Fundamental Theorem of Galois Theory PDF

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Fundamental*Theorem*of*Galois*Theory* * Introduction:* Previously*we*worked*through*an*example*illustrating*many*of*the*components*of*Galois* Theory*including*the*Fundamental*Theorem.*Here*I*will*state*and*prove*the*Fundamental* Theorem.*Supporting*definitions*and*theorems*will*be*presented*as*needed.*When* appropriate*(and*useful)*proofs*of*supporting*results*will*be*provided.*** * The$Theorem$(Dummit$&$Foote$Version):$ * Definition:*Let*K*be*a*finite*extension*of*F.*Then*K*is*said*to*be*Galois*over*F*(and*K/F*is*a* Galois'Extension)*if*|Aut(K/F)|*=*[K:*F].*In*this*case*the*group*of*automorphisms*Aut(K/F)* is*called*the*Galois'group*of*K/F,*denoted*Gal(K/F).** * Definition:*A*polynomial*over*F*is*called*separable*if*it*has*no*multiple*roots*(i.e.*all*its* roots*are*distinct).** * Theorem(s):*K$is*the*splitting*field*over*F$of*a*separable*polynomial*f(x)*if*and*only*if*K/F*is* Galois.*Further,*over*a*finite*field*or*a*field*of*characteristic*zero,*every*irreducible* polynomial*is*separable.*So,*in*those*cases,*K*is*Galois*over*F*if*and*only*if*K*is*the*splitting* field*of*an*irreducible*polynomial*over*F.*[This*is*basically*Theorem*3*found*at*the*bottom* of*this*document*(proof*omitted).]* * (Non)Example:** * Consider*the*extension*Q( 3 2 )*of*Q.**What*is*the*degree*of*this*extension?*Well,*we*know* by*closure*of*multiplication*that*Q( 3 2 )*must*contain* 3 4 .*So,*a*basis*of*this*extension* would*be*{1, 3 2 , 3 4 }.*This*means*[Q( 3 2 ):Q]*=*3.*We*can*also*see*that*[Q( 3 2 ):Q]*=*3* because*x3*–*2*is*irreducible*by*Eisenstein.** * But*if*we*factor*the*polynomial*in*Q( 3 2 )*we*see*that*x3*–*2*=* (x − 3 2 )(x 2 + 3 2 x + 3 4 ) *and*if* we*compute*the*discriminant*of*this*quadratic*we*see*that*it*is* 3 4 − 4 3 4 *which*is*negative.* So*the*other*two*roots*of*this*cubic*are*complex*and*hence*not*contained*in*Q( 3 2 ).** * This*means*that*any*automorphism*that*of*Q( 3 2 )*that*fixes*Q*must*take* 3 2 *to*itself*since* that’s*the*only*root*available.** This*completely*determines*the*map*on*Q( 3 2 )*Thus*|Aut(Q( 3 2 )\Q)| = 1. This means that Q( 3 2 )/Q is NOT Galois. This also means that Q( 3 2 )*is*not*the*splitting* field*of*an*irreducible*polynomial*over Q.* * *

Theorem:*(Fundamental*Theorem*of*Galois*Theory)*Let*K*be*a*Galois*Extension*and*set*G*=* Gal(K/F).*Then*there*is*a*bijection* * ⎧ ⎪ ⎪ Subfields E ⎪ ⎨ of K ⎪ Containing F ⎪ ⎪⎩

K ⎫ ⎧ ⎪ ⎪ | ⎪ ⎪ Subgroups H ⎪ ⎪ E ⎬ ↔ ⎨ of G | ⎪ ⎪ ⎪ ⎪ F ⎪ ⎪ ⎭ ⎩

⎫ ⎪ ⎪* ⎪ H ⎬ | ⎪ ⎪ G ⎪ ⎭

1 |

*

given*by*the*correspondences** →

E ⎧The fixed Field ⎫ ⎬ ⎨ ⎭ ⎩of H

←

⎧The elements of G ⎫ ⎬ ⎨ ⎭ ⎩Fixing E H

*

which*are*inverse*to*each*other.** *

Under*this*correspondence:* * (1) (Inclusion*Reversing)*If*E1,*E2*correspond*to*H1,*H2*respectively,*then*E1*⊂*E2*iff*H2*≤*H1.* *

