Let $Y$ be a curve, and $P\in Y$ a smooth point. So $\mathcal{O}_{P,Y}$ is a regular local ring of dimension one, therefore is a discrete valuation rings.
My question is what is the normalized valuation $v_P: K(Y)\setminus\{0\}\longrightarrow \mathbb{Z},$ and I want some concrete examplse of calculation of $v_P.$
Thanks in advance.
Being $\mathcal O_P$ a dvr, it is in particular a PID, and therefore there exists an element $t\in \mathcal O_P$ which generates the maximal ideal. In fact every element $f\in K(Y)\setminus \{0\}$ can be written uniquely as $t^n\cdot u$ for some $n\in \mathbb Z$ and some $u\in K(Y)$ such that $u$ is regular at $P$ and $u(P)\neq 0$. This already defines a valuation $K(Y)\setminus \{0\}\to \mathbb Z$ sending $f\mapsto n$.
Let's see an example. Let $E$ be the elliptic curve over $\mathbb Q$ defined by $y^2=x^3+x$ and take $P=(0,0)$. I claim that $y$ is a generator for the maximal ideal of $\mathcal O_P$. In fact, you can see easily that such an ideal is generated by $x,y$ but since $x=\frac{y^2}{x^2+1}$, it means that $x$ can be written as a power of $y$ times a function, $\frac{1}{x^2+1}$, which is regular at $P$. In particular, $v_P(x)=2$. So in general, take any $f(x,y)/g(x,y)\in K(E)\setminus \{0\}$. Now take the highest power of $y$ which divides $f$, say $y^r$, and the highest one that divides $g$, say $y^s$. Then $v_P(f/g)=r-s$.