Dimension of an analytic set

Remarks

The definition makes sense because $\operatorname{reg} A$ is dense in $A$; at a regular point $a$ it gives back the manifold dimension. Example: let $A = \set{z_1 z_3 = 0,\ z_2 z_3 = 0}$ in $\C^3$, the union of the plane $P = \set{z_3 = 0}$ and the line $L = \set{z_1 = z_2 = 0}$. The regular points are the points of $P \setminus \set{0}$, of dimension $2$, and the points of $L \setminus \set{0}$, of dimension $1$. So $\dim_0 A = 2$, $\dim_a A = 1$ for $a \in L \setminus \set{0}$, and $\dim A = 2$. Non-example of pure dimension: this set $A$ does not have pure dimension, since its local dimension takes the values $1$ and $2$. It also shows that the dimension is not the number of variables minus the number of equations, which would be $1$ here. There are equivalent forms of $\dim_a A$ when $X = \C^n$. It is the Krull dimension of $\mathcal{O}_{n,a}/\mathcal{I}(A_a)$. It is the largest of the integers $d$ given by the local parametrization theorem for the irreducible components of the germ $A_a$. It is the smallest $k$ such that some complex affine subspace of codimension $k$ through $a$ meets $A$ in the isolated point $a$.