# Equivalence of idempotents and projective modules over nonunital rings

For a nonunital ring $$R$$ (or "rng") one has to be a little bit careful when considering the category of (left or right) $$R$$-module and, furthermore, the standard equivalent definitions of projective modules are in general not equivalent anymore (see e.g.https://math.stackexchange.com/questions/120458/projective-modules-over-rings-without-unit). To fix terminology: by projective module we mean the standard categorical definition https://en.wikipedia.org/wiki/Projective_module.

Let $$M_n(R)$$ denote the set of $$(n\times n)$$-matrices with entries from $$R$$ and let $$e\in M_n(R)$$ and $$f\in M_m(R)$$ be idempotents. It is easy to show that $$M=eR^n$$ and $$N=fR^m$$ are projective modules. For a unital ring, the statement that $$M$$ is isomorphic to $$N$$ is equivalent to saying that there exists $$u\in GL_k(R)$$ such that $$e=ufu^{-1}$$ (tacitly assuming an appropriate $$k$$ and an embedding for $$e$$ and $$f$$ into $$M_k(R)$$). Now, what is the corresponding statement for nonunital rings?

The usual algebraic definition of equivalence of idempotents state that $$e\sim f$$ if there exists $$x,y$$ such that $$e=xy$$ and $$f=yx$$. It is easy to see that this implies that $$M\simeq N$$ (multiplication by $$y$$ gives the isomorphism $$M\to N$$). Now, is the converse true? That is, if $$M\simeq N$$ then there exists $$x,y$$ such that $$e=xy$$ and $$f=yx$$?

The standard proof for unital rings uses the fact that a projective module is a direct summand of a free module, which may no longer be true for nonunital rings.

• If R is a non-unital ring it is still true for idempotents e,f that $eR\cong fR$ iff e=xy and f=yx for some x,y. You just need $Hom_R(eR,fR)\cong fRe$ and dually which doesn't use identities. You don't even need addition. This is true for semigroups. – Benjamin Steinberg Oct 9 '18 at 13:07
• Thanks, I'll think about it. If you develop your comment just a little bit in an answer, I can vote it up! – Joakim Arnlind Oct 9 '18 at 15:52
• You might find sciencedirect.com/science/article/pii/002240499190096K useful for Morita equivalence of rings without unit. – Benjamin Steinberg Oct 9 '18 at 20:18

## 1 Answer

If $$R$$ is a ring (not necessarily unital) and $$e\in R$$ is an idempotent, then it is still the case that $$Hom_R(eR,M)\cong Me$$ for any right $$R$$-module $$M$$ via $$\phi\mapsto \phi(e)$$. It immediately follows that, if $$e,f\in R$$ are two idempotents, then $$eR\cong fR$$ if and only if there exists $$a\in eRf$$ and $$b\in fRe$$ with $$ab=e$$ and $$ba=f$$. This is equivalent to your condition once you notice that if $$e=xy$$ and $$f=yx$$, then putting $$a=exf$$ and $$b=fye$$ you have that $$ab=exffye=exyxye=e$$ and $$ba=fyeexf=fyxyxf=f$$.

Nothing here uses that you have a ring. You could work with semigroups and right $$S$$-sets. Then you are reducing to standard facts about Green's relations.