# A version of Hilbert's Nullstellensatz for real zeros

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

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

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 $$P$$.

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.

• 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? Apr 14 at 14:27
• @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 $Q\mid P$. Apr 14 at 14:33
• Oh, yes. Yet, can the irreducibility be also used to get the connectedness? Apr 14 at 14:44
• @IosifPinelis I amended my answer. Apr 14 at 20:50
• 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!). Apr 14 at 21:27