I would appreciate help to show this equality is valid: $\operatorname{lcm} (u, v) = \gcd (u^{- 1}, v^{- 1})^{- 1}$, where $u, v$ are elements of a field of fractions.
In the text it is stated that lcm is: there is an element $m$ in K for which $u| x$ and $v| x$ is equivalent to $m| x$
It goes on to say sending $t$ to $t^{- 1}$ in K reverses divisibility. So the proof of the existence of lcm's reduces to the proof of the existence of gcd's.
Then from the relation I am asking about above, one obtains $\operatorname{lcm} (u,v)\gcd (u,v) = uv$
I know independently how to show $\operatorname{lcm} (u,v)\gcd (u,v) = uv$. But the text I am looking at uses the first to show the second.
Thanks.
Best Answer
Below is a proof that works in any GCD domain, using the universal definitions of GCD, LCM. These ideas go back to Euclid, who defined the greatest common measure of line segments. Nowadays this can be viewed more generally in terms of fractional ideals or Krull's $v$-ideals.
Theorem $\rm\ \ \ \left(\dfrac{a}b,\,\dfrac{A}B\right)\: =\: \dfrac{(a,A)}{[b,B]}\ \ $ if $\rm\ \ (a,b) = 1 = (A,B),\ \ $ where $\rm\ \ \begin{eqnarray} (a,b) &:=&\rm\ gcd\rm(a,b)\\\ \rm [a,b]\, &:=&\rm\ lcm(a,b)\end{eqnarray}$
Proof
$\rm\begin{eqnarray} &\rm\quad c &|&\rm a/b,\,A/B \\ \quad\iff&\rm Bbc &|&\rm\ aB,\,Ab \\ \iff&\rm Bbc &|&\rm (aB,\,\color{#C00}A\color{#0A0}b) \\ \iff&\rm Bbc &|&\rm (aB,\, (\color{#C00}A,aB)\,(\color{#0A0}b,aB))\ \ &\rm by\quad (x,\color{#C00}y\color{#0A0}z) = (x,\,(\color{#C00}y,x)\,(\color{#0A0}z,x)),\ \ see\ [1] \\ \iff&\rm Bbc &|&\rm (aB,\, (A,a)\, (b,B))\ \ &\rm by\quad (a,b) = 1 = (A,B) \\ \iff&\rm Bbc &|&\rm (a,A)\,(b,B)\ \ &\rm by\quad (A,a)\ |\ a,\ (b,B)\ |\ B \\ \iff&\rm\quad c &|&\rm (a,A)/[b,B]\ \ &\rm by\quad (b,B)\:[b,B] = bB, \ \ see\ [2] \end{eqnarray}$
Here are said links to proofs of the gcd laws used: law [1] and law [2].