[Math] Equivalent definitions for projective modules

abstract-algebramodulesring-theory

Fact: Let $R$ be a ring with identity. Let $J$ be an $R$-module. Then, $J$ is injective iff for every left ideal $L$ of $R$ every $R$-module homomorphism $L\rightarrow J$ can be extended to an $R$-module homomorphism $R\rightarrow J$.

This fact provides an equivalent definition of the categorical definition of injective modules. The equivalent definition given by the fact is easier to check.

Question: Let $R$ be a ring with identity. Is there a similar fact for projective $R$-modules ? Is there an equivalent definition for projective $R$-modules such that it is much easier to check ?

Best Answer

In a Grothendieck category, i.e. an abelian category satisfying AB 5 and admitting a generator $G$, an object $M$ is injective if and only if the extension property holds for all subobjects of $G$ (see Grothendieck's Tohoku Paper I, Lemme 1). For the category of modules over a ring $R$ with $G=R$ this yields Baer's criterion for injectivity, mentioned in the question.

By duality, in an abelian category satisfying AB 5* admitting a cogenerator $Q$, an object $P$ is projective if and only if the dual extension property holds for all quotients $Q \twoheadrightarrow Q'$, that is, $\hom(P,Q) \to \hom(P,Q')$ is surjective. If the latter condition holds, let us say that $P$ is weakly projective (with respect to $Q$).

Unfortunately, the category of modules over a ring $R$ does not satisfy AB 5* (unless $R=0$). Also, the criterion fails for $Q=R$ (which is not a cogenerator, but this would be the naive analogue of Baer's criterion), for example when $R$ is a PID which is not a field, since here any nontrivial quotient is torsion and therefore all torsionfree modules are weakly projective, and the criterion also fails for the cogenerator $Q=\hom_{\mathbb{Z}}(R,\mathbb{Q}/\mathbb{Z})$: If $R$ is a finitely generated $\mathbb{Z}$-module, then $Q$ is a torsion $\mathbb{Z}$-module, and again torsionfree modules are weakly projective. Then the same holds for every quotient of $Q$.

Edit. In this new version I added the correct dualization, including cogenerators.

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