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.

  • 2
    $\begingroup$ Are you sure the functor is full? See page 256; it is just any unital subring. The whole point of relative homological algebra is the S-split but not R-split distinction. A common version is ZH ≤ ZG for finite group G and subgroup H (or replace Z by a finite field). Your question is exercise 8, page 277. Check section 8 for his explanation of relative Tor. The same idea is good for relative ext. $\endgroup$ – Jack Schmidt Mar 7 '11 at 17:47
  • $\begingroup$ Thank you for your answer! You are correct, ofcourse the functor is not full, but faithful (edited in the question) $\endgroup$ – shamovic Mar 8 '11 at 14:51
  • $\begingroup$ The following example appears in C. Weibel's book (Example 8.7.6): If $R = S/I$, then $\Psi(M) \cong M$, for every $R$-module $M$. Then $Hom_R(F(M),N) = Hom_R(M,N)$ and the bar-resolution trivializes, hence $Ext_{(R,S)}^i(M,N) = 0$ for all $i > 0$. $\endgroup$ – shamovic Mar 8 '11 at 14:58

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.


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