(2) [K*:*E]*=*|H|*and*[E*:*F]*=*|G*:*H|:* *

K | E |

}

|H |

}

|G :H |

*

F *

(3) K/E*is*always*Galois,*with*Galois*group*Gal(K/E)*=*H:* *

K | H * E *

(4) E*is*Galois*over*F*if*and*only*if*H*is*a*normal*subgroup*in*G.*If*this*is*the*case,*then*the* Galois*group*is*isomorphic*to*the*quotient*group: Gal(E / F) ≅ G / H .* * (5) If*E1,*E2*correspond*to*H1,*H2*respectively,*then*the*intersection*E1*∩*E2*corresponds*to* the*group* H 1 , H 2 generated*by*H1*and*H2*and*the*composite*field*E1E2*corresponds*to* the*intersection*H1*∩*H2.*Hence,*the*lattice*of*subfields*of*K*containing*F*and*the*lattice* of*subgroups*of*G*are*“dual”*–*the*lattice*diagram*for*one*is*the*lattice*diagram*for*the* other*turned*upside*down.** * $

$

The$Proof:$(It’s$gonna$be$a$big$one!)$ * Given*any*subgroup*H*of*G*we*obtain*a*unique*fixed*field*E*=*KH*by*Corollary*3*(below).*This* shows*that*the*correspondence*above*is*injective*from*the*right*to*left.** *

If*K*is*the*splitting*field*of*the*separable*polynomial*f(x)*∈*F[x]*then*we*may*also*view*f(x)* as*an*element*of*E[x]*for*any*subfield*E*of*K*containing*F.*Then*K*is*also*the*splitting*field* for*f(x)*over*E,*so*the*extension*K/E*is*Galois.*By*Corollary*1,*E*is*the*fixed*field*of*Aut(K/E)* ≤*G,*showing*that*every*subfield*of*K*containing*F*arises*as*the*fixed*field*for*some* subgroup*of*G.**Hence*the*correspondence*above*is*surjective*from*right*to*left*and*hence*a* bijection.*The*correspondences*are*inverse*to*each*other*since*the*automorphisms*fixing*E* are*precisely*Aut(K/E)*by*Corollary*1.* *

The*Galois*correspondence*is*inclusion*reversing*because*any*automorphism*that*fixes*a* subfield*F’$of*a*field*K’*will*also*fix*any*subfield*F1’*of*F’.**So*we*have*proved*part*(1).*[Note* that*here*I’m*using$F’,$K’,$F1’,*to*denote*an*arbitrary*sequence*of*fields.]* *

If*E*=*KH*is*the*fixed*field*of*H,*then*by*Theorem*2*(below)*[K*:*E]*=*|H|*and*[K*:*F]*=*|G|.* Taking*the*quotient*gives*[E*:*F]*=*|G*:*H|.*This*proves*part*(2).** *

Note*that*the*above*proof*uses*another*theorem*that*we*haven’t*proved.*Namely*that*if*we* have*fields*F*⊆*K*⊆*L*then*[L*:*F]*=*[L*:*K]*[K*:*F].*You*can*prove*this*simply*by*using*the* basis*for*K*over*F*and*the*basis*for*L*over*K*to*construct*a*basis*for*L*over*F*(the*same*way*I* did*when*I*constructed*the*splitting*field*for*X4*f2*over*Q).** *

Corollary*2*(below)*says*that*if*H*is*a*finite*subgroup*of*automorphism*of*a*field*K*and*E*is* the*fixed*field*then*K/E*is*Galois*with*Galois*group*H.*This*gives*us*part*(3)*immediately.** *

[Part*(4)*is*the*long*one*and*it*is*starting*now.]* *

Now*suppose*that*E*=*KH*is*the*fixed*field*of*the*subgroup*H.*Every*σ*∈G*=*Gal(K/F)*when* restricted*to*E*is*an*embedding*σ|E*of*E*with*the*subfield*σ(E)*of*K.** *

Conversely,*let*τ:*E*→*τ(E)*⊆* F be*any*embedding*of*E*(into*a*fixed*algebraic*closure* F of*F* containing*K)*which*fixes*F.** *

Then*τ(E)*is*in*fact*contained*in*K:*if*α *∈$E*has*minimal*polynomial*m α*(x)*over*F*then*τ( α)* is*another*root*of*m α*(x)*and*K*contains*all*of*these*roots*by*Theorem*3*(below).** *

As*above*K*is*the*splitting*field*of*f(x)*over*E*and*so*also*the*splitting*field*of*τ$f(x)*which*is* the*same*as*f(x)*since*f(x)*has*coefficients*in*F*over*τ(E).*Then*we*can*extend*τ*to*an* isomorphism*σ:** *

