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

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

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.

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

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

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