Galois Theory questions: Homomorphisms

[wattsup03]

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

 

[Deveno]

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 2^1/4 to one of the four complex solutions of x⁴-2 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 x⁴-2, only two are in Q(2^1/4), namely:
2^1/4 and -2^1/4.

c) this is isomorphic to Z₂, 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 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.