseparableextensions 18
§3.2. FundamentalTheoremofAlgebra 19
§3.3. Quadraticextensions 19
§3.4. Cubicextensions 19
§3.5. Galoisgroupofafinitefield 20
§3.6. Exercises 21
§4. Rootsofunity 22
§4.1. Adjoiningrootsofunity 22
§4.2. Cyclotomicfields 23
§4.3. Rationalcosines 24
§4.4. AbelianextensionsofQ 24
§4.5. Quadraticreciprocity 26
§4.6. Exercises 27
§5. Integralextensionsofrings 29
§5.1. Integralextensions 29
§5.2. Examplesoftheintegralclosure 31
§5.3. Exercises 33
§6. Cyclicextensions 33
§6.1. Cyclicextensions 33
§6.2. Artin’sLemma 35
§6.3. Lagrangeresolvents 36
1
2 JENIATEVELEV
§6.4. Exercises 37
§7. NormandTrace 37
§7.1. Exercises 38
§8. Solvableextensions 38
§8.1. Compositionseries 39
§8.2. Solvablegroups 40
§8.3. Solvableextensions: GaloisTheorem. 41
§8.4. Solvingsolvableextensions 43
§8.5. Exercises 44
§9. Samplemidterm 45
§10. Transcendentalextensions 46
§10.1. Transcendentalnumbers. Liouville’sTheorem 46
§10.2. ProofofHermite’sTheorem 47
§10.3. Transcendencedegree 48
§10.4. Exercises 50
§11. Noether’sNormalizationLemma. Nullstellensatz. 50
§11.1. Noether’sNormalizationLemma 50
§11.2. WeakNullstellensatz 51
§11.3. Algebraicsets. StrongNullstellensatz 52
§11.4. Exercises 54
§12. Localizationandlocalrings 54
§12.1. Examplesfromnumbertheoryandgeometry 54
§12.2. Localizationofrings 55
§12.3. ExtensionandcontractionofidealsinRandS−1R 56
§12.4. Nilradical 57
§12.5. Going-upTheorem 58
§12.6. Exercises 59
§13. BasicAlgebraicGeometry 60
§13.1. Irreduciblealgebraicsets 60
§13.2. Irreduciblecomponents 61
§13.3. Affinealgebraicsets. Regularfunctions. 62
§13.4. Morphismsofaffinealgebraicsets 63
§13.5. Dominantmorphisms 65
§13.6. Finitemorphisms 65
§13.7. Exercises 67
§14. Localizationofmodules 68
§14.1. Localizationasextensionofscalars 68
§14.2. Localizationisexact 69
§14.3. Nakayama’sLemma 69
§14.4. Exercises 70
§15. Rationalfunctionsonaffinevarieties 71
§15.1. Rationalfunctions 71
§15.2. Dimension 71
§15.3. DiscreteValuationRings 73
§15.4. Exercises 75
§16. Samplemidterm 75
§17. Representationsoffinitegroups 76
§17.1. DefinitionandExamples 76
GRADUATEALGEBRA:NUMBERS,EQUATIONS,SYMMETRIES 3
§17.2. G-modules 78
§17.3. Mashke’sTheorem 79
§17.4. Schur’sLemma 80
§17.5. One-dimensionalrepresentations 82
§17.6. Exercises 84
§18. Irreduciblerepresentationsoffinitegroups 85
§18.1. Characters 85
§18.2. Basicoperationsonrepresentationsandtheircharacters 85
§18.3. Schur’sorthogonalityrelations 86
§18.4. Decompositionoftheregularrepresentation 88
§18.5. Thenumberofirreduciblerepresentations 89
§18.6. Charactertableofthedihedralgroup 90
§18.7. Dimensionofanirreduciblerepresentationdivides|G| 92
§18.8. Burnside’sTheorem 93
§18.9. Exercises 95
§1. ALGEBRAIC EXTENSIONS
§1.1. Fieldextensions. Onerarelystudiesasinglefield. Atypicalsituation
is to have two fields K ⊂ F. Then F is called a field extension of K. This
willbeabasicsetupforthissection. Anothernotationforthis: F/K.
