Abstract

In a recent paper, the authors have proved that for lattices A and B with zero, the isomorphism Conc(A ⊗ B) ∼ = Conc A ⊗ Conc B, holds, provided that the tensor product satisfies a very natural condition (of being capped) implying that A ⊗ B is a lattice. In general, A ⊗ B is not a lattice; for instance, we proved that M3 ⊗ F(3) is not a lattice. In this paper, we introduce a new lattice construction, the box product for arbitrary lattices. The tensor product construction for complete lattices introduced by G. N. Raney in 1960 and by R. Wille in 1985 and the tensor product construction of A. Fraser in 1978 for semilattices bear some formal resemblance to the new construction. For lattices A and B, while their tensor product A ⊗ B (as semilattices) is not always a lattice, the box product, A □B, is always a lattice. Furthermore, the box product and some of its ideals behave like an improved tensor product. For example, if A and B are lattices with unit, then the isomorphism Conc(A □B) ∼ = Conc A ⊗ Conc B holds. There are analogous results for lattices A and B with zero and for a bounded lattice A and an arbitrary lattice B. A join-semilattice S with zero is called {0}-representable, if there exists a lattice L with zero such that Conc L ∼ = S. The above isomorphism results yield the following consequence: The tensor product of two {0}-representable semilattices is {0}-representable.

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