Is finitely generated projective module M of rank one over regular commutative notherian ring free? Bass (Illinois Math J, 1963) showed that in case M is nonfinitely generated, it is free. I am wondering about finitely generated case.


  • 3
    $\begingroup$ It's quite remarkable to see someone who knows the result for infinitely generated modules but not finitely generated ones. $\endgroup$
    – Fan Zheng
    Oct 28, 2015 at 20:41

1 Answer 1


No. Take the case of a Dedekind domain $A$ that it is not a principal ideal domain. It is a regular commutative notherian ring (of Krull dimension 1). The group of classes $\mathrm{Pic}(A)$ is not trivial, otherwise it would be a p.i.d. Therefore, there is a rank one projective module, in this case a fractional ideal (i.e. an $A$-submodule of its quotient field), that it's not free.

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    $\begingroup$ The most economical example, and an instructive one as well. $\endgroup$
    – Lubin
    Oct 28, 2015 at 13:28

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