√
Examples: R ⊂ C,Q ⊂ Q( 2),etc.
AmyfieldextensionF/K canbeviewedasavectorspaceoverK (how?)
DEFINITION 1.1.1. The dimension dimKF is called the degree of F over K.
Notation: [F : K]. If[F : K] < ∞thenF iscalledafiniteextensionofK.
For example, [C : R] = 2. The basis of C over R is given by 1 and i.
√ √
Wealsohave[Q( 2) : Q] = 2. Thebasisisgivenby1and 2. Noticethat
thisisanon-trivialstatement: oneneedstoknowthat
√ √
• 1and 2arelinearlyindependentoverQ, i.e.a+b 2 (cid:54)= 0ifaorb
√ √
isnotequaltozero. Thisisbecause 2isirrational, 2 (cid:54)= −a/b.
√ √
• 1and 2generateQ( 2)asavectorspaceoverQ. Apriori,anele-
√ √
mentofQ( 2)isafraction a+b√2 witha,b,c,s ∈ Q. However,mul-
c+d 2 √
tiplyingthenumeratoranddenominatorbyc−d 2wecanrewrite
√
thisfractionasalinearcombinationof1and 2.
Wearegoingtostudyextensionslikethisverysystematically.
DEFINITION 1.1.2. Consider a field extension K ⊂ F. An element α ∈ F
is called algebraic over K if α is a root of a non-trivial polynomial with
coefficientsinK. Anelementαiscalledtranscendentalifitisnotalgebraic.
• i ∈ Cisarootofx2−1,soiisalgebraicoverQ.
• π ∈ RistranscendentaloverQ(Lindemann’sTheorem).
• x ∈ C(x)istranscendentaloverC(obvious).
DEFINITION 1.1.3. Consider a field extension K ⊂ F. Let α ∈ F be alge-
braic. A polynomial f(x) ∈ K[x] is called a minimal polynomial of α if f(x)
isamonicpolynomialofminimaldegreesuchthatf(α) = 0.
4 JENIATEVELEV
Let us point out one persistent notational ambiguity. If x is a variable
thenK[x]denotesthealgebraofpolynomialsandK(x)denotesthefieldof
rationalfunctions(i.e.ratiosofpolynomials)invariablex. Butgivenafield
extensionK ⊂ F andanelementα ∈ F,K[α]denotestheminimalsubring
ofF generatedbyK andbyαandK(α)denotestheminimalsubfieldofF
generatedbyK andbyα. Theseobjectsarerelatedasfollows:
THEOREM 1.1.4. Consider a field extension K ⊂ F. Let α ∈ F. Consider a
uniquesurjectivehomomorphismφ : K[x] → K[α]thatsendsxtoα.
• IfαisalgebraicthenKerφisgeneratedbytheminimalpolynomialf(x),
which is irreducible and defined uniquely. Let n = degf(x). Then
[K(α) : K] = n. Moreover, K[α] = K(α). More precisely, elements
1,α,...,αn−1 formaK-basisofK(α).
• Ifαistranscendentalthenφisanisomorphism,whichinducesanisomor-
phismoffieldsK(x) (cid:39) K(α). Inparticular,[K(α) : K] = ∞.
Proof. Kerφisanidealofallpolynomialsthatvanishatα. Kerφ = 0ifand
onlyifαistranscendental(bydefinition). Inthiscase
K[x] (cid:39) K[α] ⊂ F.
Anyinjectivehomomorphismofadomainintoafieldextendstotheinhec-
tivehomomorphismofitsfieldoffractions. Soinourcaseφextendstothe
injectivehomomorphismK(x) → F,anditsimageisobviouslyK(α).
Ifαisalgebraicthen,sinceK[x]isaPID,Kerφisgeneratedbyaunique
monicpolynomialf(x),henceaminimalpolynomialisunique. Since
K[x]/Kerφ (cid:39) K[α]
injectsinF,whichisadomain,K[x]/Kerφisalsoadomain,hencef(x)is
irreducibleandKerφ = (f)isamaximalideal. HenceK[x]/Kerφisafield.
