Let $\xi$=$\ \sqrt{\vphantom{\sum}2+{\sqrt2}} $
(1) Find the minimal polynomial $r(x)$ $\in$ $\mathbb{Q}$ of $\xi$
(2)Prove that $\ \sqrt{\vphantom{\sum}2-{\sqrt2}} $ is also a root of $r(x)$
(3)Prove that $\ \sqrt{\vphantom{\sum}2-{\sqrt2}} \in \mathbb{Q}(\xi)$
(4)Find a theorem that says ; $\xi \longrightarrow \sqrt{\vphantom{\sum}2-{\sqrt2}} $ that defines an automorphism $\zeta$: $\mathbb{Q}(\xi) \longrightarrow \mathbb{Q}(\xi)$
(5) Using the basis ${1,\xi,\xi^2,\xi^3}$ of $\mathbb{Q}(\xi)$ as a vector space over $\mathbb{Q}$, find the matrix representation of $\zeta$ as a linear transformation.
(6)Show that $Aut_{\mathbb{Q}}(\mathbb{Q}(\xi))$ is cyclic of order 4
SOLUTION
(1) By the defintion og Galois I must get all the different conjugates of the generator
After much working I got $(x^2-2)^2-2=0 $ so the minimal polynomial is $x^4-4x^2+2$ and by Eisenstein's criterion it's not hard to see that it's irreducible.
(2) I reversed the process and I got the conjugates and after some working so this was easy.
(6) Using (2) the other conjugates are just additive inverses of the k=$\ \sqrt{\vphantom{\sum}2+{\sqrt2}} $ and so all the conjugates are in k. K is a Galois over $\mathbb{Q}$ and after some working we get that [k:$\mathbb{Q}$]=4 and I already showed (not here) that I get a cyclic group of order 4.
My burning question
for (5) i'm a bit confused.Normally for linear algebra related problems I would be given something like a function for $v:\mathbb{R} \longrightarrow \mathbb{R}$ which is my $\zeta $ in this case but in order to find to that I always given a matrix to begin with and then and then I use the basis given to compute.
In this question I am not given any matrix but instead asked to find it which is the reverse process and I really don't even understand how to begin this part of the problem.
Can anyone help me understand what I need to do with part (5)?
Best Answer