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.


1 Answer 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 (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 $\Bbb Z$.

  • $\begingroup$ 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. $\endgroup$ Mar 19, 2014 at 16:34
  • $\begingroup$ 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? $\endgroup$ Mar 19, 2014 at 16:48
  • $\begingroup$ 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. $\endgroup$ Mar 19, 2014 at 17:03
  • $\begingroup$ @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. $\endgroup$ Mar 19, 2014 at 18:25
  • $\begingroup$ See also this answer. $\endgroup$ Sep 6, 2014 at 5:22

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