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Let $A$ and $B$ be ideals. I want to show that there exists an element $c \in K$ (where $K$ is the quotient field of a Dedekind domain $O$) such that $cA$ is an ideal relatively prime to $B$.

As hinted in the question, I tried to apply Chinese Remainder Theorem using the fact that every prime ideal $P$ is maximal in a Dedekind domain $O$.

You may use any definition of Dedekind domains.

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1 Answer 1

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Let $p_1,\dots,p_r$ be the prime divisors of $B$. You have to find $x \in K$ such that $val_{p_1}(xA)=0$ ($i=1...r$), i.e. $val_{p_i}(x^{-1})=val_{p_i}(A)$.

For $i=1...r$, choose $a_i$ such that $val_{p_i}(a_i)=val_{p_i}(A)$. Then use CRT to choose $y \in O$ such that $y = a_i \mod p_i^{val_{p_i}(A)+1}$. This $y$ satisfies $val_{p_i}(y)=val_{p_i}(A)$ for all $i$.

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