Dimension of an algebraic variety

Remarks

For every variety this supremum is finite. If $X$ is an affine variety, then $\dim X$ is the Krull dimension of its coordinate ring $k[X]$. The reason is that irreducible closed subsets of $X$ correspond to prime ideals of $k[X]$, with inclusions reversed. For every variety, $\dim X$ equals the transcendence degree of the function field $k(X)$ over $k$. A nonempty open subset $U$ of $X$ has $\dim U = \dim X$.

Example: $\dim \mathbb{A}^n = \dim \mathbb{P}^n = n$, since $k(x_1, \dots, x_n)$ has transcendence degree $n$ over $k$. The chain $\set{0} \subsetneq V(x_2, \dots, x_n) \subsetneq \cdots \subsetneq V(x_n) \subsetneq \mathbb{A}^n$ of linear subspaces has length $n$.

Example: if $f \in k[x_1, \dots, x_n]$ is irreducible and not constant, the hypersurface $V(f) \subseteq \mathbb{A}^n$ has dimension $n - 1$.

The dimension of a reducible algebraic set is the largest dimension of its irreducible components.