Question regarding two equivalent definitions of Dedekind domains 

Theorem: If $A$ is a Noetherian integral domain, the following two properties are equivalent.
     1) $A_{\mathfrak p}$ is a DVR for every prime ideal $\mathfrak p \neq 0$;
     2) $A$ is integrally closed and of dimension $\leq 1$.


Here is the proof of 1) implies 2).


If $\mathfrak p \subset \mathfrak p^{'}$, then  $A_{\mathfrak p^{'}}$ contains the prime ideal $\mathfrak p^{'}  A_{\mathfrak p^{'}}$, which
    implies $\mathfrak p = 0$ or $\mathfrak p = \mathfrak p^{'}$. On the other hand, if $a$ is integral over $A$ it is a fortiori integral over each $A_{\mathfrak p}$ and it belongs to all the $A_{\mathfrak p}$. If one writes $a$ in the form $a = b/c$ and $c \neq 0$, and if
    $\mathfrak A$ is the ideal of those $x \in A$ such that $xb \subset cA$, the ideal $\mathfrak A$ is not contained
    in any prime ideal $\mathfrak p$, whence $\mathfrak A = A$ and $a \in A$.


Can somebody explain me that second part of the proof? How does $a \in A$? What is the purpose of $\mathfrak A$?
 A: The idea is that $A$ is the ideal of all possible denominators of the fraction $a = b/c.\,$ Thus, since $A$ is not contained in any prime ideal, $A = (1)$, so $1\in A,$ i.e. $1$ is a denominator for $a$, hence $\,a \in A.$
Indeed $\, d\in A\iff \exists e\in\! A\!:\ db = ce\!\iff\! \exists e\in\! A\!:\ b/c = e/d\!\iff\! d\,$ is a denominator for $\,b/c$
Denominator ideals play a fundamental conceptual role in number theory, e.g. conductor ideals.
A: In your proof it's used the following result

Let $A$ be an integral domain. Then $A=\bigcap A_P$, where $P$ runs over the set of all prime (actually it is enough to consider maximal) ideals of $A$. 

Obviously $A\subseteq \bigcap A_P$. Conversely, if $a\in\bigcap A_P$, then one can find for each $P$ an element $x_P\in A-P$ such that $x_Pa\in A$. The ideal $\mathfrak A$ of $A$ generated by all $x_P$ is not contained in any prime (maximal) ideal of $A$ (why?), so $\mathfrak A=A$. Thus $1\in \mathfrak A$ and we can write $1=\sum a_Px_P$, where $a_P\in A$. Multiplying this by $a$ we get $a\in A$.
