TITLE: Show that $x^{3}-3$ irreducible over $\mathbb{Q}(\sqrt{-3})$
QUESTION [9 upvotes]: Is there a slick way to show that $x^{3}-3$ is irreducible over $F= \mathbb{Q}(\sqrt{-3})$? What I did seems kind of convoluted (showing directly that there is no root in F).
Thanks

REPLY [2 votes]: Hint: $f$ in $K[x]$ irreducible of degree $m$, $K \subset L$ of degree $n$, $(m,n)=1$, implies $f$ stays irreducible in $L[x]$. 
$\bf{Added:}$
Let $\alpha$ a root of $f$.   We have 
$$[L(\alpha) \colon K]= [L(\alpha) \colon L]\cdot [L\colon K] \le m \cdot n$$
But   $[L(\alpha) \colon K]$ is divisible by both $[K(\alpha) \colon K]=m$ and  $[L\colon K]=n$ and so by $m\cdot n$. Therefore, we must have equality, $[L(\alpha) \colon L] = m$, and thus $f$ is irreducible in $L[x]$.