Dimension equals transcendence degree

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 irreducible classical $X$, the fraction fields of all nonempty affine charts identify canonically; denote the resulting field by $k(X)$. A dominant morphism $f:X\to Y$ between irreducible classical varieties induces an injection $f^{\ast }:k(Y)\hookrightarrow k(X)$. Dominant means that the image is dense. 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. (Function fields and dominant pullbacks on general varieties).

For every open cover $T={\bigcup }_{i\in I}{U}_i$ of a Noetherian space, $\dim T={\sup }_i\dim U_i$, with empty supremum $-\mathrm{\infty }$. (Dimension can be computed on an open cover).

Let $k$ be a field, let $A$ be a finite-type $k$-domain, and let $K=\mathrm{Frac}(A)$. Then $\dim A={\mathrm{trdeg}}_{k}K\mathrm{.}$ (Affine-domain dimension equals transcendence degree).