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Let $M$ be an $A$-module. Is it true that if $M/\mathfrak{m}M\simeq A/\mathfrak{m}$, for any maximal ideal $\mathfrak{m}\subset A$, then $ M\simeq A$?

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    $\begingroup$ I think this is true if there is one arrow $M\to A$ (or the other way around) inducing such isomorphisms under tensoring with $A/\mathfrak m$, but perhaps there is some exotic counterexample as with weak equivalences of topological spaces. $\endgroup$ – Pedro Tamaroff Nov 18 '17 at 19:56
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Let $A$ be a Dedekind domain, of class number greater than $1$, and let $I$ be a non-principal ideal of $A$. Then $I/\mathfrak{m}I\cong A/\mathfrak{m}$ for all $\mathfrak m$ but $A\not\cong I$ as $A$-modules.

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  • $\begingroup$ Can you remark on the isomorphism $I/\mathfrak{m} I \cong A/\mathfrak{m}$ for all $\mathfrak{m}$? I imagine it's some obvious dimension consideration but I'm not seeing it. $\endgroup$ – Mr. Chip Nov 18 '17 at 20:17
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    $\begingroup$ @Mr.Chip It's a standard fact on Dedekind domains; any reasonably comprehensive text on algebraic number theory will have it, usually in the form that $I/IJ\cong A/J$ for nonzero ideals $I$ and $J$. $\endgroup$ – Angina Seng Nov 18 '17 at 20:20

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