$X^n-Y^m$ is irreducible in $\Bbb{C}[X,Y]$ iff $\gcd(n,m)=1$

Question

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.

Answer 1

Lemma: If $f \in \mathbb{C}[X,Y]$ is irreducible polynomial over $\mathbb{C}(Y)$, then its factorization in $\mathbb{C}[X,Y]$ consists of an irreducible polynomial and zero or more univariate polynomials in $Y$.

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Theorem: If $\gcd(m,n) = 1$, then $S^m - T^n$ is an irreducible polynomial in one variable ($S$) over the field $K = \mathbb{C}(T)$.

Proof: The $m$ distinct roots of this polynomial are the values $\zeta^k T^{n/m}$ lying in the field $L = \mathbb{C}(T^{1/m})$, where $\zeta$ is a primitive $m$-th root of unity.

We can find $a,b$ so that $am + bn = 1$. This implies

$$ T^a \left(T^{n/m} \right)^b = T^{(am + bn)/m} = T^{1/m}$$

and so $L = K(T^{n/m})$. There are a number of ways to infer that $S^m - T^n$ is irreducible from this; e.g. $[L:K] = m$ along with the fact $S^m - T^n$ is degree $m$ mean that $S^m - T^n$ is the minimal polynomial of $T^{n/m}$ over $K$. $\square$

Answer 2

Assume $f(X,Y) =X^n-Y^m=g(X,Y)h(X,Y)$. Then $f(Z^m,Z^n)=0$ implies that one of $g(Z^m,Z^n)$ or $h(Z^m,Z^n)$ is the zero polynomial. Suppose that $g(Z^m,Z^n)=0$. That means that for all $k$, the monomials cancel, i.e. if $$g(X,Y)=\sum a_{i,j}X^iY^j $$ then $$\sum_{mi+nj=k}a_{i,j}=0.$$ Can we ever have $mi+nj=mi'+nj'$? That would mean $m(i-i')=n(j-j')$, hence $n\mid i-i'$ (because none of $n$'s prime factors are in $m$) and likewise $m\mid j-j'$. So if $i>i'$ this implies $j> j'$.

Answer 3

Geometric solution.

Consider the morphism of affine varieties $\mathbb{A}^1 \to V(X^n-Y^m), t \mapsto (t^m,t^n)$. It is surjective: If $x^n=y^m$ in $\mathbb{C}$, wlog $x,y \in \mathbb{C}^*$, choose some $t \in \mathbb{C}^*$ such that $x=t^m$. Then $y^m=x^n=t^{mn}$, hence $y=t^n \cdot \zeta$ for some $\zeta \in U(m)$. Since $m,n$ are coprime, $\zeta=\eta^n$ for some $\eta \in U(m)$. Then $t \eta$ is a preimage of $(x,y)$.

It follows that $V(X^n-Y^m)$ is irreducible, i.e. $\sqrt{(X^n-Y^m)}$ is a prime ideal. But the ideal $(X^n-Y^m)$ is radical, because $X^n-Y^m$ is square-free, since $\partial_X (X^n-Y^m)=n X^{n-1}$ and $X^n-Y^m$ are coprime in $\mathbb{C}(Y)[X]$.