Table Of ContentUniversitext
Editorial Board
(North America):
S. Axler
K.A. Ribet
Falko Lorenz
Algebra
Volume I:
Fields and Galois Theory
With the collaboration of the translator, Silvio Levy
FalkoLorenz
FBMathematikInstitute
UniversityMu¨nster
Mu¨nster,48149
Germany
[email protected]
EditorialBoard
(NorthAmerica):
S.Axler K.A.Ribet
MathematicsDepartment MathematicsDepartment
SanFranciscoStateUniversity UniversityofCaliforniaatBerkeley
SanFrancisco,CA94132 Berkeley,CA94720-3840
USA USA
[email protected] [email protected]
MathematicsSubjectClassification(2000):11-01,12-01,13-01
LibraryofCongressControlNumber:2005932557
ISBN-10:0-387-28930-5 Printedonacid-freepaper.
ISBN-13:978-0387-28930-4
©2006SpringerScience+BusinessMedia,Inc.
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“ThoughwhatI’msayingisperhapsnotnew,Ihave
feltitquitevividlyonthisnewoccasion.”
J.W.Goethe,inaletterfromNaples,17May1787
Foreword
Thepresenttextbookismybestefforttowritealively,problem-orientedandunder-
standable introduction to classical modern algebra. Besides careful exposition, my
goalsweretoleadthereaderrightawaytointerestingsubjectmatterandtoassume
no morebackgroundthanthatprovidedby afirstcourse inlinearalgebra.
Inkeepingwiththesegoals,theexpositionisbyandlargegearedtowardcertain
motivatingproblems;relevantconceptualtoolsareintroducedgraduallyasneeded.
This way of doing things seems more likely to hold the reader’s attention than a
moreorlesssystematicstringingtogetheroftheoremsandproofs. Thepaceismore
leisurelyandgentleinthebeginning,laterfasterandlesscautious,sothebooklends
itselfto self-study.
Thisfirstvolume,primarilyaboutfieldsandGaloistheory,inordertodealwith
thelatterintroducesjustthenecessaryamountofgrouptheory. Italsocoversbasic
applicationstonumbertheory,ringextensionsandalgebraicgeometry. Ihavefound
it advantageous for various reasons to bring into play early on the notion of the
algebraic closure of a field. Naturally, Galois’ beautiful results on solvable groups
ofprimedegreecouldnotbeleftout,norcouldDedekind’sGalois-theoreticalarith-
meticreductionprinciple. InfiniteGaloisextensionsarenotneglectedeither. Finally,
it seemedappropriateto includethefundamentalsoftranscendentalextensions.
At the end of the volume there is a collection of exercises, interspersed with
remarksthatenrichthetext. Theproblemschosenareofwidelyvaryingdegreesof
difficulty,butverymanyofthemareaccompaniedbyhints—sometimesamounting
to an outline of the solution—and in any case there are no outright riddles. These
exercisesareofcoursemeanttoallowreaderstopracticetheirgraspofthematerial,
but they serve another important purpose as well: precisely because the main text
was kept short and to the point, without lots of side-results, the appendix will give
thereaderabetterideaofthewealthofconsequencesandapplicationsderivedfrom
the theory.
Thelinearalgebrafactsused,whennottotallyelementary,areaccompaniedby
referencestomyLineareAlgebra,nowpublishedbySpektrumAkademischerVerlag
andabbreviatedLAIandLAII.Thishasnotbeentranslated,butequivalentspotsin
other linear algebratextbooks are not hard to find. Theoremsand lesser resultsare
numberedwithineachchapterinsequence,thelatterbeingmarkedF1,F2,:::—the
F is inherited from the German word Feststellung. Allusions to historical matters
aremadeonlyinfrequently(butcertainlynotatrandom). Whenatheoremorother
vi Foreword
result bears the name of a mathematician, this is sometimes a matter of tradition
morethanofaccuratehistoricalorigination.
