TITLE: $F, G \in k[X_1, \dots , X_n]$ homogeneous of degrees $r$ and $r+1$ $\implies$ $F+G$ is irreducible
QUESTION [5 upvotes]: 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.

REPLY [7 votes]: Say that a non-zero polynomial $F$ has width $w\in\mathbb N_0$ if the degree of the monomial of highest degree appearing in $F$ with non-zero coefficient minus the degree of the monomial of lowest degree appearing in $F$ with non-zero coefficient is $w$. For example, if $F=F_{n+1}+F_n$ with $F_{n+1}$ and $F_n$ non-zero and homogeneous of degrees $n+1$ and $n$, the width of $F$ is $1$; notice that the polyonomials of width $0$ are exactly the non-zero homogeneous polynomials.

Show that if $F$ and $G$ are polynomials with widths $w_F$ and $w_G$, then the product $FG$ has width $w_F+w_G$.
Use that to prove that if a polynomial $F$ of width $1$, as in your case,  is reducible it necessarily has an non-constant homogeneous factor.
Finally, use that to prove what you want.