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Galois Theory questions: Homomorphisms 
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#1
Nov211, 07:47 PM

P: 6

Let K = Q(2^(1/4))
a) Which of the morphisms from K to C are Q(2^1/2)homomorphisms b) And which are Khomomorphisms? Attempt at a solution Ok, I don't really understand this very well but for a) I know that there are 4 homomorphisms, since the minimal polynomial over C has four solutions and there is a bijection between the roots and the homomorphisms. What I don't understand is how I get from the number of homomorphisms to the homomorphisms themselves. If someone could explain that to me I think it would really help. b) I can't really do b) until I know how to get the homomorphisms I do not want to push my luck as I would be really happy if someone could give me some pointers on the previous questions, but if there was someone who didn't mind helping out a struggling student any pointers on the following would be greatly appreciated also. c) Determine the automorphism group Aut(K/Q) d) Find an element in K that is not in Q and that is fixed by every element of Aut(K/Q) e) Conclude that K/Q is not Galois 


#2
Nov411, 03:10 AM

Sci Advisor
P: 906

a) all field homomorphisms are monomorphisms. suppose φ:K→F (where K and F are fields) is a ring homomorphism.
then ker(φ) is an ideal of K (as a ring). but the only ideals K has are {0} and K. now, in a field, we insist that 0 ≠ 1, so the 0map is not a field homomorphism. that leaves just ker(φ) = {0}, so φ must be injective. it is not hard to see that for any field morphism of K into C, φ(1) = 1 = 1+0i. this, in turn implies φ maps Q into Q (as the subfield {q+0i, where q is a rational real}). furthermore φ must map 2^{1/4} to one of the four complex solutions of x^{4}2 in C. these are 2^{1/4}, 2^{1/4}, i2^{1/4}, i2^{1/4}. furthermore φ is completely determined by where it sends 2^{1/4}, so each of those 4 choices yields an injection of Q(2^{1/4}) into C. b) now, if we require that φ:K→K, then φ(2^{1/4}) has to be in Q(2^{1/4}). of the 4 roots of x^{4}2, only two are in Q(2^{1/4}), namely: 2^{1/4} and 2^{1/4}. c) this is isomorphic to Z_{2}, we have 2 automorphisms: the identity, and the automorphism that sends 2^{1/4} to 2^{1/4}, which is clearly of order 2. d) √2 will work nicely for this. e) K is clearly algebraic over Q (and thus separable), but it is not normal. from (a) we see we have embeddings of K in C which are not automorphisms of K. equivalently, note that K has a root of x^{4}2 (in fact, it has 2) but x^{4}2 does not split over K. and yet again, we see that the fixed field of Aut(K/Q) is larger than Q (in fact, it is Q(√2)), so K is not galois over Q. 


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