What is the automorphism group of the field K=Q(21/4)?

[wattsup03]

Homework Statement

Let K = Q(2^1/4)

Determine the automorphism group Aut(K/Q)

Homework Equations

An automorphism is an isomorphism from a Field to itself

Aut(K/Q) is the group of Automorphisms from k/Q to K/Q

Definition: A K-Homomorphism from L/K to L'/K is a homomorphism L---> L' that is the identity on K

The Attempt at a Solution

I am completely at a loss really. I have calculated there are four homomorphisms from K to C and think from there if I know how many are K-homomorphisms then that'll be the number of automorphisms, because a homomorphism from a field to itself is an automorphism (Please correct me if I'm wrong on this). Then that'll give me the set of Automorphisms.

My problem is that I don't know how to go from the number of homomorphisms to the actual homomorphisms. I think it has a relation to the roots of 2^{(1 /4)} in C (which I have calculated to be 2^{(1 /4)}, - 2^{(1 /4)}, i*2^{(1 /4)}, -i2^{(1 /4)} )

Please help, this lack of understanding is preventing me from moving forward with other questions and my notes from lectures completely gloss over how to do this.

 

[HallsofIvy]

(2^{1/4}^2= 2^{1/2} and (2^{1/4}^3=2^{3/4} are irrational but (2^{1/4})^4= 2 is rational so any number in [/itex]Q(2^{1/4})[/itex] is of the form a+ b2^{1/4}+ c2^{1/2}+ d2^{3/4} for rational numbers a, b, c, d. For any automorphism, f, f(a+ b2^{1/4}+ c2^{1/2}+ d2^{3/4})= a + bf(2^{1/4})+ cf(2^{1/2})+ df(2^{3/4}) so the possible values of f depend entirely upon the possible values of f(2^{1/4}), f(2^{1/2}) and f(2^{3/4}).

Further, (f(2^{1/4})^4= f(2) is rational so f(2^{1/4} must be a fourth root of 2. Also, f(2^{1/2})^2= f(2) so f(2^{1/2}) must be 2^{1/2} (or -2^{1/2} which is just a rational number times 2^{1/2}. Each f <b>permutes</b> the fourth roots of 1 while fixing 2^{1/2}.

 

[Deveno]

it is somewhat misleading to say the values of f depend on the values of f(2^1/4), f(2^1/2) and f(2^3/4).

f is a field automorphism, so (for example) f(2^3/4) = f((2^1/4)3)= (f(2^1/4)³,

so f ONLY depends on the value of f(2^1/4).

while it is true that f permutes the roots of x⁴-2, it is not true that any such permutation yields an "f". f takes conjugate pairs to conjugate pairs, as well (the square of a fourth root of 2 must be a square root of 2).

so if one regards the 4 roots of x⁴-2 as α₁,α₂,α₃,α₄, where:

α_j = αe^(j-1)πi/4

then (α₂ α₄) yield a member of Aut(K/Q), but (α₂ α₃) does not.