Let $R$ be a ring. We define  Murray Von Neumann  relation $\leq$ on $R$ as follows. 

 $a\leq b$ iff $a=xy,\;b=yx$ for some $x,y\in R$. It is reflexive  and symmetric but not transitive. 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 $p_{0}=a\simeq p_{1}\simeq p_{2}\ldots\simeq P_{n}=b$

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 bounded operator on a  Hilbert space)

So for  $R=M_{n}(\mathbb{C}),\;\;[0]$  is  an algebraic variety. 

>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 commutative unital ring with  an element $a\simeq 0$  but $a$ is  not a nilpotent element?