0
$\begingroup$

Define an algebraic curve $W(t)$ as the curve that is the solution to $G(W, t)$ for some polynomial $G$. (W(t) need not be a proper function, in which case one can, one-at-a-time, consider each branch of the function, when using it in other equations).

Given 2 algebraic curves $W_1(t), W_2(t)$ of degree bounded by $r$. And a 2-variable Polynomial $P(x_1 , x_2)$ of degree bounded by $d$, I wish to reason about the number of solutions to

$$ P(W_1 ,W_2) =0$$

I note the following facts:

if $P(x_1)$ is a one variable polynomial bounded by degree $d$ and $W_1(t)$ is an algebraic curve of degree $r$ then $P(W_1(t))=0$ has at most $rd$ solutions. This follows from the fact that,

  1. if $W_1(t)$ has degree at most $r$ points where it is equal to 0 [or any other fixed constant] Proof: given that $W_1(t)$ has degree r then $G(W,t)$ [that gives rise to $W$] is a degree r polynomial, and solving for $G(0, t) = 0$ therefore is a degree at most $r$ equation in one variable, with at most $r$ solutions.

  2. $P(x) = 0$ has at most $d$ solutions for each solution $d_i$ $W_1(t)= d_i$ has at most $r$ solutions so therefore we get a total of $rd$ solutions to $P(W_1(t)) = 0$

Furthermore, if $W_1 = W_2$ then

$$ P(W_1, W_1) = 0$$ Must have $\le rd$ solutions.This follows from the fact that since $P$ is a polynomial we can consider each of its monomial terms $A_{u,v}x_1^ux_2^v$, and fixing $P(W_1, W_1)$ will set each of these to a term of the form $A_{u,v}W_1^{u+v}$. Thus $P(W_1, W_1)$ can be viewed as a single variable polynomial $H(W_1)$ whereas the degree of $H$ is at most $d$, then the proof above yields at most $rd$ solutions.

My "gut" feeling is that

$$ P(W_1 ,W_2) =0$$

has at most $rd$ solutions, but i'm not sure how make that jump.

(I'm not sure if this has an elementary solution or requires high powered machinery, hence i tagged both algebra-precalculus, and algebraic-geometry)

$\endgroup$
  • 1
    $\begingroup$ Everything you want to know has probably falls under Bezout's theorem and it's corollaries, but your definitions are pretty opaque and possibly nonstandard. If you want people to have a conversation with you about this, you're going to need to be more specific about what you mean. For instance: where does $W$ live? Is it a subset of $\mathbb{R}^n$, $\mathbb{C}^n$, etc? $\endgroup$ – KReiser May 28 '17 at 19:24
  • $\begingroup$ @KReiser I was treating it as a subset of $\mathbb{C}^n$ . To be more specific it would be a curve in $\mathbb{C}^2$ $\endgroup$ – frogeyedpeas May 28 '17 at 21:01
  • $\begingroup$ Also Please continue listing missing details, I've mostly invented this problem using my own terminology from a different area, so I'll really appreciate being taught the proper way to go about talking about these topics $\endgroup$ – frogeyedpeas May 28 '17 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.