$(a,b)[a,b]=ab$ in non factorial monoids

Question

Do you know of a proof of $[a,b](a,b)=ab$ in $\mathbb Z$ that doesn't use prime factorization?

To be more precise let's strip all unnecessary properties and leave only the bare bones of divisibility: on a commutative monoid $M$ with the cancellation law and for which the g.c.d and l.c.m of two elements exists, does $[a,b](a,b)\sim ab$ still hold?

($\sim$ means equality up to a unit factor, also notice that this allows the existence of an infinite chain of divisors, otherwise it's known that M is factorial, see for example theorem 2.22 in Jacobson Basic Algebra I)

Edit: To clarify the notation, $[a,b]$ is the lcm of $a$ and $b$, while $(a,b)$ is the gcd.

Answer 1

We'll use $x,y$ for the $a,b$ in the post, and retain the notation $(x,y)$ for the gcd and $[x,y]$ for the lcm, which does appear in a few number theory texts. Put $(x,y)=d,\ [x,y]=m$ and write $x=da,\ y=db.$ Then $(a,b)=1$ follows (meaning it is a unit, gcd and lcm only being defined up to units). For if $k|a,\ k|b$ then also $kd|x,\ kd|y$ and then $kd|(x,y)=d$ giving $k|1$ i.e. $k$ is a unit.

We'll need a Lemma for later:

If $(a,b)=1$ then $[a,b]=ab.$

Let $e=[a,b]$, so that since $ab$ is one of the common multiples of $a,b$ we have $e|ab.$ So we may write $ek=ab.$ From $e=[a,b]$ there are $u,v$ for which $au=e,\ bv=e.$ Then $auk=ab$ gives $uk=b$ so $k|b,$ and similarly $bvk=ab$ gives $vk=a$ so $k|a.$ Then from $k|a,b$ follows $k|(a,b)=1,$ i.e. $k$ is a unit, and $[a,b]=e=ab$ up to the unit $k.$

Now define $m=[x,y]=[ad,bd]$ and consider the term $abd.$ It is a common multiple of $ad$ and $bd$ and so $m|abd.$ On the other hand, $m$ is a multiple of $d$ since each of $ad,bd$ divide $m,$ so we may write $m=dz.$ Then from $ad|m=dz$ follows $a|z,$ and from $bd|m=dz$ follows $b|z.$ So each of $a,b$ divides $z,$ hence so does $[a,b].$ We are here in the situation that $(a,b)=1,$ so by the Lemma we have $[a,b]=ab,$ and so $ab|z,$ and there is $t$ with $abt=z.$ But then $(abd)t=dz=m,$ so that $abd|m.$

We now have both $m|abd$ and $abd|m$ making these associates. Finally from that we get $md=(ad)(bd)=xy,$ up to a unit, i.e. $[x,y]\cdot(xy)=xy$ (again up to a unit).

Answer 2

Let $M$ be a commutative cancellative monoid in which gcds and lcms exist.

Lemma 0. If two elements $x,y$ of $M$ satisfy $x|t$ if and only if $y|t$, for all elements $t$, then $x$ and $y$ are associated.

Proof. For $t=x$ we see that $x|x$ implies $y|x$. By symmetry, we also get $x|y$. Hence, $x,y$ are associated. (Actually this is a special case of the Yoneda Lemma.)

Lemma 1. If $a,b,t$ are elements of $M$, then $\mathrm{gcd}(a \cdot t,b \cdot t)$ and $\mathrm{gcd}(a,b) \cdot t$ are associated.

Proof. Clearly $\mathrm{gcd}(a,b) \cdot t$ divides $a \cdot t$ and $b \cdot t$ and hence also $\mathrm{gcd}(a \cdot t,b \cdot t)$. Conversely, observe that $t$ divides $\mathrm{gcd}(a \cdot t,b \cdot t)$ since it divides $a \cdot t$ and $b \cdot t$. Write $\mathrm{gcd}(a \cdot t,b \cdot t) = c \cdot t$. Then $c$ divides $a$ and $b$, and hence $c$ divides $\mathrm{gcd}(a,b)$. So $\mathrm{gcd}(a \cdot t,b \cdot t)$ divides $\mathrm{gcd}(a,b) \cdot t$.

Proposition. If $a,b$ are elements of $M$, then $\mathrm{lcm}(a,b) \mathrm{gcd}(a,b)$ is associated to $ab$.

Proof. So let $t$ be any element of $M$. Then:

$\phantom{.--} \mathrm{lcm}(a,b)/b | t$

$\Longleftrightarrow$ $\mathrm{lcm}(a,b) | bt$

$\Longleftrightarrow$ $a | bt$ and $b | bt$

$\Longleftrightarrow$ $a | bt$

$\Longleftrightarrow$ $a | bt$ and $a|at$

$\Longleftrightarrow$ $a | \mathrm{gcd}(at,bt) \stackrel{\text{Lemma 1}}{=} \mathrm{gcd}(a,b)t$

$\Longleftrightarrow$ $a/\mathrm{gcd}(a,b) | t$

Hence, Lemma 0 implies that $\mathrm{lcm}(a,b)/b$ is associated to $a/\mathrm{gcd}(a,b)$.