Therefore,K[α]isafield. Therefore,K[α] = K(α).
Finally, we notice that 1,α,...,αn−1 are linearly independent over K
(otherwise we can find a smaller degree polynomial vanishing at α) and
span K[α] over K. Indeed, since f(x) = xn + ... vanishes at α, we can
rewrite αn as a linear combination of smaller powers of α. Then, an easy
argument by induction shows that we can rewrite any αm for m > n as as
alinearcombinationof1,α,...,αn−1. (cid:3)
Anoftenusedcorollary:
COROLLARY1.1.5.IfαisalgebraicoverK then[K(α) : K]isequaltothedegree
oftheminimalpolynomialofα.
Thenumber[K(α) : K]isalsooftencalledthedegreeofαoverK.
√ √
EXAMPLE 1.1.6.[Q( 2 + 3) : Q] = 4. This calculation involves a trick
which will become much more transparent after we discuss Galois theory.
Thepolynomialofdegree4
√ √
(cid:89)
f(x) = (x± 2± 3)
√ √
obviously has 2 + 3 as a root. To show that it is irreducible, we have
Q
toshowthatithasnolinearandquadraticfactorsover . Ifithasalinear
√ √ √ √ √
factor then 2 ± 3 ∈ Q, therefore ( 2 ± 3)2 ∈ Q, therefore 6 ∈ Q.
GRADUATEALGEBRA:NUMBERS,EQUATIONS,SYMMETRIES 5
√
Then we use the standard argument: if 6 = a/b in lowest terms then
a2 = 6b2, therefore 2|a, therefore 4|a2, therefore 2|b, contradiction. If f(x)
has a quadratic factor over Q then the sum of two roots of f(x) belongs
√ √
to Q, which implies that 2 or 3 belong to Q, unless the two roots are
√ √ √ √
opposite,say 2+ 3and− 2− 3. ButthentheirproductisinQ,which
√
impliesthat 6 ∈ Q,acontradiction.
√
EXAMPLE 1.1.7.[Q( 3 2) : Q] and the minimal polynomial is x3 − 2 (irre-
duciblebyEisenstein’scriterion).
§1.2. Multiplicativityofdegree.
LEMMA1.2.1.Consideratoweroffieldextensions
K ⊂ F ⊂ L.
If[F : K] = nand[L : F] = mthen[L : K] = nm.
Proof. Itiseasytoproveabitmore: ife ,...,e isabasisofLoverF and
1 m
f ,...,f isabasisofF overKthen{e f }isabasisofLoverK(why?) (cid:3)
1 n i j
Thisstatementhasinterestingdivisibilityconsequences. Forexample,if
√
K/Qisafinitefieldextensioncontaining 2,then[K : Q]isevenbecause
√ √ √
[K : Q] = [Q( 2) : Q][K : Q( 2)] = 2[K : Q( 2)].
§1.3. Algebraicextensions.
DEFINITION 1.3.1. AnextensionK ⊂ F iscalledalgebraicifanyelementof
F isalgebraicoverK.
THEOREM1.3.2.(a)Anyfiniteextensionisalgebraic.
(b)Letα ,...,α ⊂ F bealgebraicoverK. Then
1 r
K(α ,...,α ) = K[α ,...,α ]
1 r 1 r
isfiniteoverK. Inparticular,anyelementofK(α ,...,α )isalgebraicoverK.
1 r
Proof. If [F : K] < ∞ then 1,α,α2,... are linearly dependent over K for
anyα ∈ F. Therefore,somenon-constantpolynomialwithcoefficientinK
vanishesatα,i.e.αisalgebraic.
