Let $\phi: \mathbb{R}^m \rightarrow \mathbb{R}^n$ be a polynomial map with rational coefficients. Let $M$ be the Zariski closure of the image of $\phi$ (in the complex space $\mathbb{C}^n$). Then the following holds
$$ \dim(M) = \dim(\mathbb{R}^m) - \dim(\phi^{-1}(\phi(p))) $$ where $p$ is a generic point in $\mathbb{R}^m$ (generic means its coordinates are algebraic independent over the rational field $Q$).
My question is: why is this true? This seems to be well-known as it is given without a reference in a paper I read. Does anyone know a reference of this equality?
Expose 8 in Cartan and Chevalley, Geometrie Algebrique, Seminar Cartan-Chevalley, Secretariat Math., Paris, (1955/56) is the original reference, though this might not be the best to learn from (it is in French and uses early language).
This is also available as Hartshorne's exercise II.3.22 (one solution on MSE here) or can be found in section 11.4 of Vakil's Foundations of Algebraic Geometry.
(PS: your definition of a generic point is incorrect. What you want to interpret this statement as is that all $p$ except possibly those in some proper closed subset work. A counterexample to what you've written is $\Bbb R^2\to\Bbb R^2$ by $(x,y)\mapsto (x-x_0,(x-x_0)y)$: take $x_0$ and $y_0$ to be such that $\Bbb Q(x_0,y_0)$ is of transcendence degree 2 over $\Bbb Q$, then $\dim \varphi^{-1}(\varphi((x_0,y_0)))=1$ since $\varphi^{-1}(\varphi((x_0,y_0)))$ is $\{(x_0,y)\mid y\in\Bbb R\}$. But $M$ has dimension two.)