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Show that the number $ 2^{\frac{1}{5}} + 5^{\frac{1}{2}} $ has degree 10 as an algebraic number over $ \mathbb{Q} $.

Honestly I have found a lot of similar questions on this website, but most of the solutions involved finding the actual polynomial of which the number is a root of. I don't think it would be the best to look for a degree 10 polynomial in this case.

I saw some other hints involving finding the degree of some extension field, but I did not really understand those since my field theory knowledge is rusty at the moment. Can someone help me with a more detailed explanation, I would really appreciate it. Thanks.

irrational
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I guess it will be something like this: You know that over $\mathbb Q$, the extension $\mathbb Q(\sqrt[5]2+\sqrt5)$ and the number $\alpha:=\sqrt[5]2+\sqrt5$ have the same degree. But also, $\alpha$ is contained in the extension $\mathbb Q(\sqrt5,\sqrt[5]2)$ over $\mathbb Q$, which has degree 10. If you can show that the former extension is not just contained in the latter, but is in fact equal to it, then their degrees are equal and you are done.

To show this, you have to show that both generators of the latter extension are generated by the generator of the former extension (since then it is also a generator of the latter). That is: show that $\sqrt5$ and $\sqrt[5]2$ can be written as polynomials in $\alpha$.

Vercassivelaunos
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  • Thank you. I was able to show that $\mathbb Q(\sqrt[5]2+\sqrt5) = \mathbb Q(\sqrt[5]2, \sqrt5) $. However, I don't know why it is clear that $ \mathbb Q(\sqrt[5]2, \sqrt5) $ has degree 10. – irrational Sep 18 '22 at 07:01
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    @sneezel Note that $[\mathbb Q (\sqrt[5]{2}):\mathbb Q] = 5$, and that $[\mathbb Q (\sqrt{5}):\mathbb Q] = 2$. By multiplicity of dergees in extension towers, $[\mathbb Q (\sqrt[5]{2}, \sqrt{5}):\mathbb Q]$ is divisible by both 2 and 5, so it is divisible by 10 and thus is at least 10. It is at most 10 because clearly ${1,\sqrt[5]{2},(\sqrt[5]{2})^2,(\sqrt[5]{2})^3,(\sqrt[5]{2})^4,\sqrt{5},\sqrt{5}\sqrt[5]{2},\sqrt{5}(\sqrt[5]{2})^2,\sqrt{5}(\sqrt[5]{2})^3,\sqrt{5}(\sqrt[5]{2})^4 }$ spans it linearly. – Mor A. Sep 18 '22 at 07:09