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 ?

  • $\begingroup$ One possible definition: a module is projective if and only if it is a direct summand of a free module. $\endgroup$ – the L Mar 9 '13 at 13:15
  • $\begingroup$ @anonymous I know this. I wanted to know if there are analogues to the fact I gave . $\endgroup$ – Amr Mar 9 '13 at 13:16
  • $\begingroup$ Perhaps you should make your question ("similar fact") more precise in order to get the answers you are looking for. Or you should name the equivalent definitions you already know and want to exclude as answers. By the way, we don't need that our ring has a unit, since for a non-unital ring $R$ we have $\mathsf{Mod}(R) = \mathsf{Mod}(R^+)$, where $R^+$ is the unitalization of $R$. $\endgroup$ – Martin Brandenburg Mar 9 '13 at 13:18
  • $\begingroup$ @Marin Brandenburg You mean that the fact I gave holds for non-unital rings as well. (I don't know the definition of Mod(R) ) $\endgroup$ – Amr Mar 9 '13 at 13:26
  • $\begingroup$ So you are looking for a statement like this: $P$ is projective iff for every factor module $M$ of $R$ we have that every map $P\to M$ can be extended to a map $P\to R$? $\endgroup$ – Julian Kuelshammer Mar 9 '13 at 13:28

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.

  • $\begingroup$ From Wikipedia (article about injective module): "The dual of Baer's Criterion would give a simple test for projectivity, but even for the ring $\mathbb{Z}$ of integers, this becomes the unsolvable Whitehead problem." $\endgroup$ – Julian Kuelshammer Mar 9 '13 at 15:17
  • $\begingroup$ @Julian: This is not a comment to my answer, it is an answer on its own. Why not adding it and add some explanation? I would appreciate it. $\endgroup$ – Martin Brandenburg Mar 9 '13 at 16:07

Are you satisfied with this result?

PROPOSITION 3.1. In order that an $R$-module $A$ be projective it is necessary and sufficient that there exist a family $\{a_\alpha\}$ of elements of $A$ and a family $\{\phi_\alpha\}$ of homomorphisms $\phi_\alpha : A \to R$ such that for all $a\in A$ $$ a=\sum_\alpha (\phi_\alpha a)a_\alpha $$ where $\phi_\alpha a$ is zero for all but a finite number of indices $\alpha$.

(Cartan, Eilenberg, Homological algebra, Chapt.VII)


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