Let $R$ be a unital ring. We define the Murray Von Neumann relation $M$ on $R$ as follows:


We say $a M b$ iff $a=xy,\;b=yx$ for some $x,y\in R$. (This is inspired by the usual Murray Von Neumann equivalent relation in K theory, which is defined on the set of idempotents of a ring). The relation $M$ is a reflexive and symmetric relation but is not a transitive relation. So we consider its saturation. The saturation of this relation is an equivalent relation denoted by $\simeq$. In fact we say $a\simeq b$ if there are $p\_{i}\in R\;$ with $p\_{0} M p\_{1},\;\;\;p\_{1} M p\_{2},\ldots p\_{n-1} M p\_{n}$ where $p\_{0}=a,\;p\_{n}=b$.


The equivalent class containig $0$ is denoted by $[0]$. The set of nilpotent elements of $R$ is denoted by $N(R)$. We have $[0]\subseteq N(R)$. We are interested in the converse situation, as follows:



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> To what extent all unital rings with $[0]=N(R)$ are classified. Is there a simple unital ring $R$ which does satisfy $[0]=N(R)$?
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**Note:** If the ring $R$ is either of the following rings, then we have $[0]=N(R)$:


[Every $C^{\*}$ algebra](https://mathoverflow.net/questions/231328/the-saturation-of-murray-von-neumann-relation). 


Every $M\_{n}(F)$ where $F$ is an arbitrary field.


Every $End(V)$, the ring of endomorphisms of an arbitrary vector space $V$.



An isomorphic invariant in ring theory
If you have a finite ring $R$ that is [von Neumann regular](https://en.wikipedia.org/wiki/Von_Neumann_regular_ring), then $[0]=N(R)$ or more generally if you have a von Neumann regular ring $R$ such that each element has a power that belongs to a subsemigroup which is a group. This follows from a result on semigroups in <http://arxiv.org/abs/math/0605698>, see also <http://arxiv.org/pdf/0709.4341.pdf>. Maybe you can use some of the ideas with the ring property to get further.


