Talk:Bézout's theorem
 | This article is rated C-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects: | ||||||||||
| |||||||||||
Untitled[edit]
Who was Bezout? -- Zoe
- I think it was Etienne Bézout. -- Anon.
This article could also be made a whole lot clearer: at the moment, it's very imprecise, and confusing. -- Anon.
- What do you find imprecise and confusing? --Bernard Helmstetter 18:43, 20 Dec 2004 (UTC)
An hyperbola? Charles Matthews 12:43, 27 Oct 2004 (UTC)
- I think it is A hyperbola. --Bernard Helmstetter 18:43, 20 Dec 2004 (UTC)
why speak about mathematics in general before algebraic geometry ?[edit]
The text begins with a somewhat vague definition of the theorem in mathematics in general, and then only in algebraic geometry. I don't understand this distinction; to me, the theorem is really about algebraic geometry. --Bernard Helmstetter 18:31, 20 Dec 2004 (UTC)
Bezout's is a fundamental theorem in projective geometry, although any study of algebraic geometry requires it (since we do algebra on projective varieties mostly).
July 2013[edit]
Yo, one of the examples is wrong (the one w/ the longer ellipse). That has an intersection of multiplicity 2 at (-1,0), one of multiplicity 1 at (0,1), and one of multiplicity one at (17/8,-15/8*I). 66.71.125.103 (talk) 22:42, 10 July 2013 (UTC)
Non-intersecting Curves[edit]
What about the example with two circles centered at the origin with radius 1 and 2 respectively? I'm getting 0 intersection points instead of 4, how is this resolved? — Preceding unsigned comment added by 71.244.141.195 (talk) 20:56, 15 October 2016 (UTC)
- These two circles have two common points at infinity (the cyclic points), each of multiplicity two, giving 4, as asserted by the theorem.
- I agree that the first paragraph of the lead was confusing, and I have edit it for clarification. D.Lazard (talk) 08:42, 16 October 2016 (UTC)
Proof Sketch Confusion[edit]
The sketch begins by rewriting the 2 curves, X and Y, and somehow they're now inexplicably univariate polynomials in terms of z when they where in terms of x and y. If they're now in homogeneous coordinates then they should be in terms of x, y, and z. This, however, seems incompatible with the rest of the proof. Can someone add an example (or a link to the missing background material) for how this polynomial rewriting is supposed to happen such that x and y are no longer present, or, if such a rewriting does not exist, provide a real proof? Snydej2 (talk) 09:52, 22 April 2019 (UTC)
- I see where my confusion was (what looked like coefficients are actually polynomials in which the x's and y's are hiding. This was stated but I misunderstood it.) and I will add the example I was in need of. Once I understand the proof and have clarified any other stumbling blocks I run into I'll remove the confusion banner I added. Snydej2 (talk) 10:37, 22 April 2019 (UTC)
- I agree that to make it useful much more detail should be added, even if it's just a sketch. As of now I can't make sense of it 148.88.247.135 (talk) 09:48, 19 November 2019 (UTC)