Let $R$ be a unital ring. We define  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 $p_{0}=a M p_{1},\;\;\;p_{1}  M p_{2},\ldots p_{n-1} M p_{n}=b$

Put $R=M_{n}(\mathbb{C})$. One can show that the equivalent class containing $0$ is $$[0]=\{A\in M_{n}(\mathbb{C})\mid A^{n}=0\}$$

(In fact one can prove the following: If $A\in B(H)$ satisfy $A^{k}=0$ then there are $X,Y\in B(H)$  with $A=XY$  and $(YX)^{k-1}=0$. Here $B(H)$ is the space of bounded operators on a  Hilbert space. The same is true by replacing $B(H)$   with an  arbitrary  Von Neumann  algebra. The  same  also is true   without any topological consideration, that is by replacing $B(H)$  with $L(V)$, the space of linear endomorphisms of  a vector space $V$.)

So for  $R=M_{n}(\mathbb{C}),\;\;[0]$  is  an algebraic variety,i.e: the variety of nilpotent  matrices  $A^{n}=0$ 

>1.Is every equivalent class  of  $M_{n}(\mathbb{C})$ an  algebraic variety?(the zero set of  polynomials on $M_{n}(\mathbb{C}) \simeq \mathbb{C}^{n^{2}}$ or the zero set of  polynomials in the form $f(A)=0$ where $f\in \mathbb{C}[x]$?What is the precise description of equivalent classes?


>2. What is an example  of  a  unital ring with  a nilpotent element   $a$  which is  not equivalent to $0$?


>3. Assume that  $A$ is  a  $C^{*}$  algebra  and $a\in A$  satisfies $a^{k}=0$  for  some  $k>1$. Are there two elements $x,y \in A$  with $a=xy$  and $(yx)^{k-1}=0$?