Geodesic
Once more on the lattice of subvarieties of the wreath product of the variety of semilattices and the variety of semigroups with zero multiplication
Fundamentalʹnaâ i prikladnaâ matematika, Tome 21 (2016) no. 1, pp. 193-210 
A. V. Tishchenko. Once more on the lattice of subvarieties of the wreath product of the variety of semilattices and the variety of semigroups with zero multiplication. Fundamentalʹnaâ i prikladnaâ matematika, Tome 21 (2016) no. 1, pp. 193-210. http://geodesic.mathdoc.fr/item/FPM_2016_21_1_a15/
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Voir la notice de l'article provenant de la source Math-Net.Ru

It is known that the monoid wreath product of any two semigroup varieties that are atoms in the lattice of all semigroup varieties may have a finite as well as an infinite lattice of subvarieties. If this lattice is finite, then as a rule it has at most eleven elements. This was proved in a paper of the author in 2007. The exclusion is the monoid wreath product $\mathbf{Sl}\mathbin{\mathrm{w}} \mathbf{N}_2$ of the variety of semilattices and the variety of semigroups with zero multiplication. The number of elements of the lattice $L(\mathbf{Sl} \mathbin{\mathrm{w}} \mathbf{N}_2)$ of subvarieties of $\mathbf{Sl}\mathbin{\mathrm{w}} \mathbf{N}_2$ is still unknown. In a previous paper, we have shown that the lattice $L(\mathbf{Sl}\mathbin{\mathrm{w}} \mathbf{N}_2)$ contains a sublattice having 33 elements. In the present paper, it is proved that the lattice under consideration has exactly three maximal subvarieties. As a first application of the obtained results we calculate the finite basis of the lattice union of the variety of all semilattices and the largest variety among subvarieties of our lattice having at least one heterotypic identity. As a second application we show that the considered lattice of subvarieties has at least 39 elements.

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