Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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. –  Jack Schmidt Mar 7 '11 at 17:47
    
Thank you for your answer! You are correct, ofcourse the functor is not full, but faithful (edited in the question) –  shamovic Mar 8 '11 at 14:51
    
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 $i > 0$. –  shamovic Mar 8 '11 at 14:58
add comment

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.