Now suppose that α ,...,α ⊂ F are algebraic over K. Arguing by
1 r
induction on r (with the base of induction given by Theorem 1.1.4), let’s
assume that K(α ,...,α ) = K[α ,...,α ] is finite over K. Since α
1 r−1 1 r−1 r
is algebraic over K, it is also algebraic over K(α ,...,α ). Using Theo-
1 r−1
rem1.1.4,weseethat
K(α ,...,α )[α ] = K(α ,...,α )(α ) = K(α ,...,α ,α )
1 r−1 r 1 r−1 r 1 r−1 r
is finite-dimensional over K(α ,...,α ). Then [K(α ,...,α ) : K] < ∞
1 r−1 1 r
byLemma1.2.1. Itfollowsfrompart(a)thatanyelementofK(α ,...,α )
1 r
isalgebraicoverK. (cid:3)
Oneconsequenceof thistheoremisthat ifα andβ arealgebraicoverK
then so are α+β, αβ, and α/β. However, it can take some work to write
theirminimalpolynomials,asExample1.1.6shows.
6 JENIATEVELEV
§1.4. Adjoiningroots. Intheprevioussectionwestudiedafixedextension
K ⊂ F. Ifα ∈ F isalgebraicoverK thenK(α)isisomorphictoK[x]/(f),
wheref(x)isaminimalpolynomialofα. AnalgebraicextensionK ⊂ K(α)
generatedbyoneelementissometimescalledsimple.
Now we will start with a field K and learn how to build its extensions
andcomparethem. ThemainbuildingblockwasdiscoveredbyKronecker:
LEMMA 1.4.1.If f(x) ∈ K[x] is irreducible and monic then K[x]/(f) is a field
extensionofK generatedbyα := x+(f). Theminimalpolynomialofαisf(x).
Proof. Since f is irreducible and K[x] is a PID, K[x]/(f) is a field. It is
obviously generated by α. Since f(x) ≡ 0 mod (f), α is a root of f(x).
Sincef(x)isirreducibleoverK,f(x)istheminimalpolynomialofα. (cid:3)
WewilloftenwanttocompareextensionsF andF(cid:48) ofthesamefieldK.
We say that F and F(cid:48) are isomorphic over K if there exists an isomorphism
φ : F → F(cid:48) suchthatφ(a) = aforanya ∈ K. Hereisthebasicfact:
LEMMA1.4.2.LetK(α)andK(β)bealgebraicextensionsofK. TFAE:
• αandβ havethesameminimalpolynomial.
• ThereexistsandisomorphismofK(α)toK(β)overK thatsendsαtoβ.
Proof. If α and β have the same minimal polynomials then the analysis
above shows that both K(α) and K(β) are isomorphic to K[x]/(f), where
f(x) is the common minimal polynomial. Notice that an induced isomor-
phismbetweenK(α)andK(β)simplysendsαtoβ. (cid:3)
√ √
EXAMPLE1.4.3. FieldsQ( 3 2)andQ(ω 3 2)areisomorphic(hereωisaprim-
itivecubicrootofunity),becausetheyhavethesameminimalpolynomial
√
x3 −2. However, they are not equal inside C because Q( 3 2) is contained
R
in buttheotherfieldisnot.
EXAMPLE1.4.4. Bycontrast,itiseasytoseethat
√ √ √ √ √ √
Q( 2+ 3) = Q( 2− 3) = Q( 2, 3).
So in this case the isomorphism in the Lemma is an automorphism of the
√ √
fieldQ( 2, 3).
§1.5. Splittingfields. Wewouldliketoadjoinallrootsofapolynomial.
DEFINITION1.5.1. AfieldF ⊃ K iscalledasplittingfieldoff(x) ∈ K[x]if
• f splitsintolinearfactorsinF[x],and
• F isgeneratedbyrootsoff(x).
Inotherwords,f(x)splitsinF butnotinanypropersubfieldofF.
√ √
EXAMPLE 1.5.2. Recall that α√= 2 + 3 is a root of√f = x4√− 10x2√+ 1.
Since α2 ∈ Q(α), we see that 6 ∈ Q(α). Therefore α 6 = 2 3+3 2 ∈
√ √ √ √ √
Q(α). Therefore 2 = (2 3+3 2)−2( 3+ 2) ∈ Q(α). Thisshowsthat
√ √ √ √
Q( 3+ 2) = Q( 2, 3), whichcontainsallrootsoff(x). Sointhiscase
√ √
Q( 3+ 2)isasplittingfield.
