I am working through Stephen McAdam's Asymptotic Prime Divisors, and I've hit a snag on the proof of Lemma 3.1 with a question that basically amounts to:
Let $A \subseteq B$ be an extension of integral domains. Let $(0)\ne \mathfrak{p} \in \mathrm{Spec}(A)$ and suppose $\exists P \in \mathrm{Spec}(B)$ with $P \cap A = \mathfrak{p}$. Assume the transcendence degree of $Q(B/P)$ over $Q(A/\mathfrak{p})$ is $0$. Moreover, suppose $\mathrm{ht}(P) = 1$. Show that $P$ is isolated among prime ideals of $B$ lying over $\mathfrak{p}$.
If you are unfamiliar, to be isolated as a lying-over prime means to be both maximal and minimal with respect to this property.
So far I have used the fact that $\mathrm{ht}(P) = 1$ to show that $P$ is minimal with respect to lying over $\mathfrak{p}$. How do I show $P$ is maximal with respect to this property?
It seemingly must involve that $Q(A/\mathfrak p) \subseteq Q(B/P)$ is an algebraic extension, but I am not sure how to factor that into an argument.
Thanks.