# Show that $\operatorname{Gal}(\mathbb{Q}(\sqrt[4]{2}\,)/\mathbb{Q}) \simeq \mathbb{Z}_{4}$, or prove otherwise

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 $$\mathbb{Z}_4$$

• $\mathbb Q(\sqrt[4]{2})/\mathbb Q$ is not Galois. – Wojowu Sep 22 '18 at 20:05

roots of $$x^4-2=0$$ is $$\pm \sqrt[4]2, \pm i \sqrt[4]2$$, then $$Gal(\Bbb Q(\sqrt[4]2),\Bbb Q)=\{I,\sigma_1 \}$$; $$I(\sqrt[4]2)=\sqrt[4]2, \sigma_1(\sqrt[4]2)=-\sqrt[4]2$$
• @ruibarbosa That is because we are not talking about arbitray functions- we are discussing homomorphisms. Recall from linear algebra that a linear transformation is fully determined if we know where the elements of the vectors in some basis (any basis will do!). Similarly from group theory you should recall that a group homomorphism is fully determined when we know how it maps a set of generators. The same thing here. The element $\root4\of2$ generates the extension field $\Bbb{Q}(\root4\of2)$, so any $\Bbb{Q}$-homomorphism... – Jyrki Lahtonen Sep 24 '18 at 5:45