σ: K → | τ: E → *

K | * τ (E)

Since*σ*fixes*F*(because* τ*does)*it*follows*that*every*embedding* τ*of*E*fixing*F*is*a* restriction*to*E*of*some*automorphism*σ *of*K*fixing*F.**In*other*words,*every*embedding*of* E*is*of*the*form*σ |E*for*some*σ$∈G.* *

Two*automorphisms* σ ,$σ ′*∈G*restrict*to*the*same*embedding*of*E*if*and*only*if* σf1σ′*is*the* identity*map*on*E.*But*then* σf1σ′*∈H*since*by*Part*(3)*the*automorphisms*of*K*which*fix*E* are*precisely*the*elements*in*H.**This*means*that* σ′*is*in*the*coset*σ fH.*Thus*the*distinct* embeddings*of*E*are*in*bijection*with*the*cosets* σfH*of*H*in*G.*In*particular*this*gives* *

|Emb(E/F)|*=*[G*:*H]*=*[E*:*F]*where*Emb(E/F)*denotes*the*set*of*embeddings*of*E*(into*a* fixed*algebraic*closure*of*F)*which*fix*F.*Note*that*Emb(E/F)*contains*the*automorphisms* Aut(E/F).* *

The*extension*E/F*will*be*Galois*if*and*only*if*|Aut(E/F)|*=*[E*:*F].*By*the*equality*above,* this*will*be*the*case*if*and*only*if*each*of*the*embeddings*of*E*is*actually*an*automorphism* of*E,*i.e.*if*and*only*if*σ (E)*=*E*for*every*σ*∈G.$$ $

If* σ*∈G,*then*the*subgroup*of*G*fixing*the*field*σ(E)*is*the*group*σ Hσf1,*i.e.,* σ (E)*=*Kσ Hσ −1 .* *

To*see*this,*observe*that*if* σ( α )*∈σ(E)*then*(σh σf1)(σ( α ))*=*σ(h(α))*=* σ( α )*for*every*h*in*H,* since*h*fixes* α*∈E.***This*shows*that* σh σf1*fixes* σ(E).*On*the*other*hand,*the*group*fixing* σ (E)*has*order*equal*to*the*degree*of*K*over* σ(E).*But*this*is*the*same*as*the*degree*of*K* over*E*since*E*and*∈ σ(E)*are*isomorphic.*So,*the*order*of*this*group*is*the*same*as*H.*Hence* σ Hσf1*is*precisely*the*group*fixing* σ(E)*since*we*have*shown*one*containment*and*their* orders*are*the*same*(the*orders*of*H*and* σHσf1*are*the*same*because*they*are*conjugates).** *

Because*of*the*bijective*nature*of*the*Galois*correspondence*already*proved,*we*know*that* two*subfields*of*K*containing*F*are*equal*if*and*only*if*their*fixing*subgroups*are*equal*in*G.* Hence* σ(E)*=*E*for*all* σ *∈G$if*and*only*if*H*=*σH σf1*for*all*σ *∈G.$*In*other*words,*E*is*Galois* over*F*if*and*only*if*H*is*a*normal*subgroup*of*G.** *

We*have*already*identified*the*embeddings*of*E*over*F*as*the*set*of*cosets*of*H*in*G*and* when*H*is*normal*in*G*seen*that*the*embedings*are*automorphisms.*It*follows*that*in*this* case*the*group*of*cosets*G/H*is*identified*with*the*group*of*automorphisms*of*the*Galois* extension*E/F*by*the*definition*of*the*group*operation*(composition*of*automorphisms).* Hence*G/H*≅*Gal(E/F)*when*H*is*normal*in*G,*which*completes*the*proof*of*Part*(4).** *

Finally,*suppose*that*H1*is*the*subgroup*of*the*elements*of*G*fixing*the*subfield*E1*and*H2*is* the*subgroup*of*elements*of*G*fixing*the*subfield*E2.*Any*element*in*H1∩$H2*fixes*both*E1* and*E2,*hence*fixes*every*element*in*the*composite*E1E2,*since*the*elements*in*this*field*are* algebraic*combinations*of*the*elements*of*E1*and*E2.** *

Conversely,*if*an*automorphism* σ$fixes*the*composite*E1E2,*then*in*particular* σ$fixes*E1,*i.e.,* σ *∈$H1*and*σ$fixes*E2,*i.e.,*σ*∈$H2,*hence* σ$∈H1∩$H2.*Similarly,*the*intersection*E1∩$E2* corresponds*to*the*group* H 1 , H 2 generated*by*H1*and*H2,*completing*the*proof*of*the* Fundamental*Theorem*of*Galois*Theory!* $

