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Show that the polynomial

$$ x^3 - 3$$ Is irreducible over

$$ Q(i, \sqrt2 ) $$

I'm a little stuck as I don't think I can use Eisenstein's criterion as we're not over the rationals. Also I know that the roots of the polynomial are $ \sqrt[3] 3 $ followed by $\omega \sqrt[3] 3 $ and $\omega^2 \sqrt[3] 3 $ However I don't really know how to use this imformation to prove the polynomial is irreducible.

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If a polynomial of degree 3 or less is reducible, it means it has a linear factor in $\mathbb Q(i,\sqrt2)[x]$. Does this help you?

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  • $\begingroup$ So it will have a factor of the form $ (x-r) $ where $ r=i , \sqrt2 $ ? $\endgroup$
    – Padraic
    Mar 4, 2014 at 16:58
  • $\begingroup$ @Padraic If you assume it's reducible, then $r\in\mathbb Q(i,\sqrt2)$. So if you prove that $\sqrt[3]3,\omega\sqrt[3]3,\omega^2\sqrt[3]3\not\in \mathbb Q(i,\sqrt2)$, you are done. $\endgroup$ Mar 4, 2014 at 17:00
  • $\begingroup$ You beauty! That I can do. Thank you! $\endgroup$
    – Padraic
    Mar 4, 2014 at 17:06

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