Exercise 6.3.4

Proof. $L=\mathbb{Q}(\alpha ),\alpha =\sqrt{2+\sqrt{2}}$. We have already proved (Ex. 5.1.6) that

is the minimal polynomial of $\alpha$ over $\mathbb{Q}$, and that $L=\mathbb{Q}(\alpha )$ is the splitting field of $f$ over $\mathbb{Q}$.

$L=\mathbb{Q}(\alpha )$ is so the splitting field of the irreducible separable polynomial $f$. By theorem 6.2.1,

Write $\beta =\sqrt{2-\sqrt{2}}$. If $\sigma \in \mathrm{Gal}(L∕\mathbb{Q}),\sigma (\alpha )$ is a root of $f$, thus

Moreover, since $L=\mathbb{Q}(\alpha )$, an automorphism of $\mathrm{Gal}(L∕\mathbb{Q})$ is uniquely determined by the image of $\alpha$, and since $|\mathrm{Gal}(L∕\mathbb{Q})|=4$, all of these possibilities occur, so there exist one and only one $\sigma \in \mathrm{Gal}(L∕\mathbb{Q})$ such that $\sigma (\alpha )=\gamma$, where $\gamma \in \{\alpha ,\beta ,-\alpha ,-\beta \}$ (alternatively, since $f$ is irreducible over $\mathbb{Q}$, we can use Theorem 5.1.8).

In particular, there exists $\sigma (={\sigma }_1)\in \mathrm{Gal}(L∕\mathbb{Q})$ defined by $\sigma (\alpha )=\beta$.

Recall that $\mathit{\alpha \beta }=\sqrt{2},{\alpha }^2=2+\sqrt{2},{\beta }^2=2-\sqrt{2}$ (see Ex. 5.1.6), thus

From this equality we obtain

therefore

As $\sigma (\alpha )=\beta$,

Finally $\sigma (-\alpha )=\sigma (-1)\sigma (\alpha )=-\sigma (\alpha )=-\beta$, so

As every element in $\mathrm{Gal}(L∕\mathbb{Q})$ is uniquely determined by the image of $\alpha$,

and

So $\mathrm{Gal}(L∕\mathbb{Q})$ is cyclic, generated by $\sigma$ :

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