# Relating $\operatorname{lcm}$ and $\gcd$

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.

3
2022-07-25 20:41:44
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