# Homogeneous polynomial in $k[X,Y,Z]$ can factor into linear polynomials?

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 $$n\ge 3$$.

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 $$k[X,Y,Z]$$?

• How does it work for $X^2+Y^2-1$? You see, if $f$ factors into linear factors, then the set of solutions of $f=0$ looks like a union of hyperplanes (because the set of solution of a linear equation is a hyperplane). If the set of solutions of $f=0$ is not a union of hyperplanes then the factorization into linear factors cannot take place.
– user119256
Feb 5, 2014 at 22:07
• @karene sorry I meant $F$ homogeneous, if $F\in k[X,Y]$ is homogeneous, then $F$ have to be factored into linear polynomials? Feb 5, 2014 at 22:35
• The same ideas apply to homogeneous. The only difference is that the zero locus of homogeneous polynomials are are symmetric by homothety (i.e. they con any line passing through the origin and any of its points). The only such sets on the plane are lines (hyperplanes of the plane). That is why the factorization occurs, and that is why in higher dimensionyou can get zero loci that is not a union of hyperplanes.
– user119256
Feb 6, 2014 at 3:49

To a homogeneous polynomial $f(x,y)$ of degree $n$ we associate a polynomial $g(x)$ such that $f(x,y)=y^ng(x/y)$. Since $k$ is algebraically closed $g$ splits (in linear factors) in $k[x]$, so $g(x)=a(x-a_1)\cdots(x-a_n)$ and thus we get $f(x,y)=a(x-a_1y)\cdots(x-a_ny)$.

If $f\in k[x,y,z]$ then $f$ can be irreducible. For example, if the characteristic of $k$ is not $2$, then $x^2+y^2+z^2$ is irreducible. (For more details see $x^2 +y^2 + z^2$ is irreducible in $\mathbb C [x,y,z]$.)

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.

• Why did we have to pull out the highest possible exponents from F? Couldn't the whole argument have worked had we just stuck with F and defined f(x,y) = F(x,y,1)?
– D.R.
Aug 14, 2020 at 19:48
• @D.R. Possibly, but the idea is to reduce to the case when all variables appear. Aug 14, 2020 at 20:24