# “Classical” consequences of Bezout's theorem in dimensions $>2$

By Classical I mean something that could have been found before 1900 (say).

A well known consequence of Bezout's theorem for plane curves is Pascal's theorem http://en.wikipedia.org/wiki/Pascal's_theorem .

I am curious if there are some other statements that you find pretty that can be formulated (almost) as elementarily as Pascal's theorem and proven using higher dimensional Bezout's theorem? For example, is there some statement that involves quadrics, planes and lines (cubics?...)?

Motivation. I ask this question since I want to finish to teach my (introductory) course in algebraic geometry by higher-dimensional Bezout theorem (using Hilbert polynomials, ect), and I would be extremely happy to give some pretty application :). To give you an idea of the level of the course, it is based on some bits of Harris book "Algebraic geometry first course",

Disclaimer. I don't doubt the usefulness of Bezout theorem and am sorry if the original question sounded like I doubt it. On the contrary I based the elementary course in algebraic geometry that I teach on this theorem. Namely, the course starts with Bezout for plane curves (using resultants), intorduces projective spaces and varieties, goes through Hilbert basis theorem and Hylbert polynomials (last section of Atiyah-Macdonald) and then as an applications we get a proof of a simplest version of Bezout's theorem in high dimension.

Also, It would be difficult for me to explain what I mean by pretty in math (for myself) but still I feel that the using of this word is justified, because we, mathematicians use this word... Sometimes we disagree on what is pretty, but personally I find pretty huge amount of facts in algebraic geometry. In other words I will be happy to see any application that can be stated in the language on the level of my course.

In the comment I put the link to the question on stackexchange

• I don't like these kinds of questions. Can you be more specific about what you are looking for? For instance, do you doubt that Bezout's theorem is relevant in dimensions > 2? Are you looking for reassurance? Are you looking for applications to a specific type of problem or to a specific (allied) field? What is "pretty" to you? – Jason Starr Feb 11 '13 at 0:54
• More specifically, if one has "strong" Bezout: If E is an irreducible curve of degree d spanning Pn, and p is a singular point of E, then a general hyperplane H passing through p, meets E away from p in at most d-2 points. Cor: If D is any irreducible plane curve of degree ≥ 3, then D is isomorphic to a curve in Pn of degree d < 2n, and any such curve is birational to a non singular curve. – roy smith Feb 11 '13 at 6:46
• In "Basic algebraic geometry" by Shafarevich, the author deduces from Bezout the following fact: the dimension of a (finite-dimensional) division algebra over $\mathbb R$ must be a power of 2. – Serge Lvovski Feb 11 '13 at 10:37
• Привет Wadim, you see, I teach an algebraic geometry course and learn at the same time some slightly more advanced material. Sometimes it happen to me even to work on questions related to algebraic geometry. In any case I mostly use this account on mathoverflow to understand something in algebraic geometry. So I think that agleaner sounds more up to the point than agteacher :), also this is more modest :) – aglearner Feb 11 '13 at 13:00
• I think that aglearner should be commended for gracefully handling some off-putting comments. – Chris Brav Feb 11 '13 at 14:29

Here is an excerpt from my class notes, inspired by the discussion in Joe Harris' book on desingularizing curves as an application of Bezout in space.

The degree of a curve is the number of intersections with a general hyperplane. The projection of a spanning curve still spans. The degree of a rational map ƒ:D--->E of curves, is the common number of preimages of most points of the image curve E, and in characteristic zero, a map of geometric degree one is birational.

Strong Bezout Theorem in space: If E is an irreducible curve of degree d spanning P^n, and p is a singular point of E, then a general hyperplane H of Pn passing through p, meets E away from p in at most d-2 points.

Corollary: If an irreducible curve E of degree d spanning P^n is projected into P^(n-1) from a singular point of E, then the product of the degree of the projected curve by the degree of the projection map on E, is ≤ d-2.

Corollary: An irreducible curve in P^n of degree < n cannot span P^n. Proof: Otherwise can project down eventually to a line spanning P^m, with m ≥ 2, contradiction.

Corollary: The projection map from a point of itself, of an irreducible curve of degree < 2n from P^n to P^(n-1), is birational.

