For any \(n>e \geqslant 0\), let \(X\) be the rational scroll of degree \(d=2 n-e\) in \(\PP^{d+1}\) given by (2.19). If \(n \geqslant 2 e-2\), show that \(X\) contains a nonsingular curve \(Y\) of genus \(g=d+2\) which is a canonical curve in this embedding.
Conclude that for every \(g \geqslant 4\), there exists a nonhyperelliptic curve of genus \(g\) which has a \(g_3^1\). Cf. (V, §5).
Solution
Write \(X=\FF_e\), let \(C_0\) be its negative section, and let \(f\) be a fibre of the ruling \[ \pi:X\longrightarrow\PP^1. \] Thus \[ C_0^2=-e, \qquad C_0\cdot f=1, \qquad f^2=0, \] and \[ K_X\sim-2C_0-(e+2)f \] FE-TORHIRZ. In the embedding of (2.19), the hyperplane class is \[ H\sim C_0+nf, \] so \(H^2=2n-e=d\).
1. The divisor\[ D\coloneqq H-K_X \sim 3C_0+(n+e+2)f \]has a base-point-free complete linear system.
Put \[ b=n+e+2. \] For a divisor \(aC_0+bf\) with \(a\ge0\), the ruled-surface projection gives \[ \pi_*\OO_X(aC_0+bf) \cong \bigoplus_{i=0}^{a}\OO_{\PP^1}(b-ie). \] Hence \(\OO_X(aC_0+bf)\) is globally generated whenever \(b\ge ae\): every summand on the right is then globally generated, and the relative \(\OO_X(a)\) evaluation map is surjective.
For \(D\) one has \(a=3\), and the hypothesis \(n\ge2e-2\) gives \[ b-3e = n-2e+2 \ge0. \] Thus \(|D|\) is base-point-free.
2. The system \(|D|\) contains a nonsingular member \(Y\).
Set \[ q=n-2e+2\ge0. \] Then \[ D \sim 3(C_0+ef)+qf. \] If \(q>0\), then \[ D \sim \bigl(C_0+(e+q)f\bigr)+2(C_0+ef). \] The first summand is very ample by the rational-scroll construction (2.19), because \(e+q>e\), while step 1 with \(a=1\) shows that \(C_0+ef\) is base-point-free. The tensor product of a very ample invertible sheaf with globally generated invertible sheaves is very ample. Hence \(D\) is very ample, so a general member of \(|D|\) is nonsingular by Bertini’s theorem.
Suppose \(q=0\). The base-point-free system \(|C_0+ef|\) defines the usual morphism from \(\FF_e\) to the rational normal scroll cone: it contracts exactly \(C_0\) and is an immersion on \(X\setminus C_0\). The system
\[ |D|=|3(C_0+ef)| \]
has the same property away from \(C_0\). Moreover, \[ D\cdot C_0=0. \] Since \(|D|\) is base-point-free, a general member does not contain the contracted section \(C_0\), and hence is disjoint from it. Bertini applied on \(X\setminus C_0\), where the associated morphism is an immersion, then shows that a general member is nonsingular. Choose such a member and call it \(Y\).
3. The curve \(Y\) is connected, hence irreducible.
From \[ 0\longrightarrow\OO_X(-D) \longrightarrow\OO_X \longrightarrow\OO_Y \longrightarrow0 \] it is enough to prove \[ H^0(X,\OO_X(-D))=0, \qquad H^1(X,\OO_X(-D))=0. \] For the first vanishing, choose a fibre that is not a component of an effective divisor. Its intersection with that divisor is nonnegative, while \[ (-D)\cdot f=-3, \] so \(-D\) cannot be effective.
For the second, Serre duality and \(D=H-K_X\) give \[ H^1(X,\OO_X(-D))^\vee \cong H^1(X,\OO_X(K_X+D)) = H^1(X,\OO_X(H)). \] Since \[ \pi_*\OO_X(H) \cong \OO_{\PP^1}(n)\oplus\OO_{\PP^1}(n-e) \] and \(R^1\pi_*\OO_X(H)=0\), while \(n>e\ge0\), one obtains \[ H^1(X,\OO_X(H))=0. \] Therefore \(H^0(Y,\OO_Y)=k\), so \(Y\) is connected. By step 2 it is nonsingular; a connected nonsingular curve is irreducible.
