I'm trying to find a necessary and sufficient condition for $x^n - y^m$ to be irreducible in $\Bbb C[x,y]$.
What I have now is rather trivial sufficient conditions: if $m\mid n$ then $m=1$ and if $n\mid m$ then $n = 1$.
There are similar questions previously posted to this site. To me the answers to those questions seem overkill to this particular problem, and I am looking for an elementary solution.
I would be grateful for your help.
The polynomial $x^m-y^n$ is irreducible in ${\Bbb C}[x,y]$ iff $m$ and $n$ are coprime.
When $m$ and $n$ have a common factor $d>1$, say $m=m'd$ and $n=n'd$, then $x^m-y^n$ divisible by $x^{m'}-y^{n'}$, so $x^m-y^n$ is reducible in this case.
Now suppose that $m$ and $n$ are coprime. Then we can find $\mu,\nu\in\mathbb{Z}$ such that $\mu m + \nu n = 1$. Let $\omega$ be a root of $x^m-y^n$ in some extension $K$ of ${\Bbb C}(y)$. Then $(y^\mu \omega^\nu)^m = y$, so $K$ contains a root of $x^m - y$. But by Eisenstein's criterion $x^m-y$ is irreducible, so $[K:\mathbb{C}(y)]\geq m$, so $x^m-y^n$ must be irreducible.
EDIT: At OP's request, I am trying to translate the above into a less abstract proof.
The polynomial $x^m-y^n$ divides $x^{\nu m} - y^{\nu n}$, which is irreducible as a polynomial in $x^\nu$ and $y$: indeed, after replacing $x^\nu$ by $x^\nu y^{-\mu}$ and ignoring a factor in $\mathbb{C}(y)$, this polynomial is just $(x^\nu)^m-y$. Now I need this lemma:
Lemma: If $f(X)\in K[X]$ is an irreducible polynomial over a field $K$, then all nontrivial factors of $f(X^k)$ have degree at least $\deg f$.
But I admit I don't see how to prove this by less abstract means than this: Let $\omega$ be a root of $f(X^k)$ in some extension. Then $\omega^k$ is a root of $f(X)$, so $[K(\omega):K]\geq [K(\omega^k):K]=\deg f$.
At Pteromys's request here is a proof without field theory that $x^n-y^m$ is irreducible if $n$ is relatively prime to $m$.
Grade the polynomial ring ${\Bbb C}[x,y]$ by decreeing that $x$ has weight $m$ and $y$ has weight $n$.
The polynomial $x^n-y^m$ is then homogeneous of degree $nm$ and any factorization of it must consist of homogeneous polynomials: $$x^n-y^m=(x^r+\cdots+y^s)(x^t+\cdots-y^u)$$ with in particular $$mr=sn$$ since the first factor is homogeneous.
But since $n$ is prime to $m$ and divides $mr$, $n$ must divide $r$ so that $r\geq n$ , which successively forces $r=n$ and then $t=0$, so that the second factor is constant and irreducibility of $x^n-y^m$ is thus proved.
Here's a proof that, like the others posted so far, works over all fields $k$. It is shown in another answer why $(m,n) = 1$ is a necessary condition for irreducibility of $x^n - y^m$ in $k[x,y]$. We now show sufficiency: if $(m,n) = 1$ then $x^n - y^m$ is irreducible in $k[x,y]$.
We will look at possible factorizations of $x^n-y^m$ in $k[x,y] = k[y][x]$ (polynomials in $x$ with coefficients that are polynomials in $y$). Write such a factorization as $f(x)g(x)$ where the factors are in $k[y][x]$. Let $f(x)$ have constant term $c\in k[y]$.
Since $\deg_x f + \deg_x g = n$, $\deg_x f$ and $\deg_x g$ are at most $n$ and at least one of them is positive. Without loss of generality, say $\deg_x f > 0$. The roots of $f(x)$ in a splitting field of $x^n - y^m$ over $k(y)$ are $n$-th roots of $y^m$, and $c$ is the product of those roots, so $$c^n=y^{m\deg_x f}.$$ This is an equation in $k[y]$. Computing $y$-degrees, $n\deg_y c = m\deg_x f$, so $n\mid m\deg_x f$. Since $(m,n)=1$, we get $n\mid\deg_x f$. From the bound $\deg_f x \leq n$, we get $\deg_xf=n$. Therefore $\deg_x g = 0$, meaning $g(x)$ is constant as a polynomial in $x$: $g(x) \in k[y]$. So $g(x) = a(y)$ where the coefficients of $a(y)$ are in $k$. Comparing leading coefficients in $x$ on both sides of the equation $x^n - y^m = f(x)a(y)$, we see that $a(y)$ is invertible in $k[y]$, so $a(y) \in k^\times$.
We have shown that in a factorization of $x^n - y^m$ in $k[x,y]$, one of the two factors is in $k^\times$, so $x^n - y^m$ is irreducible in $k[x,y]$.
