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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 $f = p - q$.

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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 $$ f(x_1,\ldots,x_n)=\lim_{\epsilon\to 0}f(x_1,\ldots,x_{n-1},x_{n}+\epsilon)=\lim_{\epsilon\to 0}0=0. $$

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  • $\begingroup$ Thanks, yurnero, for your proof. Under what kind of situation can $S = 0$? $\endgroup$ – user17982 Oct 23 '16 at 13:39
  • $\begingroup$ @ColinTan $S$ is just the sum of the inputs so $S=0$ just corresponds to the case when these inputs sum to $0$. $\endgroup$ – yurnero Oct 23 '16 at 13:41
  • $\begingroup$ I understand now. Thank you for your help. $\endgroup$ – user17982 Oct 23 '16 at 13:50

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