# Automorphism group of field extension

1. Mar 7, 2006

### Pietjuh

Take $$M = Q(i, \sqrt{2} )$$. Prove that G = Aut(M) is isomorphic to $$V_4$$

I have some idea's but I don't know how to justify them:

Consider $$K(i)$$ with $$K = Q(\sqrt{2})$$. The the minimal polynomial of i over K equals to X^2 + 1. I know the fact that if x is a zero of a polynomial P and f is an automorphism, then f(x) is also a zero of P. Also, if f is in Aut(M), then it is a Q-automorphism, so it is the identity on the elements of Q. We can now construct an automorphism by sending i to -i and sqrt(2) to itself. Ofcourse we can also have the identity automorphism. Now by looking at $$L(\sqrt{2})$$ with $$L = Q(i)$$, with minimal polynomial X^2 - 2, we find an automorphism by sending i to i and sqrt(2) to - sqrt(2). If we just look at M itself, we find that the minimal polynomial equals (X^2 + 1)(X^2 - 2), and find and automorphism by sending i to -i and sqrt(2) to -sqrt(2).

Now by looking at the compositions of the automorphisms we get the structure of V_4. The only problem I have is to show that we can't have any more automorphims than the ones I found.

2. Mar 7, 2006

### AKG

You just said that if x is a zero of P, and f is an automorphism, then f(x) is a zero of P. x² + 1 only has two roots, so it must either send i to i or -i, and something similar is true for where 21/2 is sent. Now if f is any automorphism, and m is any element of M, then isn't f(m) uniquely determined by where 1, i, and 21/2 are sent? In fact, since 1 is sent to 1, f(m) is uniquely determined by where i and 21/2 are sent.