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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?

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For injectives this is just Baer's criterion. The interesting notion where the test modules M are required to be finitely presented yields Q that are called FP-injective or absolutely pure. They are the pure submodules of injective modules, and over Prüfer domains they are the divisible modules. Ext definitely does not commute will-nilly with direct limits since Q is a direct limit of Zs, but Ext(Q,Z)=R is not equal to the limit of Ext(Z,Z)=0. –  Jack Schmidt May 27 '10 at 4:11

2 Answers 2

up vote 5 down vote accepted

For the first question you already have had an answer in projective module 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.

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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.

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Are you sure you don't need the ring to be Noetherian? –  ashpool May 27 '10 at 21:51
    
Of course, I assumed that the ring is Noetherian. Otherwise it is very strange to consider the subcategory of finitely generated modules. –  Sasha May 28 '10 at 3:48

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