Let $A$ be a commutative local ring, with unique maximal ideal $\mathfrak{m}$, and residue field $k:=A/\mathfrak{m}$. Let $M$ be a faithful, finitely generated $A$-module.

If $M/\mathfrak{m}M$ is 2-dimensional over $k$, is $M$ necessarily free over $A$?


1 Answer 1


I don't think so. Let $A = \mathbf Z_{(p)}$ and consider the $A$-module $M = A \times k$.

Note that this would be true if $M/\mathfrak mM$ were $1$-dimensional, because of the fidelity.


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