Selected$Supporting$Theorems$and$Proofs$ * Theorem*1:*(Linearly*independence*of*automorphisms).*Let*σ1,*σ2,…,*σn*be*distinct* automorphisms*of*a*field*F.*Then*they*are*linearly*independent*as*functions*on*F.** * Comments:*This*is*a*special*case*of*Corollary*8*from*Section*14.2*of*Dummit*&*Foote.*In* Section*14*they*develop*theory*about*characters.$I*think*this*is*more*generality*than* needed.*So*my*proof*below*(which*is*really*for*Theorem*7)*is*specific*to*automorphisms.* This*should*be*enough*to*get*the*theorems*that*are*needed*to*support*the*Fundamental* Theorem.** * Proof:*Let*σ1,*σ2,…,*σn*be*distinct*automorphisms*of*a*field*F.*Suppose*these*automorphisms* where*linearly*dependent.*Of*all*the*possible*nonftrivial*linear*combinations*that*equal*the* zero*map,*choose*the*one*with*the*smallest*number,*m,*of*nonzero*coefficients*ci.*We*can* assume*(by*renumbering*if*needed)*that*the*m*nonzero*coefficients*are*c1,*c2,…,*cm.* *

Then*c1σ1*+*c2σ2*+…*+*cmσm*=*0*and*so*for*any*g*in*the*multiplicative*group*F×*we*have:* *

c1σ1(g)*+*c2σ2(g)*+…*+*cmσm(g)**=*0.**Now,*let*g0*∈$F×*be*such*that*σ1(g0)*≠*σm(g0).*Such*a* thing*exists*because*our*automorphisms*are*distinct.*Now,* *

c1σ1(g0g)*+*c2σ2(g0g)*+…*+*cmσm(g0g)**=*0.**Since*these*are*homomorphisms,** *

(1)**

c1σ1(g0)*σ1(g)**+*c2σ2(g0)*σ2(g)**+…*+*cmσm(g0)*σm(g)**=*0.*********

*

And*multiplying*c1σ1(g)*+*c2σ2(g)*+…*+*cmσm(g)**=*0.**,*by*σm(g0),*we*get:* *

(2)**

c1σm(g0)σ1(g)*+*c2σm(g0)σ2(g)*+…*+*cmσm(g0)σm(g)**=*0.***

*

Subtracting*(2)*from*(1)*we*get:* *

[c1σ1(g0)*σ1(g)*f*c1σm(g0)σ1(g)]+*[c2σ2(g0)*σ2(g)*f*c2σm(g0)σ2(g)]*+…*+*[cmf1σ$m=1*(g0)*σ$m=1(g)* f*c$m=1σ$m(g0)σ$m=1(g)]*=*0.* *

Doing*some*Algebra*1:* *

[c1σ1(g0)*f*c1σm(g0)]σ1(g)*+*[c2σ2(g0)*f*c2σm(g0)]σ2(g)*+…*+*[cm=1σ$m=1(g0)*f*cm=1σm(g0)]σm=1(g)*=*0.* *

Note*that*c1σ1(g0)*f*c1σm(g0)*≠*0,*and*this*equation*holds*for*any*g*in**F×.*This*means*that** * [c1σ1(g0)*f*c1σm(g0)]σ1*+*[c2σ2(g0)*f*c2σm(g0)]σ2+…*+*[c$m=1σ$m=1(g0)*f*c$m=1σm(g0)]σm*is*a*linear* combination*of*σ1,*σ2,…,*σn*that*is*equivalent*to*the*zero*map.*This,*my*friends,*is*a* contradiction*since*this*has*fewer*than*m*nonfzero*coefficients!*The*end.* * * *

Theorem*2:*(Degree*of*Extension*=*Order*of*Automorphism*Group]*Let*G*=*{σ1*=*1,*σ2,*…*,* σn}*be*a*subgroup*of*the*automorphisms*of*a*field*K*and*let*F*be*the*fixed*field*of*G.** *

Then*[K$:*F]*=*n*=*|G|.***(This*is*Theorem*9*from*14.2*of*Dummit*&*Foote.)* *

Proof:*I’ll*rule*out*both*n$>*[K$:*F]*and*n**[K$:*F].*Let*ω 1,*ω2,*…*,* ωm*be*a*basis*for*K*over*F,*where*m**equations).**Let,* β1,*…*,* βn*be*a*nonftrivial*solution.* *

