# Special nilpotent elements

Let $$R$$ be a (noncommutative, associative) ring. Set $$N_2:=\{x\in R : x^2=0\}$$, the set of nilpotent elements of degree $$2$$ (also called the square-zero elements).

If $$x,y\in R$$ satisfy $$xy=0$$, then $$yx\in N_2$$, but not every element in $$N_2$$ arises in this way. (See the example below.)

Question: Has the set $$F:=\{yx : xy=0\}$$ been studied before? Are there characterizations of these "special square-zero" elements?

Note that every element in $$F$$ is a commutator (if $$xy=0$$, then $$yx=yx-xy=[y,x]$$.) Thus, a precise question would be if $$F=N_2\cap[R,R]$$?

Example: Consider $$R=\left\{ \begin{pmatrix} a & b \\ 0 & a\end{pmatrix} : a,b\in\mathbb{R} \right\}$$. Then $$z=\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$$ belongs to $$N_2$$ but not to $$F$$, since every commutator in $$R$$ is zero.

The answer to your precise question is no: it is not always the case that $$F = N_2\cap [R,R]$$. A nice way to see this is by fixing a field $$k$$, and constructing the universal example of a $$k$$-algebra equipped with a square-zero commutator. That is, $$R = k\langle a,b\rangle/((ab - ba)^2)$$.

In that $$k$$-algebra, the element $$ab - ba$$ is certainly a square-zero commutator. The question is whether that element is equal to $$yx$$ for some pair of elements $$x,y\in R$$ such that $$xy=0$$. You can rule the existence of such a pair $$x,y\in R$$ by grading $$R$$ so that $$a,b$$ are each in degree $$1$$. Then the relation $$(ab - ba)^2$$ in $$R$$ is homogeneous of degree $$4$$. So if you have $$yx = ab- ba$$, then:

$$x(ab - ba) = xyx = 0$$, so $$x$$ must have no nonzero terms of degree $$<2$$, and

$$(ab - ba)y = yxy = 0$$, so $$y$$ must have no nonzero terms of degree $$<2$$.

But $$ab-ba$$ is homogeneous of degree $$2$$, so there's no way to multiply $$x$$ and $$y$$, each with no terms of degree $$<2$$, to yield $$ab-ba$$.

As an aside, in the case where $$k = \mathbb{F}_2$$, a certain $$k$$-algebra quotient of $$k\langle a,b\rangle/((ab - ba)^2)$$ arises quite naturally in topology: the $$k$$-algebra $$k\langle a,b\rangle/((ab - ba)^2,a^2, aba - b^2)$$ is isomorphic to the subalgebra of the Steenrod algebra generated by the Steenrod squares $$Sq^1$$ and $$Sq^2$$. The element $$ab - ba = Sq^1 Sq^2 - Sq^2 Sq^1$$ is one of the famous Milnor primitives, usually denoted $$Q_1$$. That element is also an example of a square-zero commutator which is not in your set $$F$$.

Thanks for asking an interesting question, which I enjoyed thinking about.

• Thanks for your helpful answer. May 4, 2021 at 17:58