A distributive lattice with finite length must have a finite number of join-irreducible elements.

Question

Let $L$ be a distributive lattice. Let $\mathcal{J}(L)$ be the set of all join-irreducible elements. Recall that $x$ is join-irreducible if $x \neq 0$ and $x = a \vee b$ implies that $x = a$ or $x = b$.

Claim: If $L$ has finite length, then $\mathcal{J}(L)$ is finite.

I'm struggling with the details of proving this claim. What I have is based on a hint that went along with the exercise:

Let $C$ be a maximal (finite) chain in $L$. Define $\varphi : \mathcal{J}(L) \to C$ by $\varphi(x) := \bigwedge (\uparrow x \cap C).$ This map is injective. Otherwise, if $\varphi(x) = \varphi(y) = a$, then let $b$ be the lowest cover of $a$ in $C$. Then $x\wedge(b \vee y) = x.$

Somehow this proves the claim that the map is injective, but this fact is not clear to me. I think based on this statement, I should see that both $x$ and $y$ can't both be elements in $\mathcal{J}(L)$. Can someone clarify this for me?

Reference: This is from exercise 4.19 in Introduction to Lattices and Order by Davey and Priestly.

Answer 1

It's clear that $x = x \wedge ( b \vee y)$. By distributive property, we get $x = (x \wedge b) \vee (x \wedge y)$. Since $x$ is join irreducible, we must have that $x = (x \wedge b)$ or $x = (x \wedge y)$.

1. If $x \wedge b = x$ then $\varphi(x) = b$. A contradiction.
2. If $x = x \wedge y$ then $x = y$ or $x < y$. Suppose $x < y$. Then $\uparrow y \subsetneq \uparrow x$. Let $z$ be a maximal element in $\uparrow x \setminus \uparrow y$. If $z \in C$, we have a contradiction. Otherwise, let $\hat z$ be an upper cover of $z$, which must be in $\uparrow y$ (but not $C$). Then consider $\hat z \wedge (b \vee z) = \hat z \wedge a = \hat z.$ However, $(\hat z \wedge b) \vee (\hat z \wedge z) = a \vee \hat z = a$. Thus, the lattice is not distributive. Therefore, $x = y$.