The exercise asks to find an extension $E$ of Galois of rational ones such that
$ \operatorname{Gal}(E/\mathbb{Q}) \simeq \mathbb{Z}_4$, I thought about considering $E=\mathbb{Q}(\sqrt[4]{2})$
I know that
$$\mathbb{Q}(\sqrt[4]2) \simeq \frac{\mathbb{Q}[x]}{\langle x^4 -2\rangle} \simeq \mathbb{Q}[r_i]$$
such that $r_i \; (i=1\ldots4)$ are the roots of $x^4 -2$.
Define $\phi_i: \dfrac{\mathbb{Q}[x]}{\langle x^4 -2\rangle} \to \mathbb{Q}[r_i]$ such that $\phi_{i}(g(x)+\langle x^4-2\rangle) = g(r_i)$, so we get all elements of $ \operatorname{Gal}(\mathbb{Q}(\sqrt[4]{2}\,)/\mathbb{Q}) $ and $\lvert \operatorname{Gal}(\mathbb{Q}(\sqrt[4]{2}\,)/\mathbb{Q})\rvert=4$
How can I show that $ \operatorname{Gal}(\mathbb{Q}(\sqrt[4]{2}\,)/\mathbb{Q})$ is isomorphic to $ \mathbb {Z}_4 $?
And if this is not true, how can I find $ E $ such that isomorphism is true with $\mathbb{Z}_4$
Best Answer
roots of $x^4-2=0$ is $\pm \sqrt[4]2, \pm i \sqrt[4]2$, then $Gal(\Bbb Q(\sqrt[4]2),\Bbb Q)=\{I,\sigma_1 \}$; $I(\sqrt[4]2)=\sqrt[4]2, \sigma_1(\sqrt[4]2)=-\sqrt[4]2$