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(1) Let $R=k[x_1,\ldots,x_n]$. I wish to find an example of a non-finitely generated, non-divisible, non-projective, flat $R$-module. Notice that $k(x_1,\ldots,x_n)$ is NOT an example of what I am looking for, since it is divisible.

(2) More generally, same question but now with $R=$ any integral domain.

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  • $\begingroup$ The ring of Laurent polynomials $k[x^{\pm1}]$ viewed as a$k[x]$-module satisfies your conditions in (1), no? $\endgroup$ – Mariano Suárez-Álvarez Jul 29 '15 at 1:00
  • $\begingroup$ Thanks!! I guess your example is also valid in case $n>1$? namely, $k[x_1,x_1^{-1},\ldots,x_n,x_n^{-1}]$. $\endgroup$ – user237522 Jul 29 '15 at 1:07
  • $\begingroup$ There is no need to localize all variables, in fact-. $\endgroup$ – Mariano Suárez-Álvarez Jul 29 '15 at 1:10
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Let $R=k[x]$ and $M=k[x^{\pm1}]$ viewed as an $R$-module in the obvious way. One can easily check that it is not finitely generated, not divisible and it is flat because it is a localization. To check that it is not projective, consider the map $\phi:\bigoplus_{n\in\mathbb Z}Re_n\to M$ from the free $R$-module with basis $\{e_n:n\in\mathbb Z\}$ to $M$ such that $\phi(e_n)=x^n$ for all integers $n$. The map is clearlyu surjective. If $M$ were projective, there would be a section $s:M\to \bigoplus_{n\in\mathbb Z}Re_n$, and in particular an injective homomorphism. Now the image of $1\in M$ under $s$ has to be infinitely divisible by $x$, and there are no such elements in the free module.

One can do exactly the same with more variables.

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  • $\begingroup$ Thank you very much! I wonder if there is an example which is not a localization? $\endgroup$ – user237522 Jul 29 '15 at 1:38
  • $\begingroup$ The direct sum of two copies of my M is not a localization of R. :-) $\endgroup$ – Mariano Suárez-Álvarez Jul 29 '15 at 1:56
  • $\begingroup$ :-) Nice (I should have thought of this myself..). $\endgroup$ – user237522 Jul 29 '15 at 10:50

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