---
tag: KKJ9F6JJY0
kind: definition
name: Free distributive lattice
msc: ["06B25", "06D05"]
defines: T5GK6DH8BR
uses: [0HM0HQCK6N, BNVRR925X8]
sources:
  - url: https://en.wikipedia.org/wiki/Free_lattice
---

## Definition

Let $S$ be a set.
Let $F$ be a [[distributive lattice|0HM0HQCK6N]] and let $\iota \colon S \to F$ be a map.
The pair $(F, \iota)$ is a *free distributive lattice* on $S$ if the following holds.
For every distributive lattice $L$ and every map $g \colon S \to L$, there is exactly one [[lattice homomorphism|BNVRR925X8]] $\bar g \colon F \to L$ with $\bar g \circ \iota = g$.

## Remarks

It exists for every $S$ and is unique up to a unique isomorphism compatible with $\iota$.
For nonempty $S$, every element of $F$ is a finite join of finite nonempty meets of generators $\iota(s)$.
For finite $S$ with $n \ge 1$ elements, $F$ is isomorphic to the lattice of non-constant monotone Boolean functions of $n$ variables, so it has $M(n) - 2$ elements, with $M(n)$ the [[Dedekind number|481ZPQ0GC3]].
If one also asks for $0$ and $1$ and homomorphisms that preserve them, the free bounded distributive lattice on $n$ generators has $M(n)$ elements.

Example: on $S = \set{x, y}$, the lattice $\set{x \wedge y, x, y, x \vee y}$ with $x \wedge y < x, y < x \vee y$ is free.
A map $g$ into $L$ extends by $x \wedge y \mapsto g(x) \wedge g(y)$ and $x \vee y \mapsto g(x) \vee g(y)$, and this is forced.

Non-example: the two-element chain $0 < 1$ with $\iota(x) = 0$ and $\iota(y) = 1$ is distributive and generated by the image of $\iota$, but it is not free.
The map $g(x) = 1$, $g(y) = 0$ into the same chain would need a homomorphism sending $0$ to $1$ and $1$ to $0$, which does not preserve the order.