Adjoint to multiplication in a GCD lattice

Question

Consider the lattice on the nonzero natural numbers where the meet $a \wedge b $ is defined to be the greatest common divisor of $a$ and $b$, and the join $a \vee b$ is the least common multiple. There's a partial order in that $ a \leq b \equiv a \vee b = b$.

With respect to that partial order, can we find an adjoint / to multiplication that satisfies the galois connection $ a/b \leq c \equiv a \leq b \times c$?

With the ordinary notion of order on the naturals, we get residuated division as a right adjoint to multiplication. But this doesn't work in the GCD lattice. Is there anything that does?

can you parentheses to help clarify – phdmba7of12 Oct 05 '18 at 15:46 — phdmba7of12, Oct 05 '18 at 15:46

score 5 · Accepted Answer · answered Oct 05 '18 at 17:45

$a/b$ is just $a/\mathrm{gcd}(a,b)$. Here's a perversely highbrow explanation, just for fun.

Since the lattice of natural numbers under divisibility has meets of arbitrary nonempty sets (the greatest common divisor extends fine to infinite sets) any unbounded meet-preserving functor $f$ has a left adjoint $g$ given by $g(a)=\mathrm{inf}n:a \leq f(n)$. In the case that $f$ is multiplication by $b$, we have for $a/b$ the least $n$ such that $a$ divides $n\times b$, which is as described in the first paragraph.

Of course this can also be written as $\mathrm{lcm}(a,b)/b$ which makes that last equivalence almost instantaneous. Neat answer! — Steven Stadnicki, Oct 05 '18 at 17:58

Adjoint to multiplication in a GCD lattice

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