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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.

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  • $\begingroup$ You probably came across it already, but here you might find something useful. $\endgroup$ – J.R. Jan 26 '14 at 21:11
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    $\begingroup$ @ abc: I'm not really a "show us your work" kind of guy, but I am genuinely curious in this case to see how $(m, n) \ne 1$ implies $X^n - Y^m$ is reducible. Perhaps you could show us? $\endgroup$ – Robert Lewis Jan 26 '14 at 21:13
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    $\begingroup$ @RobertLewis: One easily reduces to the observation that $X^d-Y^d$ is reducible for $d>1$. $\endgroup$ – Martin Brandenburg Jan 26 '14 at 21:13
  • $\begingroup$ Got it? Thanks! $\endgroup$ – Robert Lewis Jan 26 '14 at 21:16
  • $\begingroup$ You can do the following operation: If $m>n$ substitute $x$ by $x$ and $y$ by $xy$, and factor out $x$ as much as possible. If $m<n$, then replace $x$ by $xy$, $y$ by $y$, and factor out $y$ all you can. Observe that if the resulting polynomial factors, then so would the original polynomial. The result of doing this operation to $x^m−y^n$ is $x^{m−n}−y^n$ when $m>n$ or $x^m−y^{n−m}$ when $m<n$. As you can see these steps are doing Euclidean algorithm on the exponents. So, eventually we reach $x^d−y^d$, where $d=\gcd(m,n)$. If $d>1$, this polynomial factors and therefore the original does too. $\endgroup$ – plop Jun 13 at 12:52
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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 $Y$.


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 $K$. $\square$

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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 $\mathbb{C}(Y)[X]$.

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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 $j> j'$.

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    $\begingroup$ I did not understand the last part. $m(i-i\prime)=n(j\prime-j)$. After that where is the contradiction? $\endgroup$ – MathStudent Jan 27 '14 at 19:19
  • $\begingroup$ Your are mistaken in the last paragraph. It should be $$m(i-i')=n(j'-j).$$ $\endgroup$ – Bach Jul 27 '19 at 0:55
  • $\begingroup$ I'm sorry, could you explain why $f(Z^m, Z^n) = 0$ implies that one of the factors is zero? $\endgroup$ – Ovi Mar 27 '20 at 17:54
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I think Hagen von Eitzen's proof is good but not as explicite at the end.

I'll try to clarify:

"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 $h(Z^m,Z^n)$, wlog $g(Z^m,Z^n)$, is the zero polynomial. 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 $m(i-i')=n(j'-j)$. It follows that $i=i'(\bmod n)$ and $j=j'(\bmod m)$. But $i,i'\in {0,1,...,n-1}$ and $j,j'\in {0,1,...,m-1}$. This implies $i=i'$ and $j=j'$."

Overall, $mi+nj=mi'+nj'$ is echivalent with $i=i'$ and $j=j'$. So distinct terms from $g$ and $h$ corespond to distinct terms in $g(Z^m,Z^n)$ and $h(Z^m,Z^n)$. So when we substitute $X=Z^m$ and $Y=Z^n$ none of the terms $g(Z^m,Z^n)$ and $h(Z^m,Z^n)$ will cancel out so that $g(Z^m,Z^n)=0$ and $h(Z^m,Z^n)=0$. That means $g(Z^m,Z^n)\ne0\ne h(Z^m,Z^n)$, but $0=f(Z^m,Z^n)=g(Z^m,Z^n)h(Z^m,Z^n)\ne0$, contradiction.

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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.

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  • $\begingroup$ That's a nice an succinct solution, +1. $\endgroup$ – Hanno Apr 2 '20 at 18:47

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