# 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.

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 (or 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 $$\Bbb Z$$.

• Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab. Mar 19, 2014 at 16:34
• I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory? Mar 19, 2014 at 16:48
• I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general. Mar 19, 2014 at 17:03
• @Jacob In any domain, if $\,x\mid ab\,$ then the cofactor $\,x'\,$ such that $\,xx' = ab\,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article. Mar 19, 2014 at 18:25