Class Note for MATH 511A at UA
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Date Created: 02/06/15
1 We see that K K z 2 8 16 We deduce that the automorphisms of K are determined by 5 gt gt with 0 S j lt 8 and 2H in Note that 43 1 z By the fundamental theorem of Galois theory the sub elds and 19072 are each xed by a subgroup of order 8 In each case if we can write down 8 elements of GalKQ that x the eld these are the elements of the Galois group 0 the maps 5 gt gt qg Z i gt gt i are the ones that x Note that this map sends lt3 gt gt l 534 71 3 One checks that the order of the map with j 1 is 8 so the Galois group is cyclic of order 8 o the element is xed if and only ifj is even so the maps xing are 5 gt gt and i gt gt ii withj even One checks easily that this group of transformations is isomorphic to D3 0 note that 5 gt gt sends gt gt ily So the element H is xed by the automorphisms Si gt gt and i gt gt 71 with 0 jlt8 lfTisthemap WHisgigt gtZl thatis739hasj4 then every automorphism except the identity and 739 has square equal to 739 So the Galois group is Q3 2 The roots of this polynomial are i i V3 so the Galois group is V4 by the next problem 3 a The condition that none of D1D2D1D2 is a square implies that K F 4 The elements of the Galois group must be m gt gt im Di2 gt gt i and these are easily seen to form a group isomorphic to V4 b Suppose GalKF 2 V4 Since V4 has three subgroups of order 2 there are three quadratic elds E1 E2 E3 lying between F and K Since charF 2 each Ei has the form Ei for nonsquares Di 6 F Then K ElEg Fm D 2 Since K F we deduce that Dng is a nonsquare and the third quadratic eld inside KF is FD1D2 4 From the conditions of the problem we see that KF must be separable a We know that K has K F embeddings into L they are precisely the restrictions to K of the elements of GalLF Moreover the elements of a coset O39H C GalLF all correspond to the same embedding since elements of H act trivially on K So NKFa Ulla where 01 are my collection of left coset representatives for H in GalLF For any a E GalLF we have UNKFa 00139 a NKFa since 00139 is also a collection of left coset representatives for H in GalLF Thus NKFa is xed by GalLF and lies in F bc are trivial 1 d If F C E C K I leave it as an exercise to check that NEFNKEai Take E Fa C Ki You can check from the de nition that NFWVF a is the product of the roots of ma so NEF a 7lda0i Then NEFNKEa NEFand Dnagdv KE the rst equality because NKE ml for z E E 5 Part a is exactly like 4a For part b TrKF is the linear combination of characters EUeGalltKF a The result is immediate 6 a Follow the hint b Since a2 22 3 generates we know that E contains F By the previous part E is a quadratic extension of F so E z 8 Its easy to check that the eight given numbers are distinct and are the roots of the same polynomial with rational coef cients so they are the eight roots of the minimal polynomial of 1 c Each statement in this part is clear from the immediately preceding state ment d As noted before 2 generates over Q it is easy to see that there is only one automorphism of that sends 2 3 to 2 7 3 namely the one that xes V3 and sends 2 gt gt 72i So a is an extension to E of this map Now a 23 so 0a 723 iiei 0a 7a i Since 0a B we have 06 70A From this we see 041 a so a has order 4 e 02 7392 both send a gt gt 7a hence are equali Similarly 039739 and 73903 can both be checked to send a gt gt 7 2 7 7 so are equal f is immediate from the elements and relations constructed in e 7 Let L1iHLn be the Galois conjugates of L over F contained in some algebraic closure of L so that the Galois closure of L is the composite L1 Lni Since KF is Galois L contains K for each i and in fact LiK is Galoisi Since the extensions LiK are all Galois the degree of the composite L1 Ln K divides L1 K Ln K which is a power of pi Since K F is a power of p as well we deduce that L1 Ln F is a power of pi For a counterexample with KF not Galois take L K F 8 Suppose that the polynomial 91 z alza0 satis es gFrobp 0 Then the equation gFrobp 0 gives a linear dependence among Frobg i i Frobp 1 By linear independence of characters two of the