How do I find the orders of this rational function?

Question

How can I find the orders of $z(x)=\frac{x}{1-x}$ over $k(\mathbb P^1)$ at the zero $x=0$ and the pole $x=1$?

I saw in another question posted on MSE that the orders are both equal to $1$, but I don't know why.

Thanks.

Answer 1

Consider $z(x)$ in the local ring of $\mathbb P^1$ at $x=0$. The local ring is $k[x]_{(x)}$, with maximal ideal $(x)$. This is a DVR so every function can be written as $x^e u$, where $u$ is invertible in the local ring. In your case $e=1$.

For the local ring at $1-x$, your function can be written as $(1-x)^{-1} x$, and here $x$ is invertible. So the order is $-1$.