Alternative method : $\mathbb{Q}(\zeta_3+\sqrt[3]{7})=\mathbb{Q}(\zeta_3,\sqrt[3]{7})$

Question

Let $\zeta=\zeta_3$ be a third root of unity. I want to proof that $\mathbb{Q}(\zeta+\sqrt[3]{7})=\mathbb{Q}(\zeta,\sqrt[3]{7})$. One inclusion is clear that is :
$\mathbb{Q}(\zeta+\sqrt[3]{7})\subset\mathbb{Q}(\zeta,\sqrt[3]{7})$. For the other inclusion we can prove that either $\zeta$ or $\sqrt[3]{7}$ $\mathbb{Q}(\zeta+\sqrt[3]{7})=\mathbb{Q}(\zeta,\sqrt[3]{7})$. I want to use another approach which uses Galois theory. The field $\mathbb{Q}(\zeta,\sqrt[3]{7})$ is a splitting field of the polynomial $X^3-7\in \mathbb{Q}[X]$ and the extension $$\mathbb{Q}\subset\mathbb{Q}(\zeta,\sqrt[3]{7})$$ is a Galois extension with galois group G=Gal($\mathbb{Q}(\zeta,\sqrt[3]{7})/\mathbb{Q}$) which is isomorphic to $S_3$. The authorphisms in this group are :

1. identity

2. $\sigma: \sqrt[3]{7}\mapsto \sqrt[3]{7}, \zeta \mapsto \zeta^2$

3. $\sigma: \sqrt[3]{7}\mapsto \zeta\sqrt[3]{7}, \zeta
   \mapsto \zeta^2$

4. $\sigma: \sqrt[3]{7}\mapsto \zeta^2\sqrt[3]{7}, \zeta
   \mapsto \zeta^2$

5. $\sigma: \sqrt[3]{7}\mapsto \zeta^2\sqrt[3]{7}, \zeta
   \mapsto \zeta$

6. $\sigma: \sqrt[3]{7}\mapsto \zeta\sqrt[3]{7}, \zeta
   \mapsto \zeta$

We know that the zero's of the minimal polynomial of $\alpha=\zeta+\sqrt[3]{7}$ are its conjugates. The action of all these automorphisms on $\alpha$ give 6 different roots.
$$S=\{\zeta+\sqrt[3]{7}, \zeta^2+\sqrt[3]{7},\zeta+\zeta\sqrt[3]{7},\zeta+\zeta^2\sqrt[3]{7},\zeta^2+\zeta\sqrt[3]{7},\zeta^2+\zeta^2\sqrt[3]{7}\}$$
Does this mean that $f^\alpha_\mathbb{Q}=\prod_{\sigma\in G}(X-\sigma(\alpha))$?

We know that the minimal polynomial can not have degree greater than 6 since $\mathbb{Q}(\alpha)\subset\mathbb{Q}(\zeta,\sqrt[3]{7})$

Does having six different values for $\sigma(\alpha)$ guarantee that the product $\prod_{\sigma\in G}(X-\sigma(\alpha))$ is irreducible? Does this method also work generally?

Answer 1

This is totally fine, but you have to check $\alpha$ indeed have six distinct Galois conjugates. If you have done so, then you don't have to form the minimal polynomial. Since the only element in $\text{Gal}(\mathbb Q(\zeta, \sqrt[3]{7})\mid \mathbb Q)$ that fixes $\zeta+\sqrt 3$ is the identity element, that is $\mathbb Q(\zeta+\sqrt 3)$ is only fixed (pointwise) by the identity element, by fundamental theorem of Galois theory, $\mathbb Q(\zeta+\sqrt 3)=\mathbb Q(\zeta, \sqrt[3]{7})^{\text{id}}=\mathbb Q(\zeta, \sqrt[3]{7})$

Here we showed if $a$ is fixed only by $\text{id}\in G$, then $a$ is a primitive element. To show $\prod_{\sigma\in G}(x-\sigma(\alpha))\in\mathbb Q[x]$, we use the fact that the if $a$ is fixed by all of $G$, then $a\in\mathbb Q$.