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$\newcommand\R{\Bbb R}$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.

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    $\begingroup$ This only answers the title, not the actual question, but there is the real Nullstellensatz $\endgroup$ Apr 14 at 4:11
  • $\begingroup$ @CommandMaster : Thank you for your comment. This seems pretty close to a negative (?) answer to the question. $\endgroup$ Apr 14 at 4:39

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I think, it does.

By change of coordinates, you may suppose that $Z$ contains the origin and the tangent vector space is the hyperplane $\{x_n=0\}$. Then, by implicit function theorem, for small enough $x=(x_1,\ldots,x_{n-1})\in \mathbb{R}^{n-1}$ the polynomial $q_x(t):=Q(x,t)$ has a real root, and $p_x(t):=P(x,t)$ has the same root. So, the resultant of $p_x$ and $q_x$ is 0 for small enough $x$. Since this resultant is a polynomial in $x$, it is identical 0. Consider the polynomials $p(t)=:P(x_1,\ldots,x_{n-1},t)$, $q(t):=Q(x_1,\ldots,x_{n-1},t)$ over the field $\mathbb{R}(x_1,\ldots,x_{n-1})$. Their resultant is 0. I claim that $q$ is irreducible. Indeed, if $q(t)=a(t)b(t)$ with non-constant $a$, $b$, then multiplying by the common denominator we get a formula $Q(x_1,\ldots,x_{n-1},t)D(x_1,\ldots,x_{n-1})=A(x_1,\ldots,x_{n-1},t)B(x_1,\ldots,x_{n-1},t)$ for real polynomials $A,B$. Thus, since $Q$ is irreducible, one of $A$, $B$ must be divisible by $Q$ (in $\mathbb{R}[x_1,\ldots,x_n-1,t]$), but this is not the case as each of $A$, $B$ has smaller degree then $Q$ w.r.t. variable $t$.

So, $q(t)$ is irreducible, but the resultant of $p(t)$ and $q(t)$ is 0. This yields that $q$ divides $p$ that after multiplication by the denominator yields that $P(x_1,\ldots,x_{n-1},t)H(x_1,\ldots,x_{n-1})$ is divisible by $Q$. But if $Q$ divides $H$, then $Z$ does not have a tangent space of the form $\{x_n=0\}$ (quite the opposite), so $Q$ divides $P$.

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The argument from the link that OP included can be mimicked: reducing the problem to complex Nullstellensatz with a bit of analysis. Consider a non-singlular point $\mathbf{p}\in Z$ near which the equation $Q(x_1,\dots,x_n)=0$ can be solved for one of the coordinates, say $x_n$. Hence there exists an open neighborhood $U$ of $\mathbf{p}=(p_1,\dots,p_{n-1},p_n)$ in $\Bbb{C}^n$ and a complex analytic function $\gamma$ with
$$ \forall (x_1,\dots,x_n)\in U: Q(x_1,\dots,x_{n-1},x_n)=0 \Leftrightarrow x_n=\gamma(x_1,\dots,x_{n-1}). $$
The assumption on $P$ now shows that $P(x_1,\dots,x_{n-1},\gamma(x_1,\dots,x_{n-1}))=0$ whenever all the arguments are real numbers. But since $\mathbf{p}\in\Bbb{R}^n$ and $Q$ has real coefficients, all coefficients of the Taylor expansion of $\gamma$ at $(p_1,\dots,p_{n-1})$ are real. Therefore, all Taylor coefficients of $P(x_1,\dots,x_{n-1},\gamma(x_1,\dots,x_{n-1}))$ at $(p_1,\dots,p_{n-1})$ are real too. But the latter function is identically zero when $(x_1,\dots,x_{n-1})$ comes from a small enough neighborhood of $(p_1,\dots,p_{n-1})$ in $\Bbb{R}^{n-1}$, because then $(x_1,\dots,x_{n-1},\gamma(x_1,\dots,x_{n-1}))\in Z$. Therefore all Taylor coefficients of $P(x_1,\dots,x_{n-1},\gamma(x_1,\dots,x_{n-1}))$ at $(p_1,\dots,p_{n-1})$ are zero. In particular, this quantity is zero as $(x_1,\dots,x_{n-1})$ varies in a small enough neighborhood of $(p_1,\dots,p_{n-1})$ in $\Bbb{C}^{n-1}$. We conclude that there is a perhaps smaller open neighborhood $U'\subseteq U$ of $\mathbf{p}$ in $\Bbb{C}^n$ such that $P$ vanishes on the non-empty open subset $$ U'\cap\{(x_1,\dots,x_n)\in\Bbb{C}^n\mid Q(x_1,\dots,x_n)=0\} $$ of the complex zero locus $Z_{\Bbb{C}}(Q)=\{(x_1,\dots,x_n)\in\Bbb{C}^n\mid Q(x_1,\dots,x_n)=0\}$. Its non-singular locus $Z_{\Bbb{C}}(Q)^*$, a complex submanifold of $\Bbb{C}^n$, is dense in $Z_{\Bbb{C}}(Q)$; and is furthermore connected due to the irreducibility of $Q$ (see Chapter 0 of Griffiths & Harris for these standard facts). Now $P$ restricts to a holomorphic function on the connected complex manifold $Z_{\Bbb{C}}(Q)^*$ which is zero on the non-empty open subset $U'\cap Z_{\Bbb{C}}(Q)^*$. This holomorphic function should thus be identically zero. Consequently, $P$ vanishes on $\overline{Z_{\Bbb{C}}(Q)^*}=Z_{\Bbb{C}}(Q)$. The complex Nullstellensatz now implies that a power of $P$ is a multiple of $Q$ in $\Bbb{C}[x_1,\dots,x_n]$. But $P$ and $Q$ belong to $\Bbb{R}[x_1,\dots,x_n]$ and the latter is irreducible. We conclude that $Q\mid P$ in $\Bbb{R}[x_1,\dots,x_n]$.

