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.

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    $\begingroup$ Hartshorne, Thm I.7.7 pg 53, applied $n$ times. $\endgroup$ Mar 3, 2013 at 20:08
  • $\begingroup$ Felipe, thanks. I am afraid that what you propose would not fit into my course... I have 4 more lectures to give and finished with Hilbert basis theorem last time. It will take me more than four lectures to go in details through section 1.7 of Hartshorne and this will kill my students (and myself I guess). So I am looking for some miraculous solution. If it does not exist I'll do just like Hasset and Manin (living this statement about regular sequence as an exercise :( ...) $\endgroup$
    – aglearner
    Mar 3, 2013 at 22:03
  • $\begingroup$ It's easy to prove this if you know that every component of a hyperplane section of an irreducible variety has codimension at most 1. Obviously this is easier to show from some foundations than others, but it's probably a more important fact than Bezout. $\endgroup$ Mar 3, 2013 at 23:26
  • $\begingroup$ Jack I agree with what you say and I agree that this it is probably more important. But this is also harder to state (comparing to Bezout's theorem). $\endgroup$
    – aglearner
    Mar 3, 2013 at 23:56
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    $\begingroup$ I came here last night planning to ask exactly this question! Thanks aglearner and Sándor. My conclusion after reading the answers is that I'll probably prove the result in $\mathbb{P}^2$, explain the connection to Cohen-Macaulayness in general but not give a full general proof. $\endgroup$ Oct 7, 2014 at 14:52

3 Answers 3


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.

  • $\begingroup$ Dear Sándor, Claim 1: why does any component have same dimension? And Claim 2: for $x$ to be a NZD you need it to be outside all associated primes, not just minimal. $\endgroup$ Mar 4, 2013 at 7:10
  • $\begingroup$ Dear Hailong, I am not sure I see your problem with Claim 1. Regarding Claim 2, you are right. I was trying to avoid using "unmixedness" but it was on my mind... I meant this in the situation of the question. I guess I'll have to revise this.... $\endgroup$ Mar 4, 2013 at 7:47
  • $\begingroup$ Dear Sándor, what I did not understand in Claim 1 was that while $Z(f_1,...,f_r)$ is of dimension $n-r$, it may have component of smaller dimensions. Perhaps I was missing something? $\endgroup$ Mar 4, 2013 at 7:50
  • $\begingroup$ Hailong, that can never happen. In general, it is possible that $Z(f_1,\dots,f_r)$ has an irreducible component that has dimension larger than $n-r$, but it can never be smaller than that. $\endgroup$ Mar 4, 2013 at 7:57
  • $\begingroup$ [Hartshorne, Ex.I.1.9, page 8]. The proof is as I said above by Krull's principal ideal theorem. $\endgroup$ Mar 4, 2013 at 7:59

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


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 :)


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.


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