Is $x^a-y^b$ irreducible (over an arbitrary field $K$) if and only if $a$ and $b$ are coprime?

Question

Is $x^a-y^b\in K[x,y]$ irreducible (over an arbitrary field $K$) if and only if $a$ and $b$ are coprime? I believe this is true if $K=\mathbb{Q}$ or any algebraic extension thereof, but is it true for general $K$?

Answer 1

Yes. Clearly it is necessary for $a$ and $b$ to be coprime, since if $d$ is a common factor of $a$ and $b$ then $x^a-y^b=(x^{a/d})^d-(y^{b/d})^d$ is divisible by $x^{a/d}-y^{b/d}$. Conversely, suppose $a$ and $b$ are coprime. Since $x^a-y^b$ is clearly not divisible by $x$ or $y$, it sufficies to show that it is irreducible in the localization $K[x,y,x^{-1},y^{-1}]$. Now this localization is just the group ring $K[\mathbb{Z}^2]$, and modding out $x^a-y^b$ will give the group ring $K[A]$ where $A$ is the quotient of $\mathbb{Z}^2$ by the subgroup generated by $(a,b)$. Since $a$ and $b$ are coprime, this quotient $A$ is torsion-free and thus isomorphic to $\mathbb{Z}$ (explicitly, if $s,t\in\mathbb{Z}$ are such that $sa+tb=1$ then $\{(a,b),(-t,s)\}$ is a basis for $\mathbb{Z}^2$ so the quotient is freely generated by $(-t,s)$). That is, $K[A]\cong K[\mathbb{Z}]\cong K[t,t^{-1}]$, which is in particular a domain, so $x^a-y^b$ is irreducible.