GRADUATEALGEBRA:NUMBERS,EQUATIONS,SYMMETRIES 7
√
EXAMPLE 1.5.3. The splitting field of f(x) = x3 −2 is not eq√ual to Q( 3 2)
because not all of the roots are real. The splitting field is Q( 3 2,ω). It has
degree6overQbymultiplicativityofdegreebecauseωisarootofthequa-
√
draticpolynomialx2 +x+1. Sinceω (cid:54)∈ Q( 3 2),thisquadraticpolynomial
√
isirreducibleoverQ( 3 2).
LEMMA 1.5.4.Anypolynomialf(x) ∈ K[x]hasasplittingfield. Moreover, any
twosplittingfieldsLandL(cid:48) areisomorphicoverK.
Proof. We can of course assume that f is monic. Existence is proved by
induction on degf: if f(x) does not split then it has an irreducible factor
g(x)ofdegreegreaterthanone. Lemma1.4.1givesanextensionK ⊂ L =
K(α) such that g(x) has a root α ∈ L. Then f(x)/(x − α) ∈ L[x] has a
splittingfieldF ⊃ Lbyinductiveassumption. ThenF isasplittingfieldof
f(x) ∈ K[x]aswell.
Nowwehavetoconstructanisomorphismbetweentwosplittingfields
L and L(cid:48) over K. It is enough to construct an injective homomorphism
φ : L → L(cid:48) that preserves K. Indeed, if f(x) = (cid:81) (x − α ) in L then
i i
f(x) = (cid:81) (x−φ(α ))inφ(L),sof(x)splitsinφ(L),soφ(L) = L(cid:48).
i i
We will construct φ step-by-step. Choose a root α ∈ L of f(x). Let g(x)
betheminimalpolynomialofα. Theng(x)dividesf(x), andinparticular
g(x) splits in L(cid:48). Let β be a root of g(x) in L(cid:48). Since α and β have the
sameminimalpolynomial,K(α)andK(β)areisomorphicoverK. Fixone
isomorphism,
φ : K(α) → K(β).
0
Nowwehaveadiagramoffieldmaps
L L(cid:48)
∧ ∧
(1)
∪ ∪
K(α) > K(β)
φ
0
Notice that L is a splitting field of f(x) over K(α) and L(cid:48) is a splitting
field of f(x) over K(β). So ideally, we would like to finish by induction
bycontinuingtoaddrootsofα. However,noticethattheset-upisslightly
different: beforeweweretryingtoshowthatLandL(cid:48) areisomorphicover
K, and now we are trying to construct an isomorphism φ : L → L(cid:48) that
extendsagivenisomorphismφ : K(α) → K(β). Sothebestthingtodois
0
togeneralizeourLemmaalittlebittomakeitmoresuitableforinduction.
(cid:3)
ThisisdoneinthenextLemma.
LEMMA1.5.5. Supposewehaveadiagramofhomomorphismsoffields
L L
1 2
∧ ∧
(2)
∪ ∪
K > K
1 2
ψ
whereL isasplittingfieldoff (x) ∈ K [x],thepolynomialf (x) ∈ K [x]splits
1 1 1 2 2
inL ,andf (x)isapolynomialobtainedbyapplyingψtoallcoefficientsoff (x).
2 2 1
8 JENIATEVELEV
Then there exists a homomorphism φ : L → L such that φ| = ψ (i.e. that
1 2 K1
makesadiagramcommutative).
Proof. Choosearootα ∈ L off (x). Letg (x)betheminimalpolynomial
1 1 1
ofα. Theng (x)dividesf (x). Wehaveahomomorphism
1 1
Ψ : K [x] → K [x]
1 2
that extends ψ. Then f = Ψ(f ). Let g = Ψ(g ). Then g |f , and in
2 1 2 1 2 2
particularg (x)splitsinL . Letβ bearootofg (x)inL . Letg(cid:48)(x) ∈ K [x]
2 2 2 2 2 2
beanirreduciblefactorofg (x)withrootβ. Then
2
K (α) (cid:39) K [x]/(g ) and K (β) (cid:39) K [x]/(g(cid:48)).
