# Baer's criterion for projective modules

Let $R$ be a commutative ring. If necessary, assume that $R$ has any convenient properties you like.

• Is there some $R$-module $Q$ such that an $R$-module $P$ is projective if and only if $\hom_R(P,Q) \to \hom_R(P,Q')$ is surjective for all quotients $Q'$ of $Q$?
• If yes, can $Q$ be chosen as a cogenerator?
• What happens when we restrict to finitely generated $P$?

Neither $Q=R$ nor $Q=\hom_{\mathbb{Z}}(R,\mathbb{Q}/\mathbb{Z})$ work. This question is a follow-up of math.SE/325495.

• Hi Martin, I think there is a class $\mathcal{C}$ of $R$-modules such that it is sufficient to check the surjectivity for every quotient $Q \rightarrow Q'$ with $Q$ in $\mathcal{C}$: namely, the class of injective modules. Or more generally, any class such that any module can be embedded into some object of this class. If under non-trivial circumstances this class can be chosen to be smaller, maybe to contain only one object, I don't know. Mar 12 '13 at 11:01
• That this does not work is a reflection of the fact that the really dual notion to injectivity is not projectivity but flatness. Mar 12 '13 at 15:16
• (Notice that the result by Trlifaj mentioned by Jeremy Rickard below is precisely about rings —the perfect ones— for which projectivity and flatness coincide, so that my comment above becomes ineffective :-) ) Mar 12 '13 at 15:49
• @Mariano: Is there a $\hom$-like characterization for flat modules? Mar 12 '13 at 16:51
• Theorem (Lambek): A right $R$-module $M$ is flat iff the left $R$-module $M' := Hom_{\mathbb{Z}}(M, \mathbb{Q}/\mathbb{Z})$ is injective. Corollary: $M$ is flat iff $M \otimes I \rightarrow MI$ is an isomorphism of abelian groups for every left ideal $I$ in $R$. Stolen from Lam, Lectures on Modules and Rings, p. 125. Mar 12 '13 at 17:59

I think the paper "Whitehead Test Modules" by Jan Trlifaj (Trans. AMS 348 (1996), 1521-1554) and the references in it, answer your question positively for perfect rings, but show that a negative answer is consistent with ZFC+GCH if $R$ is not perfect.
In Lemma 2.4 he seems to prove that (if $R$ is not perfect) then for any cardinal $\kappa$, it is consistent for there to be a non-projective module $M$ such that $\operatorname{Ext}^1(M,N)=0$ for all modules $N$ of cardinality less than $\kappa$, which contradicts the property you want $Q$ to have if $\kappa$ is larger than the cardinality of $Q$. But I'm not a set theorist, and may be misunderstanding something.
For the restriction to finitely generated $P$, I assume you want $Q$ also to be finitely generated? Otherwise you could just take $Q$ to be a free module of infinite rank.
• For the restriction to finitely generated $P$, why can't we just take $Q$ to be $R$, where $R$ is our ring? The projectivity domain of a fg module is closed under arbitrary direct sums, and projectivity domains are always closed under epimorphic images, so a fg module is projective relative to $R$ iff it is projective. On the other hand, if you assume that your ring is (right, say) noetherian, it's possible to modify Torsten's initial comment to conjure up a cogenerator example that works for fg modules: choose an indecomposable injective from each isomorphism class, and take their direct sum. Mar 12 '13 at 18:27