You can apply the same argument to $\mathbb{C}[u,v]/(uv-1)$, so the relation you get is $vdu+udv$, multiply this by $v$ gives you $dv = \ldots$, which kills $dv$
But probably better is to realize that $\mathbb{C}[u,u^{-1}]$ is what you get when you invert $u$ in $\mathbb{C}[u]$, and use something about Kahler differentials and base change (which in this case is just a long way of saying you have a trivial bundle over $\mathbb{A}^1$ and you're taking an open set)
Motivation:
The (principal) value of
$$m\pmod{n}$$
for some positive integers $m> n$, might well be viewed as the value
$$m-\sum_{i=1}^{M_{m,n}}n,\tag{$\Sigma$}$$
for some $M_{m,n}\in \Bbb N$ with $M_{m,n}n\le m$ but $(M_{m,n}+1)n> m$; indeed, we have just subtracted a suitable number $M_{m,...
