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Let $S \subseteq \mathbb{R}^n$ be a subset of real points and $I(S)$ be the vanishing ideal of $S$ in $\mathbb{R}[x_1,\dotsc,x_n]$. Is $\dim V_{\mathbb{R}}(I(S)) = \dim V_{\mathbb{C}}(I(S))$? I.e., is the dimension of the real zero set of $I(S)$ equal to the dimension of the complex zero set of $I(S)$?

I think this is true, but I can't prove it rigorously. Any comments or answers are appreciated.

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    $\begingroup$ Yes, this is true. The point is that the vanishing ideal of the set $S' = V_{\mathbb{R}}(I(S))$ is also $I(S)$, so $S'$ must contain a smooth point on each irreducible component of the variety $V$ defined by $I(S)$. Now use the fact that if an irreducible real variety contains a smooth real point then the dimension of the set of real points is the same as the (algebraic) dimension of the variety. $\endgroup$
    – naf
    Commented Oct 1, 2022 at 7:37
  • $\begingroup$ I'm not sure I agree. Consider $S=\{x^2+y^2+1\}$. $\endgroup$
    – Donu Arapura
    Commented Oct 1, 2022 at 10:52
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    $\begingroup$ @DonuArapura: the ideal of vanishing of that set over R is the zero ideal. The variety of this ideal over the complex numbers is still empty. Pew is starting with a set of points, not a variety. $\endgroup$
    – Steven Gubkin
    Commented Oct 1, 2022 at 14:19
  • $\begingroup$ OK, I guess I didn't read the question carefully. $\endgroup$
    – Donu Arapura
    Commented Oct 1, 2022 at 16:58
  • $\begingroup$ @naf thank you. Do you have a reference why $S’$ must have a smooth point on each irreducible component? $\endgroup$
    – Pew
    Commented Oct 1, 2022 at 19:39

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