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• 1 University of York Heslington York, YO10 5DD Department of Mathematics UK E-mail
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## Abstract

A subsemigroup S of a semigroup Q is a straight left order in Q and Q is a semigroup of straight left quotients of S if every qQ can be written as

\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$a^\#b$$ \end{document}
for some
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$a,b \in S$$ \end{document}
with a
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathcal{R}$$ \end{document}
b in Q and if, in addition, every element of S that is square cancellable lies in a subgroup of Q. Here a denotes the group inverse of a in some (hence any) subgroup of Q. If S is a straight left order in Q, then Q is necessarily regular; the idea is that Q has a better understood structure than that of S. Necessary and sufficient conditions exist on a semigroup S for S to be a straight left order. The technique is to consider a pair
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathcal{P}$$ \end{document}
of preorders on S. If such a pair satisfies conditions mimicking those satisfied by
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$( \leqslant _\mathcal{L} , \leqslant _\mathcal{R} )$$ \end{document}
on a regular semigroup, and if certain subsemigroups of S are right reversible, then S is a straight left order. The conditions required for
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathcal{P}$$ \end{document}
to satisfy are somewhat lengthy. In this paper we aim to circumvent some of these by specialising in two ways. First we consider only fully stratified left orders, that is, the case where
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathcal{P} = ( \leqslant _{\mathcal{L}^ * } , \leqslant _{\mathcal{R}^ * } )$$ \end{document}
(certainly the most natural choice for
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathcal{P}$$ \end{document}
) and the other is to insist that S be abundant, that is, every
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathcal{R}^ *$$ \end{document}
-class and every
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathcal{L}^ *$$ \end{document}
-class of S contains an idempotent. Our results may be used to show that the monoid of endomorphisms of a hereditary basis algebra of finite rank is a fully stratified straight left order.

Author: L. Fuchs

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