Proving $x^{nm}-c$ irreducible $\leftrightarrow x^n-c,x^m-c$ irreducible

Question

I've proven the implication $\rightarrow$ so far but I'm out of ideas for $\leftarrow$. I tried to find an approach using Eisenstein's criterion but haven't had much luck.

$x^n-c,x^m-c,x^{nm}-c \in K[x]$ for arbitrary Field $K$

Answer 1

With the condition $gcd(m,n)=1$ we get following proof. Let $d\in\overline K$ such that $d^{mn}-c=0$. Then we have the field $L_{mn}:=K(d)$ and subfields $L_m:=K(d^n); L_n:=K(d^m)$. Here we get the equivalence:

$X^{mn}-c$ is irreducible iff $[L_m\cdot L_n :K]= [L_{mn}:K]=mn$ iff $[L_m:K]=m \land [L_n:K]=n$ iff $X^{m}-c$, $X^{n}-c$ are irreducible.

counterexample for gcd(m,n)>1: Choose $m=n=2$ and $K=\mathbb{R}$. Then $X^2+1$ is irreducible but $X^4+1$ is reducible.