# When are complex polynomial maps surjective?

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 $$f(z_1, z_2) = (z_1, z_1 z_2)$$.

• The Ax-Grothendieck theorem gives that when $p=q$, then injectivity implies surjectivity. Commented Jan 17, 2020 at 23:28
• Thank you. Is something known when $p > q$, e.g. something about common roots of the $f_i$?
– cs89
Commented Jan 18, 2020 at 8:52
• not that I know of, but I am far from an expert. Sorry. Commented Jan 18, 2020 at 19:02
• 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. Commented Jan 20, 2020 at 23:47