4. The canonical bundle of \(Y\) is the restriction of the hyperplanebundle:\[ \boxed{K_Y\cong\OO_Y(H)}. \]
Adjunction and the definition \(D=H-K_X\) give \[ K_Y \cong (K_X+Y)|_Y \cong (K_X+D)|_Y \cong \OO_Y(H). \]
5. The genus of \(Y\) is\[ \boxed{g(Y)=d+2}. \]
By step 4, \[ 2g(Y)-2 = \deg K_Y = H\cdot D. \] Using \(H=C_0+nf\) and \(D=3C_0+(n+e+2)f\), \[ \begin{aligned} H\cdot D &=(C_0+nf)\cdot\bigl(3C_0+(n+e+2)f\bigr)\\ &=-3e+(n+e+2)+3n\\ &=4n-2e+2\\ &=2(2n-e)+2\\ &=2d+2. \end{aligned} \] Thus \(2g(Y)-2=2d+2\), whence \(g(Y)=d+2\).
6. The restriction map\[ H^0(X,\OO_X(H)) \longrightarrow H^0(Y,K_Y) \]is an isomorphism. Consequently the given embedding of \(Y\) is itscomplete canonical embedding.
Since \(Y\sim D=H-K_X\), restriction gives \[ 0 \longrightarrow \OO_X(K_X) \longrightarrow \OO_X(H) \longrightarrow \OO_Y(H) \longrightarrow0. \] Now \(H^0(X,\OO_X(K_X))=0\): an effective canonical divisor would have nonnegative intersection with a general fibre, whereas \(K_X\cdot f=-2\). Also \[ H^1(X,\OO_X(K_X)) \cong H^1(X,\OO_X)^\vee =0 \] by Serre duality, because \(X\) is ruled over \(\PP^1\) and \(H^1(X,\OO_X)=0\). The cohomology sequence therefore makes the restriction map an isomorphism. By step 4 its target is \(H^0(Y,K_Y)\).
Moreover, \[ h^0(X,\OO_X(H)) = h^0(\PP^1,\OO(n))+h^0(\PP^1,\OO(n-e)) = 2n-e+2 = d+2. \] Thus the given embedding \(X\hookrightarrow\PP^{d+1}\) uses all sections of \(H\). Its restriction to \(Y\) is therefore defined by the complete canonical system \(|K_Y|\), as claimed.
7. For every integer \(g\ge4\), parameters \(n>e\ge0\) satisfying thehypotheses can be chosen with\[ 2n-e=g-2. \]
Put \(d=g-2\ge2\). If \(d\) is even, take \[ e=0, \qquad n=\frac d2. \] Then \(n>e\), \(2n-e=d\), and \(n\ge2e-2\).
If \(d\) is odd, then \(d\ge3\); take \[ e=1, \qquad n=\frac{d+1}{2}. \] Then \(n>1=e\), \(2n-e=d\), and \(n\ge0=2e-2\). Thus steps <1>1–<1>6 produce a canonical curve of genus \(g=d+2\) in every genus \(g\ge4\).
8. Every curve \(Y\) produced above carries a \(g^1_3\) and isnonhyperelliptic.
The ruling restricts to a morphism \[ \pi|_Y:Y\longrightarrow\PP^1 \] of degree \[ Y\cdot f = D\cdot f =3. \] Therefore the pullback of \(|\OO_{\PP^1}(1)|\) is a base-point-free \(g^1_3\) on \(Y\).
By step 6 the complete canonical map of \(Y\) is an embedding. A hyperelliptic curve of genus at least \(2\) has canonical map of degree \(2\) onto a rational normal curve, so \(Y\) is not hyperelliptic.
9. Q.E.D.
Steps <1>1–<1>6 construct the required nonsingular canonical curve of genus \(d+2\). Steps <1>7–<1>8 give, for every \(g\ge4\), a nonhyperelliptic genus-\(g\) curve carrying a \(g^1_3\).