7
$\begingroup$

How would Bezout have proved Bezout's theorem bounding the number of points in the intersection of two plane (polynomial) curves in $\mathbb{R}^2$?

I have looked at a couple of modern algebraic treatments of Bezout's theorem. For instance, I am familiar with Fulton's Algebraic Curves and his treatment of Bezout's theorem therein. While a purely algebraic approach is nice and has its uses, I feel like that proof and related modern proofs must be very different from Bezout's. I expect that Bezout's proof could not have used much beyond rudimentary calculus; in particular, I doubt he would have used the projective plane, exact sequences or local rings. Given that he was an 18th century mathematician, it is also unclear to me if he could even have used the Fundamental Theorem of Algebra which makes me uncertain that he would have proved it using resultants e.g., some version of a proof outlined on the wikipedia page: https://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem .

Alternatively, my question is what is the most classically analytic or down-to-earth proof of Bezout's theorem? Regarding analytic approaches to Bezout's theorem, I think that Griffiths--Harris proof (around page 171 therein) is analytic, but not classical.

Bonus points: If such a proof is substantially different from that in Fulton's Algebraic Curves, then can you describe how the proofs relate (assuming that they do) or how the classical proof inspires the modern proof?

$\endgroup$
12
  • 2
    $\begingroup$ The proof appears in Bézout's own book Théorie générale des équations algébriques (1779). A 2002 English translation by Eric Feron is available -- the theorem is presented in paragraph 47, which is on page 24 of the translation. Have you read this? It seems like a natural place to start... $\endgroup$ Commented Aug 18, 2021 at 12:36
  • 2
    $\begingroup$ Also, the statement "I doubt he would have used [...] exact sequences or local rings" has my vote for the understatement of the century (or two centuries, to be exact) :-) $\endgroup$ Commented Aug 18, 2021 at 12:37
  • 4
    $\begingroup$ The Fundamental Theorem of Algebra can be considered as a 1-dimensional version of Bezout, and it is absolutely essential. Without it, you may only get an inequality. Resultants and elimination theory is a natural and classical approach. $\endgroup$ Commented Aug 18, 2021 at 12:40
  • 1
    $\begingroup$ drive.google.com/file/d/1cYPvDagNHsM39ngUjr--HftY44Zznmz5/view Bezout proof was not rigorous. A simple proof can be obtained using resultants. $\endgroup$ Commented Aug 18, 2021 at 16:27
  • 1
    $\begingroup$ @Carl-FredrikNybergBrodda Thank you for pointing that out. The corrected link to Bezout's book (in French) is: gallica.bnf.fr/ark:/12148/bpt6k106053p.image I am deleting my comment to get rid of the personal link and replacing my comment here: I don't know any French, but from what I can tell Bezout proves a result about resultants which implies that the "Bezout bound" is true generically. $\endgroup$
    – K Hughes
    Commented Aug 20, 2021 at 12:14

0

You must log in to answer this question.