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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 ?

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  • $\begingroup$ A comment on style: When you throw in parenthetic remarks in the middle of an implication, it makes it very difficult to read. $\endgroup$ – RghtHndSd Mar 19 '15 at 16:06
  • $\begingroup$ @RghtHndSd You are 100% right of course, and this is pure sloppiness on my part (perhaps also because my English is not good being translated from my country language). I will edit it now. $\endgroup$ – brunoh Mar 19 '15 at 16:11
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    $\begingroup$ I gave it a go. Feel free to improve it further. $\endgroup$ – RghtHndSd Mar 19 '15 at 16:12
  • $\begingroup$ @RghtHndSd You have done a perfect job. Thank you very much for your help. I upvoted your comment. Next time I will be more careful and I have learned from you a good formatting tool ! $\endgroup$ – brunoh Mar 19 '15 at 16:23

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