Let $A$ be a Dedekind domain, $I$ be an ideal in $A$ and let $I^{-1}$ be the inverse of $I$ as a fractional ideal in $K$, where $K$ is the quotient field of $A$. It seems quite natural to have a $A-$module isomorphism between $I^{-1}/A$ and $A/I$ if $I$ is principal, but to construct this isomorphism in general appears to be hard and probably impossible. I was wondering if explicit contrexamples are available in literature? I have not found any of them so far.
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The reason that you can not find a counterexample is that the statement is true. First notice that the support of both the modules in question is just finitely many primes containing $I$ and thus the problem becomes local near these primes. But, locally $I$ is principal and thus you get what you want.
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$\begingroup$ I am sorry but I do not understand completely: I do not see any obvious isomorphism between $A/I$ and $\bigoplus_{P|I}(A_{P}/I_{P})$. Is it what it is suggested to search for? $\endgroup$– Hair80Commented Jul 6, 2017 at 14:49
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$\begingroup$ Does not apply to this case: $A$ and $A_{P}$ are different rings and $I$ is the intersection of all the $I_{P}$ for all $P$ in $A$ and not only those which divide $I$. $\endgroup$– Hair80Commented Jul 6, 2017 at 16:06
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$\begingroup$ Please read CRT carefully. It says precisely what you need and I think you have not understood what it means. For example, if $I=P_1^nP_2^m$, then $A/I=A/P_1^n\oplus A/P_2^m=A_{P_1}/{P_1^n}_{P_1}\oplus A_{P_2}/{P_2^m}_{P_2}$. $\endgroup$– MohanCommented Jul 6, 2017 at 17:09
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$\begingroup$ Dear @Mohan, I liked your argument a lot. I was wondering if you could rephrase it in terms of divisors on the corresponding arithmetic curve with a very clear and explicit geometric interpretation. Is this possible? $\endgroup$– PedroCommented Mar 13, 2018 at 10:10