The first German editionof this bookappeared in 1987. I thank my colleagues
who, already back at the writing stage, favored it with their interest and gave me
encouragement—none more than the late H.-J. Nastold, with whom I had many
fruitful conversations, W. Lütkebohmert, who once remarked that there was no
suitable textbook for the German Algebra I course, O. Willhöft, who suggested
several good problems, and H. Schulze-Relau and H. Epkenhans, whose critical
perusaloflargeportionsofthemanuscriptwasagreathelp. Thesecond(1991)and
third (1995) editions benefited from the remarks of numerous readers, to whom I
amlikewisethankful,inparticularR.Alfes,H.Coers,H.DaldropandR.Schopohl.
The response and comments on the part of students were also highly motivating.
Special thanks are due to the publisher BI-Wissenschaftsverlag (later acquired by
Spektrum)anditseditorH.Engesser, who gotmegoingin thefirstplace.
The publication of this English version gives me great pleasure. I’m grateful
to Springer-Verlag New York and its mathematics editor Mark Spencer, for their
support and competent handling of the project. And not least for seeing to it that
thetranslationbedonebySilvioLevy: Ihaveobservedtheprogressofhistaskwith
increasingappreciationandhaveincorporatedmanyofthechangeshesuggested,in
a process of collaboration that led to noticeable improvements. Further perfecting
is of course possible, and readers’ suggestions and criticism will continue to be
welcomeand relevantforfuturereprints.
Münster,July 2005 FalkoLorenz
Contents
Foreword.......................................................... v
1 ConstructibilitywithRulerandCompass ......................... 1
2 AlgebraicExtensions ........................................... 15
3 SimpleExtensions.............................................. 21
4 FundamentalsofDivisibility..................................... 33
5 PrimeFactorizationinPolynomialRings. Gauss’sTheorem......... 45
6 PolynomialSplittingFields...................................... 55
7 SeparableExtensions ........................................... 65
8 GaloisExtensions .............................................. 75
9 FiniteFields,CyclicGroupsandRootsofUnity ................... 83
10 GroupActions ................................................. 93
11 ApplicationsofGaloisTheorytoCyclotomicFields ................ 103
12 FurtherStepsintoGaloisTheory ................................ 115
13 NormandTrace ............................................... 133
14 BinomialEquations ............................................ 143
15 SolvabilityofEquations......................................... 165
16 IntegralRingExtensions........................................ 191
17 TheTranscendenceof(cid:1) ........................................ 203
18 TranscendentalFieldExtensions................................. 209
19 Hilbert’sNullstellensatz......................................... 217
Appendix:ProblemsandRemarks ................................... 231
IndexofNotation................................................... 283
Index ............................................................. 287
1
Constructibility with Ruler and Compass
1. In school one sometimes learns to solve problems where a certain geometric
figure must be constructed from given data. Such construction problems can be
quitedifficultandaffordarealchallengetothestudent’sintelligenceandingenuity.
If you have tried long and hard to solve a certain construction problem, to no
avail,youmightthenwonderwhethertherequiredconstructioncanbecarriedoutat
all. Whetheraconstructionexistsisafundamentalquestion: so muchso thatthere
are some construction problems that had already been entertained by the ancient
Greeks,and yetremainedunsolved fortwo thousandyearsandmore.
Forexample,nobodyhaseverbeenabletostateaprocedurecapableofdividing
anarbitraryangleintothreeequalangles,usingrulerandcompass. Now,ofcourse
construction problems range widely in degree of difficulty: think of the similar-
looking problem of dividing an arbitrarysegment into three equal parts using ruler
and compass—not totally trivial, but after some thought just about anyone can
carry out the construction. The problem of constructing a regularpentagon is also
solvable,butalreadysomewhatmorecomplicated. Soitiscertainlyunderstandable
that even a construction problem that has eluded would-be solvers for a long time
should leave room for hoping that success might yet be achieved through greater
ingenuity. Perhaps,then,thequestionofwhetheraparticularconstructionwithruler
and compassis possible isnotone thatcomestomindimmediately.
