A complex polynomial $\mathbb R^2 \rightarrow \mathbb R^2$ is surjective.

Question

Consider a non-constant polynomial in $\mathbb C$ and consider the map from $\mathbb R^2$ to $\mathbb R^2$ that it encodes. Is it possible to show it is a surjective function? We know it is one because of the fundamental theorem of algebra. But would we be able to prove it without knowing it is a polynomial?

As a particular example we can consider the polynomial $x^3-2x^2-4$ and see it encodes the map $f(x,y) = (x^3 + 2 x^2 - 3 x y^2 - 2 (y^2 + 2), y (3 x^2 + 4 x - y^2))$

Is there some sort of theory that encompasses this phenomenon and can be used to say functions of this kind are surjective (without knowing a-priori the expression is of this kind)? Like maybe some sort of invariant that can be calculated for the expression or something.

Do you mean maps induced by complex holomorphic functions? They are identified by satisfying the Cauchy-Riemann equations. And if they are entire (holomorphic on all of $\mathbb R^2$) then their range misses at most one point by the Little Picard Theorem. The only non-surjective ones are of the form $e^{h(z)}+c$, with $h$ entire and $c$ a constant, see Surjective entire functions on MO. — Conifold, Sep 24 '20 at 04:53
Yeah that works well, I guess it is easy to check our functions are not of that form? (polynomials in x and y) — Asinomás, Sep 24 '20 at 05:07
Not necessarily, there are some sufficient conditions on MO, but there is no constructive criterion. — Conifold, Sep 24 '20 at 05:10
but the MO question seems to assume we already know the function as a complex function — Asinomás, Sep 24 '20 at 05:13
$p(z)=w$ is a polynomial equation of degree $\ge1$ for every complex $w$, so it must have a solution. — , Sep 24 '20 at 05:28
Yes but the question is how can you take a function like that and show it is surjective without being told it is induced by a polynomial. — Asinomás, Sep 24 '20 at 05:32
Take the map $(u,v)$, check CRE, form $f=u+iv$ and apply conditions from MO to $f$. Polynomials are always surjective once they are holomorphic since any non-constant $e^{h(z)}+c$ has at least exponential growth. — Conifold, Sep 24 '20 at 05:43
hmm, but isn't there still a problem because the conditions in MO ask you to know the coefficientens in the expression $f(z)=\sum_{i=0}^{\infty}a_iz^i$ . Or how would we bound these? — Asinomás, Sep 24 '20 at 05:48
Oh I think you mean just using the expression $e^{h(z)} + c$? — Asinomás, Sep 24 '20 at 05:49
Sounds good, but how does one prove the expression $e^{h(z)} + c$ has exponential growth? I'm very rusty when it comes to this stuff. — Asinomás, Sep 24 '20 at 05:54
Entire function bounded by a polynomial is a polynomial. If $h(z)$ is non-constant it must grow at least linearly. — Conifold, Sep 24 '20 at 08:08

score 3 · Answer 1 · answered Jul 03 '23 at 14:55

I think I know exactly what you're looking for! Here's a theorem (source - example problem 4.2 in Fulton's Introduction to Algebraic Topology)

Claim: Let $g: \mathbb{C} \to \mathbb{C}$ be a continuous function such that for some integer $n \geq 1,$ $\frac{g(z)}{z^n}$ approaches a non-zero constant $a$ as $|z| \to \infty.$ Then $g$ is surjective.

Clearly polynomials fulfill this, and in some sense I believe this has what you want of "not knowing it's a polynomial".

Proof of claim: Assume there is $x \in \mathbb{C}$ with $\forall z \in \mathbb{C}: g(z) \neq x.$ We apply the Dog-on-a-Leash theorem (see Fulton for the exact details), and we try and compare $g$ and $az^n.$ WLOG $x = 0$ and $a > 0,$ and we picking an $\epsilon$ with $0<\epsilon<a$ we get that for large enough $r,$ the function $g$ restricted to the circle with radius $r$ centered at $0$ fulfills $$|g(z)-az^n| < \epsilon |z|^n < a|z|^n = |a|z|^n-0|,$$ i.e. the distance of $g$ and $az^n$ is smaller than the distance of $az^n$ and $0.$ This proves that both must have the same winding number around 0. But $az^n$ has winding number $n,$ and $g,$ as a continuous function extendable to the entire disk of radius $r,$ must have winding number 0, a contradiction as $n \geq 1.$

Thank you very much +1. Although what I was looking for was a theorem that didn't "know" that the function comes from a complex function (in other words it is just presented as a function from $\mathbb R^2$ to $\mathbb R^2$, and we are able to use some more relaxed condition to verify it is surjective). — Asinomás, Jul 04 '23 at 10:20

A complex polynomial $\mathbb R^2 \rightarrow \mathbb R^2$ is surjective.

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