Fulton algebraic curves: definition of intersection number On page 54 of Fulton's curve book the intersection number of two projective plane curves is defined. If $P\in P^2_k$, and $F$ and $G$ are two projective plane curves, then $I(P,F\cap G)$ is defined as $dim_k (O_P(P^2_k)/(F_\star,G_\star))$. In my understanding, $O_P(P^2)$ is the set (actually a local ring) of fractions $H/K$ where $H$ and $K$ are homogeneous polynomials in $k[x,y,z]$, of the same degree, and $K(P)\neq 0$. I also know that $F_\star$ is $F(x,y,1)\in k[x,y]$. Now:
Question 1: Why/in what sense can I consider $F_\star$ as element of $O_P(P^2)$? $F_\star$ is not necessarily a fraction of homogeneous polynomials of the same degree. Help me clarify the above definition.
Question 2: On page 57, in the proof of Bezout's theorem, why is true that $\sum_P I(P,F\cap G)=\sum_PI(P,F_\star\cap G_\star)$? Assuming $P=[a:b:1]\in P^2 $, in the left term of the above equation, is intendend $P=(a,b)\in A^2$ in the right term? I have understood the definition of intersection number for affine plane curves, but not for projective plane curves.
 A: In general, for affine variety $V$ and its projective completion $V^*$ and point $P\in \mathbb P^2$ such that $P\in U_3=\{[x:y:z]\in\mathbb P^2\mid z\neq 0\}$, if we write $P=[x:y:1]$ and define $P_3:=(x,y)\in\mathbb A^2$, we have the isomorphism:
$\mathcal O_P(V^*)\approx\mathcal O_{P_3}(V)$ 
by the association:
$\frac {H+I(V^*)}{G+I(V^*)}\mapsto \frac{H_*+I(V)}{G_*+I(V)}$
So we have isomorphism  $\mathcal O_P(\mathbb P^2)\approx\mathcal O_{P_3}(\mathbb A^2)$ 
Therefore we can consider projective curves $F$ and $G$ as elements in $\mathcal O_P(\mathbb P^2)$. 
So to express the statement in precise terms, define $a:=deg(F), b:=deg(G)$. 
Then we have $I(P,F\cap G):=dim_k(\mathcal O_P(\mathbb P^2)/(F/(X_3)^a),G/(X_3)^b))$ which coincides with $dim_k(\mathcal O_P(\mathbb A^2)/(F_*,G_*))$ by the ring isomorphism aforementioned.
An immediate question hence arises: If we dehomogenize with respect to $U_1$ or $U_2$ does this yield a different result? No, for suppose $P$ is simultaneously in $U_3$ and $U_2$. We have the equality:
$(F/(X_3)^a),G/(X_3)^b)=(F/(X_2)^a),G/(X_2)^b)$
as ideals in $\mathcal O_P(\mathbb P^2)$. 
