Does Ext commute with direct limit?

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 Is it true that if $\operatorname{Ext}^{1}_{A}(P,A/I)=0 $ for all $ I$ then $P$ is projective? 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

A short comment, which I can't post as a comment as I've just opened a new account (I apologize). The $R^1\varprojlim \mathrm{Hom}_R(M_{\alpha}, Q)$ contribution can be taken care by arranging the transition maps in the directed system $(M_{\alpha})$ to be injective.

Suppose we show $\mathrm{Ext}^1_R(\cdot, Q)$ vanishes on all finitely generated $R$-modules ($R$ Noetherian: we'll be using that the category of finitely generated $R$-modules is abelian when $R$ is Noetherian). We can write $M$ as the directed union of its finitely generated $R$-submodules and arrange all transition maps to be injective. Applying $\mathrm{Hom}_R(\cdot, Q)$ to the injection $M_{\alpha}\to M_{\alpha'}$, $\alpha'\ge\alpha$, we have an exact-in-the-middle seq

$$\mathrm{Hom}_R(M_{\alpha'}, Q)\to \mathrm{Hom}_R(M_{\alpha}, Q) \to \mathrm{Ext}^1_R(A, Q)$$

for $A$ a finitely generated $R$-module. That is, the inverse system $(X_{\alpha})$, $X_{\alpha} := \mathrm{Hom}_R(M_{\alpha}, Q)$, satisfies the Mittag-Leffler condition (because $\mathrm{Ext}^1_R(A, Q) = 0$) and therefore it has vanishing $R^1\varprojlim (\cdot)$.

Torsten's answer shows that in the case $R$ is Noetherian, one reduces to check injectivity of an $R$-module $Q$ to computing $\mathrm{Ext}^1_R(R/\mathfrak{p}, Q)$ to be trivial for all prime ideals of $R$ (as for $M$ finitely generated, given ses's of finitely generated $R$-modules:

$$0\to M'\to M\to M''\to 0$$

and by functoriality of Ext's, we get exact-in-the-middle sequences

$$\mathrm{Ext}^1_R(M'', Q)\to \mathrm{Ext}^1_R(M, Q)\to \mathrm{Ext}^1_R(M', Q)$$

so if we show vanishing of the outer terms, we show vanishing of the middle one. This reduces consideration to the case of simple $R$-modules (again, here $R$ is Noetherian!), ie. of the form $R/\mathfrak{p}$, $\mathfrak{p}$ a prime ideal).

Eg. Let $R = \mathbf{Z}/p^2\mathbf{Z}$. Showing $R$ is an injective $R$-module is equivalent to showing $\mathrm{Ext}^1_R(R/p, R) = 0$. More in general, can show $\mathbf{Z}/n\mathbf{Z}$ is injective as a module over itself this way.

Best

Answer 3

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.