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I need a proof of the following proposition(?). Actually I think I came up with a proof. But it's nice to confirm it and/or to know other proofs. Thanks.

Proposition Let $A$ be a Dedekind domain. Let $I$ and $J$ be non-zero ideals of $A$. Then there exist non-zero $\alpha \in I$ and an ideal $M$ such that $(\alpha) = IM, M + J = A$.

EDIT Here's my proof. Let $I = (P_1)^{e_1}...(P_n)^{e_n}$ be the prime decomposition of I. Let $Q_1, ..., Q_m$ be all the prime ideals which divide $J$, but not divide $I$. By the proposition and with its notation, there exists $\alpha \in A$ such that $v_{P_i}(\alpha) = e_i, i = 1, ..., n$. $v_{Q_j}(\alpha) = 0, j = 1, ..., m$. Since $\alpha \in I$, there exists an ideal $M$ such that $(\alpha) = IM$. Clearly $M + J = A$

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    @Dylan I'd like to know your proof.2012-06-05

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I think you are right. The essential point is the weak approximation theorem which you cite as the proposition. The weak approximation theorem is in turn essentially Chinese remainder theorem.