characters in this list must be equal and therefore n 2 Ti It follows that the characteristic polynomial of Frob is the unique monic polynomial of degree 7 which vanishes on Frobi Finally note that Froba apf a so Frobp is a root of IT 7 1 so the characteristic polynomial is IT 7 ll 9 In this problem one needs to be careful about signs it should bother you to write down an expression like an22 without making precise what the righthand side of that expression means For instance you could mean that an is de ned inductively as the image of z in the abstractly de ned eld Qan1z12 7 2 an71 but then it s not at all clear what relation is borne between an and the particular 2n2th root of unity that we started with On the other hand if you de ne an 2n2 4521 as the problem does and then derive the formula 13 an1 2 then the expression an W is meaningless the square root is not a function So it really does take some caution in order to make the statements of the problem meaningful Here is one approach Let us regard Q 2n2 as a sub eld of C and choose 2n2 to be the speci c element e so that 21 e for i S n 2 Now the expression V2V2 is perfectly wellde ned we are only taking square roots of positive numbers and we always choose the positive square root Let ai 22 Notice that 7 2m 7 2m 2w 0 C222 912 627 e W 2cos 721 So 11 and its easy enough to check that a ai712i Suppose by induction that am 2x2 with i7 1 twos in the expression The formula a 017371 2 implies that aii 2x2 with i twos in the expression But the formula ai 2 cos shows in partic ular that ai is a positive real number Therefore at 2x2 with i twos in the expression and the induction is complete Now the quadratic formula gives again there is a sign ambiguity that we must resolver Note that this may be rewritten 2x2 i 722 2 lt2n2 Since the expression inside the square root is a negative real number we may rewrite the expression again as 22 ii 272 2 27m where i is the root of 71 in the u er half lane Finall since e 2quot2 has osz39tz39ve Pp p Y7 P imaginary part we conclude that 2quot 2x2 i 27x2 42M 5W 2 This is a better theorem than a vaguer statement that the process of repeatedly adding two and taking a square root yields some 2n2th root of unity lt2n2 10 Typically in a problem like this you write down some permutations of the roots of the polynomial and prove that any automorphism must come from one of these permutationsi One subtlety to be aware of is that it s not always obvious that the permutations you7ve written down really all give you maps that is you have to give a lower bound on the size of the Galois group and not just the upper bound that you get from writing down the obvious relations among the roots of the polynomiali In this case let 2 be a root of the quadratic t2 7 4t1 for instance 2 2 V3 will do note that 2 1 is the other rooti Let y be a cube root of 2 Then the roots of 15 7 413 1 are y 3y 3 1y and y l gy l gly li Let K be the splitting eld of then an automorphism of K is determined entirely by where it sends y six choices and where it sends 3 two choices Thus S 12 We suspect that equality may hold 7 ie that we can make these two choices independently 7 but we still have to prove that the choice for y does not determine the choice for 3 In fact we also have to prove that is irreducible so that the six choices for y are all possible In this case you can see that 15 7 413 1 is irreducible for instance fz l is Eisenstein at 2 Let K be the splitting eld of over Q inside C Note that the roots of K is not contained in R eg because K must contain cube roots of unity so K K N R 2 On the other hand has two real roots letting a be one of these roots we see that Qa C K N R Since Qa 6 It follows that K K K N R K N R z is divisible by 12 in particular K 2 12 Therefore our 12 proposed automorphisms really are automorphisms Finally we can check that GalKQ 2 D12 Let a be the map sending y gt7 Lgy and 3 gt gt gill Let 739 be the map sending y gt gt y l 3 gt gt Qgi Then a has order six 739 has order two and TO39 0 17 both map y gt gt ng lt3 gt gt This gives a presentation
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