TITLE: Homogeneous polynomial in $k[X,Y,Z]$ can factor into linear polynomials?
QUESTION [7 upvotes]: My question is quite simple.

Let $k$ be an algebraically closed field and $f\in k[X,Y]$. We know that $f$ can factor into linear polynomials. I would like to know if there is some generalization of this fact to $n$ indeterminates with $n\ge 3$.

Thanks in advance
EDIT. I should have said $f\in k[X,Y]$ a homogeneous polynomial, then $f$ can be factored into linear polynomials, is that true? This fact happens for polynomials in $k[X,Y,Z]$?

REPLY [3 votes]: Basically, a homogeneous polynomial in three variables is the same as a polynomial in two variables. Collect all possible variables that appear in all the terms, with the highest possible exponent: you're left with
$$
F(X,Y,Z)=X^a Y^b Z^c G(X,Y,Z)
$$
where you can't collect any variable from $G$ (some or even all of $a,b,c$ can be zero). You have transferred the problem to $G(X,Y,Z)$. If one of the variables doesn't appear in $G$, then the polynomial is homogeneous in two variables (unless it was a monomial to begin with), so the facts you already know about it apply.
Thus you can assume all variables appear in $G$. Then you can consider
$$
g(x,y)=G(x,y,1)
$$
and you can write
$$
G(X,Y,Z)=Z^k g(X/Z,Y/Z)
$$
for some exponent $k$. Thus you see that reducibility of $G$ is equivalent to reducibility of $g$. A factorization of $g$ translates into a factorization of $G$ and conversely.
Say, for instance, that $G(X,Y,Z)=X^3-X^2Z+Y^2Z$. Then $g(x,y)=x^3-x^2+y^2$ and $k=3$. Since $g$ is irreducible, also $G$ is irreducible.