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I'm solving problems in Hartshorne. I don't know how to solve the following exercise(6.4 of Chapter 1): Let $Y$ be a nonsingular projective curve. Show that every nonconstant rational function $f$ on $Y$ defines a surjective morphism $\phi:Y\rightarrow \mathbb P^1$.

I know I can use 6.7 and 6.8 to extend the morphism $f$ to such a $\phi$. But I don't know how to prove it's surjective. I have tried something:

Through $\phi$ we get an injection $k(x)\rightarrow K(Y)$. Since all $DVR$ in $k(x)$ looks like $k[x]_{(x)}$, it's enough for us to just prove there is a DVR induced by a point in $Y$ dominates $k[x]_{(x)}$. I can find one DVR (denote it by $B$)by taking the integral closure of $k[x]$ in $K$ and do a localization at a prime that dominates $(x)$.(Just as the proof of 6.5) But I can't show it is a DVR induced by a point in Y. Could you provide some help? Thanks!

hm2020
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Richard
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  • The image of $\phi$ must have dimension $0$ or $1$, and since $f$ is nonconstant and $Y$ is irreducible, it can't be $0$. Thus the image is a closed, irreducible subset of $\mathbb{P}^1$ of dimension $1$. – Viktor Vaughn Sep 20 '21 at 20:27
  • @ViktorVaughn Thanks for answering. But how to know its image is closed? – Richard Sep 20 '21 at 23:46
  • See https://math.stackexchange.com/questions/2477575/image-of-morphism-of-projective-varieties-is-projective-variety or https://math.stackexchange.com/questions/852493/an-equivalent-statement-of-morphisms-of-projective-varieties-are-closed – Viktor Vaughn Sep 21 '21 at 00:22

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Question: "I know I can use 6.7 and 6.8 to extend the morphism f to such a ϕ. But I don't know how to prove it's surjective."

Answer: Let $C$ be an irreducible smooth projective curve over an algebraically closed field $k$ and let $K:=K(C)$ be the function field of $C$ with $x\in K$ a "non-constant" rational function. Let $P(t)\in k[t]$ and assume $P(x)=0$. Since $k$ is algebraically closed this implies $x\in k$ a contradiction, hence $x$ is "transcendental over $k$", meaning the subring $k[x] \subseteq K$ generated by $k$ and $x$ is isomorphic to a polynomial ring. You get an embedding $k \subseteq k(x) \subseteq K$ where $x$ behaves like an independent variable over $k$. This gives (by the proof of HH.I.6.12) a surjective morphism

$$\phi: C \rightarrow \mathbb{P}^1_k.$$

Since $C$ is projective and irreducible, the image is closed and irreducible and hence $Im(\phi)=\mathbb{P}^1_k$. The inverse image $\phi^{-1}(x)$ of a closed point $x\in \mathbb{P}^1_k$ is a strict closed subvariety of $C$ (since $\phi$ is non-constant) and since $C$ is irreducible it follows $\phi^{-1}(x)$ must be a finite set of points.

hm2020
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