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.


  • 2
    $\begingroup$ 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 '15 at 20:41

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ The most economical example, and an instructive one as well. $\endgroup$ – Lubin Oct 28 '15 at 13:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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