A problem concerning the lattice varietal product (Q1106869)

In general, the product of two quasivarieties of universal algebras is again a quasivariety. Thus although the product of two varieties of lattices is not necessarily a variety, it is a quasivariety and so closed under the formation of sublattices, direct products, and ultraproducts. McKenzie showed (in a forthcoming paper) that for any lattice variety \({\mathcal V}\) both \({\mathcal D}\circ {\mathcal V}\) and \({\mathcal M}\circ {\mathcal V}\) are varieties, where \({\mathcal D}\) refers to the variety of all distributive lattices and \({\mathcal M}\) refers to the variety of all modular lattices. There exist continuum many nonmodular varieties \({\mathcal V}\) such that \({\mathcal V}\circ {\mathcal D}\) is not a variety. McKenzie postulated that if \({\mathcal V}\) is a nonmodular lattice variety then \({\mathcal V}\circ {\mathcal D}\) is a variety only in the trivial case that \({\mathcal V}\) is the variety of all lattices. This paper confirms McKenzie's conjecture, showing that for any nontrivial modular lattice variety \({\mathcal P}\), if \({\mathcal V}\) is nonmodular and \({\mathcal V}\circ {\mathcal P}\) is a variety than \({\mathcal V}\) is the variety of all lattices.