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Tensor Products and Transferability of Semilattices

Published online by Cambridge University Press:  20 November 2018

G. Grätzer
Affiliation:
Department of Mathematics, University of Manitoba, Winnipeg, Manitoba, R3T 2N2 email: gratzer@cc.umanitoba.ca webiste: http://www.maths.umanitoba.ca/homepages/gratzer/
F. Wehrung
Affiliation:
C.N.R.S., E.S.A. 6081, Département de Mathématiques, Université de Caen, 14032 Caen Cedex, France email: wehrung@math.unicaen.fr website: http://www.math.unicaen.fr/˜ehrung
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Abstract

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In general, the tensor product, $A\otimes B$, of the lattices $A$ and $B$ with zero is not a lattice (it is only a join-semilattice with zero). If $A\otimes B$ is a capped tensor product, then $A\otimes B$ is a lattice (the converse is not known). In this paper, we investigate lattices $A$ with zero enjoying the property that $A\otimes B$ is a capped tensor product, for every lattice $B$ with zero; we shall call such lattices amenable.

The first author introduced in 1966 the concept of a sharply transferable lattice. In 1972, H. Gaskill defined, similarly, sharply transferable semilattices, and characterized them by a very effective condition $\left( \text{T} \right)$.

We prove that a finite lattice $A$ is amenable iff it is sharply transferable as a join-semilattice.

For a general lattice $A$ with zero, we obtain the result: $A$ is amenable iff $A$ is locally finite and every finite sublattice of $A$ is transferable as a join-semilattice.

This yields, for example, that a finite lattice $A$ is amenable iff $A\otimes \text{F}\left( 3 \right)$ is a lattice iff $A$ satisfies $\left( \text{T} \right)$, with respect to join. In particular, ${{M}_{3}}\,\otimes \,\text{F}\left( 3 \right)$ is not a lattice. This solves a problem raised by R. W. Quackenbush in 1985 whether the tensor product of lattices with zero is always a lattice.

Type
Research Article
Copyright
Copyright © Canadian Mathematical Society 1999

References

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