I have a question about Exercise 2-34 from William Fulton's Algebraic Curves book. The exercise is as follows.
Suppose that $F, G \in k[X_1, \dots , X_n]$ are forms (i.e. homogeneous polynomials) of degree $r$ and $r+1$ respectively, without common factors (where $k$ is a field). Prove that $F + G$ is irreducible.
I'm having some trouble trying to prove this. First, since $F, G$ are arbitrary I can't think of a particular irreducibility criterion to apply here, so my only idea was to try by contradiction.
Notation
I use Fulton's notation for the homogenization and dehomogenization of a polynomial with respect to the variable $X_{n+1}$. Say if $f \in k[X_1, \dots , X_n]$ has degree $d$, then its homogenization with respect to $X_{n+1}$ is denoted by $f^* = X_{n+1}^d f\left ( \frac{X_1}{X_{n+1}} , \dots , \frac{X_n}{X_{n+1}} \right)$.
Similarly, if $F \in k[X_1, \dots , X_{n+1}]$ is homogeneous, then its dehomogenization with respect to $X_{n+1}$ is denoted by $F_* = F(X_1, \dots , X_n, 1)$.
My attempt
Thus I assume by contradiction that $F + G$ is reducible, then there are polynomials $H, R \in k[X_1, \dots , X_n]$ such that $F + G = HG$ and also $ 0 < \deg{H},\deg{G} < r + 1$. Then the homogenization is
$(F + G)^* = X_{n+1} F + G = (HR)^* = H^* R^*$
Now I'm somewhat stuck. I've tried to play around with this but without luck. I suppose that maybe I should try to get a contradiction to the assumption that $F$ and $G$ have no common factors, but I don't know how.
My questions
- How can this be proved?
- If my approach by contradiction is correct, how can the argument be finished?
As always, thank you for any help.