Now,*let*a1,*a2,*…*,*am*be*m*arbitrary*elements*of*F.*These*elements*are*fixed*by*all*of*the* automorphisms*in*G.*So,*we*can*1)*multiply*the*ith*equation*by$ai,*2)*move*these*coefficients* inside*the*map,*and*3)*substitute*the*solution* β 1,*β 2,*…*,* βn**to*get:* *

σ1(a1ω 1)β1*+*σ2(a1ω 1)β2*+*…*+*σn(a1ω1)β n**=*0** *

σ1(a2ω 2)β1*+*σ2(a2ω 2)β2*+*…*+*σn(a2ω2)β n**=*0** * * *  * σ1(amωm) β1*+*σ2(am ωm) β2*+*…*+*σn(amω m) βn**=*0** *

Let’s*add*these*together*and*do*some*factoring*and*homomorphismfing:* *

σ1(a1ω 1)β1*+σ1(a2ω2) β 1*+*…*+*σ1(amωm)β 1*+*σ2(a1ω1)β 2*+*σ2(a2ω2)β 2+*σ2(amω m)β 2*+*…*+* σn(a1 ω 1)βn**+*σn(a2ω 2)β n**+*…*+*σn(amω m)β n**=*0** *

σ1(a1ω1*+*a2ω2+…+amωm) β1*+*σ2(a1 ω1+*a2ω 2*+…*+amωm)β2*+…+*σn(a1ω1*+*a2ω$+…+*am ωm) βn**=*0.* *

But*a1,*a2,*…*,*am*were*arbitrary,*so*a1ω1*+*a2ω2+…+amωm*is*an*arbitrary*element*of*K.** Thus,*the*linear*combination* β 1σ1*+*β2σ2*+…+*β nσn*is*equivalent*to*the*zero*map.*But,*the* above*theorem*assures*us*that*σ1,*σ2,…,*σn*is*an*linearly*independent*set.*This*gives*us*a*

contradiction.*So*we*cannot*have*n$>*[K$:*F].**Note$that$so$far,$we$haven’t$used$the$fact$that$ the$elements$σ1,$ σ2,…,$σ n$form$a$group!$ *

Suppose*now*that*n**equations).*Let,* β 1,* β2,*…*,*β n*be*a*nonftrivial*solution.*

If*all*of*the* β i’s$were*elements*of*F,*then*substituting*into*the*first*equation*and*bringing*the* β i’s$inside*the*maps*(they*are*fixed*by*these*automorphisms)*we*get:* ** σ1(α1 β 1)*+*σ1(α2β 2)*+*…*+*σ1(αn+1β n+1)*=*0** *

σ1(α1 β 1*+*α2β2*+*…*+*αn+1βn+1)*=*0** *

And*since*σ1*is*1f1,*this*means*that*α1 β1*+*α2β 2*+*…*+*αn+1βn+1*=*0.*But*this*contradicts*the* fact*that*α1,*α2,…,*αn+1*are*linearly*independent.*Thus*at*least*one*of*the* β i’s*is*not*in*F.** *

Now,*among*all*the*nonftrivial*solutions*to*the*system*above,*choose*the*one*with*the* fewest,*r,*nonfzero*values.*Then*by*renumbering*if*needed,*we*can*assume*that* β1,*β 2,*…*,* βr* are*nonfzero.*Note*that*a*multiple*of*any*solution*is*also*a*solution,*so*we*may*assume*that* β r*=*1.**(Note*since*at*least*one*of*the*β i’s*must*not*be*in*F,*r$>1.)*Suppose*that*β 1*is*not*in*F.* Let’s*rewrite*our*system*with*these*things*in*mind*(substituting*in*this*solution):* *

σ1(α1) β1*+*…*+*σ1(αrf1)$β r=1*+**σ1(αr)*=*0** *

σ2(α1) β1**+*…*+*σ2(αrf1)$β r=1*+*σ2(αr)*=*0**  * * * * σn(α1)β 1**+*…*+*σn(αrf1)$β r=1*+*σn(αr)*=*0** *

More*briefly:* *

(3)**σi(α1)β 1**+*…*+*σi(αrf1)$βr=1*+*σi(αr)*=*0**for*i*=*1,*2,*…,*n.** *

Now,*since*β 1*is*not*in*F,*there*is*at*least*one*1...