Let \(Y\) be a nonsingular projective curve. Show that every nonconstant rational function \(f\) on \(Y\) defines a surjective morphism \(\varphi: Y \to \PP^1\), and that for every \(P \in \PP^1\) the fibre \(\varphi^{-1}(P)\) is a finite set of points.
Solution
Let \(K=K(Y)\) and let \(f\in K\setminus k\).
1. The rational function \(f\) defines a morphism from a nonempty open subset of \(Y\) to \(\PP^1\).
There is a nonempty open subset \(U\subseteq Y\) on which \(f\) is regular. On \(U\) define \[ \varphi_U:U\longrightarrow\PP^1, \qquad Q\longmapsto[f(Q):1]. \] This is a morphism because its image lies in the standard affine chart \(\AA^1\subseteq\PP^1\) and is represented there by the regular function \(f|_U\).
2. The morphism \(\varphi_U\) extends uniquely to a morphism \(\varphi:Y\to\PP^1\).
The complement \(Y\setminus U\) is a proper closed subset of the noetherian irreducible curve \(Y\), hence is finite. Write it as \(\{P_1,\ldots,P_m\}\). Proposition I.6.8 says that a morphism from a nonsingular curve with one point removed to a projective variety extends uniquely over that point.
Starting with \(U_0=U\), put \(U_i=U\cup\{P_1,\ldots,P_i\}\). The open subset \(U_i\) of the nonsingular curve \(Y\) is itself a nonsingular curve, and \[ U_i\setminus\{P_i\}=U_{i-1}. \] Thus Proposition I.6.8 extends the morphism on \(U_{i-1}\) uniquely to \(U_i\). At each stage the target is the projective variety \(\PP^1\). After finitely many steps we obtain a morphism \[ \varphi:Y\longrightarrow\PP^1 \] whose restriction to \(U\) is \(\varphi_U\). Uniqueness at each extension step makes the final morphism unique.
3. The morphism \(\varphi\) is nonconstant and surjective.
If \(\varphi\) were constant, then its restriction to \(U\) would be constant. Since \(\varphi_U\) is given in the affine chart by \(f\), this would force \(f\) to be constant in the function field, contrary to the hypothesis.
Because \(Y\) is projective over \(k\), it is complete, so the image \(\varphi(Y)\) is closed in \(\PP^1\). The image of the irreducible space \(Y\) is irreducible. The only irreducible closed subsets of \(\PP^1\) are points and \(\PP^1\) itself. The image is not a point because \(\varphi\) is nonconstant. Therefore \[ \boxed{\varphi(Y)=\PP^1}. \]
4. Every fibre of \(\varphi\) is finite.
Fix \(P\in\PP^1\). The fibre \[ \varphi^{-1}(P) \] is closed in \(Y\) because \(P\) is closed. It is a proper subset: if it were all of \(Y\), then \(\varphi\) would be constant with value \(P\), contradicting step 3.
A proper closed subset of an irreducible noetherian curve has dimension zero. Such a closed subset has only finitely many irreducible components, each of which is a closed point. Hence \(\varphi^{-1}(P)\) is a finite set of points.
5. Q.E.D.
Steps <1>1–<1>2 construct the morphism associated to \(f\), step 3 proves that it is surjective, and step 4 proves that every fibre is finite.