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?

  • $\begingroup$ This is such a simply stated question, yet it defies any tackle with simple algebra. $\endgroup$ Mar 25, 2016 at 9:34
  • $\begingroup$ can you help in solving this then? $\endgroup$
    – Ri-Li
    Mar 25, 2016 at 9:56
  • $\begingroup$ No what I mean to say is it beats me too! $\endgroup$ Mar 25, 2016 at 10:04

3 Answers 3


This problem is easier if we use another formulation of $f$ being a homogeneous polynomial.

First I'll prove the following fact: a polynomial is homogeneous (in your definition) if and only if each monomial appearing in $f$ has total degree $n$.

Proof: First suppose that $f$ is homogeneous. Write $f$ as a sum of monomials $f_i$ of total degree $i$: $f=f_0+f_1 + \ldots+f_r$. Then $$ a^nf(x) = f(ax)=f_0(ax)+f_1(ax)+\ldots+f_r(ax)=f_0+af_1(x)+\ldots+a^rf_r(x). $$ We want to prove that $n=r$ and that $f_r$ is the only term in $f$. For the first fact, note that both sides of the equations are polynomials in $a$. The left hand side has degree $n$ and the right hand side has degree $r$. Hence $n=r$.

If we put $a=0$, we see that $f_0=0$. Now differentiate with respect to $a$ to get $$ na^{n-1}f(x) = f_1(x)+2af_2(x) + \ldots + ra^{r-1}f_r(x). $$ Put $a=0$. Then we see that $f_1=0$. Continue, to conclude that $f(x)=f_r(x)$. Hence we have shown that a homogeneous polynomial must consist of a sum of monomials of total degree $n$.

The other direction is easy and is left to the reader.


Now for your question. We want to see that the factors of a homogeneous polynomial are themselves homogeneous. We do this for two factors, since the general case follows by induction.

So suppose that $f=gh$, where $f$ is homogeneous, but one of $g,h$ is not. So suppose $g$ is not homogeneous. By the above, this means that we can write $g=g_{ih}+g_h$, where $g_{h}$ is $g$'s highest degree term, and $g_{ih}$ is its "inhomogeneous component" (=the rest). We have $$ f=gh=(g_{ih}+g_h)h=g_{ih}h+g_hh. $$ For degree reasons (the first term on the right hand side has degree less than $f$, which consist of only top degree monomials), we must have $g_{gih}h=0$, but $g_{ih}$ was nonzero, so $h=0$, but this is absurd.

  • 3
    $\begingroup$ You're of course assuming that your field has characteristic $0$. $\endgroup$
    – egreg
    Mar 25, 2016 at 12:02
  • 6
    $\begingroup$ I don't get why $g_{ih}h$ must be $0$. What if it cancels out with an "inhomogeneous" part of $g_h h$? $\endgroup$
    – User
    Nov 10, 2019 at 5:08

Let $\widehat{f}(x_1,x_2,\dots,x_n,t)=f(tx_1,\dots,tx_n)$ in the polynomial ring $k[x_1,\dots,x_n,t]$. Then $f$ is homogeneous if and only if $$ \widehat{f}(x_1,x_2,\dots,x_n,t)=t^nf(x_1,\dots,x_n) $$ This is different from your definition which is equivalent to this one only over infinite fields. For instance, with your definition, every polynomial over the two element field would be homogeneous.

Suppose $f=gh$; then we have $$ \widehat{f}(x_1,\dots,x_n,t)= f(tx_1,\dots,tx_n)= g(tx_1,\dots,tx_n)h(tx_1,\dots,tx_n) $$ Now we can write \begin{align} g(tx_1,\dots,tx_n)&=g_0+g_1t+\dots+g_at^a,\\ h(tx_1,\dots,tx_n)&=h_0+h_1t+\dots+h_bt^b, \end{align} with $g_i,h_j\in k[x_1,\dots,x_n]$, $g_a\ne0$ and $h_b\ne0$. Suppose $f$ is homogeneous; consider the least integer $c$ such that $g_i=0$ for $i<c$, $g_c\ne0$ and the least integer $d$ such that $h_j=0$ for $j<c$ and $h_d\ne0$.

Note that $a+b=n$, $c\le a$ and $d\le b$.

Then the term of degree $c+d$ in $g(tx_1,\dots,tx_n)h(tx_1,\dots,tx_n)$ is (with $g_i=0$ for $i>a$ and $h_j=0$ for $j>b$ and consider everything in the ring of polynomials in $t$ with coefficients in $k[x_1,\dots,x_n]$) $$ g_0h_{c+d}+g_1h_{c+d-1}+\dots+ g_{c-1}h_{d+1}+g_ch_d+g_{c+1}h_{d-1}+\dots+g_{c+d}h_0 =g_ch_d\ne0 $$ Since $f$ is homogeneous, $\widehat{f}$ has only the leading coefficient (at degree $n$) non zero, as a polynomial in $t$ with coefficients in $k[x_1,\dots,x_n]$, so $c+d=n$ and therefore $c=a$, $d=b$.

Therefore $$ \widehat{g}(x_1,\dots,x_n,t)=t^ag_a(x_1,\dots,x_n) $$ Evaluating at $t=1$, $$ g(x_1,\dots,x_n)=\widehat{g}(x_1,\dots,x_n,1)=g_a(x_1,\dots,x_n) $$ so we have proved that $$ \widehat{g}(x_1,\dots,x_n,t)=t^ag(x_1,\dots,x_n) $$ and therefore $g$ is homogeneous. Similarly for $h$.

  • $\begingroup$ Maybe this is a dumb question. It seems that the part showing factors are homogeneous does not require either infinite field or $char(K)\neq 0$ if I treat $t=x_0$ and $\hat(g)$ as an element of localized polynomial ring at $x_0$. $\hat(g)$ then corresponds to an element of graded polynomial ring's component. If I use the definition of homogeneous polynomial is an element of graded component of graded polynomial ring, would the proof be changed for finite/infinite/$char(K)\neq 0$? $\endgroup$
    – user45765
    Jul 16, 2017 at 15:28
  • 1
    $\begingroup$ @user45765 The key here is that $t$ is an indeterminate. $\endgroup$
    – egreg
    Jul 16, 2017 at 15:39

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.

  • $\begingroup$ Very nice! Using irreducibility to force homogeneity. $\endgroup$
    – Jose Brox
    Jul 28, 2021 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.