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I have a question about the proof that $\sqrt[12]{31}+\sqrt[31]{12}$ is irrational.

The proof should be:

Suppose $\sqrt[12]{31}+\sqrt[31]{12}$ is rational. Then

$\sqrt[12]{31} \in \mathbb{Q}(\sqrt[31]{12})$ and $\mathbb{Q}(\sqrt[12]{31}) \subset \mathbb{Q}(\sqrt[31]{12})$

As dim$_{\mathbb{Q}(\sqrt[12]{31})}=12$ over $\mathbb{Q}$, $\mathbb{Q}(\sqrt[12]{31})$ can't be a subset of $\mathbb{Q}(\sqrt[31]{12})$, because dim$_{\mathbb{Q}(\sqrt[31]{12})}=31$ over $\mathbb{Q}$.

Is that accurate?

I don't understand why $\sqrt[12]{31} \in \mathbb{Q}(\sqrt[31]{12})$, if $\sqrt[12]{31}+\sqrt[31]{12}$ is rational and how to see the conclusion that the dimension dim$_{\mathbb{Q}(\sqrt[12]{31})}=12$.

Gerturter
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    Under the assumption that $\sqrt[12]{31}+\sqrt[31]{12}=r$, for some rational $r$, then $\sqrt[12]{31}=r-\sqrt[31]{12}\in\mathbb{Q}(\sqrt[31]{12})$, since it is the difference of two elements $r$ and $\sqrt[31]{12}$ of the field $\mathbb{Q}(\sqrt[31]{12})$. – Atlantic Dec 17 '19 at 18:08
  • Is your question not answered in the linked thread? If not, please comment. – Jyrki Lahtonen Dec 17 '19 at 18:12
  • It's $\sqrt[31]{12} +\sqrt[12]{31}$ there and not $\sqrt[12]{31} +\sqrt[31]{12}$. Is it right as above then or is it $\sqrt[31]{12} \in \mathbb{Q}(\sqrt[12]{31})$ instead? Or doesn't it matter? Also, my questions weren't answered there. – Gerturter Dec 17 '19 at 18:16
  • $\sqrt[31]{12}+\sqrt[12]{31}$ and $\sqrt[12]{31}+\sqrt[31]{12}$ represent the same number. – Atlantic Dec 17 '19 at 18:27
  • So it's okay to say that $\sqrt[12]{31} \in \mathbb{Q}(\sqrt[31]{12})$? Or does it have to be $\sqrt[31]{12} \in \mathbb{Q}(\sqrt[12]{31})$? – Gerturter Dec 17 '19 at 18:34
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    Sorry, I dozed off for a while. Either of those is ok. The field $K_1=\Bbb{Q}(\root{12}\of{31})$ is a degree twelve extension of $\Bbb{Q}$. The field $K_2=\Bbb{Q}(\root{31}\of{12})$ is a degree thirty-one extension. Because $12\nmid31$ and $31\nmid 12$ neither can be a subfield of the other (by the tower law). Therefore it is impossible for $K_2$ to contain $\root{12}\of{31}$ and for $K_1$ to contain $\root{31}\of{12}$. – Jyrki Lahtonen Dec 17 '19 at 19:18

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