Transcription of Galois theory Introduction. - math.ou.edu
1 Basic idea of Galois theory is to study fieldextensions by relating them to their automorphism groups. Recall thatanF-automorphism ofE/Fis defined as an automorphism :E Ethat fixesFpointwise, that is, (a) =afor alla F. TheF-automorphisms ofE/Fform a group under composition (you can thinkof this as a subgroup ofS(E)). We call this theGalois groupofEoverFand denote it byGal(E/F) ={ :E E: is anF-automorphism}.Now consider an intermediate fieldF L E; I ll writeE/L/Ftorefer to this situation, but should issue a warning that this notation isnon-standard. Then we can similarly consider theL-automorphismsGal(E/L) ={ :E E: automorphism, (a) =afor alla L}.This is a subgroup of Gal(E/F) since any such in particular leavesF Linvariant. Conversely, if we are given a subgroupH Gal(E/F),then we can introduceInv(H) ={a E: (a) =afor all H}.
2 We call Inv(H) thefixed fieldofH. Thisisa field because ifa,b Inv(H) and H, then for example (a b) = (a) (b) =a b, soa b Inv(H), and similarlyab 1 Inv(H) ifb6= 0. It isalso clear that Inv(H) Fbecause the elements ofFare in factfixed by all automorphisms from the bigger group Gal(E/F) H. SoE/Inv(H)/F, and Inv(H) is an intermediate generally, letS Gal(E/F) be an arbitrary that Inv(S) :={a E: (a) =afor all S}is still anintermediate field. Then show that Inv(S) = Inv(H), withH= S ,the subgroup generated byS, so this doesn t really give anything given a field extensionE/F, we can pass from an intermediatefieldLto the subgroup Gal(E/L) of Gal(E/F), and also converselyfrom a subgroupHto the intermediate field Inv(H). We will be espe-cially interested in situations where these two operations Gal, Inv areinverses of each the definitions it is only clear that ifE/L/Fis a field extensionandH Gal(E/F) is a subgroup of the Galois group, then( )Inv Gal(E/L) L,Gal(E/InvH) this theory119 Definition call an algebraic extensionE/FaGalois extension(equivalently, we say thatEisGaloisoverF) if Inv Gal(E/F) = of the foundational material of this section also works for notnecessarily algebraic extensions, and I ll present it in this way.
3 How-ever, later on, we will be interested in finite extensions almost exclu-sively, so this added generality is not really ( 2) as an extension ofF=Q. Since 2 generatesE, anyQ-automorphism ofEis already determinedby what it does on 2. The minimal polynomial of 2 EoverQis given byf=x2 2 Q[x]. Since leavesQinvariant, 2can only be mapped to another zero off. This gives two potentialautomorphisms: the identity and ( 2) = 2. We observed above,after Definition , that conjugates can be mapped to each other byanF-homomorphism; this was a consequence of Lemma Moreprecisely, there is anF-homomorphism :F( 2) F( 2) thatsends 27 2. SinceF( 2) =F( 2) =E, this map is anF-automorphism Gal(E/F) ={1, }, with (a+b 2) =a b 2,a,b Q. Itnow follows thatE=Q( 2) is Galois overQbecause if (t) =tfort=a+b 2 E, thenb= 0, so no elementt / QofEis fixed by.
4 Automorphism has a simple structure from an alge-braic point of view. However, show that is discontinuous everywhereon its domainQ( 2) let s discussE=Q(21/3) in the same style. Asbefore, any Gal(E/Q) must map 21/3to one of its minimal polynomial of 21/3isf=x3 2, which has only thisone root, 21/3, inE(the other two roots inCare non-real). Thuswe must map (21/3) = 21/3; in other words Gal(E/Q) = 1. ThusInv Gal(E/Q) =E, andE/Qis not a Galois happened because the minimal polynomial of the adjoined el-ement 21/3did not split inEand we ran out of conjugates that 21/3could have been mapped to by an element of the Galois group. If weinstead consider a splitting fieldL=Q(21/3,21/3e2 i/3) off, thenLcontains (by construction) all three rootsa1= 21/3,a2= 21/3e2 i/3,a3= 21/3e4 i/3off=x3 2.
5 SinceL=Q(a1,a2), an automorphism isdetermined by what it does ona1,a2, and one can now show that all 6conceivable choicesa17 aj,a27 ak, withj6=kdrawn from 1,2,3,actually produce (a) Show that for any choice ofaj6=ak, we have thatL=Q(aj,ak),aj/ Q(ak).(b) Show that for any choice ofaj6=ak, there is a Gal(L/Q) with (a1) =aj, (a2) = :Use part (a) and apply twice, first toQ(a1) andQ(aj), and then to the full extensions.(c) Deduce thatL/Qis an example of the notions just discussed in an abstract setting, letus finally take a look at Gal(L/E). Recall thatL/E/Q, soEis an in-termediate field ofL/Q. We already know that Gal(L/E) is a subgroupof Gal(L/Q), which, as we just saw, contains the six automorphismscorresponding to the six possible choices ina17 aj,a27 a Gal(L/E) must fixE=Q(a1), so must senda17 a1;conversely, any such map fixes all ofE, of course.
