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Consider a complex polynomial map $f: \mathbb{C}^p \to \mathbb{C}^q$ for some $p \geq q \geq 1$ (not necessarily equal).

What is a sufficient condition for $f$ to be surjective?

I am aware of some necessary conditions. For example, one must of course assume that the components $f_1, \dotsc, f_q : \mathbb{C}^p \to \mathbb{C}$ are algebraically independent. (If this is not the case, then $f(\mathbb{C}^p)$ is contained in an affine algebraic variety). When they are indeed independent, one knows that $f(\mathbb{C}^p)$ is dense in $\mathbb{C}^q$ (see When are complex polynomial maps almost surjective?). However, I wonder if conditions are known ensuring that the image is equal to $\mathbb{C}^q$. The algebraic independence is not sufficient by itself, as shown by the counter-example $f(z_1, z_2) = (z_1, z_1 z_2)$.

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    $\begingroup$ The Ax-Grothendieck theorem gives that when $p=q$, then injectivity implies surjectivity. $\endgroup$
    – Noah Riggenbach
    Commented Jan 17, 2020 at 23:28
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    $\begingroup$ Thank you. Is something known when $p > q$, e.g. something about common roots of the $f_i$? $\endgroup$
    – cs89
    Commented Jan 18, 2020 at 8:52
  • $\begingroup$ not that I know of, but I am far from an expert. Sorry. $\endgroup$
    – Noah Riggenbach
    Commented Jan 18, 2020 at 19:02
  • $\begingroup$ This arises as a map of polynomial rings $F:k[x1..xq]\to k[x1..xp]$, the image scheme is defined by the kernel of $F$. For $f$ to be surjective you want $F$ to be injective. If you have concrete polynomials $f_i$, then you can compute the kernel of $F$ using computer algebra systems such as Macaulay2. $\endgroup$
    – myzhang24
    Commented Jan 20, 2020 at 23:47

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