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:

>To what extent all unital rings with  $[0]=N(R)$ are classified. Is there a simple unital ring $R$ which does satisfy $[0]=N(R)$?

**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 $V$.