Integral extensions: one prime lying over implies equal localization

Here's a problem from Matsumura's book "Commutative ring theory" page $69$.

Let $A$ be a ring and let $A \subset B$ be an integral extension, and $\mathfrak{p}$ a prime ideal of $A$. Suppose that $B$ has just one prime ideal $P$ lying over $\mathfrak{p}$. Then $B_{P}=B_{\mathfrak{p}}$.

Solution (page $290$):

$B_{\mathfrak{p}}$ is integral over $A_{\mathfrak{p}}$, so that any maximal ideal of $B_{\mathfrak{p}}$ lies over $\mathfrak{p}A_{\mathfrak{p}}$ and therefore coincides with $PB_{\mathfrak{p}}$. Hence $B_{\mathfrak{p}}$ is a local ring and the elements of $B \setminus P$ are units of $B_{\mathfrak{p}}$.

I don't understand the proof at all. Lot of questions:

1) Where it says "any maximal ideal of $B_{\mathfrak{p}}$ lies over $\mathfrak{p}A_{p}$ , why is this? And isn't $B_{\mathfrak{p}}$ a local ring? so why the expression any maximal ideal of $B_{\mathfrak{p}}$?

2) Then it says hence $B_{\mathfrak{p}}$ is a local ring. Isn't the localization always a local ring?

3) Is it possible that you can please give another proof of the exercise (or explain it in detail)? I'm really confused about this proof.

Thanks

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What happened to Arturo's answer? I was just going to accept it! –  user6495 Jul 25 '11 at 19:36
@Theo Buehler: is that OK? –  user6495 Jul 25 '11 at 20:04
@user6495: I deleted it because I was glossing over the issue of module localization vs. ring localization. –  Arturo Magidin Jul 25 '11 at 20:11
@user6495: Yes, much better! Thank you. –  t.b. Jul 25 '11 at 20:11

First of all, it should be pointed out that $B_{\mathfrak{p}}$ is the localization of the $A$ module $B$ at $\mathfrak{p}$. It is isomorphic to the ring $A_{\mathfrak{p}} \otimes_A\ B$ as an $A$ module. It need not be the case that $\mathfrak{p}$ is a prime ideal of $B$, thus to localize $B$ at $\mathfrak{p}$ as a ring doesn't make sense (for $B -\mathfrak{p}$ would not be a multiplicatively closed set). This should clear up your confusion about $B_{\mathfrak{p}}$ not a priori being a local ring.
That any maximal ideal $\mathfrak{m}$ of $B_{\mathfrak{p}}$ lies over $\mathfrak{p}A_{\mathfrak{p}}$ is a standard result used to prove the going up theorem. For since $B_{\mathfrak{p}}$ is integral over $A_{\mathfrak{p}}$, we then have that $B_{\mathfrak{p}} /\mathfrak{m}$ is integral over $A_{\mathfrak{p}}/(\mathfrak{m}\cap A_{\mathfrak{p}})$. But $B_{\mathfrak{p}} /\mathfrak{m}$ is a field, so $A_{\mathfrak{p}}/(\mathfrak{m}\cap A_{\mathfrak{p}})$ must be one as well, whence $\mathfrak{m}$ lies over $\mathfrak{p}A_{\mathfrak{p}}$ since this is the unique maximal ideal of $A_{\mathfrak{p}}$.
Why does it not make sense to localize $B$ at $p$, $S=A-p$ is a multiplicatively closed set of $A$ and hence of $B$, as $A\subset B$. The point is that $S^{-1}B=B_p$ is not a local ring. –  messi Jun 8 '12 at 19:32