If $Q \in\mathrm{Spec}(B)$ is the unique prime ideal lying over $P \in\mathrm{Spec}(A)$, then $B_P=B_Q$.

Question

Let $A \subseteq B$ be an integral extension of integral domains. Suppose $P \in \mathrm{Spec}(A)$ and $Q \in \mathrm{Spec}(B)$ is the unique prime ideal lying above $P$. Prove that $B_Q=B_P$.

My attempt: since $P=Q\cap A \subseteq Q$, we have $B_Q \subseteq B_P$. For the converse, suppose $b/s \in B_P$. Then $s \not\in P$. Since $P=Q \cap A$, we have $s \not\in Q \cap A$. That is, $s \not\in Q$ or $s \not\in A$. If it's the former, then that's what we want. If it's the latter, how should I arrive at a contradiction?

I don't know how to use the condition that $B$ is integral over $A$. If $s$ satisfies the equation $s^n+a_{n-1}s^{n-1}+\cdots+a_1s+a_0=0$ for some $a_{n-1},\ldots,a_0 \in A$. What can that do for me?

Can anyone give me a hint? Thank you!

Answer 1

Look at $B_P|_{A_P}$ which is an integral extension. The fact that only one prime lies over $PA_P$ implies $B_P$ is local. Also observe that $QB_P$ has to be the unique maximal ideal of $B_P$. Thus anything outside $Q$ is a unit in $B_P$ and hence $B_Q$ =$B_P$

Answer 2

Clearly $B_P\subseteq B_Q$. In order to prove that $B_P=B_Q$ notice that $B_Q$ is the localization of $B_P$ at $QB_P$. Now let's prove that $B_P$ is a local ring and its maximal ideal is $QB_P$. Any prime ideal of $B_P$ is of the form $Q'B_P$ with $Q'\cap A\subseteq P$. Since $Q$ is the only prime ideal lying over $P$ we have $Q'\cap A\subsetneq P$ provided $Q'\ne Q$. By Going Up there is a prime ideal $Q''$ of $B$ (strictly) containing $Q'$ and lying over $P$. By hypothesis $Q''=Q$, and thus $Q'\subsetneq Q$ proving that $QB_P$ is the only maximal ideal.