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?


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 $\mathbb{C}^n$.

  • $\begingroup$ Thanks George. It was clear that the Zariski closure of $f(\mathbb{C}^n)$ is all of $\mathbb{C}^n$. Theorem 10.2 in J.S.Milne's notes is exactly what I was looking for. Thanks! $\endgroup$
    – sreekanth
    Mar 12 '11 at 0:14
  • $\begingroup$ @Thierry : over the reals, what about $n=1$ and $f(x)=x^2$ ? $\endgroup$
    – BS.
    Mar 12 '11 at 10:13
  • $\begingroup$ @BS: I just knew my comment had to be wrong, I can't imagine why I wrote it. Thanks! $\endgroup$ Mar 12 '11 at 17:33
  • $\begingroup$ @BS: the part that does work over the reals, though, is that the $f_i$'s are algebraically independent iff $f(\mathbb{R}^n)$ is Zariski-dense. $\endgroup$ Mar 12 '11 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.