Take [tex]M = Q(i, \sqrt{2} )[/tex]. Prove that G = Aut(M) is isomorphic to [tex]V_4[/tex](adsbygoogle = window.adsbygoogle || []).push({});

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

Consider [tex]K(i)[/tex] with [tex]K = Q(\sqrt{2})[/tex]. 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 [tex]L(\sqrt{2})[/tex] with [tex]L = Q(i)[/tex], 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Automorphism group of field extension

**Physics Forums | Science Articles, Homework Help, Discussion**