Lifts for semigroups of monomorphisms of an independence algebra

João Araújo¹

¹ Universidade Aberta R. Escola Politécnica, 147 1269-001 Lisboa, Portugal and Centro de Álgebra Universidade de Lisboa Av. Gama Pinto, 2 1649-003 Lisboa, Portugal

Colloquium Mathematicum, Tome 97 (2003) no. 2, pp. 277-284.

Résumé

For a universal algebra ${\cal A}$, let $\mathop{\rm End}\nolimits ({\cal A} )$ and $\mathop{\rm Aut}\nolimits ({\cal A} )$ denote, respectively, the endomorphism monoid and the automorphism group of ${\cal A}$. Let $S$ be a semigroup and let $T$ be a characteristic subsemigroup of $S$. We say that $\phi \in \mathop{\rm Aut}\nolimits (S)$ is a lift for $\psi\in \mathop{\rm Aut}\nolimits (T)$ if $\phi|T=\psi$. For $\psi \in \mathop{\rm Aut}\nolimits (T)$ we denote by $L(\psi)$ the set of lifts of $\psi$, that is, $ L(\psi )= \{\phi \in \mathop{\rm Aut}\nolimits (S) \mid \phi|T=\psi\}. $ Let ${\cal A}$ be an independence algebra of infinite rank and let $S$ be a monoid of monomorphisms such that $G=\mathop{\rm Aut}\nolimits ({\cal A} )\leq S \leq \mathop{\rm End}\nolimits ({\cal A} )$. In  [2] it is proved that if ${\cal A}$ is a set (that is, an algebra without operations), then $|L(\phi)|= 1$. The analogous result for vector spaces does not hold. Thus the natural question is: Characterize the independence algebras in which $|L(\phi)|=1$. The aim of this note is to answer this question.

Détail

DOI : 10.4064/cm97-2-11

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