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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 \begin{equation} H(\mathbf{x})=\mathcal{E}(\mathbf{x})-\mathcal{E}(\mathbf{w}) = 0. \end{equation} 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?2012-03-29
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    A polynomial system naively implies a set of multivariate polynomial functions.2012-03-29
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    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 $F = (F_0, \underbrace{0, \dots, 0}_{(n-1)-\text{times}})$.2012-03-29
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    Are you asking what additional assumptions are necessary in order for a system to have isolated roots?2012-03-29
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    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.2012-03-29

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