TITLE: Exercise 3.18 of Fulton's Algebraic curves.
QUESTION [7 upvotes]: I'm trying to provide a proof of the following fact: If $p$ is a simple point on the curve $F$ then $I(p,F\cap G)=ord_p^F(G)$. Where $I(p,F\cap G)$ denotes the intersection number of the curves at $p$, and $ord_p^F$ is the order function in the discrete valuation ring $\mathcal O_p(F)$. The goal is use only the seven properties of the intersection numbers that him describes in the book. 
I have tried with the special case when $p$ is a simple point on $G$ and $F$, $G$ don't share any tangent at $p$; in which case the intersection number is one. But I don't know how to calculate $ord_p^F(G)$ in this simple case,  I even reduce the problem to $G$ irreducible but with no success. In fact I'm having problems with calculations of this kind, when the function $ord$ is involved; I would be very grateful with an answer that helps me in this aspect. Thank you.

REPLY [3 votes]: Fulton's axioms for the intersection number are this.
By $(3)$ we may assume $P=(0,0)$, and the tangent of $F$ at $P$ is $Y$. Then we can consider $\overline X$ as a uniformizing parameter of $\mathcal O_P(F)$. Let $n=Ord^{F}_P(G)$, then in $\mathcal O_P(F)$ there exists some unit $u$ such that $\overline G=u\overline X^n$, say $u=\overline A/\overline B$, for some $A,B\in k[X,Y]$, then $A(P)\neq 0\neq B(P)$ as $u$ is a unit. Then $GB=X^nA+CF$ for some $C\in k[X,Y]$.
We have $I(P,F\cap BG)=I(P,F\cap B)+I(P,F\cap G)$ by $(6)$, but $P\notin B$, thus by $(2)$, $I(P,F\cap B)=0$, and thus $I(P,F\cap BG)=I(P,F\cap G)$.
But $$I(P, F\cap GB)=I(P,F\cap (X^nA+CF))=I(P,F\cap X^nA),$$
by $(7)$, however as $A(P)\neq 0$, we prove $I(P,F\cap X^nA)=I(P,F\cap X^n)$ the same way we showed $I(P,F\cap BG)=I(P,F\cap G)$ above, so that $I(P,F\cap G)=I(P,F\cap X^n)$, but by $(6)$ $I(P,F\cap X^n)=nI(P,F\cap X)$, and by $(5)$, $I(P,F\cap X)=1$, thus $I(P,F\cap G)=n$, i.e., $I(P,F\cap G)=Ord_P^F(G)$.