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 $\mathbb{Z}_4$

  • 5
    $\begingroup$ $\mathbb Q(\sqrt[4]{2})/\mathbb Q$ is not Galois. $\endgroup$ – Wojowu Sep 22 '18 at 20:05

roots of $x^4-2=0$ is $\pm \sqrt[4]2, \pm i \sqrt[4]2$, then $Gal(\Bbb Q(\sqrt[4]2),\Bbb Q)=\{I,\sigma_1 \}$; $I(\sqrt[4]2)=\sqrt[4]2, \sigma_1(\sqrt[4]2)=-\sqrt[4]2$

  • $\begingroup$ What I can not understand is why Q-automorphism is defined by an element only, it would be like saying that f is a function and f (2) = 3 and automatically achieve all the values of f $\endgroup$ – rui barbosa Sep 22 '18 at 20:51
  • $\begingroup$ @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... $\endgroup$ – Jyrki Lahtonen Sep 24 '18 at 5:45

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