Affine geometric dimension equals ring dimension

Given: The objects and hypotheses in the statement.

For a classical variety $X$, let $\dim X$ be its chain dimension. If $X_1,\dots ,X_m$ are its irreducible components and $x\in X$ is a closed point, define ${\dim }_{x}X={\max }_{x\in X_i}\dim X_i$. The indexing family is nonempty. Say that $X$ has pure dimension $d$ if every irreducible component has dimension $d$; the condition on components is vacuous for the empty variety, whose dimension is nevertheless $-\mathrm{\infty }$. Work over a fixed algebraically closed field $k$, with the Axiom of Choice. Classical varieties are separated and admit finite affine covers; they may be reducible or empty unless irreducibility is specified. Irreducible means nonempty. All fibres and points below are classical closed-point fibres and points. (Global and local dimension of classical varieties).

Assume the Axiom of Choice. Let $k$ be an algebraically closed field. 1. The assignments $X⟼I(X),J⟼V(J)$ induce mutually inverse inclusion-reversing correspondences between affine algebraic sets $X\subseteq {\mathbf{A}}_k^n$ and radical ideals $J\subseteq k[x_1,\dots ,x_n]$. 2. Under this correspondence, nonempty irreducible affine algebraic sets correspond exactly to prime ideals. (Affine algebraic sets correspond to radical ideals, and irreducible ones to prime ideals).

Let $R$ be a nonzero commutative ring. A strict chain of prime ideals of length $n$ is a sequence ${\mathfrak{p}}_0⊊{\mathfrak{p}}_1⊊\cdots ⊊{\mathfrak{p}}_n$ of prime ideals of $R$. The Krull dimension of $R$ is the supremum of all integers $n\ge 0$ for which such a chain exists. This supremum is allowed to be infinite. On this page the zero ring is left outside the definition so that later chain statements do not hide that degenerate boundary. (Krull dimension of a nonzero ring).