TITLE: $X^n-Y^m$ is irreducible in $\Bbb{C}[X,Y]$ iff $\gcd(n,m)=1$
QUESTION [22 upvotes]: I am trying to show that $X^n-Y^m$ is irreducible in $\Bbb{C}[X,Y]$ iff $\gcd(n,m)=1$ where $n,m$ are positive integers. 

I showed that if $\gcd(n,m)$ is not $1$, then $X^n-Y^m$ is reducible. How to show the other direction. Please help.

REPLY [4 votes]: Let $K$ be any field, e.g. $K=\Bbb C$, let $m,n \in \Bbb N$ with $\gcd(n,m)=1$
By Gauss' lemma, it is sufficient to show that $X^n-Y^m$ is irreducible in $K(Y)[X]$, now $K(Y) / K(Y^m)$ is a degree $m$ extension and $Y^m$ is a prime element of $K[Y^m]$ since $K[Y^m]/(Y^m) \cong K$, thus $X^n-Y^m$ is irreducible over $K(Y^m)$ by the Eisenstein criterion. Now as $\gcd(\mathrm{deg}_X(X^n-Y^m);[K(Y):K(Y^m)])=1$, this implies that $X^n-Y^m$ is also irreducible over $K(Y)$, compare this question.