Higher dimensional Bezout via Hilbert polynomials: a reference

Question

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.

Answer 1

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 $\dim W\geq n-r$ 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.

Answer 2

I finally found an exposition of Bezout's theorem that I find good :) . This is by Eduard Looijenga in his course "Algebraic geometry 1":

http://www.staff.science.uu.nl/~looij101/AG2013.pdf

There is no magic of course, the proof comes after 70 pages of text. And I guess that the proof is similar to the one given in the first chapter of Hartshorne. But there is a clear (and very strong) advantage (for me): I can understand everything written in the course of Looijenga. I would advise it to and learner as myself :)

Answer 3

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.