Let $X$ be a smooth $\mathbb{R}$ irreducible algebraic set, where the closed sets zero sets in $\mathbb{R}^n$ of real polynomials.
I am trying to understand why dimension of $X$ as a manifold equals the dimension defined topologically (given as the length of a maximal chain of irreducible algebraic sets).
Any comments are appreciated. Thank you.
The tangent space $T_PX$ of a differential manifold $X$ at a point $P$ is isomorphic to the space of (real) linear derivation of the ring of germs of $X$ at $P$.
On the other hand the Zariski tangent space $T_PX$ of $X$ at a point $P$ is isomorphic to the space of $\mathbb{R}$-linear derivation of $X$ at $P$.
From all this: \begin{equation} \forall P\in X,\,\dim X=\dim_{\mathbb{R}}T_PX=\dim_{\mathbb{R}}\mathfrak{m}_{P\displaystyle/\mathfrak{m}_P^2}=\dim_{Krull}\mathcal{O}_{X,P}=\dim_{Krull}X; \end{equation} where $(\mathcal{O}_{X,P},\mathfrak{m}_P)$ is the local ring of germs of regular functions on $X$ at $P$.
Remark: This reasoning holds on complex numbers field.
A consequence. If $X$ is a semisimple, complex Lie group, then its Lie algebra $T_{1_X}X$ is its Lie algebra as semisimple, complex algebraic group.