2
$\begingroup$

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 $R \subseteq S$?

$\endgroup$
  • $\begingroup$ Every quasi-finite flat algebra $S$ over any noetherian ring $R$ that is not $R$-finite is not $R$-projective. Indeed, a separated quasi-finite flat map between noetherian schemes is finite if and only if its fiber rank is locally constant (proved in Deligne-Rapoport via Zariski's Main Theorem), so the fiber-rank of $S$ is non-constant on some connected component of Spec($R$). We can then make base change to a suitable complete dvr to reduce to the case when $R$ is a complete dvr and the special and generic fibers have different rank. Then the argument with ZMT in the answer applies. $\endgroup$ – grghxy Aug 17 '15 at 4:49
  • $\begingroup$ Also, a vast generalization is the amazing Theorem 3.3.5 in Raynaud-Gruson (see Def. 3.3.3(i) and then Ex. 3.3.4(ii), with $\mathscr{M} = O_X$ and $M=B$ in the discussion there). That is, a finitely presented flat algebra $S$ over an arbitrary ring $R$ is projective as an $R$-module if and only if it is "$R$-pure"; for $R$-smooth $S$ this has a subtle geometric meaning in terms of irreducible components of fibers (e.g., implying that if $S$ is $R$-smooth with geometrically connected non-empty fibers then it is $R$-projective) and for quasi-finite $S$ it is the equivalence with $R$-finiteness. $\endgroup$ – grghxy Aug 17 '15 at 6:45
  • $\begingroup$ Very interesting! Thank you very much! $\endgroup$ – user237522 Aug 17 '15 at 22:10
  • $\begingroup$ @grghxy Please, since the paper of Raynaud and Gruson is in French (and unfortunately I am not familiar with French), I wish to be sure about the definition of a pure module; can you write it here? $\endgroup$ – user237522 Aug 27 '15 at 16:41
  • $\begingroup$ I can't read French either. But math French is not real French. And in this part of math it is important to be able to read math French. Get English and French copies of a book of Serre and learn it. It is not as hard as you might expect, and is a very useful skill. (Context helps to fill in a lot.) $\endgroup$ – grghxy Aug 28 '15 at 14:22
6
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thank you very much for your great answer!! What you have written is actually the main theorem in sciencedirect.com/science/article/pii/0021869370900438. Especially, it says that in my situation (assuming non-divisibility is not needed): $S$ is $R$-projective iff $S$ is integral over $R$ (=the leading coefficient of $f$ is a unit in $R$). $\endgroup$ – user237522 Aug 17 '15 at 1:21
  • $\begingroup$ Your answer is not in vain; it will help me to understand the proof of the above mentioned main theorem. $\endgroup$ – user237522 Aug 17 '15 at 1:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.