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Now I try to do exercise 12.4 in the book "Commutative ring theory" by H. Matsumura.

Let $A$ be a Krull domain, $I\subseteq \mathfrak p$ and $\mathfrak p$ is a height $1$ prime ideal of $A$. I don't know how to prove the following statement:

$xI^{-1}\subseteq A_{\mathfrak p}$ is equivalent to $x\in IA_{\mathfrak p}$.

If $x\in IA_{\mathfrak p}$, then easily $xI^{-1}\subseteq A_{\mathfrak p}$. But how to prove the opposite direction of the statement above? Can someone explain it to me?

Thanks a lot!

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The title suggests that I is divisorial? –  user26857 Dec 6 '12 at 19:52
There is no assumption that $I$ is divisorial. The question I asked is one step to prove that an ideal $I$ of a Krull ring $A$ is divisorial if and only if $I$ is the intersection of a finite number of height 1 primary ideals of $A$. –  yubin Dec 7 '12 at 6:21
@YACP´╝îThank you! When $A$ is a Krull domain and I is an ideal of A, $p$ is a prime ideal of height $1$, then $(A:I)_p=(A_p:I_p)$. To prove this, we need to use the discrete valuation associated to a height 1 prime ideal, not just the ordinary arguments about ideal quotients. –  yubin Dec 11 '12 at 15:19

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