How to compute the degree of $\mathbb{Q}[\sqrt[3]{2}+\sqrt[5]{2}]$ over $\mathbb{Q}$? Here is a 2-part problem:


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*Find the number of ring homomorphisms $\mathbb Q[\sqrt[3]{2},\sqrt[5]{2}]\to \mathbb C$.

*Prove that the degree of $\mathbb{Q}[\sqrt[3]{2}+\sqrt[5]{2}]$ over $\mathbb Q$ is $15$.


Am I supposed to deduce the second part from the first? I don't see how they are related. Also, I don't know even what fact I should use to solve the first part. Any hints?
 A: You can prove that $\mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2}) = \mathbb{Q}(\sqrt[3]{2},\sqrt[5]{2})$. In fact this follows by the Primitive Element Theorem, as $\sqrt[3]{2} + \sqrt[5]{2}$ satisfies the condition to be a primitive element of $\mathbb{Q}(\sqrt[3]{2},\sqrt[5]{2})$. Nevertheless here's the explicit construction below:
Let $f(x) = x^3-2$. Now consider $g(x) = f(\sqrt[3]{2} + \sqrt[5]{2}-x) \in \mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2})[x]$. Obviously we have that $\sqrt[5]{2}$ is one of the roots ,while it's not hard to conclude that the other ones are $\sqrt[3]{2} + \sqrt[5]{2} - \zeta_3\sqrt[3]{2}$ and $\sqrt[3]{2} + \sqrt[5]{2} - \zeta_3^2\sqrt[3]{2}$, where $\zeta_3$ is the third root of unity.
On the other side $\sqrt[5]{2}$ satisfies $x^5-2$, whose roots are $\sqrt[5]{2}, \zeta_5\sqrt[5]{2}, \zeta_5^2\sqrt[5]{2}, \zeta_5^3\sqrt[5]{2}, \zeta_5^4\sqrt[5]{2}$. Now the minimal polynomial of $\sqrt[5]{2}$ over $\mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2})$ must divide both $g(x)$ and $x^5-2$ and so their greatest common divisor. But as their only common root is $\sqrt[5]{2}$, we get that it divides $x-\sqrt[5]{2}$ and as the polynomial is linear it must be itself. Hence $\min(\sqrt[5]{2},\mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2})) = x-\sqrt[5]{2}$. This yields that $\sqrt[5]{2} \in \mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2})$. Now it's not hard to conlcude that $\sqrt[3]{2} \in \mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2})$ and therefore $\mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2}) = \mathbb{Q}(\sqrt[3]{2},\sqrt[5]{2})$
As mentioned in the comments from here you can conclude that $\left[\mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2}):\mathbb{Q}\right]=15 $
