Consider a complex polynomial map $f: \mathbb{C}^n \rightarrow \mathbb{C}^n$.
For $n = 1$, the fundamental theorem of algebra says that, for any $y \in \mathbb{C}$ there exists $x \in \mathbb{C}$ such that $y = f(x)$. Thus for $n = 1$, $f(\mathbb{C}) = \mathbb{C}$ if and only if $f$ is a non-constant polynomial. Can we generalize this statement to $n > 1$?
For arbitrary $n$, what is a necessary and sufficient condition to say that closure$\left(f(\mathbb{C}^n)\right)$ $ = \mathbb{C}^n$?
Suppose the polynomials $f_1,f_2,\cdots,f_n$ are algebraically dependent, i.e., there exists an annihilating polynomial $F$ such that $F(f_1,f_2,\cdots,f_n) = 0$, then the image $f(\mathbb{C}^n)$ is a subset of the affine variety $V(F)$ of dimension $n-1$. Hence a necessary condition is that the polynomials $f_1,f_2,\cdots,f_n$ must be algebraically independent. Can we show that it a sufficient condition as well?
Best Answer
Being algebraically independent is indeed a necessary and sufficient condition for the image of $f$ to be dense.
As $f\colon\mathbb{C}^n\to\mathbb{C}^n$ is regular, its image is constructible and, in particular, contains a non-empty open subset of its closure (under the Zariski topology). See Theorem 10.2 of J.S. Milne's algebraic geometry notes. If the Zariski closure of $f(\mathbb{C}^n)$ is all of $\mathbb{C}^n$, then $f(\mathbb{C}^n)$ contains a Zariski open set and is dense in the standard topology. Otherwise, the Zariski closure of $f(\mathbb{C}^n)$ is the zero set of a nontrivial ideal $I\subset\mathbb{C}[X_1,\ldots,X_n]$ and you can take any $F\in I\setminus\{0\}$ to see that $f_i$ are algebraically dependent. In fact, this argument works for regular maps $f\colon V\to\mathbb{C}^n$ for any variety $V$. There is no need to restrict the domain to be $\mathbb{C}^n$.