**Ralph Freese and J. B. Nation**

**May 1, 1996
**

This note contains some clarifications for our paper
*Congruence lattices of semilattices*, [2].
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 [2] 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 [2]
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
[1]. 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 [2] 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 [1],
every chain is finite. Thus, if we let *b* be the
join of those *a*_{i}'s lying below *c* and
we suppose *b* < *c*, then
there is an element *r* with . By the definition
of *b*, for no *a*_{i} 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.
Let be
the lattice of all subsets of the atoms of . Map the
elements of *B* to *L* by mapping each subset to its join in
**L**.
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.

Most of the above arguments are in Crawley and Dilworth's book. We will take this opportunity to point out a small error in that book. On page 53 they state that if is a strongly atomic, compactly generated lattice then the following three conditions are equivalent:

- is locally distributive,
- is semimodular and every modular sublattice is distributive, and
- for every set of four distinct
elements
*a*,*p*_{1},*p*_{2}, for which*p*_{1},*p*_{2}, , the sublattice is an eight-element Boolean algebra.

Wed May 1 14:43:43 HST 1996