Commutator of a ring

Remarks

The commutator is biadditive, and $[x, x] = 0$ and $[y, x] = -[x, y]$ for all $x, y$. So $[x, y] = 0$ if and only if $xy = yx$. If $A$ is associative, then for all $x, y, z \in A$: $$[x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 \quad \text{and} \quad [x, yz] = [x, y]z + y[x, z].$$ The first is the Jacobi identity, so $(A, [\cdot, \cdot])$ is a Lie ring; the second says that $y \mapsto [x, y]$ is a derivation. Example: in the $2 \times 2$ real matrices, $E = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $F = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ give $EF - FE = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \neq 0$. In a commutative ring every commutator is $0$. The anticommutator is $xy + yx$.