Galois Theory - pages.uoregon.edu

1 Galois TheoryRichard KochDecember 19, 2017 Contents1 The Extension Problem; Simple Groups .. An Isomorphism Lemma .. Jordan Holder .. The Symmetric and Alternating Groups ..82 The Quadratic, Cubic, and Quartic The Quadratic Formula .. The Cubic Formula .. The Quartic Formula ..153 Field Extensions and Root Motivation for Field Theory .. Fields .. An Important Example .. Extension Fields .. Algebraic Extensions; Root Fields .. Irreducible Polynomials overQ.. The Degree of a Field Extension .. Existence of Root Fields .. Isomorphism and Uniqueness .. Putting It All Together ..294 Splitting FactoringP.. The Splitting Field.

2 Proof that Splitting Fields Are Unique .. Uniqueness of Splitting Fields, andP(X) =X3 2 ..355 Finite Finite Fields ..386 Beginning Galois Motivation for Galois Theory .. ; Putting the Ideas Together .. The Galois Group ..447 Galois Extensions and the Fundamental Galois Extensions .. Fundamental Theorem of Galois Theory .. Important Note .. An Example .. Finite Fields Again ..518 Concrete Cases of the Cyclotomic Fields .. Galois Group of a Radical Extension .. Structure of Extensions with Cyclic Galois Group ..549 Solving Polynomials by Solving Polynomial Equations with Radicals .. The Flaw .. The Fix .. Galois Theorem on Solving Via Radicals.

3 Solving Generic Equations by Radicals .. A Polynomial Equation Which Cannot Be Solved by Radicals ..6310 Straightedge and Compass Constructions With Straightedge and Compass .. Complex Extensions and Constructions .. Trisecting Angles; Doubling the Cube ..7111 Irrationality and Transcendence of Irrationalityeand .. Transcendence ofe.. Intermission: Fundamental Theorem of Symmetric Polynomials .. Transcendence of ..7812 Special The Discriminant .. The Cubic Case .. Cyclotomic Fields .. Galois Group ofXp aoverQ..8813 Constructing Regular Constructing Regular Polygons ..9114 Normal and Separable Normal Extensions, Separable Extensions, and All That ..9515 Galois Theory and Reduction Preview.

4 X 1 .. A Tricky Point Concerning Straightedge and Compass Constructions .. Polynomials with Galois GroupSn.. Every FiniteGis a Galois Group .. Proof of Dedekind s Theorem .. 106 Chapter 1 PreliminariesBefore getting to the main subject, we prove some facts about group Theory . These resultswill be used later in our study of generalizations of the quadratic chemistry, every compound is built out of atoms. Knowing the atoms is not enoughto determine the compound; for instance, graphite and diamond are both pure , we will show that every finite group is built out of simple groups . But knowingthe simple groups is not enough to determine the finall group they The Extension Problem; Simple GroupsSuppose we want to classify finite groupsGIt is natural to work by induction on the order ofG.

5 IfHis a subgroup ofG, then wealready understandH. IfHis normal, we also already in thesequence0 H G G/H 0we understand the groups at the ends and need only fill in the middle choice forGisH G/H. There are usually others. The problem of constructingsuchGis calledthe extension problemin group Theory ; it is difficult. For example, supposeH=Z4andG/H=Z2. ThenGis a group of order 8. After some work, one can show thatthere are threeGwhich fit in the sequence,Z4 Z2,D4,andQ. HereD4is the dihedralgroup of order four, that is, the group of symmetries of a square, and the groupQisthe group of unit quaternions{ 1, i, j, k}. Note that{ 1, i}is a normal subgroupisomorphic toZ4. Note also thatD4andQare not isomorphic becauseD4has five elementsof order two andQhas only one element of order 1.

