Exercise 7.1.10

Note: as a verification, note that the two parts of the definition of the Galois closure are satisfied.

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The extension $\mathbb{Q}\subset \mathbb{Q}(\omega ,\sqrt[3]{2})$ is a Galois extension, since $\mathbb{Q}(\omega ,\sqrt[3]{2})$ is the splitting field of the separable polynomial $x^3-2$.

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Let $M\supset \mathbb{Q}(\sqrt[3]{2})$ an extension such that $M$ is a Galois extension of $\mathbb{Q}$. As $\sqrt[3]{2}\in M$ and as $\mathbb{Q}\subset M$ is normal, $x^3-2$ splits completely over $M$: $$x^3-2=(x-\alpha )(x-\beta )(x-\gamma ),\alpha ,\beta ,\gamma \in M,$$

where $\alpha =\sqrt[3]{2}\in M$.

${(\beta ∕\alpha )}^3=1$, thus ${\omega }^{\prime }=\beta ∕\alpha$ is a cube root of unity in $M$, with ${\omega }^{\prime }\ne 1$ since $x^3-2$ is separable. So ${\omega }^{\prime }$ is a root in $M$ of $(x^3-1)∕(x-1)=x^2+x+1$.

$x^2+x+1$ has degree 2 and has no real root, so has no root in $\mathbb{Q}(\sqrt[3]{2})$, thus $x^2+x+1$ is irreducible over $\mathbb{Q}(\sqrt[3]{2})$. Therefore $\mathbb{Q}(\omega ,\sqrt[3]{2})\subset ℂ$ and $\mathbb{Q}({\omega }^{\prime },\sqrt[3]{2})\subset M$ are two splitting fields of $x^2+x+1$ over $\mathbb{Q}(\sqrt[3]{2})$. Therefore there exists an isomorphism $\mathbb{Q}(\omega ,\sqrt[3]{2})\simeq \mathbb{Q}({\omega }^{\prime },\sqrt[3]{2})$ which is the identity on $\mathbb{Q}(\sqrt[3]{2})$, and which sends $\omega$ on ${\omega }^{\prime }$, so there exists an embedding of $\mathbb{Q}(\omega ,\sqrt[3]{2})$ in $M$ which is the identity on $\mathbb{Q}(\sqrt[3]{2})$.