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GALOIS THEORY AT WORK: CONCRETE EXAMPLES

GALOIS THEORY AT work : CONCRETE EXAMPLESKEITH field extensionQ( 2, 3)/Qis GALOIS of degree 4, so its GALOIS grouphas order 4. The elements of the GALOIS group are determined by their values on 2 and 3. TheQ-conjugates of 2 and 3 are 2 and 3, so we get at most four possibleautomorphisms in the GALOIS group. See Table 1. Since the GALOIS group has order 4, these4 possible assignments of values to ( 2) and ( 3) all really exist. ( 2) ( 3) 2 3 2 3 2 3 2 3 Table nonidentity automorphism in Table 1 has order 2. Since Gal(Q( 2, 3)/Q) con-tains 3 elements of order 2,Q( 2, 3) has 3 subfieldsKisuch that [Q( 2, 3) :Ki] = 2,or equivalently [Ki:Q] = 4/2 = 2. Two such fields areQ( 2) andQ( 3). A third isQ( 6) and that completes the list. Here is a diagram of all the ( 2, 3)Q( 2)Q( 3)Q( 6)QIn Table 1, the subgroup fixingQ( 2) is the first and second row, the subgroup fixingQ( 3) is the first and third row, and the subgroup fixingQ( 6) is the first and fourth row(since ( 2)( 3) = 2 3).

GALOIS THEORY AT WORK: CONCRETE EXAMPLES 3 Remark 1.3. While Galois theory provides the most systematic method to nd intermedi-ate elds, it may be possible to argue in other ways. For example, suppose Q ˆFˆQ(4 p 2) with [F: Q] = 2. Then 4 p 2 has degree 2 over F. Since 4 p 2 is a root of X4 2, its minimal polynomial over Fhas to be a ...

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Transcription of GALOIS THEORY AT WORK: CONCRETE EXAMPLES

1 GALOIS THEORY AT work : CONCRETE EXAMPLESKEITH field extensionQ( 2, 3)/Qis GALOIS of degree 4, so its GALOIS grouphas order 4. The elements of the GALOIS group are determined by their values on 2 and 3. TheQ-conjugates of 2 and 3 are 2 and 3, so we get at most four possibleautomorphisms in the GALOIS group. See Table 1. Since the GALOIS group has order 4, these4 possible assignments of values to ( 2) and ( 3) all really exist. ( 2) ( 3) 2 3 2 3 2 3 2 3 Table nonidentity automorphism in Table 1 has order 2. Since Gal(Q( 2, 3)/Q) con-tains 3 elements of order 2,Q( 2, 3) has 3 subfieldsKisuch that [Q( 2, 3) :Ki] = 2,or equivalently [Ki:Q] = 4/2 = 2. Two such fields areQ( 2) andQ( 3). A third isQ( 6) and that completes the list. Here is a diagram of all the ( 2, 3)Q( 2)Q( 3)Q( 6)QIn Table 1, the subgroup fixingQ( 2) is the first and second row, the subgroup fixingQ( 3) is the first and third row, and the subgroup fixingQ( 6) is the first and fourth row(since ( 2)( 3) = 2 3).

2 The effect of Gal(Q( 2, 3)/Q) on 2 + 3 is given in Table 2. The 4 values areall different, since 2 and 3 are linearly independent overQ. ThereforeQ( 2, 3) =Q( 2 + 3). The minimal polynomial of 2 + 3 overQmust be(X ( 2 + 3))(X ( 2 + 3))(X ( 2 3))(X ( 2 3)) =X4 10X2+ particular,X4 10X2+ 1 is irreducible inQ[X] since it s a minimal polynomial CONRAD ( 2) ( 3) ( 2 + 3) 2 3 2 + 3 2 3 2 3 2 3 2 + 3 2 3 2 3 Table similar reasoning if a fieldFdoes not have characteristic 2 andaandbare nonsquaresinFsuch thatabis not a square either, then [F( a, b) :F] = 4 and all the fields betweenFandF( a, b) are as in the following ( a, b)F( a)F( b)F( ab)FFurthermore,F( a, b) =F( a+ b). The argument is identical to the special extensionQ(4 2)/Qis not GALOIS , butQ(4 2) lies inQ(4 2,i), whichis GALOIS overQ. We will use GALOIS THEORY forQ(4 2,i)/Qto find the intermediate fieldsinQ(4 2) GALOIS group ofQ(4 2,i)/Qequals r,s , wherer(4 2) =i4 2, r(i) =iands(4 2) =4 2, s(i) = i.

