On the semigroup nilpotency and the Lie nilpotency of associative algebras
This paper studies the relation between Lie nilpotence and semigroup nilpotence of associative rings. An associative ring \(R\) can be viewed as a Lie ring \(R^{(-)}\) using the multiplication \([a,b]=ab-ba\), and so the notion of the Lie nilpotence of \(R^{(-)}\) makes sense. There are two associative semigroups related to \(R\): \(M(R)\) with multiplication that of \(R\) and \(A(R)\) with multiplication \(a\cdot b=a+b+ab\). There are various notions of nilpotence for a semigroup and the author uses the following, and variants of it, due to Mal'tsev. Using noncommuting indeterminates set \(U_0=x\), \(V_0=y\), and for \(k\geq 0\), \(U_{k+1}=U_kz_kV_k\) and \(V_{k+1}=V_kz_kU_k\). Then a semigroup is nilpotent of length \(m\) if \(U_{m+1}=V_{m+1}\) is an identity but \(U_m=V_m\) is not. The main results show that: whenever \(R^{(-)}\) is nilpotent of length \(m\) then \(M(R)\) and \(A(R)\) are nilpotent of length at most \(m\); if \(R\) is an algebra over an infinite field then the nilpotence of \(R^{(-)}\) of length \(m\) and the nilpotence of \(A(R)\) of length \(m\) are equivalent; and for any \(R\), \(A(R)\) nilpotent of length 2 implies that \(R^{(-)}\) is also. In the second statement \(M(R)\) may replace \(A(R)\) when \(1\in R\) since then \(A(R)\cong M(R)\). An example is given of an algebra \(R\) without identity but over an infinite field with \(M(R)\) nilpotent of length 2 but \(R^{(-)}\) not nilpotent.
scientific article; zbMATH DE number 1257136