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.



Higher dimensional Bezout via Hilbert polynomials: a reference
This is to answer the last question on why the $F\_i$ form a regular sequence.


In my first attempt I was trying to do this without using the unmixedness of an ideal $I$ generated by $\mathrm{height } I$ number of elements, but it seems that I cannot. (Hat tip to Hailong). 


So here is the statement that one needs:


**Thm** Let $I\subseteq A$ be an ideal generated by $\mathrm{height } I$ number of elements. If $A$ is Cohen-Macaulay, then $I$ is unmixed, that is, it does not have any embedded primes. In other words, every associated prime of $I$ is a minimal prime.


I will certainly not include a proof, because there is no point repeating one of the proofs in the literature. A relatively simple one can be found in Bruns-Herzog's *Cohen-Macaulay rings* (Cambride University Press). I think if you are willing to discuss regular sequences, then this should be OK and you may have already discussed the Cohen-Macaulay property, or even this theorem, although I kind of doubt it. 


However, you don't actually need to discuss CM for this. In fact, this theorem is where the name "Cohen-Macaulay ring" comes from, because it is true that if every such ideal is unmixed, then the ring is Cohen-Macaulay and more importantly, the start of the CM property is a theorem of Macaulay from 1916 that states that the above theorem holds if $A$ is a polynomial ring. Then the next move towards creating this famous name was a theorem of Cohen from 1946 that states that the above theorem holds if $A$ is a regular local ring. Clearly you only need Macaulay's theorem and there must be a simple proof of that somewhere. Perhaps you can adopt the proof from Bruns-Herzog for that special case and make it shorter. There is also a reference to Macaulay's original paper in that book, but I doubt that that's a really good solution to go back to, but you can certainly try.


So, assuming this theorem here is how to prove that 


*Proposition* If $f\_1,\dots, f\_n$ are polynomials in $n$ variables such that $Z(f\_1,\dots,f\_n)$ is finite, then $f\_1,\dots, f\_n$ is a regular sequence. 


(this is not exactly how you stated it, but you can get to this situation easily by choosing a hyperplane that misses all the intersection points and restrict to the complement)


*Claim 1* 
If $Z(f\_1,\dots,f\_n)$ is finite, then any irreducible component of $Z(f\_1,\dots, f\_r)$ is of dimension $n-r$ for any $1\leq r\leq n$.


*Proof* Let $W$ be an irreducible component of $Z(f\_1,\dots, f\_r)$. Then by [Krull's principal ideal theorem](http://en.wikipedia.org/wiki/Krull%27s_principal_ideal_theorem) $\dim W\geq n-r$ and 
$\dim (W\cap Z(f\_{r+1},\dots, f\_n))\geq \dim W -(n-r)$. Since by assumption $\dim (W\cap Z(f\_{r+1},\dots, f\_n))=0$, it follows that $\dim W\leq n-r$ and hence has to be equal to it. $\square $


*Corollary* The ideal generated by $(f\_1,\dots, f\_r)$ has $\mathrm{height}=r$ and hence by the **Thm** it is unmixed.


*Claim 2*: Let $M$ be a finitely generated module over the ring $A$ and $x\in A$. Assume that all associated primes of $M$ have the same height and that $\dim M/xM<\dim M$. Then $x$ is *not* a zero-divisor on $M$.


*Proof*: The assumption implies that $x$ cannot be contained in any associated prime of $M$ and hence cannot be a zero divisor. $\square $


Claim 1, Corollary, and Claim 2 combined imply that $f\_1,\dots,f\_n$ is a regular sequence.