Remark) The requirement that the real zero locus of $Q$ contains an $(n-1)$-dimensional submanifold of $\Bbb{R}^n$ is necessary for having $Q\mid P$. Without that, one can merely say that $Q\mid\sum_{i=1}^sR_i^2+P^{2m}$ for suitable polynomials $R_1,\dots,R_s\in\Bbb{R}[x_1,\dots,x_n]$ and positive integer $m$ (this due to Positivstellensatz, see Theorem 5.5 here for instance). This does not imply $Q\mid P$ in general. For instance, take $Q$ to be the Motzkin polynomial $x^4y^2 + x^2y^4 + 1 − 3x^2y^2$, a famous example from the literature on Hilbert's 17th problem. Then $Q\mid x^2y^2(x^2+y^2+1)(x^2+y^2−2)^2 + (x^2−y^2)^2$. But $Q(x,y)=x^4y^2 + x^2y^4 + 1 − 3x^2y^2$ does not divide $P(x,y):=x^2-y^2$. Here, $Z=\{(\pm 1,\pm 1)\}$ is zero-dimensional.

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    $\begingroup$ Thank you for your answer. It seems you have not explicitly used the irreducibility of $Q$. Maybe, it can be used to get the connectedness? $\endgroup$ Apr 14 at 14:27
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    $\begingroup$ @IosifPinelis Isn't the irreducibility used once the problem is reduced to the complex Nullstellensatz? $Z_{\Bbb{C}}(Q)\subseteq Z_{\Bbb{C}}(P)\Rightarrow Q\mid P^r$ for some $r$. If $Q$ is irreducible, $Q\mid P^r$ implies $Q\mid P$. $\endgroup$
    – KhashF
    Apr 14 at 14:33
  • $\begingroup$ Oh, yes. Yet, can the irreducibility be also used to get the connectedness? $\endgroup$ Apr 14 at 14:44
  • $\begingroup$ @IosifPinelis I amended my answer. $\endgroup$
    – KhashF
    Apr 14 at 20:50
  • $\begingroup$ It is nice to see that my guess that the irreducibility of $Q$ can be used to get the connectedness may be correct. However, in Chapter 0 of Griffiths & Harris I have only found something only on the local irreducibility of the analytic hypersurface that is the zero set of a holomorphic function (and this local irreducibility is again based on the Nullstellensatz!). $\endgroup$ Apr 14 at 21:27

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