Transcription of GALOIS THEORY: LECTURE 18 - web.williams.edu
1 GALOIS theory : LECTURE 18 LEO GOLDMAKHER1. PROOF OF THEFUNDAMENTALTHEOREM OFGALOISTHEORYLast time we demonstrated the power of the FTGT by using it to give a short proof of the FundamentalTheorem of Algebra. Today we prove (most of) the FTGT itself. The main tools we use come from Geck sproof that|Aut(L/K)| [L:K], with equality iffL/Kis GALOIS . We recall a few particularly usefulconsequences of his proof:Proposition a finite GALOIS extensionL/K, there exists some Lsuch ( ),2. the GALOIS conjugates of are all distinct, and3. the minimal polynomial of is given bym (x) = Aut(L/K)(x ( )).With these results in hand, we re ready to prove the FTGT. Recall (from LECTURE 15) that the statement consistsof five related results; we prove them in the same order we originally listed.
2 Throughout, we assumeL/Kis agiven finite GALOIS extension, and thatG:=Aut(L/K)is the GALOIS group of this extension.(1) The GALOIS following two maps give bijections between the set of intermediatefieldsFlying betweenKandLand the set of subgroupsHofG:F7 Aut(L/F)LH7 , these maps are inverses of one plan of the proof is to reduce the problem to a simpler one, and then to apply a familiar trick in-volving minimal First reduction: it suffices to prove the maps are ( ) thatGis finite; it follows that there are only finitely many , it s a general fact (see ) that given any two functionsf:A Bandg:B Awhich are inverses of one, thenAandBhave the same cardinality andfandgmust be bijections. Thus, if we prove that the two maps given in the statement of the theoremare inverses of one another, it will immediately follow that the set of intermediate fields mustbe finite, and that each of the maps is a bijection.
3 We have thus reduced the claim to provingLAut(L/F)=FandAut(L/LH) = Second reduction: it suffices to prove|Aut(L/LH)| |H|.Recall (from LECTURE 15) thatL/Fmust be GALOIS . This implies (property (C) in the definitionof being GALOIS ) that the fixed field of Aut(L/F)is preciselyF, (L/F)=F. It thereforesuffices to prove Aut(L/LH) = thatH Aut(L/LH), since any Hfixes everything inLHby definition. Thus,it suffices to prove Aut(L/LH) H. In fact, since we knowHis a finite subset of the auto-morphism group, it s enough to prove|Aut(L/LH)| |H|.Date: April 24, on notes by Eleanor Compute the degree of the in Step 2, we know thatL/LHmust be GALOIS , whence Aut(L/LH) = [L:LH]. Thus ourgoal becomes to bound the degree of the extensionL/LHby the order ofH. Proposition us a more concrete way to think about this extension: since it s GALOIS , we deduce thatL=LH( )for some , and that the minimal polynomial of overLHis given bym (x) = Aut(L/LH)(x ( )) LH[x].
4 Since we re trying to prove that Aut(L/LH) =H, we re led to consider the polynomialf (x) := H(x ( )).We claim thatf LH[x], that its coefficients are fixed by every automorphism , applying any element Hto the set of roots{ ( ) : H}simply permutes theroots off , hence leaves the coefficients off unaffected (since they are symmetric polynomialsin the roots).1 SinceH Aut(L/LH), we deduce thatf |m . On the other hand, we knowm divides every polynomial inLH[x]with as a root, whencem |f . Since both polynomialsare monic, we deduce thatf =m . In particular,|Aut(L/LH)|= [L:LH] = degm = degf =|H|.Since we know from step 2 thatH Aut(L/LH), we conclude thatH=Aut(L/LH). It turns out this is by far the hardest part of the Fundamental Theorem to prove.(2) The GALOIS correspondence is HandF H under the Galoiscorrespondence,F F if and only ifH H.
5 F . Then any automorphisms that fixF must also fixF, whenceH =Aut(L/F ) Aut(L/F) = , supposeH H. Then the elements ofLthat are fixed by all ofHmust be also be fixed byH ,whenceF=LH LH =F . (3) Degrees are preserved under the GALOIS {e}abbaGiven thatF Hunder the GALOIS cor-respondence. Then[L:F] =|H|and[F:K] =|G/H|. that ifL/Kis GALOIS , thenL/Fis GALOIS , which implies that[L:F] =|Aut(L/F)|=|H|.Moreover, sinceL/Kis GALOIS , we know that[L:K] =|G|. The Tower Law implies[L:K] = [L:F][F:K]whence|G|=|H|[F:K].The claim immediately follows. 1 This is similar in spirit to the trick we used in LECTURE 16 to prove (C)= (B).2. MOTIVATION FOR THE NEXT STEPSG ivenL/KGalois,G=Gal(L/K), and supposeF Hunder the GALOIS can you form another intermediate field?(1) Alex: Adjoin an element ofFtoK.
6 (2) Eli: Adjoin an element ofL\FtoKorF.(3) Ben: Pick any element G,and look at (F).Question can you form another subgroup ofG?(1) Michael: Form another field by an above method, and associate a group.(2) Michael: Form a cyclic subgroup from an element ofG.(3) Isaac and Michael: We can form theconjugateofH: for any G, we have H 1 time we ll explore a connection between these: we ll show that the field and group formed in (3) of eachof the above correspond under the GALOIS correspondence.
