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Let $f:=X^5-9 \in \mathbb{Q}[X]$. Show: $f$ is irreducible. So I can't use the criterion, that f has no roots in $\mathbb{Q}[X]$, beacuse $\deg(f)>3$. Eisenstein is also not possible, because the only divider of $9$ is $3$. So I tried the reduction criterion, which doesn't really help me because $\deg(f)$ is still $5$. And in $\mathbb{Z}/7$ for example there are no prim elements, So I can't use Eisenstein. Is there another way I don't see?

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Tobi92sr
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    Well, it's irreducible $\pmod {11}$ if you want to go that route. Otherwise...it clearly has no rational roots so you just need to eliminate quadratic factors. – lulu Apr 01 '18 at 16:59
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    Alternative;: Apply Eisenstein to $(x-1)^5-9$. – lulu Apr 01 '18 at 17:00
  • Eisenstein can also be used to determine if $p(aX+b)$ is irreducible for any $p$ which will then imply that $p$ is irreducible as well. There are automorphisms of $\mathbb{Q}$ that drag $X$ to $aX+b$. And autmorphisms permute irreducible elements. –  Apr 01 '18 at 17:05
  • @lulu I don't see what's the point to apply Eisenstein on $(x-1)^5-9$. How does that help me with $f$? – Tobi92sr Apr 01 '18 at 17:18
  • Knowing that $f(x)$ is irreducible is equivalent to knowing that $f(x-1)$ is irreducible. – lulu Apr 01 '18 at 17:21

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The roots are $\sqrt[5]9\left(\cos(\frac {2k\pi}5+i\sin(\frac {2k \pi}5)\right)$ for $k \in [0,4]$ The trig functions have known values that involve $\sqrt 5$. Any product of two of them will still have the $\sqrt[5]{81}$ factor out front, so will not have rational coefficients.

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