Let be the extension of fields $\mathbb{Q}(\alpha,\sqrt{2})/\mathbb{Q}$ with $\alpha=\sqrt{5+2\sqrt{5}}$. If I haven't miscalculated, the degree of the extension is 8 and the statement of my problem says that the extension is galois and to calculate its elements. However I only find 4 $\mathbb{Q}$-automorphisms of $\mathbb{Q}(\alpha,\sqrt{2})$ which are $\sigma_1$=Identity, $\sigma_2$ is determined by $\sigma_2$($\alpha$)=$\alpha$,
$\sigma_2$($\sqrt{2}$)=$-\sqrt{2}$,
$\sigma_3$ is determined by
$\sigma_3$($\alpha$) =-$\alpha$,
$\sigma_3$($\sqrt{2}$)=$\sqrt{2}$ and finally $\sigma_4$($\alpha$)=-$\alpha$ and
$\sigma_4$($\sqrt{2}$)=$-\sqrt{2}$. I don't know what's wrong, because according to my calculations the extension is of degree 8 and the statement says that it is galois so there must be 4 more $\mathbb{Q}$-automorphisms but I don't know what they are. If someone could write them explicitly I would greatly appreciate it.
Galois Group of a Extension
galois-theory
Best Answer
$\alpha=\sqrt{5+2\sqrt{5}}$
$\alpha^2=5+2\sqrt{5}$
$\alpha^2-5=2\sqrt{5}$
$\alpha^4-10\alpha^2+5=0$
Note that $f(x)=x^4-10x^2+5$ is irreducible by Eisenstein criteria with $p=5$, and thus is the minimal polynomial for $\alpha$ over $\mathbb{Q}.$ Thus $[\mathbb{Q}(\alpha):\mathbb{Q}]=4.$
Now, the four roots of the minimal polynomial of $f(x)$ are $\pm\sqrt{5 \pm 2\sqrt{5}}$ and it can be shown that all of these roots are in $Q(\alpha)$, and thus $f(x)$ splits in $\mathbb{Q(\alpha)}$. Thus $\mathbb{Q(\alpha)}$ is a splitting field for the minimal polynomial of $\alpha$ over $\mathbb{Q}$, and the roots of this minimal polynomial are all distinct. Thus $\mathbb{Q} \subseteq \mathbb{Q(\alpha)}$ is a Galois extension.
Now, since $\mathbb{Q} \subseteq \mathbb{Q(\alpha)}$ is a simple extension and the minimal polynomial of $\alpha$ splits in $\mathbb{Q}(\alpha)$, we have that $gal(\mathbb{Q(\alpha):\mathbb{Q})} = \{e,\gamma_1,\gamma_2,\gamma_3\}$ where:
$e(\alpha)=\alpha$
$\gamma_1(\alpha)=-\sqrt{5 + 2\sqrt{5}}$
$\gamma_2(\alpha)=\sqrt{5 - 2\sqrt{5}}$
$\gamma_3(\alpha)=-\sqrt{5 - 2\sqrt{5}}$
$gal(\mathbb{Q(\alpha):\mathbb{Q})}$ is a group of order $4$ and so is isomorphic to either $C_2 \times C_2$ or $C_4$. I leave it as an exercise to you to figure which one it is.