# How to show $\dim_{\mathbb{A}_{\mathbb{R}}^n} V= \dim_{\mathbb{A}_{\mathbb{C}}^n} V$?

Suppose $V$ is an affine algebraic set defined by real polynomials. Let $\mathbb{A}_{\mathbb{R}}^n$ be $\mathbb{R}^n$ endowed with Zariski topology where the closed sets are algebraic sets (in $\mathbb{R}^n$) defined by real polynomials. Let $\mathbb{A}_{\mathbb{C}}^n$ be the usual affine $n$ space. Suppose $V(\mathbb{R})$ has at least one non-singular point. Then I think it should be that $$\dim_{\mathbb{A}_{\mathbb{R}}^n} V(\mathbb{R}) = \dim_{\mathbb{A}_{\mathbb{C}}^n} V(\mathbb{C}).$$

I was wondering how I can prove this... Any comments are appreciated. Thank you very much.

The upper bound on the real Dimension needs no assumption on the non-singular point. This frees us to perform induction on $n$. With finitely many exceptions, the fibers of the projection to the first coordinate have complex dimension one less than $\dim V$, and the remainder have dimension $\dim V$. Hence by induction their real dimensions are bounded by the same. Whatever your definition of dimension, it should be easy to conclude.