GCD and LCM via groups and rings

Let $a$ and $b$ be integers. Consider the ring

$G=\{ax+by:x,y\in\mathbb Z\}$.

This is a well-ordered group. So by a result in this post it is infinite cyclic. We call the positive generator $(a,b)$ the greatest common divisor (GCD) of $a$ and $b$.

Consider now the ring

$L=\{m\in\mathbb Z:m$ is divisible by both $a$ and $b\}$.

It is also a well-ordered group. Hence it is generated by a single positive element $[a,b]$, called the least common multiple (LCM) of $a$ and $b$.

Let

$M_a=\{ax:x\in\mathbb Z\}$ and $M_b=\{by:y\in\mathbb Z\}$.

$M_a$, $M_b$ and $L$ are subings of $G$. Moreover, $G=M_a+M_b$ and $L=M_a\cap M_b$. So by the Chinese remainder theorem,

$G/L\cong G/M_a\times G/M_b$,

which can be written as

$(\mathbb Z/L)/(\mathbb Z/G)\cong(\mathbb Z/M_a)/(\mathbb Z/G)\times(\mathbb Z/M_b)/(\mathbb Z/G)$

by the third isomorphism theorem. The groups in brackets are all finite groups of orders $[a,b]$, $(a,b)$, $a$ and $b$. Hence $[a,b]/(a,b)=ab/(a,b)^2$, i.e.,

$\boxed{ab=(a,b)[a,b]}$.

In general, let $a_1,\dots,a_n\in\mathbb Z$. As before, we can define

$G=\{a_1x_1+\cdots+a_nx_n:x_i\in\mathbb Z\ \forall i\}$

$L=\{m\in\mathbb Z:m$ is divisible by $a_i\ \forall i\}$

$M_i=\{a_ix:x\in\mathbb Z\}$ for $i=1,\dots,n$.

Then $G=M_1+\cdots+M_n$ and $L=M_1\cap\cdots\cap M_n$. As before,

$G/L\cong\displaystyle\prod_{i=1}^n(G/M_i)$,

i.e.,

$(*)\qquad\qquad\qquad (\mathbb Z/L)/(\mathbb Z/G)\cong\displaystyle\prod_{i=1}^n((\mathbb Z/M_i)/(\mathbb Z/G))$.

So $\displaystyle [a_1,\dots,a_n]/(a_1,\dots,a_n)=a_1\cdots a_n/(a_1,\dots,a_n)^n$, i.e.

$\boxed{a_1\cdots a_n=(a_1,\dots,a_n)^{n-1}[a_1,\dots,a_n]}$.

If we replace $\mathbb Z$ by $\mathcal O_K$ for any number field $K$, then $(*)$ takes the form

$(\mathcal O_K/L)/(\mathcal O_K/G)\cong\displaystyle\prod_{i=1}^n((\mathcal O_K/M_i)/(\mathcal O_K/G))$.

Taking cardinalities gives the following equation in terms of ideal norms

$N(L)/N(G)=\displaystyle\prod_{i=1}^n(N(M_i)/N(G))$.

Thus

$\boxed{N_{K/\mathbb Q}(a_1\cdots a_n)=N(G)^{n-1}N(L)}$.