Polynomials of a fixed degree have a nonzero partial at all points of their vanishing set? I'm having difficulty with the second half of a question from an old homework assignment (for a differential geometry class I am currently taking).  
The first half of the question asked me to assume that $f:\mathbb{C}^n \to \mathbb{C}$ was a polynomial function (with complex coefficients) and to find a condition (in terms of the partial derivatives of $f$) for the vanishing set of $f$ to be a smooth submanifold of $\mathbb{C}^n$ (of complex dimension $n-1$).  The correct condition seemed to me to be that $df_x : \mathbb{C}^n \to \mathbb{C}$ is surjective for 
every $x$ in the vanishing set of $f$.  But since $\mathbb{C}$ has complex dimension $1$, this just means that for every $x$ in the vanishing set there is an $i$ such that $\partial f/\partial z_i(x) \neq 0$.
The second half of the question asked me to show that these conditions are satisfied for "almost all" polynomials of a fixed degree.  I'm assuming that the "almost all" means everywhere except on a set of measure zero, but this is difficult for me to understand.  Should I give the set of polynomials of fixed degree the structure of a smooth manifold?  What is the topology and what are the charts?  Presumably I should then consider the subset of polynomials for which every partial vanishes at every root and show that its intersection with each chart has measure zero.  Does anyone have any other ideas?  Maybe involving Sard's theorem (which we had just learned)? 
Thank you for any help.      
 A: A polynomial of degree $\le d$ is $P_a(z)=\sum_\nu a_\nu z_1^{\nu_1}\cdots z_n^{\nu_n}$,where $\nu=(\nu_1,\dots,\nu_n)$, $|\nu|=\sum_i\nu_i\le d$. Hence $P_a$ is represented by the tuple $a=(a_\nu)_\nu\in\mathbb C^N$ for suitable $N$. Then the map $F:\mathbb C^N\times\mathbb C^n\to\mathbb C$ given by $F(a,z)=P_a(z)$ is a submersion: $\partial F/\partial a_0\equiv1$. Thus $0$ is a regular value of $F$ and $F^{-1}(0)=M$ is a smooth manifold of dimension $d=2(N+n-1)$ of $\mathbb C^N\times\mathbb C^n\equiv\mathbb R^{2(N+n)}$; the tangent space is
$$
T_{(a,z)}M=\ker(d_{(a,z)}F).
$$
From now on we consider always real dimensions. Now restrict to $M$ the linear projection $\pi:\mathbb C^N\times\mathbb C^n\to\mathbb C^N$ to get a smooth mapping $g:M\to\mathbb C^N$. By the Sard theorem the set $R\subset\mathbb C^N$ of regular values of $g$ has complement of measure zero. This set $R$ gives what wanted.
Indeed, let $a\in R$ be a regular value of $g$. Its inverse image 
$$
A=g^{-1}(a)=\{a\}\times\{x\in\mathbb C^n: P_a(z)=0\}
$$
is a smooth manifold of dimension $d-2N=2(n-1)$ with tangent space
$$
T_{(a,z)}A=\ker(d_{(a,z)}g)=(\{0\}\times\mathbb C^n)\cap\ker(d_{(a,z)}F)=\{0\}\times\ker d_zP_a.
$$
Thus $\dim(\ker d_zP_a)=2(n-1)$, which means the image of $d_zP_a:\mathbb C^n\to\mathbb C$ has dimension $2$. In other words, $d_zP_a$ is onto for $(a,z)\in A$, and fixed $a$, for every $z\in\mathbb C^n$ with $P_a(z)=0$. We are done.
