Commutative Algebra – Flatness Over Regular Local Rings of Dimension 3

ac.commutative-algebraflatnessprojective-modules

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?

Best Answer

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".

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