PORTUGALIAE MATHEMATICA Vol. 52, No. 3, pp. 279287 (1995) 

Equations Linéaires dans les Anneaux Nilpotents au Sens de LieGérard EndimioniUniversité de Provence, UFRMIM, Unité de Recherche Associée au CNRS \no225,3, place Victor Hugo, F13331 Marseille Cedex 3  FRANCE Abstract: {\sl Linear equations in Lie nilpotent rings}. Let $A$ be a ring and let $(\calc{L}_{n}(A))_{n\ge1}$ be the family of twosided ideals defined by $\calc{L}_{1}(A)=A$ and $\calc{L}_{n}(A)= A[\calc{L}_{n1}(A),A]$ with $[x,y]=xyyx$. Assume that $\bigcap_{n\ge1}\calc{L}_{n}(A)=0$. In the usual way, we regard the family $(\calc{L}_{n}(A))_{n\ge1}$ as a fondamental system of neighbourhoods of 0 for a separated topology such that $A$ is a topological ring. Let $\widehat A=\limgets(A/\calc{L}_{n}(A))$ the completion of $A$. If $a_{1},...,a_{k},b_{1},...,b_{k}\in A$, we prove that the linear equation $a_{1}xb_{1}+...+a_{k}xb_{k}=c$ is soluble in $\widehat A$ for all $c\in\widehat A$ if and only if $a_{1}b_{1}+...+a_{k}b_{k}$ is invertible in $\widehat A$. Moreover, solution is unique and is given by the limit of a certain sequence. If $A$ is strongly Lie nilpotent (i.e. $\calc{L}_{n}(A)=0$ for some $n$), we can take $\widehat A=A$ in the previous result. We give an application to linear equations in an arbitrary ring. In particular, we prove that in a ring $A$, if $a_{1}b_{1}+...+a_{k}b_{k}$ is invertible, then all the solutions of the equation $a_{1}xb_{1}+...+a_{k}xb_{k}=0$ are in $\bigcap_{n\ge1}L_{n}(A)=0$, where $L_{n}(A)$ is the twosided ideal generated by elements of the form $[x_{1},...,x_{n}]$. Full text of the article:
Electronic version published on: 29 Mar 2001. This page was last modified: 27 Nov 2007.
© 1995 Sociedade Portuguesa de Matemática
