Closed-point dimension lemma for affine domains ($\dim A_{\mathfrak{m}}=\dim A$)

If $A$ is a finite-type integral domain over an algebraically closed field and $\dim A=d$, every maximal ideal $\mathfrak{m}$ has height of a prime ideal $d$. Choose a Noether normalization $k[z_1,\dots ,z_d]\subset A$. The contraction of $\mathfrak{m}$ is a maximal ideal of the polynomial ring and has height $d$; going-down theorem lifts its full chain since the polynomial ring is integrally closed domain. The upper bound is $\dim A$. A principal open subset containing a closed point consequently has dimension $d$, because the whole chain under that point survives localization.

For an irreducible variety which is an affine variety of dimension $d$ and a nonzero nonunit regular function $f$, each irreducible component of $V(f)$ has dimension $d-1$. Choose a closed point on just the component in question. In its local ring $R$, $\sqrt{(f)}$ is that component's prime ideal. A system of parameters of $R/(f)$, lifted and supplemented by $f$, generates a maximal-primary ideal; the Krull height theorem gives $\dim R/(f)\ge d-1$. Extending any chain above $(f)$ by the zero prime of the domain gives the reverse inequality. This proof does not identify dimension with transcendence degree.