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Let $R$ be a ring. Consider the inclusion functor from the category of finitely generated, projective $R$-modules to the category of all finitely generated $R$-modules. For which rings does it have a left adjoint.

For example for any principal ideal domain $R$, we have the structure theorem of f.g. $R$-modules. The upper left adjoint is given by $M\mapsto M/torsion$.

In general there is no reason, why such a construction should exist. I tried to check the assumptions for Freyds adjoint functor theorem. But it seems quite hard to check whether, they are satisfied for a given ring $R$.

So does anyone know a counterexample, where such a adjoint doesn't exist or a weaker property than PID, that implies the existence ?

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up vote 6 down vote accepted

A necessary condition is that, for any finitely generated R-module M, the Hom-object $Hom_R(M,R)$ is a finitely generated projective right R-module. This is because if a left adjoint existed, this would be isomorphic to the set $Hom_R("Proj(M)",R)$, and this is a summand of a free right module.

Consider the case $R = \mathbb{Z}/p^2$, $M = \mathbb{Z}/p$, $P = R$. Then $$Hom_{\mathbb{Z}/p^2}(\mathbb{Z}/p, \mathbb{Z}/p^2) \cong \mathbb{Z}/p$$ and so this ring admits no projectivization.

Conversely, if this condition is satisfied then we can define $D_R M = Hom_R(M,R)$, and $DM$ is always finitely generated projective with a natural double-duality map $M \to D_{R^{op}} D_R M$ which is an isomorphism for projective modules. This provides the desired adjoint, because any map $M \to P$ produces a natural factorization $$M \to DDM \to DDP \leftarrow^{\sim} P$$ which is the desired adjoint.

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Don't you need for $\mathrm{Proj}(M)$ to be a finitely generated projective for your argument for necessity? – Mariano Suárez-Alvarez May 20 '10 at 17:33
Ach, yes, you're correct. – Tyler Lawson May 20 '10 at 17:59
However, the proof of sufficiency is fine, at least assuming that the ring is noetherian. Since your condition is satisfied for regular 2-dimensional rings, this gives another class of examples. – Angelo May 20 '10 at 18:15

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