Dimension zero for the empty set loses the empty-fibre distinction

Given: The objects and hypotheses in the statement refuted, for the explicit witness below.

For a Noetherian topological space $T$, define $\dim T$ as the supremum of the lengths $s$ of strict chains $Z_0⊊\cdots ⊊Z_s$ of nonempty irreducible closed subsets of $T$. Thus a one-member chain has length zero. Set $\dim \mathrm{\varnothing }=-\mathrm{\infty }$, and allow $\dim T=+\mathrm{\infty }$. The supremum of an empty family of dimensions is $-\mathrm{\infty }$. (Chain dimension and the empty-space convention).

For a morphism $f:X\to Y$ of classical varieties and a closed point $y\in Y$, let $X_y=f^{-1}(y)$ have its reduced closed-subvariety structure. Its dimension is the chain dimension, with $\dim X_y=-\mathrm{\infty }$ if the fibre is empty. On affine charts $V\subseteq Y$ containing $y$ and $U\subseteq f^{-1}(V)$, writing $A=k[V]$ and $B=k[U]$, the fibre chart has coordinate ring $B\mathrm{/}\sqrt{{\mathfrak{m}}_{y}B}$. Here general morphisms have the locally ringed-space meaning; the earlier affine morphism definition applies to the restrictions $U\to V$. 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. (Reduced closed-point fibres and their dimension).