Construct a morphism from a nonsigular projective curve onto P1 from a rational function, and its quasi-finiteness (II)

Question

I'm currently doing exercise I.6.2 in Hartshorne's Algebraic Geometry. I've tried to prove this and get a so-called "proof" which full of gaps and uncertainty.

The exercise goes like this:

Let $Y$ be a nonsingular projective curve. Show that: (1) every nonconstant rational function $f$ defined on $Y$ defines a surjective morphism $\phi: Y \rightarrow \mathbb{P}^1$, (2) and that for every $P \in \mathbb{P}^1$, $\phi^{-1}(P)$ is a finite set of points.

Questions on (1) has already been posted in the question: Construct a morphism from a nonsigular projective curve onto $\mathbb{P}^1$ from a rational function, and its quasi-finiteness (I)

In this post, I'm going to raise some questions on the proof of (2). This proof is adapted from an online solution manual of Hartshorne. Here $k$ is the base algebraically closed field.

"Proof" of (2): Since $\phi$ is dominant, it incudes an inclusion $K(Y) \rightarrow K(\mathbb{P}^1)$. Since both fields are finitely generated extension fields of transcendence degree 1 of $k$, $K(\mathbb{P}^1)$ must be a finite algebraic extension of $K(Y)$. To show that $\phi$ is quasi-finite (i.e. $\phi(P)$ is a finite set), look at any open affine set $V$ in $P$. Its coordinate ring is $A(V)$, and by "Finiteness of Integral Closure" (i.e. Thm (I.3.9) of Hartshorne), $A(V)$ is a finite $A(V)$-module. The corresponding affine set to $A(V)$ is isomorphic to an open subset $U$ of $Y$. Clearly $U = \phi^{-1}(V)$, and thus $\phi$ is a quasi-finite morphism, as desired.

I feel quite confused following the above proof. Here are my questions:

Question 1: Is the induced inclusion map obtained by [Hartshorne's Corollary I.6.12]? (I will list the corollary below.) If so, it seems that the inclusion obtained should be $K(\mathbb{P}^1) \rightarrow K(Y)$ instead of claimed in the "proof". Am I getting something wrong when reading Hartshorne.

Question 2: I'm not quite familiar with the concept and properties of quasi-finite maps. I have only read the first chapter of Shafarevich's Algebraic Geometry Volume 1 besides Chapter 1 of Hartshorne. A similar concept is a finite map, which is defined as:

Defn: Let $f:X \rightarrow Y$ be a morphism between affine varieties. We call $f$ a finite morphism, if the coordinate ring $A(X)$ is integral over $A(Y)$.

And such a morphsim is called finite, probably because the following property holds:

FACT 1: Using the same notations as Defn, if $f$ is a finite map, then any point $y \in Y$ has at most a finite number of inverse images.

And may be in the proof above, we also use the fact that finiteness is a local property, namely,

FACT 2[Shafarevich Vol.1 Thm 1.13]: If $F: X \rightarrow Y$ is a morphism of affine variety, and every point $x \in Y$ has an affine neighborhood $x \in U$, such that $V=f^{-1}(U)$ is affine and $f: V \rightarrow U$ is finite, then $f$ itself is finite.

HOWEVER, the above seemingly satisfactory facts cannot apply here, since here we are talking about PROJECTIVE curves and PROJECTIVE line $\mathbb{P}^1$. So I'm asking whether such demanding results hold in projective case? (Is "quasi-finiteness" a corresponding notion in projective case to "finiteness" in affine case?) I'm hoping to get some reference on that. (I have looked up Milne's Algebraic Geometry notes, yet the materials there seems inadequate to solve the above exercise.)

OR, I'm trying to convert this exercise into an affine version. For example, fix a point $P$ and take an open affine chart containing it and so on.... Yet I feel it quite challenging to me to fill up all the gaps and details. If such ideas works, could anyone help me clean up the details?

IF the ideas above does not work, could anyone help me to clear up the details in the original proof listed above?

Question 3: It seems that the proof above (after filling up the details) is quite complicated and involves so many concepts and tools not mentioned in Hartshorne. Does Hartshorne really hope us to give such a proof? Is there any simple ones that Hartshorne is demanding?

Thank you all in advance! :)

P.S. [Hartshorne's Corollary I.6.12]:

The following three categories are equivalent: (1) Nonsigular projective curves, with dominant morphisms; (2) Quasi-projective curves, and dominant rational maps; (3) Function fields of dimension one over $k$, and $k$-homomorphisms.

Answer 1

Question 1: Yes. You are right. The arrow should be reversed. And the whole "proof" is just nonsense.

To prove this exercise, you only need to do a dimension count.

An "algebraic" proof: Take affine chart $V \in \mathbb{P}^1$ and $U := \varphi^{-1}(V) \in Y$. We have an inclusion $A(V) \to A(U)$. We need to prove $A(U)$ is integral over $A(V)$. And then the theorem you mentioned would tell us that the fiber is finite.

Now because $A(V)$ and $A(U)$ are both (Krull) dimension $1$ ($U$, $V$ are nonempty open sets), by Noether normalization lemma (the wiki of the lemma explained this fact well, we need the refined version https://en.wikipedia.org/wiki/Noether_normalization_lemma), $A(U)$ is indeed integral over $A(V)$.

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However, Hartshorne did not include Noether normalization lemma in his book, so I guess the proof in his mind should be the following "geometrical" one:

Because the preimage of a closed set is again a closed set. $\varphi^{-1}(P)$ must be closed. However, it cannot be the empty set or the whole $Y$, so it must be a set consists of finite many points.

Answer 2

There are number of issues here. First, this is problem I.6.4, not I.6.2, but that's relatively minor. Next, this "proof" kind of stinks: there's an immediate error in the first line, in that the inclusion of function fields ought to the other way, $K(\Bbb P^1)\to K(Y)$, and it only goes downhill from there.

Here's a better approach: Consider the map of function fields $k(t)\to k(Y)$ by $t\mapsto f$. This is a $k$-homomorphism between two function fields of dimension one, so by corollary I.6.12 it gives us a dominant (in particular, nonconstant) morphism $\varphi:Y\to \Bbb P^1$ which is given by $y\mapsto [1:f(y)]$ where $f(y)\neq \infty$. If $\varphi(Y)$ misses a point, then by taking the copy of $\Bbb A^1$ which is the complement of that point, we can find a nonconstant regular function on $\Bbb A^1$ which pulls back to a nonconstant regular function on $Y$. This is impossible by theorem I.3.4, so $\varphi(Y)=\Bbb P^1$ and $\varphi$ is surjective.

Note that since any point $P$ in $\Bbb P^1$ is closed, $\varphi^{-1}(P)$ is a closed set since morphisms are continuous. On the other hand, the fiber over any point must be a proper subset of $Y$, since $\varphi$ is nonconstant. Since all proper closed subsets of a curve are finite (exercise I.4.8, for instance), we are done.

Answer 3

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.