Showing a polynomial is irreducible

[kalish1]

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 $y-1 \not | y^{n-1}+\cdots+1$.*

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[jvicens]

To show that $x^m+y^n-1$ is irreducible in $\mathbb{C}[x,y]$, we can use the following adaptation of Eisenstein's Criterion:

If there exists an irreducible polynomial $p(y) \in \mathbb{C}[y]$ such that $p(y)$ divides each coefficient of $y^{n-1}+\cdots+1$ except the leading coefficient, and $p(y)$ does not divide the constant term, then $x^m+y^n-1$ is irreducible in $\mathbb{C}[x,y]$.

In this case, we can choose $p(y)=y-1$. Since $p(y)$ does not divide the constant term $-1$, we just need to show that $p(y)$ divides each coefficient of $y^{n-1}+\cdots+1$.

Notice that when we expand $y^{n-1}+\cdots+1$, each term has a factor of $y-1$ except for the constant term $1$. Therefore, $p(y)$ divides each coefficient of $y^{n-1}+\cdots+1$ and thus $x^m+y^n-1$ is irreducible in $\mathbb{C}[x,y]$.

This is because if $y-1$ divides each coefficient of $y^{n-1}+\cdots+1$, then each term can be written as $a(y-1)$ for some $a \in \mathbb{C}$. This means that when we expand the polynomial, each term will have a factor of $y-1$ and we will be left with only the constant term $1$, which $p(y)$ does not divide. Therefore, $p(y)$ must divide each coefficient of $y^{n-1}+\cdots+1$.

Hence, $x^m+y^n-1$ is irreducible in $\mathbb{C}[x,y]$ by our adapted version of Eisenstein's Criterion.