Closed-subset and specialization chains compute the same scheme codimension

In an irreducible space, any two nonempty open subsets meet: otherwise their complements would be two proper closed subsets covering the space. It follows that a nonempty open subset is dense, since a proper closed superset would have a nonempty open complement disjoint from it. Such an open subset is also irreducible: two of its nonempty relatively open subsets are nonempty opens of the original space by [F2], so meet, and taking complements gives the irreducibility criterion. Now take a nonempty irreducible closed subset $Z\subseteq X$, and choose an affine open $W$ meeting it by [F1]. The subset $Z\cap W$ is closed in $W$ and is a nonempty irreducible open subset of $Z$, dense in $Z$. Choose an affine presentation of $W$ as in [F1], with underlying homeomorphism $h:W\to \mathrm{Spec}A$. It bijects the closed subsets of the two spaces and therefore carries smallest closed supersets to smallest closed supersets: $h({\bar{B}}^W)={\overline{h(B)}}^{\mathrm{Spec}A}$ for $B\subseteq W$. It also identifies relatively closed covers of subsets, so $h(Z\cap W)$ is nonempty, irreducible and closed. By the affine-spectrum theorem in [F1], this image has a unique generic point $p$. The closure identity shows that $\eta =h^{-1}(p)$ is the unique generic point of $Z\cap W$ in $W$. Its closure in $X$ lies in $Z$ and, by [F3], contains $Z\cap W$. Density then forces ${\overline{\{\eta \}}}^X=Z$. Any other generic point ${\eta }^{\mathrm{\prime }}$ of $Z$ must belong to $W$: otherwise the closed set $X\setminus W$ would contain ${\eta }^{\mathrm{\prime }}$ and hence all of $Z$, contradicting $Z\cap W\mathrm{\ne }\mathrm{\varnothing }$. By [F3], its closure in $W$ is also $Z\cap W$, so affine uniqueness gives ${\eta }^{\mathrm{\prime }}=\eta$. This proves existence and uniqueness in $X$.

The closure of any nonempty irreducible subset is irreducible. Indeed, two relatively closed subsets covering that closure are closed in $X$ by [F2]; their traces cover the original irreducible subset, which lies in one of them, and then its closure lies there by [F3]. In particular the closure of a point is irreducible, since a singleton is irreducible. Also each component $C$ is closed: its irreducible closure contains it, so maximality in [F2] makes that closure equal to $C$. By step 1.1, taking the generic point of a nonempty irreducible closed subset is inverse to taking the closure of a point. For points $a,b$, the relation $b\in \overline{\{a\}}$ implies $\overline{\{b\}}\subseteq \overline{\{a\}}$ by the smallest-closed-superset property; the converse follows from $b\in \overline{\{b\}}$. Equality of these two closures forces $a=b$ by generic-point uniqueness in step 1.1. Thus strict specialization is exactly strict containment of point closures in the indicated direction. Applying the inverse assignments term by term now gives the asserted bijection of chains, with endpoints $C,\overline{\{x\}}$ corresponding to ${\eta }_C,x$. This preserves each finite length, including zero. The sets of possible lengths are identical, so their suprema agree, also when unbounded. The empty scheme has neither points nor nonempty irreducible closed subsets, and all assertions are then vacuous.