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. 



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


The lower bound follows from considering the tangent space at the non-singular point.


