# $M/\mathfrak{m}M\simeq A/\mathfrak{m}$, for any maximal ideal $\mathfrak{m}\subset A$

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 $M\simeq A$?

• 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. – Pedro Tamaroff Nov 18 '17 at 19:56

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.
• 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. – Mr. Chip Nov 18 '17 at 20:17
• @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 $J$. – Angina Seng Nov 18 '17 at 20:20