3
$\begingroup$

Let $R$ be a regular local ring of dimension $n$. Let $i:\text{Flat}\to\text{Fin}$ be the fully faithful inclusion of the category of flat finitely generated type $R$-modules into all finitely generated modules.

  • If $n=1$ then $M$ is flat iff $M$ is torsionfree, and the functor $M\mapsto M/(torsion)$ is left adjoint to $i$.
  • If $n=2$ then $M$ is flat iff $M$ is reflexive, and the functor $M\mapsto M^{\vee\vee}$ is left adjoint to $i$.

If $n=3$ (or in fact $n\ge 3$), is there a nice characterization of flat modules? Is there a left adjoint to $i$, providing a flatification functor? Or are there reasons why such a left adjoint can't exist?

$\endgroup$
2
  • $\begingroup$ Do you really mean "local"? If $R$ is local, finite flat = free. $\endgroup$ Commented Mar 21, 2023 at 8:56
  • $\begingroup$ Ah - thank you Laurent! Yes, we can remove "local". Alternatively, keep the ring "local" and call the modules "free". In fact, I guess that the kind of flatification functor that I would like would be compatible with localization... $\endgroup$ Commented Mar 21, 2023 at 13:35

1 Answer 1

1
$\begingroup$

Fix $R$ (not necessarily local) and $M$. Let us call a map $u:M\to \overline{M}$ a free hull if $\overline{M}$ is finite free and for every finite free $F$ the induced map $\mathrm{Hom}(\overline{M},F)\to \mathrm{Hom}({M},F)$ is bijective.

Claim: $M$ has a free hull if and only if $M^\vee$ is free, and in that case $\overline{M}=M^{\vee\vee}$ (with the obvious map $v_M:M\to M^{\vee\vee})$.

Namely, assume $u:M\to \overline{M}$ is a free hull. Then (taking $F=R$) we conclude that $\overline{M}^\vee\cong M^{\vee}$, hence $M^{\vee}$ is free since $\overline{M}$ is.

Conversely, for any $M$, any map $\varphi:M\to F$ with $F$ free (or just reflexive) factors as $$M\xrightarrow{v_M} M^{\vee\vee} \xrightarrow{\varphi^{\vee\vee}} F^{\vee\vee} \xrightarrow{v_F^{-1}}F$$ and if $M^\vee$ is free so is $M^{\vee\vee}$, whence the claim.

EDIT: As Matthieu observes, everything also works with "free" replaced by "locally free".

$\endgroup$
3
  • $\begingroup$ This is a nice observation! It seems that you can define similarly a locally free hull and obtain that such a hull exists iff $M^\vee$ is locally free, right? $\endgroup$ Commented Mar 21, 2023 at 20:40
  • $\begingroup$ Sounds right, with the same argument. $\endgroup$ Commented Mar 21, 2023 at 21:59
  • $\begingroup$ Ok, then I guess that it is enough to provide a module over a regular ring of dimension 3 whose dual is not locally free, and there are such examples here mathoverflow.net/questions/4590/when-are-dual-modules-free. Thanks! $\endgroup$ Commented Mar 22, 2023 at 12:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .