Solutions Manual
Galois Theory 5th Editon by Ian Stewart
All Chapters Covered
, Introduction 1
Introduction
ThisdSolutionsdManualdcontainsdsolutionsdtodalldofdthedexercisesdindthedFifth
dEdi-dtiondofdGaloisdTheory.
Manydofdthedexercisesdhavedseveralddifferentdsolutions,dordcandbedsolveddu
singdseveralddifferentdmethods.dIfdyourdsolutiondisddifferentdfromdthedonedpresentedd
here,ditdmaydstilldbedcorrectd—
dunlessditdisdthedkinddofdquestiondthatdhasdonlydonedanswer.
Thedwrittendstyledisdinformal,danddthedmaindaimdisdtodillustratedthedkeydideas
din-
dvolveddindansweringdthedquestions.dInstructorsdmaydneeddtodfilldindadditionaldde
tailsdwheredthesedaredstraightforward,dordexplaindassumeddbackgrounddmaterial
.dOndthedwhole,dIdhavedemphasisedd‘baredhands’dmethodsdwheneverdpossible,d
sodsomedofdthedexercisesdmaydhavedmoredelegantdsolutionsdthatdusedhigher-
powereddmethods.
IandStewart
CoventrydJanuaryd2022
1 ClassicaldAlgebra
1.1 Letdud=dxd+diyd≡d(x,dy),dvd=dad+dibd≡d(a,db),dwd=dpd+diqd≡d(p,dq).dThen
uvd =d (x,dy)(a,db)
=d(xad—dyb,dxbd+dya)
=d(axd—dby,dbxd+day)
=d (a,db)(x,dy)
=d vu
(uv)wd =d [(x,dy)(a,db)](p,dq)
=d(xad—dyb,dxbd+dya)(p,dq)
=d (xapd—dybpd—dxbqd—dyaq,dxaqd—dybqd+dxbpd+dyap)
=d(x,dy)(apd—dbq,daqd+dbp)
=d (x,dy)[(a,db)(p,dq)]
=d(uv)w
1.2 (1)dChangingdthedsignsdofda,dbddoesdnotdaffectd(a/b)2,dsodwedmaydassumeda,dbd>d0.
,(2) Anydnon-emptydsetdofdpositivedintegersdhasdadminimaldelement.dSincedbd>d0dis
andinteger,dthedsetdofdpossibledelementsdbdhasdadminimaldelement.
, 2
(3) Wedknowdthatda2d=d2b2.dThen
(2bd—da)2d—d2(ad—db)2d =d 4b2d—d4abd+da2d—d2(a2d—d2abd+db2)
=d2b2d—da2d=d0
(4) Ifd2b ≤ adthend4b2≤ a2d=d2b2,dadcontradiction.dIfd≤
a bdthend2a2≤
2b d=da ,dadcontradiction.
2 2
(5) Ifda —bd ≥ bdthenda≥ 2bdsoda2≥
4b2d=d2a2,dadcontradiction.dNowd(3)dcontra-ddictsdthedminimalitydofdb.
NotedondthedGreekdapproach.
ThedancientdGreeksddiddnotdusedalgebra.dTheydexpresseddthemdsamedunderl
yingdideadindtermsdofdadgeometricdfigure,dFigured1.
√
FIGUREd1:dGreekdproofdthatd 2disdirrational.
StartdwithdsquaredABCDdanddletdCEd=dAB.dCompletedsquaredAEFG.dThedr
estdofdthedfiguredleadsdtodadpointdHdondAF.dClearlydAC/ABd=dAF/AE.dIndmodern
dnotation,dletd ABd =d b ,d ACd =d a .d Sinced ABd =d HFd =d ABd andd BHd =d AC,d wed have
′ ′
d AEd =d a d+db d =db,
′ ′
′d
√ dAFd=da′d+d2b′d=da,dsay.dThereforeda′d+db′d=db,db′d=dad—db,dandd ad =bda . b′
say,dand
Ifd 2disdrational,dwedcandmakedad bdintegers,dindwhichdcaseda′d b′
, √ , aredalsodintegers,
andd thed samed processd ofd constructingd rationalsd equald to 2d withd ever-decreasing
numeratorsdandddenominatorsdcoulddbedcarrieddout.dThedGreeksddidn’tdarguedthedpr
oofdquitedthatdway:dtheydobserveddthatdthed‘anthyphaeresis’dofdAFdanddAEdgoesdon
dforever.dThisdprocessdwasdtheirdversiondofdwhatdwednowdcalldthedcontinueddfractio
ndexpansiond(ordthedEuclideandalgorithm,dwhichdisdequivalent).dItdstopsdafterdfinite
lydmanydstepsdifdanddonlydifdthedinitialdratiodliesdindQ.dSeedFowlerd(1987)dpagesd
33–35.
1.3 Adnonzerodrationaldcandbedwrittenduniquely,dupdtodorder,d asd adproducedofdprime
powersd (withd ad signd ±):
m m
rd=d±p1d1d ·d·d·dpkd k
wheredthedmjd aredintegers.dSo
r2d=dp2m1d
1
·d·d·dp
k
2mk
Galois Theory 5th Editon by Ian Stewart
All Chapters Covered
, Introduction 1
Introduction
ThisdSolutionsdManualdcontainsdsolutionsdtodalldofdthedexercisesdindthedFifth
dEdi-dtiondofdGaloisdTheory.
