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Let $X \subseteq \mathbf{R}^n$ be Zariski closed and absolutely irreducible, and let $U \subseteq \mathbf{R}^n$ be Euclidean open. My guess is that $X$ is the Zariski closure of $X \cap U$, if the latter is nonempty.

In the case of $X = \mathbf{R}^n$, the proof as given in this answer can be easily modified to work here: if $f$ is a polynomial that vanishes on $U$ ($= X \cap U$), then it is zero on all of $\mathbf{R}^n$ by analyticity. Then $X$ is the Zariski closure of $X \cap U$.

In the general case, it seems like you would want to choose some analytic local coordinates, but this is not easy for at least two reasons:

  • The statement is clearly false if $X$ is allowed to be reducible, and when $U$ has empty intersection with any of its irreducible components. This shows how hard it is to make a global statement about the zeros of some function $f$ that vanishes on $X \cap U$.
  • Maybe $X \cap U$ contains singularities.

In the case of $X \subseteq \mathbf{C}^n$, the analogous statement is implied by the theorem in Mumford's book as cited by this answer.

I tried the following: Suppose that $f$ vanishes on $X \cap U$; we want that $f$ vanishes on $X$. Define $V = X \cap V(f)$ and $W = \overline{X \setminus V}$ (Zariski closure). Since $X$ is irreducible, it is sufficient if we can prove that $W \neq X$, because then $V=X$. I'm not sure if this is very helpful: At this point, I've basically just postponed comparing a Zariski closure with a Euclidean open set.

Thanks in advance for any insight!

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  • $\begingroup$ Let $X$ be an irreducible affine scheme of finite type over $\mathbf{R}$ with positive dimension. To study $X(\mathbf{R})$ one has to be careful about isolated points, so it seems safer to assume $X$ is absolutely irreducible over $\mathbf{R}$. The smooth locus is a Zariski-dense open, and we want $\mathbf{R}$-points of that arbitrarily close to a chosen point of an open subset of $X(\mathbf{R})$. Perhaps you can use Bertini to slice down to a curve and exploit features of curves? But that will only give Zariski density in the scheme $X$, perhaps weaker than in $X(\mathbf{R})$... $\endgroup$
    – user29283
    Commented May 13, 2013 at 16:52
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    $\begingroup$ To give a counterexample without absolute irreducibility, let $Y$ be an affine curve over $\mathbf{C}$ viewed as an $\mathbf{R}$-scheme and "pinch" $Y$ at a couple of $\mathbf{C}$-points to artificially lower the residue field there to be $\mathbf{R}$. Then that gives you an irreducible $\mathbf{R}$-scheme $X$ which is not absolutely irreducible, and $X(\mathbf{R})$ is just a couple of points, so you get a counterexample by choosing an open around just one of those points. $\endgroup$
    – user29283
    Commented May 13, 2013 at 16:54
  • $\begingroup$ e.g. $x^2+y^2+x^4=0$ or some similar equation. $\endgroup$
    – Will Sawin
    Commented May 13, 2013 at 22:30
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    $\begingroup$ Probably the equation of $x^2+y^2+y^3+x^3=0$ should be what you are looking for. This is a curve with an isolated singular point, and the intersection of any small ball around the origin only consist of the point. If the variety is smooth and contains real points, then I think that the answer is yes and it just follows from dimension arguments: the intersection $X\cap U$ has the same dimension as $X$ and is therefore Zariski dense. $\endgroup$
    – Jérémy Blanc
    Commented May 14, 2013 at 7:53
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    $\begingroup$ I have never seen the notion of absolutely irreducible, but on the web it seems to mean irreducible over $\mathbb{C}$, so $x^2+y^2+y^3+x^3$ is again a counterexample, and the same with any variety with isolated singularities. $\endgroup$
    – Jérémy Blanc
    Commented May 14, 2013 at 8:01

1 Answer 1

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This answer is distilled from the comments by Jérémy Blanc, François Brunault and xuhan.

The guess that $X$ is the Zariski closure of $X \cap U$, if the latter is nonempty, is false in general. As a counterexample, take the Zariski closed set in $\mathbf{R}^2$ given by the zero set of the equation

$x^2 + x^3 + y^2 + y^3 = 0$

It consists of a curve that asymptotically approached the line $x + y = 0$, together with the isolated point $(x,y)=(0,0)$. Taking any small Euclidean neighbourhood of this isolated point for $U$ is a counterexample to this guess.

However, if $X$ is in addition assumed to be a connected submanifold of $\mathbf{R}^n$, then the guess is true. Namely, if $X$ is a submanifold and Zariski closed, then it actually is a (real) analytic submanifold, meaning that analytic functions that vanish locally vanish, by analytic continuation, on the entire connected component. This implies that the Zariski closure of $X \cap U$ is the entire connected component containing $X \cap U$. Assuming $X$ to be connected, this was the 'guess' in the question.

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  • $\begingroup$ The equation seems wrong. It is not an irreducible curve: it factors into the line $x=y$ and the ellipse $x^2+xy+y^2+x+y=0$. If I am not mistaken, though, $y^2 + x^2+x^3 = 0$ does work. It is irreducible, and its real zero set is the union of a curve and of the isolated point $(0,0)$. $\endgroup$
    – Ilia Smilga
    Commented Mar 3, 2019 at 11:01
  • $\begingroup$ Ah, good point! The error is mine: I missed a sign when copying Jérémy Blanc's example. Thanks for the trip down memory lane. $\endgroup$
    – tkluck
    Commented Mar 3, 2019 at 13:37

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