If $f_n, f_{n-1}$ are irreducible homogenous polynomials, then $I := \langle f_n + f_{n-1}\rangle$ is prime.

Question

Assume that $f_n$ and $f_{n-1}$ are irreducible and homogenous polynomials in two variables $x,y$. I am trying to show that the algebraic variety
$$X : f_n + f_{n-1} = 0$$
is irreducible.

To do this, the natural way is to show that the ideal $I = \langle f_n + f_{n-1}\rangle$ is prime. This is done if I show that $f_n + f_{n-1}$ is irreducible in $\mathbb{C}[x,y]$ provided $f_n, f_{n-1}$ are, but I was not able to prove it.

Any hint?

Answer 1

The $\Bbb C[x,y]$ case is a bit silly, because in that ring the only irreducible homogeneous polynomials are the ones of degree $1$.

The true version of the statement, as opposed to the current one which would be false with more variables because it allows $f_n$ and $f_{n-1}$ to have the same degree, is the following (henceforth, I'll indicate by $p_s$ the homogeneous component of degree $s$ of the polynomial $p$):

Let $R$ be an integral domain and let $f\in R[x_1,\cdots,x_r]$ be a polynomial such that $f_{\deg f}$ is irreducible. Then, $f$ is irreducible.

This is an obvious consequence of the identity $(pq)_{\deg(pq)}=p_{\deg p}q_{\deg q}$: on one hand, $f$ isn't a unit because $f_{\deg f}\notin R^*$ prevents it from being an invertible constant; on the other hand, if $f=pq$, then $f_{\deg f}=p_{\deg p}q_{\deg q}$ and therefore either $p_{\deg p}\in R^*$ or $q_{\deg q}\in R^*$. Either way, one of the factors $p,q$ is an invertible constant.