Extending valuation to function field

Question

I'm reading Serre's Algebraic groups and I was not able to understand a definition he gave there. Let $(X, \mathscr{O}_X)$ be a smooth projective curve. One defines $$ \mathscr{O}_{X,p} = \varinjlim_{p\in U \subseteq X} \mathscr{O}_X(U) $$ and $$ k(X) = \varinjlim_{U \subseteq X} \mathscr{O}_X(U) $$ which induces injections $$ \mathscr{O}_X(X) \to \mathscr{O}_{X,p} \to k(X). $$ Since $X$ is smooth $\mathscr{O}_{X,p}$ is a DVR. Let $\nu_p$ be its associated valuation. How is this valuation extended to $k(X)$? Serre says that if $f = ut^n$ where $u$ is an invertible element in $\mathscr{O}_{X,p}$ and $t$ a uniformizing parameter of $\mathscr{O}_{X,p}$ then $\nu_p(f) = n$. However, this is only defining $\nu_p$ for elements in $\mathscr{O}_{X,p}$ not $k(X)$. Am I missing something?

Answer 1

$\mathcal O_{X, p}$ is an integral domain, and $k(X)$ is its field of fractions.

Therefore, seeing that all elements in $\mathcal O_{X, p}$ can be written in the form $u t^n$ with $u$ a unit and $n \geq 0$, it must be the case that all elements in $k(X)$ can be written in the form $u t^n$ with $n$ positive or negative.

So for a given $f \in k(X)$, $v_p(f)$ can take any value in $\mathbb Z \cup \{ \infty \}$. Such an $f$ is contained in $\mathcal O_{X, p}$ if and only if $v_p(f) \geq 0$.