Surjective Closed Map from Affine Plane to Affine Line

Question

I was wondering if there is a way to prove that there does not exist a surjective, closed map $f \colon \mathbb{A}^2 \to \mathbb{A}^1$ where closed is in terms of the Zariski topology. Intuitively, it seems that collapsing a variety in the plane to a line should result in a singularity resulting in some open set somewhere. For example, if we have the natural projection $\pi \colon \mathbb{A}^2 \to \mathbb{A}^1$ then it cannot be universally closed as $Z(xy-1)$ maps to $D(0)$, which is what motivated my intuition.

Answer 1

I'll assume that the field is the complex numbers.

Let $ \phi: \mathbb{A}^2 \rightarrow \mathbb{A}^1 $ be surjective and closed. The map $ \phi $ is given by a nonconstant polynomial $ f = f(x,y) $.

For a general choice of $ g = g(x,y) $ of large enough degree, the polynomial $ h = fg-1 $ is irreducible. ($\bullet $)

Then $ Z = \mathbb{V}(h) $ is an irreducible affine curve, so $ f(Z) $ is irreducible and closed (by assumption), so $ f(Z) $ is either a point or $ \mathbb{A}^1 $. But clearly $ f(Z) $ misses $ 0 $, so $ f(Z) = c $ for some constant $ c $. Then $ f-c $ vanishes on $ Z $, i.e. $ f-c \in I(Z) = (h) $. Writing $ f-c = qh= q(fg-1) $ for some $ q $ gives $ (qg-1)f = q-c $, a contradiction because the degree of $ g $ was assumed large enough.

I only haven't justified $ (\bullet) $ fully for lack of time, I'll add it later.