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  "body": "\n## Definition\n\nLet $S$ be a set.\nLet $F$ be a [[distributive lattice|0HM0HQCK6N]] and let $\\iota \\colon S \\to F$ be a map.\nThe pair $(F, \\iota)$ is a *free distributive lattice* on $S$ if the following holds.\nFor 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 $\\bar g \\circ \\iota = g$.\n\n## Remarks\n\nIt exists for every $S$ and is unique up to a unique isomorphism compatible with $\\iota$.\nFor nonempty $S$, every element of $F$ is a finite join of finite nonempty meets of generators $\\iota(s)$.\nFor 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]].\nIf one also asks for $0$ and $1$ and homomorphisms that preserve them, the free bounded distributive lattice on $n$ generators has $M(n)$ elements.\n\nExample: on $S = \\set{x, y}$, the lattice $\\set{x \\wedge y, x, y, x \\vee y}$ with $x \\wedge y \u003c x, y \u003c x \\vee y$ is free.\nA 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.\n\nNon-example: the two-element chain $0 \u003c 1$ with $\\iota(x) = 0$ and $\\iota(y) = 1$ is distributive and generated by the image of $\\iota$, but it is not free.\nThe 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.\n",
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