# Polynomials (or analytic functions) vanishing on a real algebraic set

I have seen the following result stated several times in the literature, without proof:

Let $$\mathbb{K}$$ be $$\mathbb{R}$$ or $$\mathbb{C}$$, and assume $$P\in\mathbb{K}[X_{1},\ldots,X_{n}]$$ is an irreducible polynomial of $$n$$ variables, and let

$$V = \{z\in\mathbb{C}^{n},~P(z)=0\},\quad I(V)=\{Q\in\mathbb{C}[X_{1},\ldots,X_{n}],~Q=0\text{ on }V\}.$$

Assume that $$V$$ contains a real point $$a\in\mathbb{R}^{n}$$ which is regular, that is

$$\text{dim }V=n-\text{rank }J_{a}(V),$$

where $$J_{a}(V)$$ denotes the Jacobian of a family of generators of $$I(V)$$, evaluated at $$a$$.

Then the set $$V_{\mathbb{R}}=V\cap\mathbb{R}^{n}$$ of real points of $$V$$ is Zariski dense in $$V$$, or equivalently any polynomial that vanishes on $$V_{\mathbb{R}}$$ must vanish on $$V$$.

I am also interested by the analytic version of this result (if true) where $$V$$ is still the zero set of a polynomial $$P$$ but $$I(V)$$ is the ideal of analytic functions vanishing on $$V$$, and the vanishing of an analytic function $$f$$ on $$V_{\mathbb{R}}$$ implies the vanishing of $$f$$ on $$V$$.

Could someone provide a proof for the algebraic or analytic case?

• what do you mean by "(or complex ?)" - naturally, if $P\in\mathbb{R}[X]$ then the coefficients are real. May 10, 2019 at 9:29
• I mean $P\in\mathbb{C}[X]$. May 10, 2019 at 9:31

The more general statement that is true is the following:

Let $$X \subset \mathbb{C}^d$$ be an irreducible affine variety defined by real polynomials. If $$X$$ has a smooth real point, then $$X(\mathbb{R})$$ is Zariski dense in $$X.$$

A sketch of the proof and some good examples are given here: Real Algebraic Geometry for Geometric Constraints by Frank Sottile (page 8).

The main fact used is, under the assumption there is a real smooth point, the smooth points of the $$\mathbb{R}$$-locus is a real manifold of the same dimension.

The main reference is: Real Algebraic Geometry by Bochnak, Coste, Roy.

Fun facts:

1. If you ask instead for the density of $$\overline{\mathbb{Q}}$$-points then that is true more generally for any $$d$$–dimensional affine variety $$V$$ defined over $$\mathbb{Q}$$ by Noether Normalization.
2. If you remove the assumption that there is a smooth point, then the statement is false as $$V(x^2+y^2)$$ has only one $$\mathbb{R}$$-point which is not smooth, namely $$(0,0)$$, and $$x^2+y^2$$ is irreducible over $$\mathbb{R}$$ and the $$\mathbb{C}$$-locus is dimension 1 (so the $$\mathbb{R}$$-locus not Zariski dense).