Evenifsomeoneasksthisquestionofprinciple,itisnotclearapriorithatthere
is a promising way to tackle it. Yet there is, as the development of algebra since
Gauss(1777–1855)hasshown. Iwouldliketoexplainnow,atthebeginningofour
introduction to algebra, how one can arrive at broad statements about the general
constructibilityproblem,bytranslatingthisgeometricproblemintoanalgebraicone.
Asweelaborateonthis,wewillhavethechancetomotivatequitenaturallycertain
fundamentalalgebraic concepts. Moreover the subsequent treatmentof the derived
algebraic problem will require many of the tools usually treated in an Algebra I
course. Thisprocedurehastheadvantagethatonestartsfromaconcreteandeasily
understoodquestionandkeeps thegoalofsolving itin mindasonegoesalong.
Letitbesaid,however,thattheproblemofconstructibilitywithrulerandcom-
pass by no means played a central role in the development of algebra. In this
2 1 ConstructibilitywithRulerandCompass
regard the problem of solving algebraic equations by means of radicals was surely
moresignificant,nottomentionothermotivationsandstimulicomingfromoutside
algebra—fromnumbertheoryand analysis, forexample. Incidentally,in due time
we will makeprecise the problemofsolubilityofequationsbyradicalsand keep it
in viewas theexpositionunfolds.
2. First let’s describe properly what is to be understood by constructibility with
rulerand compass. For thiswe start fromthe plane(cid:1)2 of elementarygeometry. A
constructionproblemaskswhetheracertainpointP oftheplanecanbeconstructed
with ruler and compass, starting from a given initial set M of points. Thus, let a
subsetM of(cid:1)2 begiven(wemayaswellassumeithasatleasttwopoints). Then
lookat theset
˚ ˇ (cid:1)
M D P 2(cid:1)2ˇP isconstructiblefromM withrulerand compass ;
to bedefinedmorepreciselyasfollows. Let
Li.M/Dset ofstraightlinesjoiningtwodistinctpointsofM,
Ci.M/Dset ofcircleswhosecenterbelongstoM
and whoseradiusequalsthe distancebetween twopointsofM.
Thenconsiderthefollowingelementarystepsfortheconstructionof“new”points:
(i) intersectingtwo distinctlinesinLi.M/;
(ii) intersectinga linein Li.M/witha circleinCi.M/;
(iii) intersectingtwo distinctcirclesinCi.M/.
LetM0 betheunionofM withthesetofpointsobtainedbytheapplicationof
one of these steps. The points of (cid:1)2 that can be obtained by repeated application
ofsteps(i)–(iii),startingfromM andreplacingM byM0 eachtime,aresaidtobe
constructiblefromM withrulerandcompass. They formthe set M.
Wejustmentionrightnowfourwellknownconstructibilityproblemsthatwere
posed alreadybythe ancientGreeks.
˛: Trisectionoftheangle. Q0
Givenanangleofmeasure',constructanangleof
measure '=3 with ruler and compass. We regard
the given angle as determined by its vertex S and
X
points Q;Q0 on each of its sides; one may as well
assume that Q and Q0 are equidistant from S. Let
'
X bethepointindicatedinthefigure. Thequestion
'=3
theniswhetherX 2 fS;Q;Q0g.
S Q
ˇ: Doublingofthecube(Delianproblem).
Givenacubeofsidelengtha,findacubeoftwicethevolume. Thesidelengthx of
p
thedesiredcubesatisfiesx3D2a3,soxDa3 2. ThusletP;Q;X bepointsonthe
p
reallinesuchthatPQDaandPX Da3 2;thequestioniswhetherX 2 fP;Qg.
Classicalconstructionproblems 3
(cid:1): Quadratureofthecircle.
