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
 A: 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 $A_{\mathfrak{p}}$.
