Galois group of $\mathbb{Q}(\sqrt[3]{5}, e^{2\pi i/3})$ over $\mathbb{Q}$ and $\mathbb{Q}(e^{2\pi i/3})$

Question

I want to find $Gal(\mathbb{Q}(\sqrt[3]{5}, e^{2\pi i/3}) | \mathbb{Q})$ and $Gal(\mathbb{Q}(\sqrt[3]{5}, e^{2\pi i/3})| \mathbb{Q}(e^{2\pi i/3}))$

Let $w = e^{2\pi i/3}, \alpha = \sqrt[3]{5}, L = \mathbb{Q}(\sqrt[3]{5}, w)$.

Since $L$ is the splitting field of $f(x) = x^3 - 5$, it is a finite, normal, seperable extension, and therefore $|Gal(L|\mathbb{Q})| = [L:\mathbb{Q}] = [\mathbb{Q}(w, \alpha):\mathbb{Q}(w)][\mathbb{Q}(w):\mathbb{Q}]$.

The minimal polynomial for $w$ over $\mathbb{Q}$ is $x^3 - 1$, so $[\mathbb{Q}(w):\mathbb{Q}] = 3$

I thought that since the minimum polynomial for $\alpha$ over $\mathbb{Q}(w)$ is $x^3 - 5$, we would have $ [\mathbb{Q}(w, \alpha):\mathbb{Q}(w)] = 3 \implies |Gal(L|\mathbb{Q})| = 9$, but I have been told that $|Gal(L|\mathbb{Q})| = 6$.

Where is the problem in my reasoning above? Also, using the same logic, I found $Gal(L: \mathbb{Q}(w))$ to be $\{id, \sigma_1, \sigma_2\}$, where $\sigma_1(\alpha) = w\alpha, \sigma_2(\alpha) = w^2\alpha$. Is this wrong as well?

Answer 1

The polynomial $x^3-5$ actually gives you all of $\Bbb Q(\sqrt[3]5, e^{2\pi i/3})$ by itself. It is irreducible, by Eisenstein, so $[\Bbb Q(\sqrt[3]5),\Bbb Q]=3$. However, the polynomial factors over $\Bbb Q(\sqrt[3]5)$ as $$ (x-\sqrt[3]5)(x^2+\sqrt[3]5x+\sqrt[3]5^2) $$ And it is the second-degree factor on the right that adjoins $\sqrt[3]5e^{2\pi i/3}$ and thus $e^{2\pi i/3}$ to $\Bbb Q(\sqrt[3]5)$. Which means that $[Q(\sqrt[3]5, e^{2\pi i/3}), Q(\sqrt[3]5)]=2$ (or technically, by its own, it only means ${}\leq 2$, but the equality is easily proven as we move from a field contained in $\Bbb R$ to one that isn't).

This happens in general: given any irreducible degree-$n$ polynomial, if you extend your field by adjoining one root of that polynomial, the polynomial will now factor into one linear term (given by the root you just adjoined) and one term of degree $n-1$ (which might even be reducible over the extended field). So the next root you adjoin will have strictly smaller degree. The Galois extension given by a degree $n$ polynomial can therefore at most have degree $n!$.

Answer 2

Since $x^3-1 = (x-1)(x^2+x+1)$ and $e^{2\pi i/3} \neq 1$, the minimal polynomial of $e^{2\pi i/3}$ over $\mathbb Q$ is $x^2+x+1$.