Manydofdthedexercisesdhavedseveralddifferentdsolutions,dordcandbedsolveddu
singdseveralddifferentdmethods.dIfdyourdsolutiondisddifferentdfromdthedonedpresentedd
here,ditdmaydstilldbedcorrectd—
dunlessditdisdthedkinddofdquestiondthatdhasdonlydonedanswer.
Thedwrittendstyledisdinformal,danddthedmaindaimdisdtodillustratedthedkeydideas
din-
dvolveddindansweringdthedquestions.dInstructorsdmaydneeddtodfilldindadditionaldde
tailsdwheredthesedaredstraightforward,dordexplaindassumeddbackgrounddmaterial
.dOndthedwhole,dIdhavedemphasisedd‘baredhands’dmethodsdwheneverdpossible,d
sodsomedofdthedexercisesdmaydhavedmoredelegantdsolutionsdthatdusedhigher-
powereddmethods.
IandStewart
CoventrydJanuaryd2022
1 ClassicaldAlgebra
1.1 Letdud=dxd+diyd≡d(x,dy),dvd=dad+dibd≡d(a,db),dwd=dpd+diqd≡d(p,dq).dThen
uvd =d (x,dy)(a,db)
=d(xad—dyb,dxbd+dya)
=d(axd—dby,dbxd+day)
=d (a,db)(x,dy)
=d vu
(uv)wd =d [(x,dy)(a,db)](p,dq)
=d(xad—dyb,dxbd+dya)(p,dq)
=d (xapd—dybpd—dxbqd—dyaq,dxaqd—dybqd+dxbpd+dyap)
=d(x,dy)(apd—dbq,daqd+dbp)
=d (x,dy)[(a,db)(p,dq)]
=d(uv)w
1.2 (1)dChangingdthedsignsdofda,dbddoesdnotdaffectd(a/b)2,dsodwedmaydassumeda,dbd>d0.
,(2) Anydnon-emptydsetdofdpositivedintegersdhasdadminimaldelement.dSincedbd>d0dis
andinteger,dthedsetdofdpossibledelementsdbdhasdadminimaldelement.
, 2
(3) Wedknowdthatda2d=d2b2.dThen
(2bd—da)2d—d2(ad—db)2d =d 4b2d—d4abd+da2d—d2(a2d—d2abd+db2)
=d2b2d—da2d=d0
(4) Ifd2b ≤ adthend4b2≤ a2d=d2b2,dadcontradiction.dIfd≤
a bdthend2a2≤
2b d=da ,dadcontradiction.
2 2
(5) Ifda —bd ≥ bdthenda≥ 2bdsoda2≥
4b2d=d2a2,dadcontradiction.dNowd(3)dcontra-ddictsdthedminimalitydofdb.
NotedondthedGreekdapproach.
ThedancientdGreeksddiddnotdusedalgebra.dTheydexpresseddthemdsamedunderl
yingdideadindtermsdofdadgeometricdfigure,dFigured1.
√
FIGUREd1:dGreekdproofdthatd 2disdirrational.
StartdwithdsquaredABCDdanddletdCEd=dAB.dCompletedsquaredAEFG.dThedr
estdofdthedfiguredleadsdtodadpointdHdondAF.dClearlydAC/ABd=dAF/AE.dIndmodern
dnotation,dletd ABd =d b ,d ACd =d a .d Sinced ABd =d HFd =d ABd andd BHd =d AC,d wed have
′ ′
d AEd =d a d+db d =db,
′ ′
′d
√ dAFd=da′d+d2b′d=da,dsay.dThereforeda′d+db′d=db,db′d=dad—db,dandd ad =bda . b′
say,dand
Ifd 2disdrational,dwedcandmakedad bdintegers,dindwhichdcaseda′d b′
, √ , aredalsodintegers,
andd thed samed processd ofd constructingd rationalsd equald to 2d withd ever-decreasing
numeratorsdandddenominatorsdcoulddbedcarrieddout.dThedGreeksddidn’tdarguedthedpr
oofdquitedthatdway:dtheydobserveddthatdthed‘anthyphaeresis’dofdAFdanddAEdgoesdon
dforever.dThisdprocessdwasdtheirdversiondofdwhatdwednowdcalldthedcontinueddfractio
ndexpansiond(ordthedEuclideandalgorithm,dwhichdisdequivalent).dItdstopsdafterdfinite
lydmanydstepsdifdanddonlydifdthedinitialdratiodliesdindQ.dSeedFowlerd(1987)dpagesd
33–35.
1.3 Adnonzerodrationaldcandbedwrittenduniquely,dupdtodorder,d asd adproducedofdprime
powersd (withd ad signd ±):
m m
rd=d±p1d1d ·d·d·dpkd k
wheredthedmjd aredintegers.dSo
r2d=dp2m1d
1
·d·d·dp
k
2mk