3 (Viewing elements ofQ(4 2,i) as complex numbers,sacts on them like complex conjuga-tion.) The group r,s is isomorphic toD4, wherercorresponds to a 90 degree rotation ofthe square andscorresponds to a reflection across a diagonal. What is the subgroupHofGal(Q(4 2,i)/Q) corresponding toQ(4 2)?( )Q(4 2,i){1} 2 2Q(4 2)H 4 4QD4 Sincesis a nontrivial element of the GALOIS group that fixesQ(4 2),s H. The sizeofHis [Q(4 2,i) :Q(4 2)] = 2, soH={1,s}= s . By the GALOIS correspondence forQ(4 2,i)/Q, fields strictly betweenQ(4 2) andQcorrespond to subgroups of the Galoisgroup strictly between s and r,s . From the known subgroup structure ofD4, the onlysubgroup lying strictly between s and r,s is r2,s . Therefore only one field lies strictlybetweenQ(4 2) andQ. SinceQ( 2) is such a field it is the only THEORY AT work : CONCRETE EXAMPLES3 Remark GALOIS THEORY provides the most systematic method to find intermedi-ate fields, it may be possible to argue in other ways.

4 For example, supposeQ F Q(4 2)with [F:Q] = 2. Then4 2 has degree 2 overF. Since4 2 is a root ofX4 2, its minimalpolynomial overFhas to be a quadratic factor ofX4 2. There are three monic quadraticfactors with4 2 as a root, but only one of them,X2 2, has coefficients inQ(4 2) (letalone inR). ThereforeX2 2 must be the minimal polynomial of4 2 overF, so 2 [F:Q] = 2,F=Q( 2) by counting s exploreQ(4 2, 8), where 8=e2 i/8is a root of unity of order 8,whose minimal polynomial overQisX4+ 1. BothQ(4 2) andQ( 8) have degree 4 overQ. Since 28=i,Q(4 2, 8) is a splitting field overQof (X4 2)(X4+ 1) and therefore isGalois overQ. What is its GALOIS group? We have the following field (4 2, 8) 4 4Q(4 2)4Q( 8)4 QThus [Q(4 2, 8) :Q] is at most 16. We will see the degree isnot16: there are some hiddenalgebraic relations between4 2 and Gal(Q(4 2, 8)/Q) is determined by its values( ) ( 8) = a8(a (Z/8Z) ) and (4 2) =ib4 2 (b Z/4Z).

5 There are 4 choices each foraandb. Taking independent choices ofaandb, there are atmost 16 automorphisms in the GALOIS group. But the choices ofaandbcannotbe madeindependently because 8and4 2 are linked to each other:( ) 8+ 18=e2 i/8+e 2 i/8= 2 cos( 4)= 2 =4 says 2 belongs to bothQ( 8) andQ(4 2). Here is a field diagram that emphasizesthe common subfieldQ( 2) inQ(4 2) andQ( 8). This subfield is the source of ( ).Q(4 2, 8) 4 4Q(4 2)2Q( 8)22Q( 2)2Q(i)2 QRewriting 8+ 18= 2 as 28 2 8+ 1 = 0, 8has degree at most 2 overQ(4 2).Since 8is not real, it isn t insideQ(4 2), so it has degree 2 overQ(4 2). Therefore4 KEITH CONRAD[Q(4 2, 8) :Q] = 2 4 = 8 and the degrees marked as 4 in the diagram both equal rewrite the field diagram (4 2, 8)22Q(4 2)2Q( 8)22Q( 2)2Q(i)2 QReturning to the GALOIS group, ( ) tells us the effect of Gal(Q(4 2, 8)/Q) on4 2partially determines it on 8, and conversely: ( (4 2))2= ( 8) + ( 8) 1, which in thenotation of ( ) is the same as( )( 1)b= a8+ a8 8=e2 i/8= (1 +i)/ 2, a calculation shows a8+ a8= 2 ifa 1,7 mod 8 and a8+ a8= 2 ifa 3,5 mod 8 (note 78= 18and 58= 38).

6 Thus ifa 1,7 mod 8we have ( 1)b= 1, sob 0,2 mod 4, while ifa 3,5 mod 8 we have ( 1)b= 1, sob 1,3 mod 4. For example, can t both fix4 2 (b= 0) and send 8to 38(a= 3) because( ) would not simplest way to understandQ(4 2, 8) is to use a different set of generators. Since 8=e2 i/8=e i/4= (1 +i)/ 2,Q(4 2, 8) =Q(4 2,i),and from the second representation we know its GALOIS group overQis isomorphic toD4with independent choices of where to send4 2 (to any fourth root of 2) andi(to anysquare root of 1) rather than4 2 and 8. A different choice of field generators can makeit easier to see what the GALOIS group looks like. We also see immediately from the secondrepresentation that [Q(4 2, 8) :Q] = s consider the splitting field ofX8 2 overQ, which isQ(8 2, 8),where 8=e2 i/8= (1 +i)/ 2. While [Q(8 2) :Q] = 8 and [Q( 8) :Q] = 4, we ll see that[Q(8 2, 8) :Q] is 16, not 32, and determine Gal(Q(8 2, 8)/Q).

