I am studying divisors from Shafarevich. The book is available as a free download at http://userpage.fu-berlin.de/aconstant/Alg2/Bib/Shafarevich.pdf. My issue comes at page 164 of the pdf.
Assume we have a smooth variety $X$ (the book assumes only that $X$ is nonsingular in codimension $1$, but I only care about the smooth case right now). Let $C$ be an irreducible closed subvariety of codimension $1$. A theorem from the books tells us that there exists an open set $U$ in $X$ such that in $k[U]$, we have that the ideal of $U\cap C$ is $(\pi)$ for some $\pi\in k[U]$. Now let $f\in k[U]$ nonzero. We want to prove that there exists some $n\in\mathbb Z_{\ge0}$ such that $f\in(\pi^n)$, but $f\notin(\pi^{n+1})$. Clearly $f\in(\pi^0)$, so if such an $n$ doesn't exist, then for all $n\in\mathbb Z_{\ge0}$, $f\in(\pi^n)$. To show that this can't happen we need to show that $f=0$ in $k[U]$.
This is where I get stuck. I do not understand the argument from here. If we restrict to $\mathcal O_C$, then we get that $f=0$ on $C$, but we already knew this since $f\in(\pi)$ and $\pi$ vanishes on $C$. I also don't understand how to use theorem 2.8 which is about having a unique Taylor series (since if we restrict to some $\mathcal O_x$, with $x\notin C$, then we don't know what the equations are there, do we?)
Any help would be much appreciated.
(Depending on what you mean by variety, shrink $U$ to make it irreducible.)
The localization $k[U]_{(\pi)}$ of the domain $k[U]$ at the prime ideal $(\pi)$ is a discrete valuation ring (by the nonsingular in codimension $1$ hypothesis: being a discrete valuation ring is the same as being a Noetherian regular local domain of dimension $1$).
Every ideal in the discrete valuation ring $k[U]_{(\pi)}$ is of the form $(\pi^n)$ for some positive integer $n$. In particular, $(f) = (\pi^n)$ where $n = \nu_C(f)$.
Alternately, note that if $A$ is a Noetherian domain and $\mathfrak{a} \neq (1)$ is an ideal in $A$, we have $\cap_{n \geq 0} \mathfrak{a}^n = 0$ by the Krull intersection theorem.