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I think the following proposition is likely to be true. I'd like to know a proof of it if any.

Proposition Let $A$ be an integral domain, $K$ its field of fractions. Let $P_1, ..., P_n$ be prime ideals of $A$. Let $S = (A - P_1)\cap\cdots\cap(A - P_n)$. If we regard $A_S$ and $A_{P_1}, \ldots, A_{P_n}$ as subrings of K, then $A_S = A_{P_1}\cap \cdots\cap A_{P_n}$.

EDIT I came up with a proof thanks to the Bill's hint. Let $\alpha \in A_{P_1}\cap \cdots\cap A_{P_n}$. Let $I$ = {$x \in A; x\alpha \in A$}. $I$ is an ideal of $A$. Since $I$ is not contained in any $P_i$, it is not contained in $P_1\cup\cdots\cup P_n$ by Proposition 1.11 of Atiyah-MacDonald. Hence $\alpha \in A_S$. Therefore $A_{P_1}\cap \cdots\cap A_{P_n} ⊂ A_S$. The other inculsion is obvious.

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Certainly, one direction is clear: if $x\in A_S$, so that $x=\frac{a}{s}$ for some $a\in A$ and $s\in S=(A-P_1)\cap \cdots\cap (A-P_n)$, then certainly $s\in A-P_i$ for all $i$, so that $x\in A_{P_i}$ for all $i$, so that $x\in A_{P_1}\cap\cdots\cap A_{P_n}$. Thus $A_S\subseteq A_{P_1}\cap\cdots\cap A_{P_n}$. – Zev Chonoles Jun 5 '12 at 20:09
up vote 2 down vote accepted

Hint $\ $ An integral domain $\displaystyle \rm\:D = \bigcap_{max\ M}\! D_{\:\!M}.\ $ Now put $\rm\:D = A_S.\:$

Proof $\ $ If fraction $\rm\:f\not\in D\:$ then its denominator ideal $\rm\: I = \{ d\mid d\:\!f\in D\} \ne (1),\:$ so $\rm\!\: I\!\:$ is contained in a maximal ideal $\rm\:M,\:$ so $\rm\:f\not\in D_M,\:$ thus $\rm\:f\:$ is not in the intersection. The converse is clear. $ \ $ QED

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Thanks. I think we need the following result to apply your hint. Let $I$ be an ideal of A. If $I ⊂ P_1\cup\cdots\cup P_n$, then $I ⊂ P_i$ for some $i$. The proof can be found in Atiyah-MacDonald, Proposition 1.11. – Makoto Kato Jun 6 '12 at 10:52
@Makoto Kato I dont think we need that. Your result directly follows from Bill's hint. But I am curious and cant see why $I$ is not contained in any $P_i$ in your solution above. – user114539 Nov 20 '15 at 20:41

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