Discrete set of zeroes of polynomials must be finite? Let  $F:\mathbb C^n\to\mathbb C^n$ be a polynomial mapping (i.e. $n$ polynomials in $n$ variables).
Suppose that $Z = \left\{z \in \mathbb C^n : F(z) = 0\right\}$ is a discrete set (all points are isolated).


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*I think $Z$ must be finite. How to prove this?

*Also, is it true that a sufficient condition for $Z$ to be discrete is that $\det \left(\frac {\partial F_i}{\partial z_j}(z)\right) \ne 0$ for all $z \in Z$? I think this is true by the complex version of the Implicit Function Theorem (for instance, Proposition 1.1.11 in Complex Geometry - An Introduction by D. Huybrechts).
 A: @Yoni
Dear Yoni.
Because your claim deals with polynomial maps but would fail for holomorphic maps, I consider your claim a problem from algebraic geometry.
ad 1) Your assumption implies that the only irreducible components of the algebraic variety $Z \subset \mathbb A_{\mathbb C}^n$ are single points. But any algebraic variety has only finitely many irreducible components.
ad 2) Because of the condition on the functional matrix the algebraic variety $Z$ is non-singular and has dimension $=0$ at each of its points. Hence each irreducible component of $Z$ is $0-$dimensional. Finiteness of $Z$ follows from part 1).
A: This is a more ring-theoretic alternative to jo wehler's geometric answer. Let $F : \mathbb{C}^n \to \mathbb{C}^n$ be given by $F = (f_1, \ldots, f_n)$, where $f_i \in \mathbb{C}[z_1, \ldots, z_n]$. 
1) The zero set $Z$ is in bijection with the set of maximal ideals of $R := \mathbb{C}[z_1, \ldots, z_n]/(f_1, \ldots, f_n)$, and $Z$ is discrete iff $R$ has dimension $0$ (i.e. every point of $\text{Spec}(R)$ is closed). But $R$ is Noetherian, so $\dim R = 0$ iff $R$ is Artinian, and an Artinian ring has finitely many prime ideals.
2) Your condition on the Jacobian is equivalent to requiring the variety $Z$ to be nonsingular, which implies that it is (locally) a complete intersection, and thus has codimension $n$ ($= \#$ of equations), so $\dim Z = 0$ is discrete. 
