# Showing a polynomial is irreducible over an extension field.

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.

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?
• So it will have a factor of the form $(x-r)$ where $r=i , \sqrt2$ ? Mar 4, 2014 at 16:58
• @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. Mar 4, 2014 at 17:00