Show that $\Bbb Q(α)/\Bbb Q$ is not a normal field extension.

Question

Let $f=x^{6}-11$ in $\mathbb{Q}[x]$ and let $\alpha \in \mathbb{C}$ be a root of $f$.

Show that $\mathbb{Q}(\alpha)/\mathbb{Q}$ is not a normal field extension.

Attempt:

We have that $f=x^{6}-11$ is irreducible in $\mathbb{Q}[x]$ by a simple application of Eisenstein's criterion. Clearly $\alpha \in \mathbb{Q}(\alpha)$ so $f$ has a root in $\mathbb{Q}(\alpha)$.

Then to show that $\mathbb{Q}(\alpha)/\mathbb{Q}$ is not a normal field extension we must show that $f$ does not split over $\mathbb{Q}(\alpha)$. In other words, there is some root of $f$ that is not in $\mathbb{Q}(\alpha)$. As the roots of $f$ are precisely $\{\alpha, \alpha \omega, \alpha \omega^{2}, \alpha \omega^{3},\alpha \omega^{4},\alpha \omega^{5}\}$ where $\omega=e^{\frac{2\pi i}{6}}$, this is equivalent to saying that $\omega \notin \mathbb{Q}(\alpha)$.

At this point I'm not sure how to proceed and would appreciate a hint to push me in the right direction. I think it may be related to the degrees of the extensions, for example we have:

$[\mathbb{Q}(\alpha,\omega)/\mathbb{Q}(\alpha)]=1$ if $\omega \in \mathbb{Q}(\alpha)$

$[\mathbb{Q}(\omega)/\mathbb{Q}] = 2$ as $x^{2}-x+1$ is the minimal polynomial for $\omega$ in $\mathbb{Q}$

$[\mathbb{Q}(\alpha)/\mathbb{Q}]=6$ as $x^{6}-11$ is the minimal polynomial of $\alpha$ in $\mathbb{Q}$

$[\mathbb{Q}(\alpha,\omega)/\mathbb{Q}]=[\mathbb{Q}(\alpha,\omega)/\mathbb{Q}(\omega)][\mathbb{Q}(\omega)/\mathbb{Q}]=[\mathbb{Q}(\alpha,\omega)/\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha)/\mathbb{Q}]$

I think these pieces may combine to give a contradiction showing that $\omega \notin \mathbb{Q}(\alpha)$ but I can't quite get it.

Answer 1

Hint: Prove that $\mathbb{Q}(\alpha)\cong \mathbb{Q}[X]/(X^6-11)\cong \mathbb{Q}(\sqrt[6]{11})$. Conclude that $\mathbb{Q}(\alpha)$ contains exactly two roots of $X^6-11$ and so is not normal.

Answer 2

Note that if $\alpha$ is chosen to be a real root of $f$, then $\omega\notin \Bbb R$ gives $\omega\notin \Bbb Q(\alpha)$, which gives us $[\Bbb Q(\alpha, \omega):\Bbb Q]=12$. So you need a degree-12 extension of $\Bbb Q$ for $f$ to split.

Answer 3

If the extension were normal, it would contain all the 6-th roots of $11$, hence also a primive 6-th root $\zeta_6$ of $1$. But $[\mathbf Q(\zeta_6):\mathbf Q]=\phi(6)=3$

Warning. I made a stupid mistake (confusion between 6 and $11$). Shouldn't work when not well awake. The correct argument is that of @Arthur.