Exercise 7.2.4

Proof.

Here $\sigma ,\tau$ are the elements of $G=\mathrm{Gal}(L∕\mathbb{Q})$, where $L=\mathbb{Q}(\omega ,\sqrt[3]{2})$, determined by

We show that the map $K\mapsto \mathrm{Gal}(L∕\mathbb{Q})$ applies the left diagram on the right diagram, the inclusion arrows are opposite by Exercise 3.

$\text{∙}$ If $K=L$, $\mathrm{Gal}(L∕L)=\{e\}$, and if $K=\mathbb{Q}$, $\mathrm{Gal}(L∕K)=\mathrm{Gal}(L∕\mathbb{Q})=G$.

$\text{∙}$ If $K=\mathbb{Q}(\omega )$, note that $\sigma (\omega )=\omega$, thus $\sigma (\alpha )=\alpha$ for all $\alpha \in \mathbb{Q}(\omega )$, so $\sigma \in \mathrm{Gal}(L∕\mathbb{Q}(\omega ))$. Therefore

Moreover, as $\mathbb{Q}\subset L$ is a Galois extension, then $K\subset L$ is also Galois for all intermediate fields $K$, therefore $|\mathrm{Gal}(L∕\mathbb{Q}(\omega ))|=[L:\mathbb{Q}(\omega )]=3$. Consequently

$\text{∙}$ If $K=\mathbb{Q}(\sqrt[3]{2})$, then $[L:K]=2=|\mathrm{Gal}(L∕K)|$, and $\tau \in \mathrm{Gal}(L∕K)$, thus

$\text{∙}$ If $K=\mathbb{Q}(\omega \sqrt[3]{2})$, with the same reasoning, as ${\sigma }^2\tau$ has order 2 and $({\sigma }^2\tau )(\omega \sqrt[3]{2})={\sigma }^2({\omega }^2\sqrt[3]{2})={\omega }^4\sqrt[3]{2}=\omega \sqrt[3]{2}$,

$\text{∙}$ If $K=\mathbb{Q}(\omega \sqrt[3]{2})$, we have a similar result, by exchanging $\omega$ with $\bar{\omega }={\omega }^2$:

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