7 It is a group of order 16,but is not isomorphic to the dihedral group of order we do will build on the previous example, where [Q(4 2, 8) :Q] is 8, not 16 (theproduct of [Q(4 2) :Q] and [Q( 8) :Q]), sinceQ(4 2) andQ( 8) are not independent overQ: the fields both contain 2 Since 86 Q(8 2) but 28 2 8+ 1 = 0 by the algebraic relation ( ), the degree of 8overQ(8 2) is 2. Thus [Q(8 2, 8) :Q] = [Q(8 2, 8) :Q(8 2)][Q(8 2) :Q] = 2 8 = 16, so|Gal(Q(8 2, 8)/Q)|= Gal(Q(4 2, 8)/Q) is determined by the values( ) ( 8) = a8(a (Z/8Z) ) and (8 2) = b88 2 (b Z/8Z). GALOIS THEORY AT work : CONCRETE EXAMPLES5 There is a condition linkingaandb, from ( ):( ) a8+ a8= ( 8+ 18) = ( 2) = (8 24) = (8 2)4= ( 1)b 2.(This is essentially the same as ( ), excepts nowbcomes fromZ/8 Zinstead ofZ/4Z.) Itfollows from ( ), by reasoning like that in the previous example, that ifa 1,7 mod 8thenbis even and ifa 3,5 mod 8 thenbis odd. Therefore, sincebis inZ/8Z, there areat most 4 values ofballowed for each value ofa.

8 That means Gal(Q(8 2, 8)/Q) has orderat most 16. Since we earlier showed the GALOIS group has order 16, the parity conditiononbfor eachais the only constraint: Gal(Q(8 2, 8)/Q) consists of the automorphismsdescribed by ( ) where (i)bis even ifa 1,7 mod 8 and (ii)bis odd ifa 3,5 mod is a field diagram, which is very far from showing all intermediate (8 2, 8)42Q(8 2)4Q( 8)22Q( 2)2Q(i)2 QEach automorphism in Gal(Q(8 2, 8)/Q) can be written by ( ) as a,bwhere a,b( 8) = a8and a,b(8 2) = b88 2. Two automorphisms a,band c,dcompose as follows: a,b( c,d( 8)) = a,b( c8) = ac8, a,b( c,d(8 2)) = a,b( d88 2) = a,b( 8)d a,b(8 2) = ad+b88 is how matrices (a b0 1) multiply: (a b0 1)(c d0 1) = (ac ad+b0 1). Therefore Gal(Q(8 2, 8)/Q)can be viewed as 2 2 mod 8 matrices by a,b7 (a b0 1)wherea (Z/8Z) andb Z/8 Zsubject to the conditions bis even ifa 1,7 mod 8, bis odd ifa 3,5 mod constraint onbgivena, or onagivenb, is important to remember, , the matrix(1 10 1) isnotin the GALOIS group: no has the effect ( 8) = 8and (8 2) = 88 2!

9 What group of order 16 is the GALOIS group? It isnotisomorphic to the dihedral groupof order 16 since that dihedral group has 9 elements of order 2 (all 8 reflections and therotationr4) while the GALOIS group has 5 elements of order 2: (1 40 1), (7 00 1), (7 20 1), (7 40 1), and(7 60 1). (The mod 8 matrices (3 00 1) and (5 00 1) have order 2, but they are not in the Galoisgroup since a matrix of the form (a00 1) in the GALOIS group hasa 1,7 mod 8.)The GALOIS group contains := (5 10 1), which has order 8. Its successive powers are inthe table below. An element of the GALOIS group with order 2 that is not a power of is1A concise way of describing the two parity constraints onbmod 8 givenamod 8 isb a2 18mod CONRAD := (7 00 1). Since the GALOIS group has order 16, it is , . By an explicit mod 8 matrixcalculation, 1= = (5 70 1) = k(5 10 1) (1 60 1) (5 70 1) (1 40 1) (5 50 1) (1 20 1) (5 30 1) (1 00 1)The group with two generatorsxandywherexhas order 8,yhas order 2, andyxy 1=x3is called thesemidihedral groupof order 16.

10 You can look up information about it on :SD16. For example,that webpage says this group has 3 subgroups of order 8, so by GALOIS theoryQ(8 2, 8) has3 quadratic fields. There are 3 quadratic fields inQ( 8):Q( 2),Q(i), andQ(i 2). Thusthese are all the quadratic fields ofQ(8 2, 8).2 The webpage also says the 3 subgroups oforder 8 are isomorphic toZ/8Z,D4,3andQ8, so the GALOIS group has a subgroup isomorphicto each of these groups. For example, check fixesi= 48, and since has order 8 we haveGal(Q(8 2, 8)/Q(i)) = = determining which elements of the GALOIS group fix 2 =8 24andi 2, it is left as anexercise to the reader to showGal(Q(8 2, 8)/Q( 2)) =D4,Gal(Q(8 2, 8)/Q(i 2)) = concludes our look atQ(8 2, 8) and its GALOIS group GALOIS extension is said to have a given group-theoretic property (being abelian, non-abelian, cyclic,etc.) when its GALOIS group has that quadratic extension ofQis an abelian extension since its GALOIS grouphas order 2.


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