A primitive element for the extension $\mathbb{Q}(\sqrt{7}, \sqrt[5]{2}) / \mathbb{Q}$ I'm trying to represent the field extension $\mathbb{Q}(\sqrt{7}, \sqrt[5]{2}) / \mathbb{Q}$ as a simple extension over $\mathbb{Q}$.
I know the degree of the extension is $10$ and a $\mathbb{Q}$-basis for $\mathbb{Q}(\sqrt{7}, \sqrt[5]{2})$ is $$\{ 1, 2^{\frac{1}{5}},2^{\frac{2}{5}},2^{\frac{3}{5}},2^{\frac{4}{5}},7^{\frac{1}{2}}, 7^\frac{1}{2}2^{\frac{1}{5}},7^\frac{1}{2}2^{\frac{2}{5}},7^\frac{1}{2}2^{\frac{3}{5}},7^\frac{1}{2}2^{\frac{4}{5}}\}$$
Normally I would find something that is in the basis that is also in another extension field and that would be my primitive element. However, because 7 and 2 are relatively prime, it is not like other simple extension fields I have found in other examples.  How would I start to find this, or is it already in its simplest form?
 A: Consider $\alpha =\sqrt{7} \sqrt[5]{2}\,$. It is clear that $\mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\sqrt{7}, \sqrt[5]{2})\,$. Now, $\alpha^5 = 2\cdot49\cdot\sqrt{7}$ and this tells you that $\sqrt{7}\in\mathbb{Q}(\alpha)\,$. But then $\sqrt[5]{2} = \frac{\alpha}{\sqrt{7}}$ is in $\mathbb{Q}(\alpha)$. This proves that $\mathbb{Q}(\sqrt{7}, \sqrt[5]{2}) \subseteq \mathbb{Q}(\alpha)$.
You have proved that $\mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt{7}, \sqrt[5]{2})$. 
You can get a little bit more out of this. As you have mentioned in your question, $[\mathbb{Q}(\sqrt{7}, \sqrt[5]{2}):\mathbb{Q}]=10$. The polynomial $f=X^{10} - 7^5\cdot2^2$ is in $\mathbb{Q}[x]$, is monic, has degree $10=[\mathbb{Q}(\sqrt{7}, \sqrt[5]{2}):\mathbb{Q}]=[\mathbb{Q}(\alpha):\mathbb{Q}]$, and $f(\alpha)=0$. So $f$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$ and that says $f$ is irreducible over $\mathbb{Q}$. Note that it's not immediate from Eisenstein or considerations for low degree polynomials that $X^{10} - 7^5\cdot2^2$ must be irreducible.
