LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-07-31

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Every join-irreducible element of a distributive lattice is join-prime

Statement

Let $L$ be a finite distributive lattice and let $j \in L$ be join-irreducible. If $j \leq a \lor b$ , then $j \leq a$ or $j \leq b$ . Thus $j$ is join-prime.

Facts & Assumptions

Given: A finite distributive lattice $L$ , a join-irreducible $j \in L$ , and elements $a, b \in L$ with $j \leq a \lor b$ .

[F1]

In a lattice, $x \leq y$ exactly when $x \land y = x$ ; distributivity gives $x \land (y \lor z) = (x \land y) \lor (x \land z)$ (Lattices, distributive lattices, and order ideals).

[F2]

If $j = u \lor v$ and $j$ is join-irreducible, then $j = u$ or $j = v$ (Join-irreducible elements of a nonempty finite lattice).

Proof

technique · direct

1.1

Since $j \leq a \lor b$ , one has $j = j \land (a \lor b)$ . Distributivity rewrites this as $j = (j \land a) \lor (j \land b)$ .

givenF1

2.1

Join-irreducibility applied to step 1.1 gives $j = j \land a$ or $j = j \land b$ . These equalities are respectively equivalent to $j \leq a$ or $j \leq b$ .

step 1.1F1F2

3.1

Hence every join-irreducible element of a distributive lattice is join-prime.

step 2.1∎

Depends on

Used by

Birkhoff representation theorem: every finite distributive lattice is isomorphic to the lattice of order ideals of its join-irreducible poset Theorem

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

MIT OpenCourseWare 18.212, Lecture 16: Distributive lattices (standard reference, not scraped)