1 1 1 2 2 2
Notice that Ψ induces a homomorphism K [x]/(g ) → K [x]/(g(cid:48)). This
1 1 2 2
givesanhomomorphism
φ : K (α) → K (β)
0 1 2
that sends α to β and such that φ | = ψ. Now we have a commutative
0 K1
diagramoffieldmaps
L L
1 2
∧ ∧
∪ ∪
K (α) > K (β) (3)
1 2
φ
∧ 0 ∧
∪ ∪
K > K
1 2
ψ
NoticethatL isasplittingfieldoff (x)overK (α)andf (x)splitsinL .
1 1 1 2 2
So we are in the same set-up as in the statement of the Lemma, but now
[L : K (α)] < [L : K ]. Sowecanfinishbyinductionon[L : K ]. (cid:3)
1 1 1 1 1 1
§1.6. Algebraicclosure.
LEMMA1.6.1. LetK beafield. Thefollowingareequivalent:
• anypolynomialf ∈ K[x]hasarootinK.
• anypolynomialf ∈ K[x]splitsinK.
• TheonlyalgebraicextensionofK isK itself.
Proof. Easy. (cid:3)
IfanyoftheseconditionsaresatisfiedthenK iscalledalgebraicallyclosed.
DEFINITION 1.6.2. Let K be a field. A field K¯ containing K is called an
algebraicclosureofK if
• K¯ isalgebraicallyclosed.
• K ⊂ K¯ isanalgebraicextension.
C R Q
Forexample, isanalgebraicclosureof butnotof (why?)
LEMMA1.6.3.LetK ⊂ F beafieldextensionwithF algebraicallyclosed. Then
K¯ = {a ∈ F |a is algebraic over F}
isanalgebraicclosureofK.
GRADUATEALGEBRA:NUMBERS,EQUATIONS,SYMMETRIES 9
Proof. Ifα,β ∈ K¯ thenK(α,β)isfinitealgebraicoverK. Inparticular,K¯ is
afield(obviouslyalgebraicoverK). Anypolynomialf(x) = xn+a xn−1+
1
... + a in K¯[x] has a root α ∈ F. Then K(a ,...,a ) is finite over K,
n 1 n
K(a ,...,a )(α) is finite over K(a ,...,a ). Therefore, [K(α) : K] < ∞
1 n 1 n
andαisalgebraic. (cid:3)
Forexample,ifK = QandF = CthenK¯ = Q¯ isthefieldofallalgebraic
C
numbersin (rootsofpolynomialswithrationalcoefficients).
LEMMA 1.6.4.AnyfieldK iscontainedinafieldF suchthatanypolynomialin
K[x]hasarootinF.
Proof. The idea is to use a Kronecker-type construction to adjoin roots of
all polynomials at once. Let K[x ] be the algebra of polynomials in vari-
f
ables x , one variable for each irreducible polynomial with coefficients
f
inK. Considertheideal
I = (cid:104)f(x )(cid:105) ⊂ K[x ]
f f
with one generator for each irreducible polynomial f. Notice that each
polynomialisapolynomialinitsownvariable. WeclaimthatI isaproper
ideal. Ifnot,thenwecanwrite
s
(cid:88)
1 = g f (x ),
i i fi
i=1
whereg aresomepolynomialsthatinvolveonlyfinitelymanyvariablesx .
i f
LetLbeasplittingfieldoftheproductf ...f . Theformulaaboveremains
1 s
validinL[x ]. However,ifweplug-inanyrootoff forx (foreachf),we
f f
willget1 = 0,acontradiction. ItfollowsthatI isaproperideal.
Let M be a maximal ideal containing I. Then F := K[x ]/M is a field
f
that contains K. We claim that any irreducible polynomial f ∈ K[x] has a
rootinF. Indeed,f(x ) ∈ M,andthereforex +M isarootoff inF. (cid:3)
f f
THEOREM 1.6.5.Any field K has an algebraic closure. It is unique up to an
isomorphismoverK.
