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I have a question about example 6.11.4 (page 142-143) of Hartshorne's Algebraic Geometry.

In this example we are trying to determine the Cartier divisor class group of the cuspidal cubic curve $y^2 z = x^3,$ let's call it $X$. Let $Z$ be the singular point $(0,0,1).$ To each closed point of $X-Z$ we associate a Cartier divisor and we want to prove (by contradiction) that this map is injective.

At some point in this proof Hartshorne deduces that there exists an $f\in K^*,$ which is invertible at $Z,$ and such that $(f) = P - Q$ on $X-Z$ for distinct points $P$ and $Q$ of $X-Z$.

He continues: "Then $f$ gives a morphism of $X$ to $\mathbb{P}^1$, which must be birational." Why must this morphism be birational?

Then he goes on by saying: "But then the local ring of $Z$ on $X$ would dominate some discrete valuation ring of $\mathbb{P}^1$." Why is this?

It's quite possible both of these remarks are rather straight forward, but I just don't see why they are true at the moment, so any help is much appreciated!

EDIT: I found this previous post (Question about cusp cubic example in Hartshorne) about the same example, which however doesn't answer my question.

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  • $\begingroup$ It's barational because it has one zero and one pole, i.e., it's generically one-to-one. Then whatever point maps to Z has its local ring (a DVR) injecting into the local ring at Z. $\endgroup$ – John Brevik Mar 17 '15 at 18:56
  • $\begingroup$ Excellent, thank you! $\endgroup$ – Misja Mar 17 '15 at 19:29

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