homological algebra - Does Ext commute with direct limit?

A short comment, which I can't post as a comment as I've just opened a new account (I apologize).
The $R^1varprojlim mathrm{Hom}_R(M_{alpha}, Q)$ 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 $M_{alpha}to M_{alpha'}$, $alpha'gealpha$, 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 $R^1varprojlim (cdot)$.

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:

$$0to M'to Mto M''to 0$$

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

Best