Galois Theory questions: Homomorphisms

[wattsup03]

Homework Statement

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?

Homework Equations

Theorem: There is a bijection between roots of minimal polynomial and number of homomorphisms

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

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

 

[mighty2000]

in a) so if someone could please help me with that first, that would be great.

Thank you for your question. Let me explain the process for finding the homomorphisms from K to C that are Q(2^(1/2))-homomorphisms.

First, we know that there are four homomorphisms from K to C, as you correctly stated. This is because the minimal polynomial of K over C has four roots, which correspond to the four homomorphisms.

To find the specific homomorphisms, we need to consider the definition of a Q(2^(1/2))-homomorphism. This means that the homomorphism must fix the elements of Q(2^(1/2)), which is the same as saying that it must be the identity on Q(2^(1/2)). So, we need to find the elements of K that are fixed by the homomorphisms.

Recall that K is the field extension of Q(2^(1/4)), which means that any element in K can be written as a linear combination of powers of 2^(1/4). So, we can write an element in K as a + b*2^(1/4) + c*2^(1/2) + d*2^(3/4) where a, b, c, and d are rational numbers.

Now, let's consider the four homomorphisms. We know that each of these homomorphisms must send 2^(1/4) to one of the four roots of the minimal polynomial. So, for example, one of the homomorphisms might send 2^(1/4) to the root 2^(1/4), which means that it will also send 2^(3/4) to 2^(3/4). This means that the homomorphism will fix the element 2^(1/4) + 2^(3/4) in K.

Similarly, we can consider the other three homomorphisms and find the elements in K that are fixed by each of them. These fixed elements will give us the four homomorphisms from K to C that are Q(2^(1/2))-homomorphisms.

Once you have determined the four homomorphisms, you can use the same process to find the K-homomorphisms. These are the