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Suppose we have a group $G$ then one can construct $BG$ and one of the essential part of the construction is the co-unit map. Now suppose we have a ring spectrum $R$, then having a co-unit splits the ring spectrum as $R = S \vee R'$, which is never an interesting case.

So my question is what is the best one can do? I know its a vague question. But is there an alternative thing that one can do, which closely resembles bar construction that one does for a group?

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I don't see an analogy in your proposal since you don't take any bar construction on $R$. The 'bar construction' on $R$ would be the topological Hochschild homology spectrum $THH(R)$, which has a 'counit' $TTH(R)\rightarrow R$. – Fernando Muro Feb 19 '13 at 14:48
Prasit: Such a thing does work in a good category of spectra, and in fact it is the case that you can split of the unit in a similar way to get a reduced bar construction. I will look around for the right diagram, I can't recall it off the top of my head. – Sean Tilson Feb 19 '13 at 15:06
To add to Sean's comments, the two-sided bar construction B(M,A,N) can be formed in myriads of contexts in which one has a monoid A in a monoidal category with right action on M and left action on N. Topological groups G and ring spectra R are just two examples. The fact that you can't generally take M=S in general in the ring spectra example is not so important. The precise relationship with THH is on pp 173-174 of EKMM. – Peter May Feb 19 '13 at 15:25
Edit: Fernando: There is a bar construction for spectra and it does not necessarily give you $THH$. The naive two sided bar construction $B(M,A,N)$ frequently models $M \wedge_A N$, there are some cellularity (is that a word) hypotheses. The cyclic bar construction is an example of the two sided bar construction. This can all be found in EKMM. – Sean Tilson Feb 19 '13 at 17:27
Why is it uninteresting for $S$ to split off of $R$? Doesn't this happen in the case that $R = \Sigma^\infty G_+$ for a group $G$? The associated Tor spectrum $Tor^R(S, S)$ (or derived smash product of $S$ with itself over $R$) will then compute the suspension spectrum of $BG$, certainly an interesting object. – Craig Westerland Feb 20 '13 at 11:29

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