Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $A$?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.