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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.

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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.

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    $\begingroup$ Thanks for your helpful answer. $\endgroup$ May 4, 2021 at 17:58

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