9. Let Kn=Q(ζ2n+2) for n≥0.
(a) Prove that K+n/Q is cyclic of degree 2n.
(b) Prove Kn/K+n−1 is biquadratic and that two of the three intermediate subfields are Kn−1 and K+. Prove the remaining intermediate field is a cyclic extension of Q of degree 2n.
Proof: (7) Note that complex conjugation is an automorphism of C, and also restricts to an automorphism of K since ¯ζn=ζ−1n (|ζnn|=|ζn|n=1 implies for ζn=a+bi that |ζn|=a2+b2=ζn¯ζn=1). Now, when n>2 we note ζn∉R so Kn/K+ is of degree greater than or equal to 2. As well, 12(ζn+ζ−1n)=a and 1−a2=b2, so when α is a root of the equation x2+b2 we observe K+(α)=Kn and hence Kn/K+ is of degree exactly 2. Since the maximal real subfield is necessarily properly contained in Kn, we must have K+ is the maximal real subfield.
(9)(a) The degree of the extension follows from above. Here we may observe the isomorphism Gal(K/ℚ)≅(ℤ/2n+2ℤ)×≅ℤ/2ℤ×ℤ/2nℤ shown in Exercises 2.3.22-23. We show (ℤ/2n+2ℤ)× is generated by 5 and −1. This is clearly the case when n=0 so assume n≥1. Now 5 is of order 2n by the mentioned exercises and it suffices to show 5 doesn't generate −1; we observe this by demonstrating kn such that 5kn≡2n+1+1 mod 2n+2, a different element of order 2. First, we see k1=1. Now, given kn, we have kn+1=2kn since (5k1)2=(2n+1+1+m2n+2)2≡2(n+1)+1+1 mod 2(n+1)+2 when n≥1. Thus −1 and 5 generate (ℤ/2n+2ℤ)×, and since we notice the quotient group of (ℤ/2n+2ℤ)× over the subgroup generated by −1 is the cyclic group generated by the image of 5, we conclude Gal(K+/ℚ)≅Gal(K/ℚ)/Gal(K/K+) is cyclic.
(b) We note that since Kn/ℚ is Galois necessarily Kn/K+n−1 is Galois. This Galois extension by degree must be either cyclic or biquadratic, and since it cannot be cyclic by the presence of two distinct subfields of degree two, it must be biquadratic. We see that Kn−1, K+n, and the third subfield manifest as the fixed fields of the (isomorphic pullbacks) of the elements of order 2 in (ℤ/2n+2ℤ)×, and viewing their Galois groups as the quotient of this group with the subgroups generated by those elements, we see as a result of only one of them intersecting nontrivialy with the order-2n subgroup generated by 5 (that is, 2n+1+1 as shown above, which corresponds to Kn−1/ℚ which we see is not cyclic) that the third subfield's Galois group's representation as a quotient is cyclic. ◻
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