If $B$ is integral over $A$ and there is only one $P$ over $\mathfrak p$, then $B_P=B_\mathfrak p$?

Question

I am stuck in the Exercise 5.3 of Matsumura's Commutative Algebra: $\newcommand{\p}{\mathfrak p}$ $\newcommand{\sp}{\operatorname{Spec}}$

Let $B$ be a ring, $A$ be a subring and $\p\in\sp(A)$. Suppose that $B$ is integral over $A$ and that there is only one prime ideal $P$ of $B$ over $\p$. Then $B_P=B_\p$, where $B_\p=B\otimes_A A_\p$.

In addition, this book gives a hint: show that $B_\p$ is a local ring with maximal ideal $PB_\p$. I am really confused about this problem, especially in the following points:

• What is the exact meaning of $B_P=B_\p$? Can I say $B_P\subset B_\p$ since $\p\subset P$?

• Even if I have showed that $B_\p$ has a unique maximal ideal $PB_\p$, how does this fact imply $B_P=B_\p$?

• I know that the going-up theorem holds for $A\hookrightarrow B$ since $B$ is integral over $A$, but how can the condition that there is only one prime ideal $P$ such that $P\cap A=\p$ can be used to show that $(B_\p, PB_\p)$ is a local ring?

Thus I would like to ask for some explanation and further hints. Thanks in advance...

Answer 1

I hope my thinking is right. Here is what I have come up with.

• $(B_\mathfrak{p},PB_\mathfrak{p})$ is local: If we rewrite $B_\mathfrak{p} = S^{-1}B$ and $A_\mathfrak{p} = S^{-1}A$ with $S = A\setminus \mathfrak{p}$, we see that since $A \subseteq B$ is integral, $A_\mathfrak{p} = S^{-1}A \subseteq S^{-1}B = B_\mathfrak{p}$ is therefore also integral (and it is still an inclusion!). Let $Q \in \operatorname{Spec} B_\mathfrak p$ be a prime ideal. Then, since $(A_\mathfrak p , \mathfrak p A_\mathfrak p)$ is local, we have $Q \cap A_\mathfrak p \subseteq \mathfrak p A_\mathfrak p$, and by going-up, there exists a $Q^\prime \in \operatorname{Spec} B_\mathfrak p$ such that $Q \subseteq Q^\prime$ and $Q^\prime \cap A_\mathfrak p = \mathfrak p A_\mathfrak p$. By ideal correspondence, $Q^\prime$ must be of the form $Q^\prime = P^\prime B_\mathfrak p $ with $P^\prime \in \operatorname{Spec} B$ and $P^\prime \cap A = \mathfrak p$. By our assumption, we must have $P^\prime = P$. This concludes that $(B_\mathfrak p , PB_\mathfrak p)$ is local.
• Since $P \cap A = \mathfrak p$, we have an inclusion $A\setminus \mathfrak p \subseteq B \setminus P$ which should imply $\require{enclose}\enclose{horizontalstrike}{B_\mathfrak p = (A\setminus \mathfrak p)^{-1}B \subseteq (B \setminus P)^{-1}B = B_P}$ EDIT: which gives us a homomorphism of rings $B_\mathfrak p = (A\setminus \mathfrak p)^{-1}B \rightarrow (B \setminus P)^{-1}B = B_P$ (a priori it is not injective, since we could have zero-divisors in our multiplicative subsets).

I think $B_P = B_\mathfrak p$ shortly follows. I hope I didn't overlook anything. Good luck!