# A proposition on a Dedekind domain

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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Since $M$ is completely determined by $I$ and $\alpha$, perhaps you might want to state the result only in terms of $\alpha$: there is a nonzero $\alpha \in A$ such that $(\alpha) + IJ = I$, or equivalently ${\rm gcd}((\alpha),IJ) = I$. (An $\alpha$ satisfying this equation is automatically in $I$.) Does your proof use the Chinese remainder theorem? –  KCd Jun 5 '12 at 1:47
There's probably a pithy way of stating this in terms of the class group, right? –  Dylan Moreland Jun 5 '12 at 2:17
@KCd Yes, it does. –  Makoto Kato Jun 5 '12 at 8:52
The proof I'm thinking of uses the CRT as well. I can reproduce it below, but it seems likely that it's identical to yours. –  Dylan Moreland Jun 5 '12 at 20:07
@Dylan I'd like to know your proof. –  Makoto Kato Jun 5 '12 at 21:02