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When I learned abstract algebra many years ago,I noticed the author deals with commutative ring say,$A$ has the proposition:$A^2=A$(without assuming it has identity).It seems that many proposition of commutative ring and module theory can be carried to the ring satisfying the condition above.Is there a name for this kind of ring? It is well known that we have the following fact:

The category of semi-modules(or non-unital modules) over $A$(commutative ring without identity,semi-modules means the action of $A$ is non-unital) is equivalent to the category of modules over $A$.I wonder know what is the motivation to understand the ring with property that $A^2=A$.It seems that this kind of ring appears in theory of operator algebras,noncommutative differential geometry naturally.Can anyone give some short introduction to these stuff?

There is another proposition of ring called firm ring and firm module and I noticed that Quillen and Suslin ever considered High algebraic K-theory on this kind of ring.It seems that High algebraic K-theory on this kind of ring has many properties similar(or even the same)to that of unital ring.I wonder what is the motivation to develop the theory on it.

The motivation to ask this question is that there is a preprint by Alexander Rosenberg and Alain Connes on kind of non-unital algebraic geometry,they developed Non-unital Barr-Beck's theorem,higher algebraic K-theory on firm module over firm semimonad(non unital monad).It seems that they were trying to build a foundation to connect noncommutative differential geometry and noncommutative algebraic geometry,however they did not provide any motivation to study this kind of objects,they just mentioned that this work grew out from a talk given by Quillen and I can hardly find out any expository references talking about them.

I believe this might be a soft question.

Thanks in advance

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  • $\begingroup$ What do you mean by $A^2=A$? That every element is a square root? $\endgroup$
    – Denis Nardin
    Commented Jan 9, 2014 at 8:35
  • $\begingroup$ $A^2=span\{xy,x,y\in A\}$ $\endgroup$
    – user41650
    Commented Jan 9, 2014 at 8:47
  • $\begingroup$ As for the name, these rings are called "idempotent rings". $\endgroup$
    – Name
    Commented Jan 9, 2014 at 9:25
  • $\begingroup$ a reference is ci.nii.ac.jp/naid/110000027349 (not available to me). $\endgroup$
    – Name
    Commented Jan 9, 2014 at 9:28
  • $\begingroup$ If for every ideal J of R, $J^2=J$, then these rings are called fully idempotent rings, and are more studied in the literature. $\endgroup$
    – Name
    Commented Jan 9, 2014 at 9:30

1 Answer 1

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Hope this helps

http://webs.um.es/leandro/miwiki/lib/exe/fetch.php?id=curriculum&cache=cache&media=dssrttn.pdf

If you remove the condition A^2=A things are much more complicated, but there is some literature about that.

Leandro

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