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I am looking for a reference for examples showing the following phenomena: Let $A$ be a commutative noetherian ring, and let $F$ be an $A$-module such that for all $p \in Spec(A)$ it holds that $F_p$ is a free $A_p$-module, but $F$ it not a projective $A$-module.

Of course, such an $F$ must be infinitely generated and flat.

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  • $\begingroup$ Are you sure you need not finitely generated? What if you take a non free but torsion free module over a Dedekind domain? I mean, take $K$ a number field of class number grater than $1$ and take a non principal ideal of $\mathcal{O}_K$. $\endgroup$
    – Ricky
    Commented Apr 6, 2022 at 14:19
  • $\begingroup$ @Ricky: In a Dedekind ring any ideal is projective. As pointed out in the question, $F$ is flat, hence projective if finitely generated. $\endgroup$
    – abx
    Commented Apr 6, 2022 at 14:28
  • $\begingroup$ Sorry, I read "... that is not free"! $\endgroup$
    – Ricky
    Commented Apr 6, 2022 at 14:29

1 Answer 1

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I found an answer in Exercise 19.12 of Eisenbud's book "Commutative algebra with a view...". It says the following: we are working over the ring of integers Z, and take the Z-module M of all rational numbers with square-free denominators. Then every localization of M as a prime is free, but M is not projective.

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    $\begingroup$ Could you expand on the answer you found? It may be of interest to those who do not have access to the book. $\endgroup$
    – Michael Albanese
    Commented Apr 7, 2022 at 13:28
  • $\begingroup$ @MichaelAlbanese done. $\endgroup$
    – Flat but not projective
    Commented Apr 7, 2022 at 14:08

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