Proving $\mathbb{Q}(r)$ is Galois over $\mathbb{Q}$ ?

Question

Problem: Let $r = \sqrt{2 + \sqrt{2}}$.

1. Calculate the minimal polynomial of $r$ and $r / \sqrt{2}$ over $\mathbb{Q}$.

2. Prove that $\mathbb{Q}(r)$ is Galois over $\mathbb{Q}$.

Attempt: I calculated the minimal polynomial in both cases, and found the same polynomial $p(x) = x^4 - 4x^2 + 2$ which is irreducible by Eisenstein criterion with prime $2$.

The roots of $p(x)$ are $\pm \sqrt{2 \pm \sqrt{2}}$. My attempt consisted in showing that $\mathbb{Q}(r)$ is the splitting field of $p(x)$. So I wanted to show that the other root $\sqrt{2 - \sqrt{2}} \in \mathbb{Q}(r)$, and the result would follow. But I couldn't write $\sqrt{2 - \sqrt{2}}$ in terms of $r$ by the usual operations. I know that $1/r = \frac{2 - \sqrt{2}}{2}$ and so $2 - \sqrt{2} \in \mathbb{Q}(r)$, but I cannot take the square root of this.

I think I have to use the fact somehow that $r$ and $r/\sqrt{2}$ have the same minimal polynomial and that splitting fields of the same polynomial are isomorphic, but not sure how.

Thanks for any help.

Answer 1

Set $s=\sqrt{2-\sqrt{2}}$. Then $rs=\sqrt{2}=r^2-2$, so $s=r-2r^{-1}\in\mathbb{Q}(r)$.

Now, the roots of $p$ are $\pm r,\pm s$, so the splitting field of $p$ is $\mathbb{Q}(\pm r,\pm s)=\mathbb{Q}(r)$.

Answer 2

$p$ is irreducible and has $4$ distinct roots. $r$ and $\sqrt{2}$ are clearly in $\mathbb{Q}(r)$. So $r/\sqrt{2}$ is as well so all 4 roots of $p$ lie in the extension. Sending $r$ to any of the other roots gives the automorphisms, so there are $4$, which is the degree of the extension, so the extension is Galois.