# Showing a polynomial is irreducible over $\mathbb{C}[x,y]$

Given $m,n \in \mathbb{N},$ how can I show that the polynomial $x^m+y^n-1$ is irreducible in $\mathbb C[x,y]$?

I'm given the following hint, but I don't follow. Note: I know Eisenstein's Criterion.

Adapt Eisenstein's Criterion to work in $\mathbb C[x,y]$ by using irreducibles in $\mathbb C[y]$ instead of primes in $\mathbb Z$, namely $y-1$ in this case - it needs to be shown that $y-1 \nmid y^{n-1}+\cdots+1$.

We apply the Eisenstein Criterion as follows. We view $\mathbb{C}[x, y]$ as a polynomial ring $R[x]$ in one variable, where we set $R=\mathbb{C}[y]$.
So the polynomial $f(x)\in R[x]$ is given by $f(x) = x^{m} + y^{n}-1$. In order to apply the Eisenstein Criterion to the element $y-1\in R$, we need check that $y-1$ does not divide the leading coefficient of $f(x)$, which is $1$ in this case; and $y-1$ divides $y^{n}-1$ to the first power, that is, $y-1 \mid y^{n}-1$ but $(y-1)^2 \not\mid y^{n}-1$.
Since $y^{n}-1=(y-1)(y^{n-1}+y^{n-2}+\cdots + y+1)$, it is clear that $y-1\mid y^{n}-1$. In order to show $(y-1)^{2}\not\mid y^{n}-1$, we need that $y-1\not\mid y^{n-1}+y^{n-2}+\cdots + y + 1$. One way to see this is to note that $1$ is a root of $y-1$, but not a root of $y^{n-1}+y^{n-2}+\cdots+y+1$.
• I think the first thing to check is that $y-1$ is prime. Jun 23 '15 at 7:45
• @user26857: You are absolutely right: $\mathbb{C}[y]/(y-1)\cong \mathbb{C}$ is a domain, so $y-1$ is prime. Jun 23 '15 at 19:24