TITLE: constructing a projection onto a variety
QUESTION [9 upvotes]: Consider the vector space $\mathbb{C}^n$. Given any linear subspace $S$ we can choose a complement of $T$ in $V$, i.e. $\mathbb{C}^n=S \oplus T$ and we can subsequently define a projection $\pi_S:\mathbb{C}^n \rightarrow S$ given by $x=x_S+x_T \mapsto x_S$, where $x_S,x_T$ are the unique components of $x$ in $S,T$ respectively.
Now let $f_1,\cdots,f_k$ be elements of $\mathbb{C}[y_1,\cdots,y_n]$. These polynomials define a variety $V$, i.e. the zero set of the ideal that they generate. Question: is there a way to define a projection $\mathbb{C}^n \rightarrow V$ in a similar manner as we did for linear subspaces? Alternatively, is there any way of defining a surjective map $\mathbb{C}^n \rightarrow V$ given the polynomials $f_1,\cdots,f_k$?
Edit: From the comments, i understand that this is not possible in general. Then, under what conditions can we obtain such a surjection? Is there an approximation theory of how close a point of $\mathbb{C}^n$ is to the variety of interest?

REPLY [8 votes]: Let me just explain why the question is subtle and interesting as said Georges. 
An obvious necessary condition to have a surjective map $\mathbb C^n \to V$ is $V$ is irreducible because the image of an irreducible topological space is irreducible. Also you can replace $V$ by the associated reduced variety, this doesn't change the existence of a surjective map from $\mathbb C^n$. So we suppose $V$ is an integral affine algebraic variety. 
Now another more serious necessary condition is that the function field of $V$ must be the function field of a (smooth projective) unirational variety $\overline{V}$ (by definition this means there exists a dominant rational map $\mathbb P^n --\to \overline{V}$, or equivalently, a dominant morphism from an open subset of $\mathbb P^n$ to $\overline{V}$. This is because the surjevtive map $\mathbb C^n\to V$ induces a dominant rational map from $\mathbb P^n$ to any smooth projective variety birational to $V$. Unirational varieties are  rationally connected. There is a vast litterature on this subject (a good starting point is Kollar's "Rational curves on algebraic varieties" or Debarre's "Higher-dimensional algebraic geometry"). 
Suppose $\mathbb C(V)$ is the function field of a unirational smooth projective variety $\overline{V}$, can one decide whether there is a surjective map from $\mathbb C^n$ ? This in my opinion is a hard question. 
If $\dim V=1$, then unirational implies rational by Lüroth's theorem. And it is then easy to see that there exists a surjective map $\mathbb C^n\to V$ if and only if $V$ is a (possibly singular, but integral) rational projective curve minus a smooth point. But in higher dimension, even when $V$ is a smooth surface, I don't know a characterization (though $V$ then must be rational). 
Remark: the usual definition of unirational requires a dominant rational map from $\mathbb P^{\dim V}$. One can replace $\dim V$ by any integer $n$ by  Debarre, 4.2(2).  
A weaker and purely algebraic question is when a $\mathbb C$-algebra of finite type is contained in a polynomial ring $\mathbb C[y_1, \dots, y_n]$.