Maximal Sublattices of Finite Distributive Lattices. III: A Conjecture from the 1984 Banff Conference on Graphs and Order

Let L be a finite distributive lattice. Let Sub0(L) be the lattice {S | S is a sublattice of L} ∪ {∅} and let ∗[Sub0(L)] be the length of the shortest maximal chain in Sub0(L). It is proved that if K and L are non-trivial finite distributive lattices, then ∗[Sub0(K × L)] = ∗[Sub0(K)] + ∗[Sub0(L)]. A conjecture from the 1984 Banff Conference on Graphs and Order is thus proved. 1 Motivation Let L be a finite lattice. Let Sub0(L) denote the lattice {S | S is a sublattice of L} ∪ {∅} ordered by inclusion. (Recall that a lattice or sublattice is by definition non-empty; if |L| = 1, we say L is trivial.) Let ∗[Sub0(L)] be the length of the shortest maximal chain in this lattice. Figures 1 through 4 illustrate maximal chains in Sub0(L) where L equals 3, 2 × 2, and 3 × 3. (For n ≥ 0, n is the n-element chain.) We exhibit two maximal chains of Sub0(3) of different lengths, one of length 9, one of length 6. How do we know there are not maximal chains that are shorter still? In [3, Theorem 2(i)], Chen, Koh, and Lee proved the following. Theorem 1.1 Let m ≥ 1; let n1, . . . , nm ≥ 2. Then ∗[Sub0(n1 × · · · × nm)] = m ∑