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I am wondering how should I go about to prove that the polynomial $x^3-5$ is irreducible over $\Bbb Q(\sqrt2)?$ The reason why I do this is that I need to find a minimal polynomial of $\sqrt[3]{5}$ over the field $\Bbb Q(\sqrt2).$ Is it true that the degree of the minimal polynomial of $\sqrt[3]{5}$ must divide the degree of any polynomial that has $\sqrt[3]{5}$ as a root? If this is true, how to prove that ?

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A cubic polynomial is irreducible over a field if and only if it doesn't have a root in the field, so it's enough to show that $\mathbb{Q}(\sqrt{2})$ doesn't contain a cube root of $5$.

Since $x^3-5$ is irreducible over $\mathbb{Q}$, any extension of $\mathbb{Q}$ containing a cube root of $5$ has degree at least $3$ over $\mathbb{Q}$. So since $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$, $\mathbb{Q}(\sqrt{2})$ doesn't contain a cube root of $5$.

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  • $\begingroup$ Thanks for your answer. Is it true that the degree of the minimal polynomial of $\sqrt[3]{5}$ must divide the degree of any polynomial that has $\sqrt[3]{5}$ as a root? $\endgroup$
    – Y.X.
    Mar 26, 2017 at 2:20
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    $\begingroup$ No. For example, $x^4-5x$ has $\sqrt[3]{5}$ as a root, but $4$ isn't divisible by $3$. $\endgroup$ Mar 26, 2017 at 2:23
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How about the following argument: As pointed out by carmichael561, it is enough to check that the polynomial has no root. If it did, it would factor mod $p$ such that $2$ is a quadratic residue mod $p.$ The smallest prime with that property is $7.$ (OK, there is $2,$ also). Now, is $5$ a cube mod $7?$ The cubes are $0, 1, -1,$ so $5$ is not among them.

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