Transfer Between LCM, GCD for Rings?

I am starting a chapter on divisibility in commutative rings, and I was wondering if there was a way to translate theorems about gcd to lcm and vice versa. I know the concepts are considered "dual" in some sense, so perhaps the theorems relating to them are also dual.

Solution 1:

A prototypical example is proving $\ \rm gcd(a,b)\:lcm(a,b) = ab,\ $ using the $\,\overbrace{{\rm involution}\,\ x'\! =\, ab/x}^{\rm\large cofactor\ duality\ \ }\ $ on the divisors of $\rm\:ab.\ $ Notice that $\rm\ x\mid y\color{#c00}\iff y'\mid x',\ $ by ${\,\ \rm\dfrac{y}x = \dfrac{x'}{y'} \ }$ by $\rm\, \ yy'\! = ab = xx'.\, $ Thus

$$\begin{align}\rm c\mid\gcd(a,b)\!\iff&\rm\ c\mid a,b\\[2px] \color{#c00}\iff&\ \rm a',b'\mid c'\\[2px] \iff &\ \rm lcm(a',b')\mid c'\\ \color{#c00}\iff &\ \rm c\mid lcm(a',b')'\\ {\rm Thus}\rm\quad \gcd(a,b)\, \ =\ &\rm \, lcm(a',b')'=\ \dfrac{ab}{lcm(b,a)} \end{align}\quad $$

The black arrows above are the universal property (definition) of gcd and lcm, and the red arrows follow by cofactor duality.

Notational abuse alert: the gcd, lcm "equalities" are up to unit factors (i.e. "equal" if associate), to keep the post accessible to beginners working in $\Bbb Z$.