Given a circle, construct a square of the same surface area. The given circle of
radiusr isdeterminedbypointsP andQadistancerDPQapart. Thesidelength
p
xofthedesiredsquaremustsatisfyx2D(cid:2)r2,soxDr (cid:2). Whatmustbedecided,
p
then,iswhethersome pointX such that PX Dr (cid:2) belongsto fP;Qg.
ı: Constructionofaregularn-gon(n-sectionofthecircle).
As before, we think of the circle as being given by X
twopointsP andQ,thecenterandapointonthe
circumference. LetX bethepointshownonthe
right. ForwhatnaturalnumbersndoesX liein 2(cid:2)=n
fP;Qg? Q
P
Thisisthecase,forexample,fornD6(and
thereforeforallnumbersoftheformnD3(cid:1)2m):
it is enough to draw a circle of radius PQ with
center Q. By successively repeating the procedure
withthe newlyfoundpointsonegetsthewell knownrosette:
3. To make the problem of constructibilitywith ruler and compass accessible, one
must first“algebraize”it. To that endit isusefultoemploythe identification
(cid:1)2D(cid:2);
thatis,toregardpointsintheplaneascomplexnumbers,andsotakeadvantageof
the possibility not only of (vector) addition but also of multiplication. Assuming
thebasicpropertiesofthefield(cid:2) ofcomplexnumbers,theproblemofdividingthe
circle into n parts (see ı above) amounts to the following question: Is it the case
that
e2(cid:1)i=n2 f0;1g ?
The next statement points out that the fundamental algebraic operations of (cid:2)
can bedescribedconstructively.
F1. LetM beanysubsetof(cid:2) containingthenumbers0and1. Then:
(1) i 2 M;
4 1 ConstructibilitywithRulerandCompass
(2) z2 M ) z2 M;
(3) z2 M ) Rez; Imz2 M;
(4) z2 M ) (cid:2)z2 M;
(5) z ;z 2 M ) z Cz 2 M;
1 2 1 2
(6) z ;z 2 M ) z z 2 M;
1 2 1 2
(7) z2 M; z¤0 ) 1=z2 M.
Proof. (1)Thelineconnecting0and1,thatis,therealline(cid:1),belongstoLi.M/by
definition. Intersecting(cid:1) withtheunitcircle,whichbelongstoCi.M/,weseethat
(cid:2)12 M. IfwenowconstructtheperpendicularbisectoroftheintervalŒ(cid:2)1;1(cid:3)in
the well-knownway and intersectitwiththe unitcircle,weobtaini 2 M.
(2) Drop a perpendicular from z to (cid:1). From the foot of this perpendicular, say a,
draw acircle whoseradiusisthedistance fromato z. Itssecondintersectionwith
the straightline throughz anda givesz2 M.
(3) As just verified, we have a D Rez 2 M. To obtain b D Imz, draw the
perpendicular to the imaginary axis through z, and then transfer to (cid:1) the absolute
valueofthefootbi oftheperpendicular.
(4)Intersectthe linethrough0and z withthe circleofradiusjzjandcenter0.
(5) Intersect the circle of center z and radius jz j with the circle of center z and
1 2 2
radius jz j. One of the intersections is the vertex z Cz of the parallelogram
1 1 2
determinedby z ;z .
1 2
(6)Ifz Da Cib and z Da Cib we have
1 1 1 2 2 2
z z D.a a (cid:2)b b /C.a b Ca b /i:
1 2 1 2 1 2 1 2 2 1
Now z ;z 2 M implies a ;b ;a ;b 2 M, by (3). If we assume the claim is
1 2 1 1 2 2
trueforrealnumbers,itwillalsobetrueforarbitrarycomplexnumbers,becauseof
(4)and(5). Thereforewemust provethat givenrealnumbersr and r ,
1 2
r ;r 2 M ) r r 2 M:
1 2 1 2
Clearlyone canassumer ;r >0. To completetheproof,considerthisdiagram:
1 2
z
0 1 r r x
1 2
