I'm studying polynomials and wondering if it is true that the roots of a multivariate polynomial depend continuously on its coefficients?

I have seen this kind of result for "univariate" polynomials with "complex" coefficients but not aware of any generalization to the multivariate and real cases.

Thanks for any help,

  • 3
    $\begingroup$ What exactly do you mean by "depend continuously?" The set of roots of a multivariate polynomial is usually uncountable, so it's not so clear how to identify a root in a perturbation with a root in the original. $\endgroup$ – Matt Samuel Dec 27 '16 at 22:28
  • $\begingroup$ what i meant is that the roots can be shown as smooth functions of the coefficients. in particular, if the coefficients are slightly perturbed then the roots should vary slightly. $\endgroup$ – jayki Dec 28 '16 at 0:09
  • $\begingroup$ What are the domain and codomain of the function? $\endgroup$ – Matt Samuel Dec 28 '16 at 0:17
  • $\begingroup$ for univariate case, the polynomial $P:\mathbb{C}\to\mathbb{C}$ and $\endgroup$ – jayki Dec 28 '16 at 0:28
  • $\begingroup$ for multivariate case, the polynomial $P(x_1,\ldots,x_n)$ goes from $\mathbb{C}^n$ to $\mathbb{C}$ with complex coefficients. In case of real polynomials, both coefficients and roots are real numbers. $\endgroup$ – jayki Dec 28 '16 at 0:34

You can't really make much sense of this over $\mathbb{R}$ even in the univariate case because there may not even be roots. For instance, consider the polynomial $$x^2 + C$$ For $C \leq 0$ there are roots; for $C > 0$ there are not. What would it mean for the roots to "vary continuously" here?

Also, multivariate polynomials don't have a discrete set of "roots" the way univariate polynomials do. They cut out an algebraic hypersurface. For instance, consider the zeroes of the polynomial $x^3 - y^3 + x^2 + C$ for various values of C:

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(images courtesy Wolfram|Alpha)

What does it mean for these to "vary continuously"?

  • 4
    $\begingroup$ The question is not completely nonsensical, and the answer is probably yes, at least over the complex numbers. The question is: what is the right question? $\endgroup$ – Matt Samuel Dec 27 '16 at 22:57
  • $\begingroup$ @Daniel McLaury You are perfectly right. Sorry for the inapropriate comment that I erase for future readers. $\endgroup$ – Jean Marie Dec 27 '16 at 23:04

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