fulton "algebraic curves" intersection numbers and bezout' s theorem proofs I've used fulton's "algebraic curves" (here is the link http://xahlee.info/math/i/algebraic_geometry_by_william_fulton_73219.pdf) to prepare my exam and i had just 2 issues with proofs, 2 details i didn't understand:
1)in the lemma a) at page 39 (when proving the property 5) of intesection numbers) it proves the existence of an integer N for which $I^{2N} ⊂ (F,G)O_P(A_2)$ but why does this imply that $A_{ij}$ belongs to $(F,G)O_P(A_2)$? I cannot figure out why  BF' belongs to $(F,G)O_P(A_2)$.enter image description here 
2) in the third step of the proof of the bezout theorem (page 58), when proving the independence of the $a_i$, I cannot figure out how it gets  $$z^r Σλ_i A_i = z^s B^*F+z^t C^*G$$ Using  the proposition 5 in chapter 2 as suggested by the book i should have every $A_i$  multiplied by different powers of z as i do not know whether the $A_i$  are all divided by the same powers of z or not. Are all the $A_i$ actually divided by the same powers of z and why? Or is there another reason why the $A_i$ are all multiplied by the same factor $z^r$ ?enter image description here
 A: Hello. I a moment ago just worked through the proof on page 39, and let's fill in the gap.
If we read the line above carefully, you will see the expression 

$ A_{ij} = BF - BF^{'} $, where each term of $BF^{'}$ has degree $ \geq i+j
 +1$

Look closely: $BF \in (F,G)\mathcal O$ doubtlessly, so in order to express $A_{ij}$, all we need to do now is to express $BF^{'}$, whose degree is one higher than that of $A_{ij}$.
What Fulton did here, is to transform the problem of expressing elements of degree $i+j$, where $i + j \geq m+n-1$, into the degree $i+j+1$ counterpart of the same problem. That is, all the work we have done so far accumulates to the fact that, if we can express everything of degree $d+1$, where $d \geq m+n-1$, we can express everything of degree $d$. 
So if the whole of all polynomials of a certain VERY HIGH DEGREE can be expressed, any polynomial below that degree and above $m+n-1$ can be expressed, by inducting downwards. If there is some (possibly very big) $t$ such that $I^t \subset (F,G)\mathcal O$ , we can induct from $t$ downwards to patch up the gap between $t$ and $m+n-1$, and anything above $t$ is automatically in the ideal anyway.
Now we can continue reading the next paragraph.
