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Let $R=k[x_1,\ldots,x_n]$. According to the first answer, every finitely generated flat module over an integral domain is necessarily projective.

Therefore, the only hope to find a flat non-projective $R$-module $M$ is when $M$ is not finitely generated. Can anyone please suggest such modules?

Actually, here there are some examples of flat non-projective modules (but not over a polynomial ring), and the following general claim, due to Bass: Flat modules are projective iff the ring is perfect.

(A somewhat relevant question can be found here).

EDIT: I have posted here my last comment as a question, which already has an answer.

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migrated from mathoverflow.net Jul 28 '15 at 20:27

This question came from our site for professional mathematicians.

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    $\begingroup$ Projective $R$-modules are not divisible. Every projective $R$-module is a direct summand of a free $R$-module, and every nonzero element in a free $R$-module obviously fails to be divisible. Thus, a divisible, flat $R$-module, e.g., the fraction field, is not projective. $\endgroup$ – Jason Starr Jul 28 '15 at 0:40
  • $\begingroup$ Thanks!! So given any integral domain $D$, the field of fractions of $D$ is a a non-f.g. non-projective flat module over $D$? I am curious if there exist other examples? (I guess yes). BTW, if you would like to post your comment as an answer, I will accept it. $\endgroup$ – user237522 Jul 28 '15 at 1:34
  • $\begingroup$ How did we end up here, user237522? Somebody out there does not appreciate your question (or maybe they do not appreciate my answer). $\endgroup$ – Jason Starr Jul 29 '15 at 3:07
  • $\begingroup$ @JasonStarr I do not know :-) I think it is more plausible that someone did not appreciate my question (than your answer). It seems to me that the claim in your answer ("divisible implies non-projective") is not so known (at least not to me, though I am not an expert) and hence appropriate for MO. $\endgroup$ – user237522 Jul 29 '15 at 10:42
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    $\begingroup$ Well, it is better to learn, to discover, and to have fun than it is to be appreciated. I hope that you have fun learning about some algebra. $\endgroup$ – Jason Starr Jul 29 '15 at 13:11
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Let $R$ be an integral domain that is not a field. Then $R$ contains a nonzero element that is not invertible. For every nonzero element $r$ that is not invertible, $r$ is not a multiple of $r^2$: if it were, then $r=r^2s$ for some $s\in R$. Since $R$ is an integral domain, $1=rs$, contradicting that $r$ is not invertible.

Now, for $I$ any nonzero ideal in $R$, if $I$ equals $R$, then $I$ is only divisible by invertible elements of $R$. By hypothesis, there exists a nonzero, noninvertible element $q$ of $R$, and $I$ is not divisible by $q$. On the other hand, if $I$ does not equal $R$, then $I$ contains a nonzero element $r$ that is not invertible. Thus $r$ is not divisible by $q=r^2$. Since $r$ is not divisible by $q$, also $I$ is not divisible by $q$. Thus, for every nonzero ideal $I$ in $R$, there exists a nonzero, noninvertible element $q$ of $R$ such that $I$ is not divisible by $q$.

Finally, for any free $R$-module $M$, say free on a basis $S$, for every nonzero element $m$ of $M$, the ideal $I$ generated by the finitely many nonzero $S$-coefficients of $m$ is nonzero. Thus there exists a nonzero element $q$ of $R$ such that $I$ is not divisible by $q$. Therefore also $m$ is not divisible by $q$. Therefore, also, for every nonzero $R$-submodule $P$ of $M$, for every nonzero element $m$ of $P$, there exists nonzero, noninvertible $q$ in $R$ such that $m$ is not divisible by $q$. In particular, if $P$ is a direct summand of $M$, then $P$ is not divisible.

Thus, every divisible $R$-module is not projective. In particular, the fraction field of $R$ is divisible. Therefore the fraction field of $R$ is a flat $R$-module that is not projective.

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  • $\begingroup$ Thank you very much for your answer; the elaboration really helped me. I wonder if there exists an example of a non-finitely generated non-projective flat $R$-module which is not divisible, where $R$ is an integral domain. I guess there is such an example? and what about the special case $R=k[x_1,\ldots,x_n]$?. (Should I post this as a separate question?). $\endgroup$ – user237522 Jul 28 '15 at 23:43

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