TITLE: Higher dimensional Bezout via Hilbert polynomials: a reference
QUESTION [7 upvotes]: For the purposes of teaching my elementary course in algebraic geometry I am looking for a reference (or notes) that contains a complete proof of a higher-dimensional weak Bezout theorem. I only want to learn about a proof that is based on the approach when dimensions and degrees of projective varieties are introduced via Hilbert polynomials.
Weak Bezout. Let $F_1,...,F_n$ be homogeneous polynomials of degrees $d_1,...,d_n$
such that hypersurfaces $F_1=0$,..., $F_n=0$ have only finite number of intersections in
$\mathbb P^n_K$ (with $K$ algebraically closed). Then the number of intersections is 
at most $d_1\cdot...\cdot d_n$.
Justification of the question. Of course there is a large amount of proofs of this statement in many books on algebraic geometry
(Shafarevich, Harris, Hasset, ect.) but these proofs usually come after 200 pages of text, while 
I want an honest proof that is contained in a complete set of notes of 40 pages (this will be the maximal 
length of my notes and I don't want to spend  more than 10 pages on higher dimensional Bezout). Also these proofs often develop dimension theory basing on transcendence
degree of the field of rational functions and I don't want to use this approach, (since it will require 
2-3 additional lectures which I don't have time to give). 
I know one place where the approach that I want to use (namely everything is based on Hilbert polynomials)
is taken, these are the notes of Manin (end of 60ties). Unfortunately it seems to me that there is 
a problem with the proof he proposes. Basically everything works if $F_1,...,F_n$
form a regular sequence, but to understand why in the condition of the theorem $F_1,...,F_n$
do form a regular sequence is left in the notes as something "not hard to do"...
In the book of Hasset, it seems to me there is a similar problem (i.e. it is not explained 
why $F_1,...,F_n$ form a regular sequence if the number of intersections of hypersurfaces is finite). I hope there is a nice complete proof somewhere... 
In other words, what will completely satisfy me is a short proof of the following statement:
If the number of intersections of $F_i=0$ is finite, then $F_i$ form a regular sequence.
Is there a short proof of this statement?
PS. I think that the answer of Sandor shows that probably there is no "magic easy" solution to the question (if one sticks to Hilbert polynomial approach instead of using higher-dimensional resultants), so I decided to accept it.

REPLY [2 votes]: I'm writing out what is essentially Looijenga's proof for my own reference. Let $Z_r = \{ F_1 = F_2 = \cdots = F_r = 0 \}$. From basic dimension theory (eg Shavarevich Theorem I.6.2.5, much easier than Macaulay's result), all components of $Z_r$ have dimension $\geq n-r$. If any of them had dimension $>n-r$, then $Z_n$ would have a component of dimension $>0$, a contradiction. So all components of $Z_r$ have dimension $n-r$.
Let $A_r = k[x_0, x_1, \ldots, x_n]/\langle f_1, f_2, \ldots, f_r \rangle$. Look at the exact sequence of graded modules:
$$0 \to K_r \to A_r \overset{\cdot f_{r+1}}{\longrightarrow} A_{r}[d_r] \to A_{r+1}[d_r] \to 0$$
where $K_r$ is defined to be the kernel of the map it precedes. 
If we knew unmixedness, we would know that $K_r=0$. Although we don't know that, we do know that the support of $K_r$ does not contain any irreducible component of $Z_r$. That is because, if $W \subset Z_r$ were an irreducible component of $Z_r$ on which $f_{r+1}$ was $0$, then $W$ would be an $n-r$ dimensional component of $Z_{r+1}$, contradiction. 
So the Hilbert polynomial of $K_r$ has degree $\leq n-r-1$. (Of course, in reality, it is zero.) We now inductively prove that the Hilbert polynomial of $A_r$ has leading term $d_1 d_2 \cdots d_r \frac{x^{n-r}}{(n-r)!}$, as desired.