A curve \(X\) is called hyperelliptic if \(g \geq 2\) and there exists a finite morphism \(f: X \to \PP^1\) of degree 2.
If \(X\) is a curve of genus \(g=2\), show that the canonical divisor defines a complete linear system \(\abs{K}\) of degree 2 and dimension 1, without base points. Use (II, 7.8.1) to conclude that \(X\) is hyperelliptic.
Show that the curves constructed in (1.1.1) all admit a morphism of degree 2 to \(\PP^1\). Thus there exist hyperelliptic curves of any genus \(g \geq 2\).
Note: we will see later (Ex. 3.2) that there exist non-hyperelliptic curves. See also (V, Ex. 2.10).
Solution
1. If \(g=2\), then the canonical system has degree \(2\) and dimension \(1\):\[ \deg K=2, \qquad \dim\abs{K}=1. \]
For a smooth projective curve of genus \(g\), \[ \deg K=2g-2 \] and \[ \ell(K)=g. \] Thus for \(g=2\), \[ \deg K=2 \] and \[ \dim\abs{K} = \ell(K)-1 = 1. \]
2. The canonical system \(\abs{K}\) has no base point.
Suppose that \(P\in X\) were a base point of \(\abs{K}\). Then every canonical section vanishes at \(P\), so \[ H^0(X,\mco_X(K-P)) = H^0(X,\mco_X(K)). \] Hence \[ \ell(K-P)=\ell(K)=2. \]
Riemann–Roch for the degree-one divisor \(P\) gives \[ \ell(P)-\ell(K-P) = 1+1-g = 0, \] so \[ \ell(P)=2. \] Therefore \[ \dim\abs{P} = \ell(P)-1 = 1 = \deg P. \] Since \(P\ne0\), the equality criterion in Exercise IV.1.5 would force \(g=0\), contradicting \(g=2\). Thus \(\abs{K}\) has no base point.
3. Every genus-\(2\) curve is hyperelliptic.
By steps <1>1–<1>2, the complete linear system \(\abs{K}\) is base-point free of dimension \(1\). Hence it defines a morphism \[ \varphi_{\abs{K}}:X\longrightarrow\PP^1 \] with \[ \varphi_{\abs{K}}^*\mco_{\PP^1}(1) \cong \mco_X(K). \] The map is nonconstant, hence finite. Taking degrees gives \[ \deg\varphi_{\abs{K}} = \deg K = 2. \] Thus \(X\) is hyperelliptic. This proves part (a).
4. The curves constructed in (IV.1.1.1) admit a finite morphism of degree \(2\) to \(\PP^1\).
In that construction, for every \(g\ge2\) one obtains a smooth curve \[ X\subseteq Q, \qquad Q\cong\PP^1\times\PP^1, \] whose divisor class has type \((g+1,2)\).
Take the ruling projection \[ p:Q\longrightarrow\PP^1 \] whose fibre class meets a divisor of type \((g+1,2)\) in degree \(2\). The restriction \[ p|_X:X\longrightarrow\PP^1 \] is nonconstant, hence finite. For any point \(a\in\PP^1\), the fibre divisor is \[ (p|_X)^*(a)=X\cap p^{-1}(a), \] and its degree is the intersection number \[ X\cdot p^{-1}(a)=2. \] Therefore \[ \deg(p|_X)=2. \] Hence every curve in the construction is hyperelliptic.
5. Hyperelliptic curves exist in every genus \(g\ge2\).
The construction in (IV.1.1.1) produces such a smooth curve for every \(g\ge2\), and step 4 gives a degree-two morphism from each one to \(\PP^1\). This proves part (b).
6. Q.E.D.
Step 3 proves part (a), and steps <1>4–<1>5 prove part (b).