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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 $\mathfrak A$?

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    $\begingroup$ @Downvoter Can you please explain the downvote? $\endgroup$ – Mohan Mar 2 '13 at 3:00
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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.

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  • $\begingroup$ Why $A$ is not contained in any prime ideal? $\endgroup$ – Mohan Mar 1 '13 at 17:04
  • $\begingroup$ Think about what it means for $\mathfrak p \supseteq \mathfrak A.\:$ This says that every possible denominator for $a$ is in $\mathfrak p$ (is "divisible" by $\mathfrak p$) which contradicts $a$ is $\mathfrak p$-integral (which means $a$ can be written as a fraction with denominator not in $\mathfrak p$, i.e. not "divisible" by $\mathfrak p$) $\endgroup$ – Math Gems Mar 1 '13 at 17:13
  • $\begingroup$ Thanks for the answer. I get it now. $\endgroup$ – Mohan Mar 2 '13 at 3:33
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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 $a\in A$.

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  • $\begingroup$ Actually it is enough to utilize any set S of primes such that the denominator ideals of proper fractions over $A$ are contained in some prime P in S. Thus associated to every proper fraction is a witness P in S, which asserts that all denominators of the fraction lie in P (are "divisible" by P), therefore $1$ is not a denominator, i.e. the fraction is proper. In other words, S must be large enough to contain a witness for every denominator ideal $\neq (1).$ $\endgroup$ – Math Gems Mar 1 '13 at 17:55
  • $\begingroup$ The above comment (after being stripped of the explicit denominator language) is essentially Exercise 1-6-20, p. 42 in Kaplansky's Commutative Rings. $\endgroup$ – Math Gems Mar 1 '13 at 18:16

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