TITLE: Example of relative Ext functor
QUESTION [8 upvotes]: Greetings,
I've been reading Maclane's "Homology" and ran into the following question:
Let $(R,S)$ be a resolvent pair of ring, i.e $R$ is an $S$-algebra and we have a functor $\Psi \colon \operatorname{R-Mod} \to \operatorname{S-Mod}$, that is additive exact and faithful. We also have an left adjoint functor of $\Psi$, namely $F \colon \operatorname{S-Mod} \to \operatorname{R-Mod}$.
One defines then relative $Ext_{(R,S)}$ functor, using the bar-resolution. Can you please help me find a concrete example of a case when $Ext^1_{(R,S)} \neq Ext^1_R$ (where $Ext^1_R$ is the regular $Ext$ functor).
My thought on the matter. One can identify $Ext^1_{(R,S)}(A,B)$ with the set of extensions of $B$ by $A$, that are $S$-split. One must then find an extension in $R$ that is not $S$-split to do the trick. But this argument feels a bit like cheating.
Any help will be appreciated.

REPLY [4 votes]: Well, if $R=S$, then an $S$-split extension is obviously also $R$-split, so $\operatorname{Ext}^1_{(R,S)}$ is always zero. On the other hand, $\operatorname{Ext}^1_{R}$ can be pretty much anything you like.