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.

Every $M_{n}(F)$ where $F$ is an arbitrary field.

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

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.

Every $M_{n}(F)$ where $F$ is an arbitrary field.

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

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 is denoted by NIL. We have $[0]\subseteq NIL$. We are interested in the converse situation, as follows:

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

Note: If the ring $R$ is either of the following rings, then we have $[0]=NIL$:

Every $C^{*}$ algebra. Every $M_{n}(F)$ where $F$ is an arbitrary field. Every $End(V)$, the ring of endomorphisms of an arbitrary vector space $V$.

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. 

Every $M_{n}(F)$ where $F$ is an arbitrary field. 

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