Survey 1979: Equational Logic - Department of Mathematics ...

36 EQUATIONAL LOGIC
dual covers: one is M 5 ^ N 5 and each of the other fifteen (-.... in our picture) is
generated by a single subdirectly irreducible lattice.
We close this section with some remarkable general results of McKenzie on the
full lattice A = A(nt: t C T), where, temporarily, we make the type explicit.
THEOREM. (McKenzie [292] ). From the isomorphism type of A(nt: t C T)one
can recover the function
(1 (t T: n t = m} I: m co).
(In other words, one can recover the type (nt: t G T) up to renaming of all the
operations.)
THEOREM. (McKenzie [292]). (Appropriate (nt: t G T).). There exists a first
order formula 0(x) with one free variable in the language of lattice theory such that
the unique element of A satisfying 0(x) is the equational theory of groups. (Resp.
semigroups, lattices, distributive lattices, commutative semigroups, Boolean algebras.)
This last theorem had a precursor in Jeek [209]' the variety of commutative
semigroups obeying x2y = xy is definable (in a similar fashion).
PROBLEM. (McKenzie [292]). Does A possess any non-obvious
automorphisms 9.
PROBLEM. (McKenzie and Maltsev). Which lattices L are isomorphic to some
A(Z)? McKenzie has conjectured that the following easy necessary condition is also
sufficient: L is algebraic and its largest element is compact.
(Je,ek has proved [211 ] that L is isomorphic to an interval in A((2)) iff L is
algebraic and has only countably many compact elements.)
PROBLEM. (McKenzie [292]). When does an element tx C A(Z) have a "dual
cover," i.e., 13 < tx such that there exists no 3' with 13 < 3' ix? (E.g. the dual covers of
(x = y) are the equationally complete theories.) The answer is yes for semigroups
[430] and lattices [211 ]; see [225] for a dual cover of modular lattices. The result of
Je,ek just cited implies that there exists b tx such that there is no dual cover t3 with
b 13
ix. In fact and tx can be taken to be equational theories of modal algebras (W.
J. Blok, Notices Amer. Math. Soc. abstract 77T-A232).
ring"9,
PROBLEM. (Pigozzi [354] - q.v. for details.). Is A(Z) always a "partial Boolean