TITLE: How to prove that factors of homogeneous polynomial are homogeneous?
QUESTION [17 upvotes]: How to prove that factors of homogeneous polynomial are homogeneous?
I was thinking that for a homogeneous polynomial of degree $n$, 
$f(ax_1,....,ax_n)=a^nf(x_1,....,x_n)$ where $a\in k$
Now if $f=f_1...f_r$ and atleast one $f_i$ is not homogeneous then we'll not get $a^n$ but one flaw in this arguement is that this might possible that one $f_i$ and another $f_j$ both non-homogeneous gives rise to a homogeneous polynomial so how to fix this problem?

REPLY [2 votes]: Say $f\in \mathbb{K}[x_1,\dots , x_n]$ a homogeneous polynomial with degree $d>0$.
Write his factorization $f=g_1\dots g_h$ with $g_i$ irreducible with degree $d_i>0$ for all $i=1,\dots ,h$.
Write each $g_i$ as sum of his homogeneous components, say $g_i=\sum_{j=0}^{d_i}g_{i,j}$ with $g_{i,j}$ homogeneous of degree $j$, so that $g_{i,d_i}\neq 0$ for all $i=1,\dots,h$.
Of course $\sum_i d_i=d$ and $f=g_1\dots g_h=\text{terms of degree}< d+\prod_{i=1}^{h}g_{i,d_i}$.
By $\prod_{i=1}^{h}g_{i,d_i}\neq 0$ and because of homogeneity of $f$, the terms of degree lower than $d$ must vanish and we must have $f=\prod_{i=1}^{h}g_{i,d_i}$.
Now we write the factorization of $g_{i,d_i}=\prod_{j=1}^{k_i}r_{i,j}$ with each $r_{i,j}$ irreducible of positive degree.
So we have $g_1\dots g_h=f=(\prod_{j=1}^{k_1}r_{1,j})\dots(\prod_{j=1}^{k_h}r_{h,j})$. Now because of uniqueness of decomposition, we must have $k_1=\dots =k_h=1$ (otherwise we would have a factorization at left shorter than that at right) so $r_{i,1}=g_{i,d_i}$ is homogeneous for all $i=1,\dots, h$.
So we have $g_1\dots g_h=g_{1,d_1}\dots g_{h,d_h}$ and again because of uniqueness of decomposition we have $g_{i,d_i}=g_i$ for all $i=1,\dots,h$ up to rienumeration and multiplication by a costant, so each $g_i$ is homogeneous.