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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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    @YACP,Thank 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.2012-12-11

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