Under going down and incomparability, lying-over primes have the same finite height

By [L5], strict prime chains below $\mathfrak{p}$ in $A$ correspond to strict prime chains in $A_{\mathfrak{p}}$, and strict prime chains below $\mathfrak{q}$ in $B$ correspond to strict prime chains in $B_{\mathfrak{q}}$. Since [L1] defines height as the dimension of these local rings and [L2] defines dimension as the supremum of the lengths of strict prime chains, it is enough to compare finite strict chains below $\mathfrak{p}$ and $\mathfrak{q}$ in the original rings.

Let ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_n=\mathfrak{p}$ be any finite strict prime chain in $A$. Repeatedly applying [L3] from the top prime $\mathfrak{q}$ downward produces primes ${\mathfrak{q}}_0\subseteq \cdots \subseteq {\mathfrak{q}}_n=\mathfrak{q}$ with ${\mathfrak{q}}_i\cap A={\mathfrak{p}}_i$. These inclusions are strict, because ${\mathfrak{q}}_i={\mathfrak{q}}_{i+1}$ would force ${\mathfrak{p}}_i={\mathfrak{p}}_{i+1}$. Therefore step 1.1 gives $\mathrm{ht}(\mathfrak{q})\ge \mathrm{ht}(\mathfrak{p})$.

Conversely, let ${\mathfrak{q}}_0⊊\cdots ⊊{\mathfrak{q}}_m=\mathfrak{q}$ be any finite strict prime chain in $B$. The contractions form a chain ending at $\mathfrak{p}$, and [L4] makes each adjacent contraction strict. Hence step 1.1 gives $\mathrm{ht}(\mathfrak{p})\ge \mathrm{ht}(\mathfrak{q})$.

If one of the two heights is finite, the inequalities from steps 2.1 and 2.2 force the other to be finite and equal to it. Therefore $\mathrm{ht}(\mathfrak{q})=\mathrm{ht}(\mathfrak{p})$ whenever one of them is finite.