# Polynomial System has only isolated solutions

How does one show that the polynomial system $F(x)=0,$ where $F:\mathbb{C}^n \rightarrow \mathbb{C}^n,$ has only isolated roots?

As an example, let $\mathcal{E}(\mathbf{x})\equiv(e_{1}(\mathbf{x}),e_{2}(\mathbf{x}),e_{3}(\mathbf{x}),e_{4}(\mathbf{x})),$ where $\mathbf{x}\equiv(x_{1},x_{2},x_{3},x_{4})$ and

$e_{1}(\mathbf{x}) = x_{1}+x_{3}+5(x_{1}x_{4}+x_{2}x_{3})$

$e_{2}(\mathbf{x}) = x_{1}x_{3};$

$e_{3}(\mathbf{x}) = x_{2}+x_{4}-6(x_{1}x_{4}+x_{2}x_{3});$

$e_{4}(\mathbf{x}) = x_{2}x_{4}$

An easy result to show is that the set $\mathcal{S}=\left\{ \mathbf{x}\in\mathbb{C}^{4}:\mathcal{E}^{\prime}(\mathbf{x})\mbox{ is nonsingular}\right\}$ has measure zero. With $\mathbf{w}\equiv(w_{1},w_{2},w_{3},w_{4})\in\mathcal{S},$ consider the polynomial system given by $$H(\mathbf{x})=\mathcal{E}(\mathbf{x})-\mathcal{E}(\mathbf{w}) = 0.$$ For a family of such systems, I wish to know whether there are finite roots. I would like to determine properties that ensure the roots are finite.

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Suresh, could you explain please what is meant by a "polynomial system", and whether by $C^n$ you mean $\mathbb{C}^n$ - the $n$-dimensional complex space? – William Mar 29 '12 at 6:32
A polynomial system naively implies a set of multivariate polynomial functions. – Suresh Mar 29 '12 at 6:34
Then I am even more confused now. The zeros locus of a polynomial $F_0: \mathbb{C}^n\rightarrow \mathbb{C}$ is not (most often) a discrete set, but rather an algebraic surface. Now, take such an $F_0$, and let $F = (F_0, \underbrace{0, \dots, 0}_{(n-1)-\text{times}})$. – William Mar 29 '12 at 6:38
Are you asking what additional assumptions are necessary in order for a system to have isolated roots? – bgins Mar 29 '12 at 6:49
I am aware that the zeros of a system are not often a discrete set. I wish to know what conditions does F need to satisfy to ensure this. – Suresh Mar 29 '12 at 6:51

Example: $F(z_1,z_2)=(z_1-z_2, z_1^2-z_2^2)$; many roots, not isolated.

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If this is a particular example of what Suresh means, then my general example above shows that the zeros are not necessarily isolated. – William Mar 29 '12 at 6:41
@WNY: It is a counterexample. – André Nicolas Mar 29 '12 at 6:43
I wasn't careful with wording in my comment above, sorry. Yes, it is a particular counterexample. I am just confused on the definition of a polynomial system, since the original question seems to strongly insist that zeros are isolated (one just needs to prove it). – William Mar 29 '12 at 6:44
@WNY: The statement $F: \mathbb{C}^n \to \mathbb{C}^n$ seemed pretty definite. – André Nicolas Mar 29 '12 at 6:48
It wasn't, because as soon as I read "isolated zeros", I thought of a system $F(x_1, \dots, x_n) = (f_1(x_1), \dots, f_n(x_n))$, where each $f_i$ is a polynomial (not identically equal to zero). – William Mar 29 '12 at 6:51

I believe you have a specific system in mind. What you seek is to show that your system of polynomials is a Complete intersection, http://en.wikipedia.org/wiki/Complete_intersection

There are various computer methods for doing this. I am not an expert in this field, but there are sort of generalized Betti numbers, http://en.wikipedia.org/wiki/Betti_number that you may compute, and from those, see how your variety looks like.

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What you are asking about is whether the ideal generated by the polynomials of the system is zero dimensional. This is a difficult condition to test, which requires the computation of a Gröbner basis for the ideal. Certainly the fact of having as many polynomials as indeterminates, as your question suggest, does not suffice, even if the set of polynomial equations is independent (none of the is implied by the others).

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