Question

Is it true that if $\mbox{Ext}^{1}(P,M)=0$ for every finitely generated module $M$ then $P$ is projective? Or that if $\mbox{Ext}^{1}(M,Q)=0$ for every finitely generated module $M$ then $Q$ is injective?

Answer 1

For the first question you already have had an answer in http://mathoverflow.net/questions/25687/projective-module/25698 if $\mathrm{Ext}^1_{\mathbb Z}(P,M)=0$, then it depends on the axioms of set-theory whether the conclusion is true or not. The answer to the second question is yes, it is one of the basic characterisation of injective modules that $\mathrm{Ext}^1(A/I,Q)=0$ for all ideals $I$ iff $Q$ is injective. As for the question in your title, the answer should be no for the second variable (irrespective of the axioms of set theory, but I am too lazy to try to come up with an example). For the first variable things are a little bit more interesting: If $M$ is the direct limit of ${M_\alpha}$, then we have a spectral sequence with $E_2$-term $lim^i\mathrm{Ext}^j(M_{\alpha},Q)$ ("lim" means inverse limit, there is some strange problem with using "varprojlim" which sometimes works and sometimes doesn't) and converging to $\mathrm{Ext}^{i+j}(M,Q)$. Somewhat strangely this spectral sequence does not seem to formally give the above characterisation of injective modules as there is a potential $lim^1\mathrm{Hom}(M_{\alpha},Q)$ contribution.

Answer 2

The answer to the first question is positive, if $P$ is itself finitely generated. Indeed, then $P$ is a direct summand of a free module, hence projective. In general, $Hom(P,-)$ does not commute with colimits. Moreover, it commutes only if $P$ is compact (this is the definition of compactness), so I don't think that in general such $P$ will be projective.