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.
Best Answer
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.