Non-finitely generated, non-projective flat module, over a polynomial ring 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.
 A: 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.
