Deciding whether a non-f.g. non-divisible flat module is projective or not Assume $S= R[T]/(f)= R[w]$ is a flat non-divisible $R$-module, where $R$ is a noetherian UFD, $T$ is an indeterminate over $R$, and $f\in R[T]$ is a non-monic polynomial of positive degree.
Can we say if $S$ is $R$-projective or not?
I guess there is not enough information to answer this question. If so, is there an additional condition which guarantees the projectivity of $S$ as an $R$-module?
This question is somewhat relevant. Originally, I have posted this question here.
Any comment is welcome.
Edit: After receiving an answer: Let $R$ be a noetherian UFD, $S$ a non-finitely generated non-projective flat $R$-module. For example, $S$ can be the field of fractions of $R$ or $S=R[T]/(f)$ with non-monic $f$. I wonder if there exist other "types" of examples for such an extension $R \subseteq S$?
 A: There's more than enough information: the  answer is that $S$ is never projective when it isn't "obviously" projective (i.e., never happens when the leading coefficient of $f$ is a non-unit).  This is an application of Zariski's Main Theorem (in EGA formulation). There is no need for hypotheses about non-divisibility, so let's forget about that. 
An obvious necessary condition for $S$ to be $R$-flat is that the coefficients of $f$ have total gcd equal to 1 (indeed, otherwise $f$ is divisible by some nonzero non-unit $r \in R$, so $S$ would have nonzero $r$-torsion, contradicting $R$-flatness of $S$).  But this condition is also sufficient.  Indeed, to check sufficiency we may assume $R$ is local (that step preserves the gcd hypothesis), and then some coefficient of $f$ must be a unit in $R$.  Hence, $f_0 := f \bmod \mathfrak{m}_R$ is nonzero in $k[T]$ with $k = R/\mathfrak{m}_R$, so $f_0$-multiplication on $k[T]$ is injective and thus $f$-multiplication on the $R$-flat $R[T]$ has $R$-flat cokernel $S$ by the usual flatness criteria (see Cor. to Thm. 22.6 in Matsumura's "Commutative Ring Theory").
That being said, now let's assume (as we have seen we must) that the coefficients of $f$ have total gcd equal to 1 in $R$, so $S$ is a quasi-finite flat $R$-algebra.  I claim that $S$ is projective if and only if the leading coefficient of $f$ is a unit in $R$ (in which case clearly it is finite flat and finitely presented as an $R$-module). The "if" direction is obvious.  Now suppose the leading coefficient is not a unit, so $R$ is not a field and 
we aim to show that $S$ is not projective as an $R$-module. 
Note that if $S$ were projective as an $R$-module then the same would hold after localizing at any prime, so it is harmless to localize at a height-1 prime containing the non-unit leading coefficient of $f$. We assume $S$ is $R$-projective and seek a contradiction.
Now $R$ is a dvr and some lower-degree coefficient is a unit.  It is likewise harmless to extend scalars to the completion of $R$ so that $R$ is complete.  But now we can apply Zariski's Main Theorem in the EGA formulation to the quasi-finite $R$-algebra $S$ to get an $R$-algebra decomposition  $S = S_{\rm{fin}} \times S'$ where $S_{\rm{fin}}$ is finite over $R$ and $S'$ has empty special fiber. If $S'$ is nonzero then it is a nonzero finite-dimensional vector space over the fraction field $K$ of $R$, so then $K$ would occur as an $R$-module direct summand of $S$. But $S$ is $R$-projective, so $K$ is $R$-projective, an absurdity (as $K$ is divisible, unlike any projective $R$-module).  Thus, $S'=0$, so $S$ is $R$-finite, and hence by projectivity it is a free $R$-module.  But the generic and special fibers of $S$ over $R$ have unequal ranks since the leading coefficient of $f$ is a non-unit and some lower-degree coefficient is a unit, contradicting $R$-freeness. Thus, $S$ wasn't $R$-projective after all.  
QED
