TITLE: Converse to a proposition on homogeneous polynomials
QUESTION [7 upvotes]: I know that for a homogeneous polynomial $P$, if $P(x_1, ... , x_n) = 0$, then $P(ax_1, ..., ax_n) = 0$ for every $a$ in the field of $P$. Is the converse of this proposition true? That is, if $P(x_1, ... , x_n) =0$ implies $P(ax_1, ..., ax_n) = 0$ for every $a$ in the field of $P$, is $P$ homogeneous?

REPLY [2 votes]: We are assuming that we are working on field of characteristic $0$.
Let $(x_1,x_2,\ldots,x_n)$ be a root of $P$. Then consider the one variable polynomial $Q(t)=P(tx_1,tx_2,\ldots,tx_n)$. Hence $Q(a)=0$ for any $a$ in the field implies all the coefficients of $Q$ are $0$. What are the coefficients of $Q$ ? Write $P(x_1,x_2,\ldots,x_n)$ as sum of monomials and proceed. I think it will work.
The coefficient of $X^d$ in $Q(X)$ is infact a homogeneous polynomial of degree $d$ in the variables $X_1,X_2,\ldots,X_n$.
$P(X_1,X_2,\ldots,X_n)$ can be represented as follows,
$$\sum_{j=1}^{m}a_jR_j(X_1,X_2,\ldots,X_n)$$
Where $m$ is the degree of $P$ and each $R_j$ is a homogeneous polynomial of degree $j$ with $R_j(x_1,x_2,\dots,x_n)=0$ for all $1\leq j\leq m$. Moreover $a_m\neq0$.