Let $\mathbb{Q} \subset \mathbb{Q}(\sqrt[5]{3}, \sqrt{2}i)$ be a field extension. I computed that $[\mathbb{Q}(\sqrt[5]{3}, \sqrt{2}i) : \mathbb{Q}] = 10$.
I now want to show that $\mathbb{Q}(\sqrt[5]{3} \sqrt{2}i) = \mathbb{Q}(\sqrt[5]{3}, \sqrt{2}i)$.
The first inclusion seems trivial to me. Since $\mathbb{Q}(\sqrt[5]{3}, \sqrt{2}i)$ is the smallest subfield of $\mathbb{C}$ that contains $\mathbb{Q}, \sqrt[5]{3}$ and $\sqrt{2}i$, it must also contain their product $\sqrt[5]{3}\sqrt{2}i$. This shows that $\mathbb{Q}(\sqrt[5]{3} \sqrt{2}i) \subseteq \mathbb{Q}(\sqrt[5]{3} ,\sqrt{2}i)$ (is this argument correct)?
But the other inclusion is harder. I want to take the product $\sqrt[5]{3}\sqrt{2}i$ and then by applying field operations on this, show that this field must also contain the separate factors. Obviously it must also contain the $5$th power, that is $(\sqrt[5]{3}\sqrt{2}i)^5 = 3 \sqrt[5]{2} i.$ Now, can I conclude from this that it must also contain $\sqrt{2}i$? and how to show it contains $\sqrt[5]{3}$?
Also, Let's say I would want to find the minimal polynomial of $\sqrt[5]{3}\sqrt{2}i$ over $\mathbb{Q}$. I let $x = \sqrt[5]{3}\sqrt{2}i$. Then $x^5 = 3 \sqrt{32}i$ and so $x^{10} + 288 = 0$. This is the minimal polynomial I think. I want to prove this, by proving it is irreducible over $\mathbb{Q}$. I wanted to show this using Eisenstein criterion, but couldn't find the right prime number. So how would I prove it is the minimal polynomial? Should I assume it is not, and then derive a contradiction somehow?
Help/suggestions are appreciated.
To show that $\Bbb{Q}(\sqrt[5]{3},\sqrt{2}i)$ is a subset of $\Bbb{Q}(\sqrt[5]{3}\sqrt{2}i)$ it suffices to show that both $\sqrt[5]{3}$ and $\sqrt{2}i$ can be formed as a rational expression involving only rational numbers and $\sqrt[5]{3}\sqrt{2}i$.
Well, $$ \frac{(\sqrt[5]{3}\sqrt{2}i)^5}{12} = \sqrt{2}i \\ \frac{(\sqrt[5]{3}\sqrt{2}i)^6}{-72} = \sqrt[5]{3} $$