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). It 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?
- What is an example of a unital ring with a nilpotent element $a$ which is not equivalent to $0$?
- 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 $(yx)^{k-1}=0$?