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?

Many thanks in advance.

  • $\begingroup$ what do you mean by "(or complex ?)" - naturally, if $P\in\mathbb{R}[X]$ then the coefficients are real. $\endgroup$ May 10, 2019 at 9:29
  • $\begingroup$ I mean $P\in\mathbb{C}[X]$. $\endgroup$
    – user111
    May 10, 2019 at 9:31

1 Answer 1


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).

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.