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Aug 31, 2023 at 9:09 comment added Neil Strickland I think that if $R$ is any proper subring of $\mathbb{Q}$ then we can choose a prime $p$ that is not invertible in $R$ and use the above argument to show that $\chi\colon F\to\text{Hom}(\text{Hom}(F,R),R)$ is an isomorphism whenever $F$ is free and countable. Essentially the same argument shows that $\chi$ is a split monomorphism whenever $F$ is free, even if it is uncountable. If I understand correctly, there is some story relating the cokernel to ultrafilters and measurable cardinals.
Aug 31, 2023 at 8:24 history edited Jeremy Rickard CC BY-SA 4.0
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Aug 31, 2023 at 7:52 vote accept Neil Strickland
Aug 31, 2023 at 7:28 history answered Jeremy Rickard CC BY-SA 4.0