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A classical consequence of Bezout's theorem for plane curves is 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?...)?

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, the course is very close to some bits of Harris book "algebraic geometry first course", and covers some bits of it)

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The Fundamental Theorem of Algebra is generalized by Bezout. (Doesn't appear to be mentioned in the Wikipedia entry).

"The roots of $f(x)$ correspond to the points at which the zero set of the polynomial $y- f(x)$ and the zero set of the polynomial $y$ intersect." -- Stephanie Fitchett "Bezout's theorem: a taste of algebraic geometry"

She goes on to write: "Based on our experience with the fundamental theorem of algebra, we would like to replace the inequality with an equality, and in fact, this is exactly what Bezout’s Theorem claims..."

The curves $f(x)$ and $y=0$ are replaced by arbitrary curves in projective space.

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  • $\begingroup$ Thank you for this answer an for the link to the article, it looks nice. Though this is not exactly what I am looking for, namely I look for consequences of Bezout's theorem. $\endgroup$
    – agleaner
    Feb 11, 2013 at 15:35
  • $\begingroup$ What do you mean by consequence? If BT is a generalization of FTA, then FTA is a consequence of BT, even though FTA was clearly worked out earlier. $\endgroup$ Feb 11, 2013 at 16:19
  • $\begingroup$ You are right of course, FTA is a partial case of BT. But it is in dimension 1. I am interested in high dimension, i.e., >2 $\endgroup$
    – agleaner
    Feb 11, 2013 at 16:40
  • $\begingroup$ I see. I thought you might consider a polynomial of degree $n$ as a higher dimensional object (vector space of). $\endgroup$ Feb 11, 2013 at 16:54

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