Non-singular zeros are isolated. Suppose that $f: \mathbb{C}^N \to \mathbb{C}^n$ is a function with components $(f_1, \ldots, f_n)$ being polynomials in $N$ variables (so $f_i \in \mathbb{C}[X_1, \ldots, X_N]$). A zero $z^\ast$ is called singular if the Jacobian of $f$ at $z^\ast$
$$J(f)(z^\ast) = \begin{pmatrix}
\frac{\partial f_1(z^\ast)}{\partial X_1} & \frac{\partial f_1(z^\ast)}{\partial X_2} & \ldots &\frac{\partial f_1(z^\ast)}{\partial X_N}\\
\frac{\partial f_2(z^\ast)}{\partial X_1} & \frac{\partial f_2(z^\ast)}{\partial X_2} & \ldots &\frac{\partial f_2(z^\ast)}{\partial X_N}\\
\vdots & \vdots & & \vdots\\
\frac{\partial f_n(z^\ast)}{\partial X_1} & \frac{\partial f_n(z^\ast)}{\partial X_2} & \ldots & \frac{\partial f_n(z^\ast)}{\partial X_N}
\end{pmatrix}$$
is rank-deficient (does not have full rank). The book I am reading is 'Numerically Solving Polynomial Systems with Bertini' by Bates, Hauenstein, Sommese and Wampler and it states that 'non-singular zeros are isolated', where $z^\ast$ is an isolated zero if there exists some $r > 0$ such that $B(z^\ast, r)$ (the ball with center $z^\ast$ and radius $r$) does not contain any other zeros. 
In case $n = N$, this follows from the Inverse Function Theorem. However, I have trouble with the case where $n \neq N$. 
For the case where $n \neq N$, I have tried using a contradiction: suppose $z^\ast$ is not isolated, then for the open sets $B(z^\ast, \frac{1}{n})$ there must be some other zero, say $z_n$. Hence $z_n \to z^\ast$ as $n \to \infty$. I have tried using this to show that $z^\ast$ can not be a non-singular zero, but I am not able to do so. 
Question: How can I prove this or could anyone give a reference to some proof?
Remark: The book I use is mostly on the numerical part of finding zeros, but it has some sections which go in more detail, however these details consist mostely out of facts which are stated with little to no proof. This question is just such a statement, without proof.
 A: The statement in your book is partly erroneous. 
The key thing to know is the constant rank theorem: If $U$ is a domain in $C^N$ where a complex-analytic  map $F: U\to C^n$ satisfies
$$
rank(dF)_z=k
$$
for all $z\in U$, then near each $\zeta\in U$, after holomorphic change of coordinates near $\zeta$ and near ${\mathbf 0}\in C^n$, the map $F$ has the form
$$
(x_1,...,x_N)\mapsto (x_1,...,x_k,0,...0).
$$ 
If $\zeta$ is a nonsingular zero of $F: C^N\to C^n$ then, by continuity of the determinant, there is a neighborhood of $\zeta$ where $dF$ has constant rank $k$, equal $rank(dF_\zeta)$. Now consider the following special cases:


*

*If $N\le n$ and $\zeta\in C^N$ is a nonsingular zero, then $k=N$ and, hence, locally, near $\zeta$, the map is 1-1. Hence, $\zeta$ is indeed an isolated zero of $F$ in this case.

*If $N>n$ and $\zeta\in C^N$ is a nonsingular zero, then $k=n$ and the map $F$ (at $\zeta$, after a local change of coordinates) becomes a linear projection. Hence, $\zeta$ is not an isolated zero in this case: The set of zeroes near $\zeta$ has dimension $N-n$. 
(In fact, one can prove more in your situation: If $F: C^N\to C^n$ is a holomorphic map and $N>n$ then $F^{-1}({\mathbf 0})$ has no isolated points regardless of whether they are singular or not.) 
The conclusion is that the statement in your book only holds if $N\le n$. 
