A nontrivial principal section has pure codimension one

Given: The objects and hypotheses in the statement.

For a nonempty affine algebraic set $X$, $\dim X=\dim k[X]$, where the right side is Krull dimension. For this comparison only, extend ring dimension to the zero ring by $\dim (0)=-\mathrm{\infty }$; then the equality also holds for $X=\mathrm{\varnothing }$. 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. (Affine geometric dimension equals ring dimension).

For a nonempty irreducible closed subvariety $Z$ of an irreducible classical variety $X$, define ${\mathrm{codim}}_{X}Z=\dim X-\dim Z$. These are finite integers. In a reducible ambient variety a difference of global dimensions must not be substituted for the height of a local prime; the containing component matters. 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. (Codimension of an irreducible closed subvariety).

Let $R$ be a Noetherian commutative ring, let $x\in R$, and let $\mathfrak{p}$ be a prime ideal minimal over $(x)$. Then $\mathrm{ht}(\mathfrak{p})\le 1$. (Krull's principal ideal theorem).

Let $k$ be a field, let $A$ be a finite-type $k$-domain, and let $\mathfrak{p}\in \mathrm{Spec}(A)$. Then $\mathrm{ht}(\mathfrak{p})+\dim (A\mathrm{/}\mathfrak{p})=\dim A\mathrm{.}$ (Height plus quotient dimension equals ambient dimension in an affine domain).

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).