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