# If $x_1+ \cdots + x_n - 1$ divides a homogeneous polynomial $f$, then is $f$ necessarily the zero polynomial?

Let $f \in \mathbb{R}[x_1,\ldots, x_n]$ be a homogeneous real polynomial. Suppose that there exists a polynomial $h\in \mathbb{R}[x_1,\ldots, x_n]$ such that $$f = h \cdot (x_1 + \cdots + x_n - 1).$$ Is it true that $f = 0$?

Motivation. Suppose $p$ and $q$ are homogeneous real polynomials such that $p \equiv q \pmod{x_1 + \cdots + x_n - 1}$. Is it true that $p = q$? This is equivalent to the above question by setting $f = p - q$.

Consider $f(x_1,\ldots,x_n)$ and let $S=\sum_ix_i$. If $S\neq 0$, then by homogeneity, there is some $k\geq 1$ such that $$f(x_1,\ldots,x_n)=S^kf(x_1/S,\ldots,x_n/S)=S^kh(x_1/S,\ldots,x_n/S)\cdot\left(\sum_ix_i/S-1\right)=0$$ because $\sum_ix_i/S=1$. If $S=0$, then we also have $f(x_1,\ldots,x_n)=0$ due to continuity: if $\sum_ix_i=0$ then $\sum_ix_i+\epsilon\neq 0$ for $\epsilon\neq 0$ and so $$f(x_1,\ldots,x_n)=\lim_{\epsilon\to 0}f(x_1,\ldots,x_{n-1},x_{n}+\epsilon)=\lim_{\epsilon\to 0}0=0.$$
• Thanks, yurnero, for your proof. Under what kind of situation can $S = 0$? – user17982 Oct 23 '16 at 13:39
• @ColinTan $S$ is just the sum of the inputs so $S=0$ just corresponds to the case when these inputs sum to $0$. – yurnero Oct 23 '16 at 13:41