$\mathbb{Q}(\sqrt[3]{2}, \zeta_{9})$ Galois group

Question

How do I calculate the degree of $\mathbb{Q}(\sqrt[3]{2}, \zeta_{9})$ over $\mathbb{Q}$. Should it be 18, as $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$, and $[\mathbb{Q}(\zeta_{9}):\mathbb{Q}] = 6$?

However $(\sqrt[3]{2})^{3} \in \mathbb{Q}(\zeta_{9})$, how this affect the calculation?

Thanks

Answer 1

$\newcommand{\Q}{\mathbb{Q}}\newcommand{\Size}[1]{\lvert #1 \rvert}$$\sqrt[2]{2}$ has minimal polynomial  $f = x^{3}-2$ over $\Q$. You have to show that $f$ is also the minimal polynomial over $F = \Q(\zeta_{9})$, that is, that $f$ is irreducible in $F[x]$, and since $f$ has degree $3$, it is enough to show that $f$ has no roots in $F$. From this it will follow that $$ \Size{\Q(\sqrt[3]{2}, \zeta_{9}) : \Q} = \Size{F(\sqrt[3]{2}) : F} \cdot \Size{F : \Q} = 3 \cdot 6 = 18. $$

Since $F/\Q$ is Galois, if it contains a root of the polynomial $f \in \Q[x]$, which is irreducible over $\Q$, then it contains all the roots of $f$, and thus it contains a splitting field $E$ of $f$ over $\Q$.

But the Galois group of $E/\Q$ is nonabelian, while that of $F/\Q$ is abelian, so this rules out the possibility that $E \subseteq F$.

Alternatively, since the Galois group of $F/\Q$ is abelian of order $6$, there is only one intermediate field $\Q \subset L  \subset F$ with $\Size{L : \Q} = 3$, and so in $F$ you cannot find the three distinct extensions $\Q(\sqrt[3]{2}), \Q(\omega \sqrt[3]{2}), \Q(\omega^{2} \sqrt[3]{2})$ of degree $3$ over $\Q$. Here $\omega$ is a primitive third root of unity, and $\sqrt[3]{2}, \omega \sqrt[3]{2}, \omega^{2} \sqrt[3]{2}$ are the three roots of $f$.

Answer 2

How about doing it the other way around:

Certainly $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]$ is $3$, so we aim to compute $n=[L:K]=|Gal(L/K)|$, where $L = \mathbb{Q}(\sqrt[3]{2},\zeta_9)$ and $K=\mathbb{Q}(\sqrt[3]{2})$. The extension is cyclotomic, so certainly is Galois.

The minimal polynomial of $\zeta_9/K$ divides $\varphi_9(X)$, so we know that $n \leq6$. Now note that $L$ contains the subextension $\mathbb{Q}(\zeta_9)/\mathbb{Q}$ of degree $6$, so $3$ divides $n$. But since $K$ is real and $L/K$ is Galois, complex conjugation gives an order $2$ element inside $Gal(L/K)$, so $2$ divides $n$. Thus $n = 6$.

Answer 3

If $X^3-2$ is not irreducible over $\mathbb{Q}(\zeta_9)$, it must have a root $\alpha \in \mathbb{Q}(\zeta_9)$ (this holds only because the degree of the polynomial is $3$). Now $\mathbb{Q}(\zeta_9)/\mathbb{Q}$ has a subextension $\mathbb{Q}(\alpha)/\mathbb{Q}$ which is not normal, since we know that the splitting field of $X^3-2$ over $\mathbb{Q}$ has degree $6$. This is not possible, because the Galois group of $\mathbb{Q}(\zeta_9)/\mathbb{Q}$ is abelian (generally, $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \simeq (\mathbb{Z}/n\mathbb{Z})^{\times}$), therefore every subgroup is normal, and hence (using the Galois correspondence) every subextension of $\mathbb{Q}(\zeta_9)/\mathbb{Q}$ is normal.