I read the following in Eisenbud's Commutative Algebra with a view .... Let $k$ be an algebraically closed field. Recall that a retract is a morphism which is a retraction of the inclusion, and $X$ is a rational curve if its function field is a purely transcendental extension of the ground field $k$.
A retract of a principal open subset $D(f)$ of $\mathbb{A}^n(k)$ on an irreducible algebraic curve $X$ of $\mathbb{A}^n(k)$ exists if and only if $X$ is rational.
There seems to be an easy proof using the Riemann-Hurwitz formula which could go like this : taking any irreducible rational curve contained in $D(f)$, the Hurwitz formula implies that its genus is greater than the one of $X$ which therefore should be $0$.
I have three questions :
1- Is the proof using the Hurwitz formula correct ?
2- Is there another one more elementary (i.e. not using Riemann-Hurwitz) ?
3- How to interpret this fact intuitively ?