6 PRELIMINARIES5 Luckily, we don t need to solve the extension problem for Galois Theory . But supposewe completely understood this problem. What more would be needed to classify finitegroups?Some groups have no non-trivial normal subgroups. We call such groupssimple groupsandwe would need to construct them another way. This has been one of the greatest researchtopics of the twentieth century, and there is now a complete list of all finite simple knowledge of this list, and a complete solution of the extension problem (which willnever happen!) would produce a complete classification of all finite the list we only need theabeliansimple groups:Theorem 1A finite abelian group is simple if and only if it equalsZpfor a :All subgroups of an abelian group are normal, so it suffices to list all groups with nonon-trivial subgroups.

7 CertainlyZphas no non-trivial subgroups, since every subgroup hasorder dividingpand thus equals{e}orZp. Conversely ifGis has no non-trivial subgroupsandg6=eis inG, then the cyclic subgroup generated bygmust be all ofG, soGis cyclicof some ordern. Ifnis not prime then it has non-trivial cyclic An Isomorphism LemmaWe are about to prove theJordan Holder theorem, which says that every finite group isbuilt from a uniquely determined collection of simple groups. In the midst of the proof,we need a little lemma, so we prove it a group with subgroupsAandB. By definitionABis the set of all elements inGof the varyingk. This set is clearly a subgroup ofG. IfAandBare normal, so isABbecauseg( )g 1=(ga1g 1)(gb1g 1).

8 (gakg 1)(gbkg 1)Lemma 1 IfAandBare normal,AB/B =A/(A B)Proof:We have a natural group homomorphismA AB AB/Band clearly this map sendsA Bto the identity. So it inducesA/(A B) AB/BCHAPTER 1. PRELIMINARIES6 The kernel of this map isebecause ifa Amaps toe AB/B, thena A mapis onto becauseaibi aiinAB/B, Jordan HolderContinuing with the idealistic program of the previous sections, supposeGis an arbitraryfinite group. Find a normal subgroupH1 Gunequal toGand as large as is easy to see thatG/H1must be simple, since ifK G/H1is a non-trivial normalsubgroup, the inverse image ofKinGwill be a normal subgroup KwithH1 K G. Ifwe understood the extension problem, we could constructGfromH1 the process.

9 Find a normal subgroupH2 H1unequal toH1and as largeas possible. ThenH1/H2is simple. If we understood the extension problem, we couldconstructH1fromH2andH1 in this vein, we eventually construct acomplete composition series, that is, achain of subgroups{e}=Hn Hn 1 .. H1 Gwith eachHinormal inHi 1and as large as possible, and eachHi 1/Hisimple. The groupGis constructed from the simple groupsHi 1/Hiby a series of group , theHiare not unique. For example, letG=Z6. LetZ2 Z6be thesubgroup{0,3}and letZ3 Z6be the subgroup{0,2,4}.Then we obtain two compositionseries{e} Z2 Z6{e} Z3 Z6 However, both series have length 2 ; the simple quotients in the first case areZ2,Z3=Z6/Z2and the simple quotients in the second case areZ3,Z2=Z6/Z3and thus the sameup to , this holds in generalTheorem 2 (Jordan-Holder)Any two complete composition series for a finite grouphave the same length, and their simple quotients are isomorphic up to :This theorem is a generalization of the unique factorization theorem for , ifn= , it is easy to find a composition series forZnwith simple quotientsZpi, each :Sometimes progress is made in mathematics by throwing information away untilonly the crucial information remains.

10 Composition series allow us to throw away theCHAPTER 1. PRELIMINARIES7intricate extension information until only the simple quotient information remains. It isthis quotient information which is important in Galois the previous section, we listed the three groups of order four obtained by extendingZ4byZ2. Notice that the simple quotients of all three groups areZ2,Z2,Z2. So in this case,extension information is definitely thrown :We prove the theorem by induction on the order ofG; the result is trivial for groupsof order less than or equal to the induction step, supposeGhas two composition A1 A B1 B GIfA=B, the theorem holds by induction, so supposeA6= a normalsubgroup larger thanA, and soAB=G.