Let \(X\) be a curve of genus \(g\). Show that there is a finite morphism \(f: X \to \PP^1\) of degree \(\leq g+1\).
Recall that the degree of a finite morphism of curves \(f: X \to Y\) is defined as the degree of the field extension \([K(X): K(Y)]\) (II.6).
Solution
Choose a point \(P\in X\) and put \[ D=(g+1)P. \]
1. One has\[ \ell(D)\ge2. \]
By Riemann--Roch, \[ \ell(D)-\ell(K-D) = \deg D+1-g = 2. \] Since \(\ell(K-D)\ge0\), it follows that \[ \ell(D)=2+\ell(K-D)\ge2. \]
2. There is a nonconstant rational function \(f\in K(X)\) whose pole divisor satisfies\[ (f)_\infty\le D. \]
By definition, \[ L(D)=H^0(X,\mco_X(D)) = \{f\in K(X):\operatorname{div}(f)+D\ge0\}\cup\{0\}. \] The constants form a one-dimensional subspace of \(L(D)\), while step 1 gives \(\dim_kL(D)\ge2\). Hence there is a nonconstant \(f\in L(D)\).
The inequality \[ \operatorname{div}(f)+D\ge0 \] says precisely that the polar divisor of \(f\) is bounded by \(D\): \[ (f)_\infty\le D. \]
3. The function \(f\) defines a finite morphism\[ f:X\longrightarrow\PP^1 \]of degree at most \(g+1\).
A nonconstant rational function on the smooth projective curve \(X\) defines a nonconstant morphism to \(\PP^1\). A nonconstant morphism between projective integral curves is finite.
Its degree equals the degree of its polar divisor: \[ \deg f=\deg (f)_\infty \] by the degree formula for a rational function on a projective curve. By step 2, \[ \deg (f)_\infty \le \deg D = g+1. \] Therefore \[ \boxed{\deg f\le g+1}. \]
4. Q.E.D.
Step 3 gives the required finite morphism.