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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.

user26857
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beanshadow
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    Suppose that there is $Q\supsetneq P$ another prime lying over $\mathfrak p$. Let $b\in Q\setminus P$. Since $\bar b\in B/P$ is algebraic over $Q(A/\mathfrak p)$, there is a non-zero multiple of $\bar b$ in $A/\mathfrak p$. But this is a contradiction with $(Q/P)\cap (A/\mathfrak p)=(0)$. – user26857 Feb 18 '19 at 10:21
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    As a side remark, I always found that book as unreadable. – user26857 Feb 18 '19 at 10:23
  • Thanks 26857! Always you appear in my fortunes on stack exchange. – beanshadow Feb 18 '19 at 18:39

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