Theorem. If D is any irreducible plane curve of degree ≥ 3, then D is isomorphic to a curve in P^n of degree d < 2n, and any such curve is birational to a non singular curve. Proof: The Veronese map :D-->P^N by homogeneous polynomials of degree d-2, where N = [d(d-1)/2] - 1, is an embedding of D as a curve spanning P^N and having degree d(d-2). But 2N = > d(d-2), precisely when d > 2, which holds by hypothesis. If the re - embedded curve D has a singular point, we project from it and lower the degree by at least 2, and the dimension of the ambient space by exactly one. We continue as long as the curve has singularities. Since the projected curve always spans, the degree can never drop below the dimension of the space, so the projection map always has degree one on the curve, i.e. the projected curve is always birational to the original curve. As long as there are singularities, the degree goes down at least twice as fast as the dimension, so the projected curve always has degree less than twice the dimension of the ambient space. Eventually then we either get to a point where the projected curve has no more singularities, or we get to a singular curve of degree 3 spanning P^2, and one more projection is birational to P^1. QED.

• Dear Roy, thank you for the answer, it is very nice! – aglearner Feb 12 '13 at 10:07
• Similar arguments with only weak Bezout, as you wanted, show that an irreducible curve of degree m in P^m is rational, and then again using strong version, you get Theorem: An irreducible curve D in P^2 of degree d ≥ 3 with (d-1)(d-2)/2 singular points is birational to an irreducible curve of degree d-2 spanning P^(d-2), hence is rational. (map by polynomials of degree d-2 through the singular points.) E.g. a nice special case is to map a plane quartic with three singular points say at [1,0,0], [0,1,0], [0,0,1], to a plane conic via conics through the singularities e.g. by [yz,xz,xy]. – roy smith Feb 12 '13 at 14:42

Two useful consequences of Bézout's Theorem in $\mathbb{P}^n$ are the following

Proposition 1. Let $X \subset \mathbb{P}^n$ be a hypersurface of degree $d+1$ with only isolated singularities. Then every irreducible curve $C \subset \mathbb{P}^n$ of degree $\mu$ contains at most $\mu d$ singular points of $X$.

Proof. Let $F=0$ be the equation of $X$ and assume that there is an irreducible curve of degree $\mu$ containing $\mu d +1$ singular points of $X$. Each partial derivative of $F$ is a polynomial of degree $d$, hence by Bézout theorem either its zero locus contains at most $\mu d$ points of $C$ or it contains the whole of $C$.

On the other hand, all singular points of $X$ are contained in the zero locus of each partial derivative, so by our assumption each partial derivative contains the whole of $C$.

This means that all the points of $C$ are singular points of $X$, contradicting the fact that $X$ has only isolated singularities.

Proposition 2. Let $C \subset \mathbb{P}^{g-1}$ be the canonical model of a smooth, irreducible, non hyperelliptic curve $C$ of genus $g$. If $C$ is trigonal, then its homogeneous ideal $I(C)$ is not generated by quadrics.

Proof. By the geometric version of Riemann Roch theorem, a $g^1_3$ on $C$ gives a trisecant line $L$ of the canonical model. By Bézout Theorem, such a $L$ must be contained in any quadric $Q$ in the homogeneous ideal $I(C)$ (since it has at least three intersection point with $Q$). Thus, if $I(C)$ were generated by quadrics, $L$ would be a component of $C$, contradiction.

• Dear Francesco, thank you, this is cute! – aglearner Feb 12 '13 at 10:08
• Francesco, thanks for adding the second statement! – aglearner Feb 17 '13 at 13:31
• Dear aglearner, you are welcome – Francesco Polizzi Feb 17 '13 at 16:49

A small remark: Bezout's theorem is not just about plane curves but includes the higher-dimensional version. See: General Theory of Algebraic Equations. The French original is freely available here.

• Abdelmalek, thank you for this book and for its French version! In fact I never really saw this book (not so surprisingly) and now I am curious if the answer to my question is not in this book... In fact I started to read the introduction (in French) ans it is very interesting, at least from human point of view :) – aglearner Feb 11 '13 at 15:18

I just found one more classical consequence which is quite nice. It is described in the book of Harris "First course in algebraic geometry".

Namely, one can prove that all projective automorphisms of $\mathbb CP^n$ are projective transformations. In order to prove this one shows that any hyperplane in $\mathbb CP^n$ is sent by any automorphism to a hyperplane (otherwise it would intersect the image of the line in more than one point). From this the statement follows rather directly.