A bivariate polynomial of degree $m+n$ is,

$ p(x,y) = \sum_{k=1}^n\sum_{j=1}^m a_{jk}x^ky^j$

where $a_{mn}\neq0$ and $a_{jk}\in\mathbb{R}$ for $1\leq j\leq m$, $1\leq k\leq n$.

I would like to understand how the roots of a bivariate polynomial behave. It is clear that the roots cannot form patches (unless $p$ is the zero polynomial) but I'm sure about the following:

  1. If some of the roots form a curve, what properties does this curve have? Can it bifurcate? Can the curve be parameterised to a univariate polynomial? Can the curve have end points (which are not $\pm\infty$)?

  2. What is the maximum number of isolated roots of $p(x,y)$? What is the maximum number of zero curves?

  3. How badly ill-conditioned is bivariate polynomial root finding? For example the polynomial $p(x,y) = x^2+2x+1$ has one zero curve but $p(x,y) = x^2 + (2-\epsilon)x+1$ has two zero curves. Is there a polynomial $p(x,y)$ with a zero curve but with a small perturbation of the coefficients has only isolated zeros?

  4. I guess the fundamental theorem of algebra does not hold. The polynomial $p(x,y) = x^n-y$ seems to be a counterexample. Is there a multivariate function theorem?

This post is a barrage of questions, but I feel they are all intrinsically related and a person who can answer one of them is likely to be able to give answers to them all.

Thank you.


I don't think most of your questions are appropriate for MO, you should try math.stackexchange.com if you have any follow-up questions, as it sounds as if your knowledge and questions are at advanced undergraduate/beginning graduate level, but I'll try to answer some of them. A good reference is Fulton "Algebraic Curves" or perhaps some older books, like Walker's which go deeper into real points.

First a degree $d=m+n$ polynomial is usually $\sum_{j+k \le d} a_{jk}x^jy^k$, which is a little more general than what you wrote.

  1. The zero set is a union of finitely many (real smooth) curves and points and has a finite number of singularities (nodes, cusps and higher) which look like multiple crosses and pointy bits (you have to look at a picture). It cannot end abruptly and, if by bifurcation you mean one curve separate into two, then I don't think you can have that. The curve can be parametrized by univariate polynomials only in very special cases (irreducible, genus zero and only one point at infinity).

  2. At most $d^2$ isolated points (can improve bound slightly) because they are singular. There is also at most some quadratic function of $d$ number of "zero curves", I forget the exact formula. This is known as Harnack's theorem.

  3. I don't know, ask a numerical analyst. Edit: Re last part of question. Having a curve of solutions is an open condition on the coefficients. An useful example is $x^2+y^2=a$, and look at $a$ small, positive, negative or zero.

  4. Question doesn't make sense.

  • $\begingroup$ Question 4. I guess the OP asks for the algebraic closure of the field R(y) of rational functions (ie, quotients of polynomials)? That would be the only way to make some sense of it if you want to find x (a function of y) in some field as a root of a polynomial with coefficients in R[y]. If that is the point, the subjects to learn are Riemann Surfaces and Puiseux series. (Real Puiseux series parameterize these curves, and they are to Puiseux series like real numbers to complex numbers). $\endgroup$ – quim Oct 6 '11 at 11:47
  • $\begingroup$ Question 4. He should work over C rather than R. There is always a complex zero, whether the polynomial has one variable or several variables. $\endgroup$ – KConrad Oct 6 '11 at 13:30
  • $\begingroup$ I don't think he wants a nullstellensatz, but he thinks of functions as roots (otherwise his counterexample doesn't make sense). Of couse the algebraically closed field of Puiseux series contains C (and R). $\endgroup$ – quim Oct 6 '11 at 14:04
  • 3
    $\begingroup$ Trying to figure out what the question was supposed to mean is pointless. If the OP wants an answer, he should come here and explain his question. The burden of effort should be his, not ours. $\endgroup$ – Felipe Voloch Oct 6 '11 at 15:39

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