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The divisors of $60$ form a finite distributive lattice and realize Birkhoff's representation concretely

Example

Order the positive divisors of $60$ by divisibility. Since $60=2^2\cdot3\cdot5$ , every divisor has a unique form

$2^\alpha3^\beta5^\gamma,\qquad 0\le\alpha\le2,\quad0\le\beta,\gamma\le1.$

Divisibility is componentwise comparison of the exponent triples. Meet and join are componentwise minimum and maximum, so this is a finite distributive lattice.

Facts & Assumptions

Given: The positive divisors of $60$ , ordered by divisibility.

[L1]

Every positive integer has a prime factorization unique up to order (The fundamental theorem of arithmetic: every integer $n \ge 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i<r} p_i = \prod_{j<s} q_j$ with every $p_i$ and $q_j$ prime, then $r = s$ and $q_i = p_{\pi(i)}$ for some $\pi \in \operatorname{Sym}(r)$ ).

[L2]

Every finite distributive lattice is isomorphic to the order-ideal lattice of its join-irreducible poset (Birkhoff representation theorem: every finite distributive lattice is isomorphic to the lattice of order ideals of its join-irreducible poset).

Verification

technique · direct

1.1

By [L1], the exponent-triple description is unique, and $d\mid e$ exactly when every exponent of $d$ is at most the corresponding exponent of $e$ .

givenL1

2.1

Componentwise minimum and maximum give the greatest common divisor and least common multiple, and the distributive identities hold coordinatewise for minimum and maximum on chains. Thus the divisor poset is a finite distributive lattice.

step 1.1algebra

2.2

Its join-irreducibles are $2,4,3,5$ . In their inherited order, $2<4$ and $3,5$ are incomparable with these and with each other.

step 1.1

3.1

A divisor $d$ maps to the order ideal of join-irreducibles dividing it: its $2$ -exponent chooses $\varnothing$ , $\{2\}$ , or $\{2,4\}$ , while its $3$ - and $5$ -exponents independently choose whether to include $3$ and $5$ . This is exactly the Birkhoff map of [L2].

step 2.2L2

4.1

Hence the divisors of $60$ concretely realize Birkhoff's representation as the order ideals of the poset $2<4$ with isolated elements $3$ and $5$ .

step 2.1step 3.1∎

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Sources

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

The divisors of 606060 form a finite distributive lattice and realize Birkhoff's representation concretely