6 So Gal(L/E) containsthe following two automorphism: (1)a17 a1,a27 a2, and this is justthe identity; (2)a17 a1,a27 observed above that Gal(L/E) is always a subgroup ofGal(L/F) for an intermediate fieldL/E/F. Please verify that the twomaps from Gal(L/E) that we just obtained indeed form a is Gal(L/Q) in this example (please find a familiargroup that this Galois group is isomorphic to)? , whereE=Q(21/4,i) is the splitting fieldoff=x4 2. Find [E:Q] and show that|Gal(E/Q)|= [E:Q] andthatEis Galois (x) =x3+x2 2x 1 Q[x]. (a) Showthatfis irreducible; (b) show that ifr Cis a root off, then so isr2 2; (c) conclude thatQ(r) is a splitting field for any suchr; (d)find Gal(Q(r)/Q). that every homomorphism :F Ffixes theprime fieldPofFpointwise. Conclude that Aut(F) = Gal(F/P), ifFis viewed as an extension of its prime thatRdoes not have any non-trivial (that is, (a)6=afor somea R) :Show that ifa < b,then (a)< (b) for any homomorphism.
7 For this, start out with thecasea= 0 and try to characterize the condition thatb >0 Galois now return to the general situ-ation and explore theGalois connectionbetween (sub)groups ofF-automorphisms and fixed (sub)fields in more detail. As we alreadydiscussed, we can move back and forth between these objects with theGalois theory121help of the operations Gal and Inv. It will be useful to temporarilysimplify the notation, as follows: letE/Fbe a field extension, andletG= Gal(E/F). Then, ifLis an intermediate field,E/L/F, wewriteL := Gal(E/L); as we observed above,L is a subgroup , ifH Gis a subgroup, then we writeH := InvH; this isan intermediate fieldE/H can mean either Gal or Inv, and which operation is meantdepends on the context. There is no danger of confusion, however,because Gal is applied to intermediate fields while Inv must be ap-plied to subgroups of automorphisms, so only one of the two possibleinterpretations of makes sense in any given Kbe intermediate fields of a field extensionE/K/L/F, and letJ H G:= Gal(E/F)be subgroups.
8 Then:(a)K L andH J ;(b)L LandH H;(c)L =L andH =H . (a) is clear from the definitions: for example, if K =Gal(E/K), then (a) =afor alla K, so in particular this holds foralla L K, and thus L = Gal(E/L). Part (b) is ( ) restatedin our new notation and was discussed in Exercise for part (c), notice thatL = (L ) L by (b), but alsoL Lby (b) again and thusL L by (a). The proof ofH =H iscompletely analogous. Lemma a field extension, and suppose thatK/Lis of finite degree. Then[L :K ] [K:L].In particular, this can be applied toK=E,L=FifE/Fis finite,and sinceE = Gal(E/E) = 1, so [L :E ] =|L |, it then says that|Gal(E/F)| [E:F].This statement is immediately plausible because ifa E\F, then weknow that an automorphism can mapaonly to one of its conjugates,and there are at most degfaof these (if we are unlucky, there are fewer,iffadoesn t split inEor has multiple roots).
9 This already settles thecaseE=F(a), and in general, we would hope to be able to apply thisstep several times to deduce the organize the formal proof as an induction onn= [K:L].Ifn= 1, thenK=L, soL =K and the claim becomes assume thatn >1 and that the inequality holds for extensionsK /L of with [K :L ]< n. If there is an intermediate fieldMprop-erly betweenKandL, then [K:M],[M:L]< n, so the induction122 Christian Remlinghypothesis can be applied to these extensions and we obtain that[L :K ] = [L :M ][M :K ] [M:L][K:M] = [K:L].Here, we use Exercise for the first equality and Theorem for thesecond one. (In fact, we use a version of Exercise for potentiallyinfinite groups.)If there are no such intermediate fieldsM, then we can take anya K\L, and we will have thatK=L(a) (otherwiseM=L(a)would be a proper intermediate field).
10 Letfa L[x] be the minimalpolynomial ofaoverL. We then know that degfa=n. Now considera coset K = Gal(E/K) L /K , L = Gal(E/L), and let , K , be an arbitrary element of this coset (also recall that thegroup operation inL is composition, so this is the composition of theautomorphisms , ). Since Gal(E/K) fixesK=L(a), we havethat ( )(a) = (a). So all automorphisms from a fixed coset K sendato the same image. This image (a) must be another root offabecause the coefficients offaare inLand are thus fixed by .Moreover, if we now consider two distinct cosets jK , then 1, 2donotsendato the same image: if they did, it would follow that 1(b) = 2(b) for allb K=L[a], so 12 1 Gal(E/K) =K andhence 1K = 2K . So the representatives of a given coset can bedescribed asexactlythose automorphisms fromL that sendato acertain fixed image.