Proof. ApplyingLemma1.6.4inductivelygivesaninfinitetoweroffields
K = F ⊂ F ⊂ F ⊂ ...
0 1 2
such that any polynomial in F [x] has a root in F . Then F = ∪ F is
k k+1 i i
algebraicallyclosedasanypolynomialinF[x]infactbelongstosomeF [x],
k
and therefore has a root in F. Applying Lemma 1.6.3 gives an algebraic
closureK¯.
Let K¯, K¯ be two algebraic closures of K. It suffices to show that there
1
existsahomomorphismφ : K¯ → K¯ overK. Indeed,φ(K¯)isthenanother
1
algebraicclosureofK containedinK¯ . SinceK¯ isalgebraicoverφ(K¯), it
1 1
mustbeequaltoit.
Finally, we construct φ using Zorn’s lemma. Consider a poset of pairs
(F,φ), where K ⊂ F ⊂ K¯ and φ : F → K¯ is a homomorphism over K.
1
We say that (F,φ) ≤ (F ,φ ) if F ⊂ F and φ is the restriction of φ to F.
1 1 1 1
Then any chain has a maximal element (F,φ) (take the union of fields in
thechainandthemapφinducedbymapsinthechain). Thismaximalfield
mustbeequaltoK¯: ifF isproperlycontainedinK¯ thentakeanyα ∈ K¯\F.
10 JENIATEVELEV
ByLemma1.5.5,wecanextendφtoahomomorphismF(α) → K¯ . (cid:3)
1
Thesameargumentshowsthefollowingslightlymoreusefulstatement:
PROPOSITION1.6.6.Supposewehaveadiagramofhomomorphismsoffields
L
1
∧
(4)
∪
K > K
1 2
ψ
where L is algebraic over K and K is algebraically closed. Then there exists
1 1 2
a homomorphism φ : L → K such that φ| = ψ (i.e. that makes a diagram
1 2 K1
commutative).
§1.7. Finitefields.
THEOREM 1.7.1.For any prime p and positive integer n, there exists a field Fpn
with pn elements. Moreover, any two such fields are isomorphic. We can embed
Fpm inFpn ifandonlyifmdividesn.
Proof. Let K be a splitting field of the polynomial f(x) = xpn −x ∈ F [x].
p
Since f(cid:48)(x) = −1 is coprime to f(x), there are exactly pn roots. Recall that
F : K → K, F(x) = xp is a Frobenius homomorphism. In particular, if α
and β are roots of f(x) then ±α±β and αβ, and α/β are roots as well. It
followsthatK haspn elementsandallofthemarerootsoff(x).
SupposeK isafieldwithpn elements. ThegroupofunitsK∗ isAbelian
of order pn −1, and therefore xpn−1 = 1 for any x ∈ K∗.1 It follows that
K is a splitting field of xpn − x. But any two splitting fields of the same
polynomialareisomorphicbyLemma1.5.4.
IfFpm ⊂ Fpn thenthelatterfieldisavectorspace(ofsomedimensionr)
overtheformer. Itfollowsthat
pn = (pm)r = pmr.
Itfollowsthatmdividesn.
Finally,supposethatmdividesn. Thenpm−1|pn−1(easy),andtherefore
xpm−1−1|xpn−1−1(equallyeasy). Itfollowsthatthesplittingfieldofxpm−x
iscontainedinthesplittingfieldofxpn −x. (cid:3)
§1.8. Compositefield.
DEFINITION 1.8.1. Let K1K2 be subfields of a field K. Then the composite
fieldofK andK ,denotedbyK K ,isthesmallestsubfieldofK contain-
1 2 1 2
ingK andK .
1 2
√ √ √ √
Forexample,Q( 2)Q( 3) = Q( 2, 3) ⊂ C.
1RecallthatinfactthisanalysisimpliesthatK∗isacyclicgroup. Indeed,otherwisewe
wouldhavexr =1foranyx∈K∗andr<pn−1.However,thepolynomialcannothave
morerootsthanitsdegree.