$\newcommand\R{\Bbb R}$Let $Q(x_1,\dots,x_n)\in\R[x_1,\dots,x_n]$ be an irreducible polynomial such that the dimension of the set $Z:=\{(x_1,\dots,x_n)\in\R^n\colon Q(x_1,\dots,x_n)=0\}$ (defined, say, as the maximal dimension of the tangent vector spaces at the nonsingular points of $Z$) is $n-1$. Let $P(x_1,\dots,x_n)\in\R[x_1,\dots,x_n]$ be another polynomial such that $P(x_1,\dots,x_n)=0$ for all $(x_1,\dots,x_n)\in Z$.

Does it then necessarily follow that $Q(x_1,\dots,x_n)$ divides $P(x_1,\dots,x_n)$?

For $n=2$, the answer to this question is affirmative.

$\newcommand\R{\Bbb R}$Let $Q(x_1,\dots,x_n)\in\R[x_1,\dots,x_n]$ be an irreducible polynomial such that the dimension of the set $Z:=\{(x_1,\dots,x_n)\in\R^n\colon Q(x_1,\dots,x_n)=0\}$ (defined, say, as the maximal dimension of the tangent vector spaces at the nonsingular points of $Z$) is $n-1$. Let $P(x_1,\dots,x_n)\in\R[x_1,\dots,x_n]$ be another polynomial such that $P(x_1,\dots,x_n)=0$ for all $(x_1,\dots,x_n)\in Z$.

Does it then necessarily follow that $Q(x_1,\dots,x_n)$ divides $P(x_1,\dots,x_n)$?

When $n=2$ and the total degree of $Q(x_1,x_2)$ is $2$, the answer to this question is affirmative.