Number of distributive lattices with exactly one atom and four irreducible elements in total

Question

I was given the following problem:

How many distributive lattices are there with exactly one atom and four irreducible elements in total?

An irreducible element, in this context, is an $x$ in the lattice set $L$ such that $x = y \lor z \to x =y $ or $x = z$ for any $y, z \in L$, where $\lor$ is the supremum operator.

There clearly are distributive lattices satisfying this property. Take, for example, the Hasse diagram of $P(\{a, b, c\}, \subseteq)$ with its typical cube form. Add to this diagram a minimum node $0$, such that it looks like this:

Here, $I_1, ... I_3$ are irreducible elements, and so is the atom $a$, and the graph is a distributive lattice. In fact, my intuition is that one can "extend" elementary distributive lattices satisfying the constraint$-$such as the one whose Hasse diagram is above$-$ indefinitely while preserving the number of irreducible elements and the unique atom. For example,

is still distributive and satisfies the constraints. All I did was "add a floor" to the cube, and I could repeat that step indefinitely.

So far my intuition goes. But I am unable to prove this at all. I was requested to give a formal proof or justification of whatever answer I give, and here's where I'm falling short. How could one tackle this?

Thanks.

Answer 1

We can find exactly 5 such distributive lattices, the ones depicted on the Hasse diagrams below. The atom is highlighted in blue, the other irreducibles in red. Textual descriptions are provided below.

Our proof uses the following well-known result:

Theorem. (Birkhoff) Every finite distributive lattice $H$ arises as the the lattice of lower sets of the poset of irreducible elements of $H$.

Since the distributive lattices of interest have exactly 4 irreducibles, one could feasibly start by enumerating all 16 posets on 4 elements and proceed from there. But we know the one-atom restriction, which means we can start by fixing one element as that atom $0$: all other elements $x$ must satisfy $0 \leq x$, so it suffices to enumerate the posets with 3 elements. You can make quick work of these by drawing Hasse diagrams, or by considering maximal chains. Either way, you'll find five such posets:

• $P_1$: $1;\: 2;\: 3$
• $P_2$: $1 \leq 2;\: 3$
• $P_3$: $1 \leq 2;\: 1 \leq 3$
• $P_4$: $1 \leq 3;\: 2 \leq 3$
• $P_5$: $1 \leq 2 \leq 3$

This means that there are 5 distributive lattices with exactly one atom and four irreducibles. While you don't have to actually construct them to answer the question as stated, I'll describe which ones they are:

• $P_1$ gives rise to your example above, the 8-element Boolean algebra with a bottom element adjoined (so 9 elements in total).
• $P_2$ gives rise to the product of the 2-element Boolean algebra and the 3-chain, with a bottom element adjoined (so 7 elements in total).
• $P_3$ gives rise to the 4-element Boolean algebra with a bottom element adjoined twice (so 6 elements in total).
• $P_4$ gives rise to the 4-element Boolean algebra with a bottom and top element adjoined (so 6 elements in total).
• $P_5$ gives rise to the 5-chain.

And that's all of them!

Your strategy had a few problems:

• Your definition of irreducible does not quite correspond to the usual one: the bottom element has to be excluded (since one could regard it as the join of no elements).

• You drew the Hasse diagram of the 8-element Boolean algebra incorrectly (compare with mine above, you have an extra edge).

• Finally, the 12-element poset you constructed never satisfied the definition of distributive lattice (what would be the least upper bound of $I_1$, $I_2$ and $I_3$?). This led you down the wrong path, where you tried to argue that we can find infinitely many such lattices.

edit: an upscaled version of the diagrams is now available.