Check that $\sqrt{7}(\sqrt[3]{5} - \sqrt[5]{3})$ is not rational

Question

How to prove that $\sqrt{7}(\sqrt[3]{5} - \sqrt[5]{3})$ is not rational.

I will appreciate any proof, but I had such exercise during lecture in field theory.

Thanks.

Answer 1

Suppose $\alpha = \sqrt{7}(\sqrt[3]{5}-\sqrt[5]{3})$ is rational. Then $\mathbb{Q}(\sqrt{7})=\mathbb{Q}(\sqrt{7}\alpha)=\mathbb{Q}(\sqrt[3]{5}-\sqrt[5]{3})$. This extension is then degree $2$ over $\mathbb{Q}$. However, it is also a subextension of $\mathbb{Q}(\sqrt[3]{5},\sqrt[5]{3})$ which has degree $15$ over $\mathbb{Q}$*, contradicting the tower law.

*Can you prove this?

Answer 2

Let $b=\sqrt[3]5-\sqrt[3]3$. Using a computer program, you can obtain:

$$b^2=\sqrt[3]{25}-2\sqrt[3]{15}+\sqrt[3]9$$ $$b^8=-563\sqrt[3]{25}+778\sqrt[3]{15}-131\sqrt[3]9$$ $$b^{14}=276901\sqrt[3]{25}-271574\sqrt[3]{15}-67259\sqrt[3]9$$

Inverting the matrix you can find that $$87903696b^2+406092b^8+516b^{14}=2154816\sqrt[3]{25}$$

This proves that $b^2$ is irrational and so is $7b^2$ and hence, $\sqrt 7b$.

Answer 3

Hint: If $\sqrt{7}(\sqrt[3]{5} - \sqrt[5]{3})$ is rational, then so is $\sqrt{7}(\sqrt[3]{5} + \sqrt[5]{3})$.