Showing that $\mathbb{Q}(\sqrt[3]{5}, \sqrt[3]{7}) = \mathbb{Q}(\sqrt[3]{5}+\sqrt[3]{7})$

Asked Jan 03 '20 at 16:17

Active Jan 03 '20 at 19:00

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How can I show that $\mathbb{Q}(\sqrt[3]{5}, \sqrt[3]{7}) = \mathbb{Q}(\sqrt[3]{5}+\sqrt[3]{7})$?

I tried:

$\sqrt[3]{5}+\sqrt[3]{7} \in \mathbb{Q}(\sqrt[3]{5}, \sqrt[3]{7}) \Rightarrow \mathbb{Q}(\sqrt[3]{5}+\sqrt[3]{7}) \subset \mathbb{Q}(\sqrt[3]{5}, \sqrt[3]{7})$

For the other direction I get:

$(\sqrt[3]{5}+\sqrt[3]{7})^3=12+(\sqrt[3]{5})^2\sqrt[3]{7}+2(\sqrt[3]{5})^2\sqrt[3]{7}+2\sqrt[3]{5}(\sqrt[3]{7})^2+(\sqrt[3]{7})^2\sqrt[3]{5} \in \mathbb{Q}(\sqrt[3]{5}+\sqrt[3]{7})$

Now I don't know what to do next.

How to continue to show that $\mathbb{Q}(\sqrt[3]{5},\sqrt[3]{7}) \subset \mathbb{Q}(\sqrt[3]{5} + \sqrt[3]{7})$?

edited Jan 03 '20 at 19:00

asked Jan 03 '20 at 16:17

Gerturter

This holds the same way as for square roots in general, see this post. See also this post. – Dietrich Burde Jan 03 '20 at 17:43
I don't see how to transform this to get the inclusion here. My idea was to write the equation in terms of $\sqrt[3]{5}$, but it doesn't work properly. – Gerturter Jan 04 '20 at 11:57

Showing that $\mathbb{Q}(\sqrt[3]{5}, \sqrt[3]{7}) = \mathbb{Q}(\sqrt[3]{5}+\sqrt[3]{7})$

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