A question of the uniqueness of the prime ideal lying over in an integral extension (Corollary 5.9 in Atiyah's Commutative Algebra) I am reading Atiyah's Commutative Algebra.
I can understand all the proof of Corollary 5.9 except the sentence $\mathfrak{n}^c=\mathfrak{n}'^c=\mathfrak{m}$.
Question 1.
Why $\mathfrak{n}^c=\mathfrak{n}'^c=\mathfrak{m}$?
My attempt: Since $\mathfrak{p}\subseteq \mathfrak{q}$, 
we have $\mathfrak{m}=S^{-1}\mathfrak{p}\subseteq S^{-1}\mathfrak{q}=\mathfrak{n}\subseteq B_{\mathfrak{p}}$.
But $\mathfrak{m}$ is maximal in $A_{\mathfrak{p}}$, 
which is not necessarily maximal in $B_{\mathfrak{p}}$. 
I can't get $\mathfrak{m}=\mathfrak{n}$ by this.
Question 2.
When we use that notation $A_{\mathfrak{p}}$, 
which means the localization $S^{-1}A$ of $A$ at the prime ideal $\mathfrak{p}$ of $A$.
But in this corollary, $\mathfrak{p}$ doesn't necessarily be a prime ideal of $B$. Why can he write $B_{\mathfrak{p}}$?
Should we write $S^{-1}B$ rigorously?

Corollary 5.9.
  Let $A\subseteq B$ be rings, 
  $B$ integral over $A$;
  let $\mathfrak{q}, \mathfrak{q}'$ be prime ideals of $B$ such that $\mathfrak{q}\subseteq\mathfrak{q}'$ and $\mathfrak{q}^c=\mathfrak{q}'^c=\mathfrak{p}$ say.
  Then $\mathfrak{q}=\mathfrak{q}'$.
Proof. By (5.6), 
  $B_{\mathfrak{p}}$ is integral over $A_{\mathfrak{p}}$.
  Let $\mathfrak{m}$ be the extension of $\mathfrak{p}$ in $A_{\mathfrak{p}}$ and let $\mathfrak{n}, \mathfrak{n}'$ be the extensions of $\mathfrak{q}, \mathfrak{q}'$ respectively in $B_{\mathfrak{p}}$.
  Then $\mathfrak{m}$ is the maximal ideal of $A_{\mathfrak{p}}$;
  $\mathfrak{n}\subseteq \mathfrak{n}'$, 
  and $\mathfrak{n}^c=\mathfrak{n}'^c=\mathfrak{m}$.
  By (5.8) it follows that $\mathfrak{n}, \mathfrak{n}'$ are maximal, 
  hence $\mathfrak{n}=\mathfrak{n}'$, 
  hence by (3.11)(iv) $\mathfrak{q}=\mathfrak{q}'$.
Corollary 5.6.
  Let $A\subseteq B$ be rings, 
  $B$ integral over $A$.
  i) If $\mathfrak{b}$ is an ideal of $B$ and $\mathfrak{a}=\mathfrak{b}^c=A\cap \mathfrak{b}$, 
  then $B/\mathfrak{b}$ is integral over $A/\mathfrak{a}$.
  ii) If $S$ is a multiplicatively colsed subset of $A$, 
  then $S^{-1}B$ is integral over $S^{-1}A$.
Corollary 5.7.
  Let $A\subseteq B$ be integral domains, 
  $B$ integral over $A$.
  Then $B$ is a field if and only if $A$ is a field.
Corollary 5.8.
  Let $A\subseteq B$ be rings, 
  $B$ integral over $A$;
  let $\mathfrak{q}$ be a prime ideal of $B$
  and let $\mathfrak{p}=\mathfrak{q}^c=\mathfrak{q}\cap A$.
  Then $\mathfrak{q}$ is maximal if and only if $\mathfrak{p}$ is maximal.

 A: To answer your first question, look at the commutative diagram
$$\require{AMScd}
\begin{CD}
A @>{}>> B\\
@VVV @VVV \\
A_\mathfrak p @>{}>> B_\mathfrak p
\end{CD}$$
Going the right way, $\mathfrak n$ is contracted to $\mathfrak p$ by assumption. Hence the same holds for the left way. But the map $\operatorname{Spec} A_\mathfrak p \to \operatorname{Spec} A$ is well known to be injective and the sole pre-image of $\mathfrak p$ is $\mathfrak m$. Thus $\mathfrak n$ is contracted to $\mathfrak m$ by the map $A_\mathfrak p \to B_\mathfrak p$. Of course the same arguments works for $\mathfrak n'$.
For your second question, note that $B$ is an $A$-module, so this is just the usual notation $M_\mathfrak p$, when $M$ is an $A$-module and $\mathfrak p$ a prime of $A$.
A: Since $\mathfrak{p}\subseteq \mathfrak{q}$,
we have $\mathfrak{m}=\mathfrak{p}^{e}\subseteq \mathfrak{q}^{e}\cap A_{\mathfrak{p}}=\mathfrak{n}\cap A_{\mathfrak{p}}$.
Since $\mathfrak{q}\cap A=\mathfrak{p}$,
we have $\mathfrak{q}\cap (A-\mathfrak{p})=\emptyset$.
By Lemma III.4.9.(iii) in Hungerford's Algebra,
$\mathfrak{n}=\mathfrak{q}^{e}$ is prime in $B_{\mathfrak{q}}$.
It follows that $\mathfrak{n}\cap A_{\mathfrak{p}}$ is prime in $A_{\mathfrak{p}}$ ($\mathfrak{n}\cap A_{\mathfrak{p}}$ is the inverse image of $\mathfrak{n}$ under the inclusive mapping from $A_{\mathfrak{p}}$ to $B_{\mathfrak{q}}$)
and $\mathfrak{n}\cap A_{\mathfrak{p}}\neq A_{\mathfrak{p}}$.
Since $\mathfrak{m}$ is maximal in $A_{\mathfrak{p}}$ and $\mathfrak{m}\subseteq \mathfrak{n}\cap A_{\mathfrak{p}}\neq A_{\mathfrak{p}}$,
we have $\mathfrak{n}\cap A_{\mathfrak{p}}=\mathfrak{m}$.
