Structure of étale maps

Question

Is it true that any étale map of affine schemes is a composition of finitely many finite étale maps and Zariski localizations?

Answer 1

This is a good question! The answer is, I think, no. As étale maps are open, one can assume the map $f: X \to Y$ to be surjective. One can then construct an affine scheme $\bar{X}$ containing $X$ as open subscheme together with a finite morphism $\bar{f}: \bar{X} \to Y$ which prolongs $f$ - however there is no reason for $\bar{f}$ to be étale at $\bar{X}\setminus X$.

This guides us in finding a counterexample: Consider the map of rings $$\mathbb{C}[T] \to \mathbb{C}[T,T^{-1}], \ T \mapsto T^2(T-1)$$ and the corresponding scheme morphism $$f:\mathbb{A}^1 \setminus \{0\} \to \mathbb{A}^1.$$ It is easy to see that this morphism is étale but not finite.

I'm a bit sketchy on proving precisely that this can't be factored, but I think one can make the following precise: The morphism has prime degree, so we should factor it either as $\bar{f} \circ g$ or $g \circ \bar{f}$ where $\bar{f}$ is finite étale and $g$ is a Zariski-immersion. As $f$ is surjective, the latter can't happen, so we need to exclude the first possibility. This can be done analytically, where standard arguments about Riemann surfaces show that the only possible factorization corresponds to $$\mathbb{A}^1 \setminus \{0\} \to \mathbb{A}^1 \to \mathbb{A}^1$$ where the first map is the inclusion, the second given by $z \mapsto z(z^2-1)$. However the second map is not étale at the origin.