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Find the minimal polynomial of $\sqrt[3]{5}+\sqrt{2}$ over $\mathbb{Q}(\sqrt[3]{5})$

Attempt:

Let $u:=\sqrt[3]{5}+\sqrt{2}\\ u-\sqrt[3]{5}=\sqrt 2\\ (u-\sqrt[3]{5})^2=2\\ \boxed{u^2-2\sqrt[3]{5}u+5^{2/3}-2=0}$

Is this correct? the "over $\mathbb{Q}(\sqrt[3]{5})$" confuses me

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    $\begingroup$ Yes, this is correct. $\endgroup$ – José Carlos Santos May 29 '17 at 14:07
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Yes, it's correct. For simplicity, set $a=\sqrt[3]{5}$. Then $u=a+\sqrt{2}$ becomes $$ u-a=\sqrt{2} $$ so $$ u^2-2au+a^2=2 $$ and the element $u$ is a root of $X^2-2aX+a^2-2$ which is a polynomial with coefficients in $\mathbb{Q}(a)$.

However, you should prove also that $u=a+\sqrt{2}\notin\mathbb{Q}(a)$, which is equivalent to $\sqrt{2}\notin\mathbb{Q}(a)$.

This can be tackled by considering that $[\mathbb{Q}(a):\mathbb{Q}]=3$, so the field $\mathbb{Q}(a)$ cannot contain a quadratic extension of $\mathbb{Q}$.

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