Ralph Freese and J. B. Nation
May 1, 1996
This note contains some clarifications for our paper Congruence lattices of semilattices, . In that paper we use the term relatively pseudo-complemented for the condition:
Unfortunately, (1) and (2) are not the same; in fact, (2) implies distributivity. However, (1) and (2) are equivalent for distributive lattices. Thus (1) could have been used to define relatively pseudo-complemented without changing the meaning for distributive lattices.
Of course, our statement (3) of  that the congruence lattice, Con(L), of a semilattice S, is relatively pseudo-complemented means that it satisfies (1). It is worth pointing out that for compactly generated lattice condition (1) is equivalent to meet semidistributivity:
In statement (4) of  we claim that Con(L) is locally distributive. This is true, but our argument stated that a compactly generated lattice is locally distributive if and only if it is semimodular and satisfies (1). We referred to Crawley and Dilworth's book . However, Crawley and Dilworth prove this under the assumption that is strongly atomic. The lattice diagrammed in Figure 1 is compactly generated and locally distributive but is not semimodular. Thus one direction of the above equivalence does not hold without strong atomicity. However, the following theorem shows that the other direction does hold and thus our statement (4) of  is correct.
Theorem 1. If a compactly generated, semimodular lattice satisfies (1) then
it is locally distributive.
Proof. Let be a compactly generated, semimodular lattice satisfying (1). Let and let u be the join of the covers of z. We need to show the interval sublattice u/z is distributive. Since every interval of also is compactly generated, semimodular and satisfies (1), we may assume z=0 and u=1, i.e., 1 is the join of the atoms.
Note that (1) implies that any set of atoms of is independent. Let be a compact element. Since the atoms join to 1, there is a finite set of atoms which join above c. Then, by semimodularity and independence,
is a maximal chain in and so, by (3.8) of , every chain is finite. Thus, if we let b be the join of those ai's lying below c and we suppose b < c, then there is an element r with . By the definition of b, for no ai is and obviously and by semimodularity each either equals or covers b. But it is easy to see that this violates (1).
We conclude that every compact element, and thus every element,
of is the join of atoms.
the lattice of all subsets of the atoms of . Map the
elements of B to L by mapping each subset to its join in
By what we have just shown, this map is onto and it
clearly preserves joins. Let B and C be sets of atoms. If
then there is an atom with . But since sets of
atoms are independent, implies ; similarly,
. Thus , a contradiction.
Hence is isomorphic to the lattice of all subsets of its
atoms and thus is certainly distributive.