Portugaliæ Mathematica   EMIS ELibM Electronic Journals PORTUGALIAE
Vol. 52, No. 3, pp. 279-287 (1995)

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Equations Linéaires dans les Anneaux Nilpotents au Sens de Lie

Gérard Endimioni

Université de Provence, UFR-MIM, Unité de Recherche Associée au CNRS \no225,
3, place Victor Hugo, F-13331 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 two-sided ideals defined by $\calc{L}_{1}(A)=A$ and $\calc{L}_{n}(A)= A[\calc{L}_{n-1}(A),A]$ with $[x,y]=xy-yx$. 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 two-sided ideal generated by elements of the form $[x_{1},...,x_{n}]$.

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