Local fibre dimension equals local ring dimension plus residue transcendence degree

By [F1], the minimal primes ${\mathfrak{a}}_1,\dots ,{\mathfrak{a}}_r$ of $A$ contained in $\mathfrak{q}$ form a nonempty finite list. The components of $\mathrm{Spec}A$ through $\mathfrak{q}$ are $V({\mathfrak{a}}_j)$. By [F3], ${\dim }_{\mathfrak{q}}\mathrm{Spec}A={\max }_j\dim (A\mathrm{/}{\mathfrak{a}}_j)\mathrm{.}$ Every prime chain of $A_{\mathfrak{q}}$ starts above one of these minimal primes, so $\dim A_{\mathfrak{q}}={\max }_j{\mathrm{ht}}_{A\mathrm{/}{\mathfrak{a}}_j}(\mathfrak{q}\mathrm{/}{\mathfrak{a}}_j)$.

Apply [F2] to each domain $A\mathrm{/}{\mathfrak{a}}_j$ and its prime $\mathfrak{q}\mathrm{/}{\mathfrak{a}}_j$. The residue field at that prime is $\kappa (\mathfrak{q})$, independent of $j$, and [F2] gives $\dim (A\mathrm{/}{\mathfrak{a}}_j)={\mathrm{ht}}_{A\mathrm{/}{\mathfrak{a}}_j}(\mathfrak{q}\mathrm{/}{\mathfrak{a}}_j)+{\mathrm{trdeg}}_k\kappa (\mathfrak{q})\mathrm{.}$ Taking maxima over $j$ and using step 1.1 yields the displayed equality. AC is inherited by the affine-domain dimension theorem; only finitely many components are compared.