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

The natural candidate is $x^5-2$. But the problem is to prove it is irreducible. In theory, I can write down the roots explicitly and by squaring multiple times somehow show that no root lies in the field. But even then there may be a quadratic factor, and using the method of undetermined coefficients is way too long. Can I prove irreducibility easier?

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    $\begingroup$ If you already know that $X^5 - 2$ is irreducible in $\mathbb{Q}[X]$, it suffices to note that $2$ and $5$ are coprime. $\endgroup$ – Daniel Fischer Jul 15 '18 at 21:04
  • $\begingroup$ @DanielFischer Why is it sufficient? $\endgroup$ – user437309 Jul 15 '18 at 21:13
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    $\begingroup$ @user437307 We know that $[Q(\sqrt{3},\sqrt[5]{2}):Q(\sqrt{3})]*[Q(\sqrt{3}):Q]=[Q(\sqrt{3},\sqrt[5]{2}):Q]$ and $[Q(\sqrt{3},\sqrt[5]{2}):Q(\sqrt[5]{2})]*[Q(\sqrt[5]{2}):Q]=[Q(\sqrt{3},\sqrt[5]{2}):Q]$ Now use the fact that $2$ and $5$ are coprime $\endgroup$ – Sorfosh Jul 15 '18 at 21:54
  • $\begingroup$ @Sorfosh Do you mean I should use it to conclude that since $5k=2n$, $2n$ is divisible by $5$, so $n$ is divisible by $5$. And since $n\le 5$ and it's divisible by $5$, $n=5$? Here $n=[\mathbb Q(\sqrt[5]{2},\sqrt{3}):\mathbb Q(\sqrt 3)]$. But I'm not sure how this is related to irreducibility. $\endgroup$ – user437309 Jul 15 '18 at 22:28
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    $\begingroup$ @user437307 Yes, the idea is to use it to conclude $n = 5$. And since $n$ is the degree of the minimal polynomial of $\sqrt[5]{2}$ over $\mathbb{Q}(\sqrt{3})$ it follows that $X^5 - 2$ is that minimal polynomial, hence irreducible. $\endgroup$ – Daniel Fischer Jul 16 '18 at 9:45

$\Bbb Z[\sqrt3]$ is UFD, $a^2-3b^2=2$ has no solution (since it has no solution mod $3$), so $2$ is prime in $\Bbb Z[\sqrt3]$. Now use Eisenstein.


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