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Galois Theory questions: Homomorphisms

  1. Nov 2, 2011 #1
    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 K-homomorphisms?

    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. jcsd
  3. Nov 4, 2011 #2


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    Science Advisor

    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 0-map 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 21/4 to one of the four complex solutions of x4-2 in C. these are 21/4, -21/4, i21/4, -i21/4.

    furthermore φ is completely determined by where it sends 21/4, so each of those 4 choices yields an injection of Q(21/4) into C.

    b) now, if we require that φ:K→K, then φ(21/4) has to be in Q(21/4). of the 4 roots of x4-2, only two are in Q(21/4), namely:
    21/4 and -21/4.

    c) this is isomorphic to Z2, we have 2 automorphisms: the identity, and the automorphism that sends 21/4 to -21/4, 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 x4-2 (in fact, it has 2) but x4-2 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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