For complex polynomials of many variables the following theorem holds:

If $p(x_1,...,x_n), q(x_1,...,x_n)\in \mathbb K[x_1,...,x_n]$, where $\mathbb K=\mathbb C$, are polynomials such that $p(x_1,...,x_n)$ is irreducible and for all $a_1,...,a_n \in \mathbb K$ $$ p(a_1,...,a_n)=0 \Rightarrow q(a_1,...,a_n)=0, $$ then $p(x_1,...,x_n) | q(x_1,...,x_n)$.

I look for a counterpart of this theorem in the case $\mathbb K=\mathbb R$ for polynomials of the second order. Some additional assumptions are necessary, because for example for $p(x_1,x_2)=x_1^2+x_2^2$ and $q(x_1,x_2)=2x_1^2+x_2$ the above theorem is not true.

Is it maybe true under additional assumption that $p,q$ are the second order and that

there exists a $y=(y_1,...y_n)\in \mathbb R^n$ such that $p(y)=0, grad f(y)\neq 0$ ?


Yes. If there is $y\in\mathbf R^n$ such that $p(y)=0$ and $\mathop{grad} p(y)\neq0$ then the set $\{p=0\}$ is a smooth real hypersurface of $\mathbf R^n$ in a neighborhood of the point $y$. This implies that $\{p=0\}$ is Zariski dense in the set $\{z\in\mathbf C^n\mid p(z)=0\}$. Hence, if $q$ vanishes on the real zero set $\{p=0\}$, it also vanishes on the complex zero set of $p$. It also implies that $p$ is irreducible in $\mathbf C[x_1,\ldots,x_n]$. Therefore, $p$ divides $q$ in $\mathbf C[x_1,\ldots,x_n]$ and hence in $\mathbf R[x_1,\ldots,x_n]$.

Note that the conditions on the degree of $p$ and $q$ are superfluous.

  • $\begingroup$ Hi, I am interested by the result you mention, the fact that the real points of $\{p=0\}$ are Zariski dense in the complex points of $\{p=0\}$, but have difficulty to locate a proof in the literature (I am not an expert on this subject). Could you possibly provide a reference with a proof ? Many thanks in advance. $\endgroup$ – user111 May 8 '19 at 10:43
  • $\begingroup$ @user111 I don't know for a reference and I was typing a proof as a comment, but it would be much nicer if you just ask the question as a new question so that I, or somebody else, can write a neat answer to it! $\endgroup$ – Johannes Huisman May 10 '19 at 7:41
  • $\begingroup$ See here $\endgroup$ – user111 May 10 '19 at 9:10

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