# On Lie nilpotent associative algebras

Research paper by Claud W. G. Dias, Alexei Krasilnikov

Indexed on: 17 Sep '17Published on: 17 Sep '17Published in: arXiv - Mathematics - Rings and 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 $y_1, y_n \in X$.