Let $u\in \mathbb{C}$ such that $u^3+2=0$. Is the extension $\mathbb{Q}(u):\mathbb{Q}$ Galois?

Question

Let $u\in \mathbb{C}$ such that $u^3+2=0$, I'm asked to determine whether the extension $\mathbb{Q}(u):\mathbb{Q}$ is Galois.

Now, the polynomial $p(t)=t^3+2\in \mathbb{Q}[t]$ can be decomposed into linear factors in $\mathbb{C}$ as $$(t+\sqrt[3]{2})(t+\sqrt[3]{2}e^{2\pi /3})(t+\sqrt[3]{2}e^{4\pi /3})$$ where $e^{2\pi /3}=\frac{-1+i\sqrt{3}}{2}$. The splitting field of $p$ over $\mathbb{Q}$ is then $\mathbb{Q} (\sqrt[3]{2}, i\sqrt{3})$. The issue is that $\mathbb{Q}(u)$ is probably not equal to $\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})$, and I'm having issue finding another approach besides (dis)proving the extension is a splitting field. I would appreciate any help.

Answer 1

One of the extensions has order $3$ ($u$ has minimal polynomial of degree $3$ over $\Bbb Q$) and the other has order $6$ (extending by $\sqrt[3]2$ has order $3$, and you are still within the reals, so extending by $i\sqrt3$ is a proper extension), so they are clearly not equal.

Answer 2

$\mathbb Q(u)/\mathbb Q$ is not Galois. We can prove this using the following lemma

Lemma $L/K$ is Galois if and only if $[L:K] = |Aut(L/K)|$

On the LHS we have $[\mathbb Q(u) : \mathbb Q] = 3$ because our field is spanned by the basis $\{1, u, u^2\}$.

On the RHS we actually have 1, the only automorphism of our field is the identity*.

This proves that the extension is not Galois.

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How do we show that there is only the identity automorphism?

Suppose $\sigma$ was a non identity automorphism, then $\sigma u = u'$ would be another root of the polynomial. $X^3 - 2$ would divide by $(X - u)(X - u')$ to give us a linear factor containing the 3rd root. This would imply that we have all 3 roots in our field.

In that case we could do $u/u'$ (or $u''/u'$ or similar) to retrieve the quadratic irrational $\zeta_3$. This would give us $2 | [\mathbb Q(u) : \mathbb Q]$ which is not contradictory.