On Lie nilpotent associative algebras

Abstract

Let $G$ be a group generated by a set $X$. It is well known and easy to check that \[ [g_1, g_2, \dots ,g_n] = 1 \mbox{ for all } g_i \in G \qquad \iff \qquad [x_1, x_2, \dots , x_n] =1 \mbox{ for all } x_i \in X. \] Let $L$ be a Lie algebra generated by a set $X$. Then it is also well known and easy to check that \[ [h_1, h_2, \dots , h_n] = 0 \mbox{ for all } h_i \in L \qquad \iff \qquad [x_1, x_2, \dots ,x_n] = 0 \mbox{ for all } x_i \in X. \] Now let $A$ be a unital associative algebra generated by a set $X$. Then the assertion similar to the above does not hold: for $n>2$, it is easy to find an algebra $A$ with a generating set $X$ such that $[x_1,x_2,\dots ,x_n]=0$ for all $x_i\in X$ but $[a_1,a_2,\dots ,a_n]\ne 0$ for some $a_i\in A$. However, we prove the following result. Let $R$ be a unital associative and commutative ring such that $\frac{1}{3}\in R$. Let $A$ be a unital associative $R$-algebra generated by a set $X$. Let $X^2=\{x_1{x}_2\mid x_i\in X\}$ be the set of all products of $2$ elements of $X$. Then \[ [a_1, a_2, \dots ,a_n] = 0 \mbox{ for all } a_i \in A \qquad \iff \qquad [y_1, y_2, \dots , y_n] =0 \mbox{ for all } y_i \in X \cup X^2. \] Moreover, one can assume that in the commutator $[y_1,y_2,\dots ,y_n]$ above $y_1,y_n\in X$.

18 p

References in corpus (2)

• On $[A,A]/[A,[A,A]]$ and on a $W_n$-action on the consecutive commutators of free associative algebra
• On universal Lie nilpotent associative algebras