Let \(Y\) be a nonsingular curve on a quadric cone \(X_0\) in \(\PP^3\). Show that either
\(Y\) is a complete intersection of \(X_0\) with a surface of degree \(a \geqslant 1\), in which case \(\deg Y=2 a\), \(g(Y)=(a-1)^2\), or,
\(\deg Y\) is odd, say \(2 a+1\), and \(g(Y)=a^2-a\).
Cf. (V, 6.4.1). Hint: Use (2.11.4).
Solution
Let \[ v\in X_0 \] be the vertex, and let \[ \pi:\widetilde X_0\longrightarrow X_0 \] be the blowup of \(v\).
1. The resolution \(\widetilde X_0\) is the Hirzebruch surface \(\FF_2\).If \(C_0\) is its exceptional section and \(f\) a ruling fibre, then\[ C_0^2=-2, \qquad C_0\cdot f=1, \qquad f^2=0, \]and\[ K_{\FF_2}=-2C_0-4f. \]Moreover, if \(H\) denotes the pullback of the hyperplane class of\(X_0\subseteq\PP^3\), then\[ \boxed{H\sim C_0+2f}. \]
The ruled-surface model referred to in the hint identifies the blowup of the vertex of the quadric cone with \(\FF_2\); equivalently, the quadric cone is obtained by contracting the negative section of \(\FF_2\). The intersection numbers and \[ K_{\FF_2}=-2C_0-4f \] are the standard Hirzebruch-surface formulas recorded in FE-TORHIRZ.
Write \[ H\sim rC_0+sf. \] A ruling fibre maps isomorphically to a ruling line of the cone, so \[ H\cdot f=1. \] Hence \(r=1\). Since \(C_0\) is contracted to the vertex, \[ H\cdot C_0=0. \] Thus \[ 0=(C_0+sf)\cdot C_0=-2+s, \] so \(s=2\) and \(H\sim C_0+2f\).
2. Let \(D\subseteq\FF_2\) be the strict transform of \(Y\). Then\[ D\cong Y, \]and\[ D\cdot C_0= \begin{cases} 0,&v\notin Y,\\ 1,&v\in Y. \end{cases} \]
Away from \(v\), the morphism \(\pi\) is an isomorphism, so if \(v\notin Y\) the strict transform is isomorphic to \(Y\) and is disjoint from \(C_0\).
Assume now that \(v\in Y\). Work in an affine chart of \(\PP^3\) centered at \(v\). Since the curve \(Y\) is nonsingular at \(v\), choose a local parameter \(t\) on \(Y\) at \(v\). Among the ambient affine coordinates, at least one, say \(x\), restricts to \[ x=t u \] with \(u\) a unit at \(v\); every other ambient coordinate restricts to a multiple of \(t\). In the \(x\)-chart of the blowup, the strict transform of \(Y\) is therefore still parametrized by \(t\), and the exceptional divisor is cut on it by \(t=0\). Hence the strict transform is nonsingular, the map \(D\to Y\) is an isomorphism, and \(D\) meets the exceptional curve once and transversely. Thus \[ D\cdot C_0=1. \]
3. Write\[ D\sim aC_0+bf. \]Then\[ \boxed{ D\sim \begin{cases} aH,&v\notin Y,\\ aH+f,&v\in Y, \end{cases}} \]for an integer \(a\ge0\); in the first case \(a\ge1\).
Since \[ \Pic(\FF_2)=\ZZ C_0\oplus\ZZ f, \] such integers \(a,b\) exist. Intersecting with a fibre gives \[ D\cdot f=a, \] so \(a\ge0\) because \(D\) is an effective curve distinct from \(C_0\).
By step 2, \[ b-2a = D\cdot C_0 \in\{0,1\}. \] Thus \[ b=2a \] when \(Y\) avoids the vertex, and \[ b=2a+1 \] when \(Y\) passes through it. Using \(H=C_0+2f\) from step 1 gives the two displayed classes. In the first case \(a=0\) would make \(D\sim0\), impossible for a nonempty effective curve, so \(a\ge1\) there.
4. The degree of \(Y\) is\[ \boxed{ \deg Y= \begin{cases} 2a,&v\notin Y,\\ 2a+1,&v\in Y. \end{cases}} \]
The degree is intersection with a hyperplane. Pulling the hyperplane class back to \(\FF_2\) therefore gives \[ \deg Y=D\cdot H. \] From step 1, \[ H^2=(C_0+2f)^2=-2+4=2, \qquad H\cdot f=1. \] Using the two classes from step 3 gives \[ (aH)\cdot H=2a \] and \[ (aH+f)\cdot H=2a+1. \]
5. The genus of \(Y\) is\[ \boxed{ g(Y)= \begin{cases} (a-1)^2,&v\notin Y,\\ a^2-a,&v\in Y. \end{cases}} \]
By step 2, \(D\cong Y\), so it is enough to compute the genus of \(D\) on the nonsingular surface \(\FF_2\). Step 1 gives \[ K_{\FF_2}=-2C_0-4f=-2H. \]
If \(D=aH\), then \[ D^2=2a^2, \qquad D\cdot K_{\FF_2}=-4a. \] Adjunction gives \[ g(Y) = 1+\frac12D\cdot(D+K_{\FF_2}) = 1+a^2-2a = (a-1)^2. \]
If \(D=aH+f\), then \[ D^2=2a^2+2a, \] while \[ D\cdot K_{\FF_2} = -2D\cdot H = -4a-2. \] Hence \[ g(Y) = 1+\frac12(2a^2+2a-4a-2) = a^2-a. \]
6. If \(v\notin Y\), then \(Y\) is the scheme-theoretic completeintersection of \(X_0\) with a surface of degree \(a\).
In this case \(Y\) lies in the nonsingular locus of the normal surface \(X_0\), so it is an effective Cartier divisor on \(X_0\). Since it avoids the center of the blowup, its total transform is its strict transform: \[ \pi^*Y=D. \] By step 3, \[ \OO_{\FF_2}(D) \cong \OO_{\FF_2}(aH) \cong \pi^*\OO_{X_0}(a). \] Thus \[ \pi^*\OO_{X_0}(Y) \cong \pi^*\OO_{X_0}(a). \]
The pullback \[ \pi^*:\Pic(X_0)\longrightarrow\Pic(\FF_2) \] is injective. Indeed, \(X_0\) is normal and \(\pi\) is proper birational, so \[ \pi_*\OO_{\FF_2}=\OO_{X_0}. \] If \(\pi^*L\) is trivial, the projection formula gives \[ L \cong L\tensor\pi_*\OO_{\FF_2} \cong \pi_*\pi^*L \cong \OO_{X_0}. \] Therefore \[ \OO_{X_0}(Y)\cong\OO_{X_0}(a). \]
The canonical section cutting out \(Y\) is consequently a section of \(\OO_{X_0}(a)\). From the hypersurface sequence of the quadric cone, \[ 0 \longrightarrow \OO_{\PP^3}(a-2) \longrightarrow \OO_{\PP^3}(a) \longrightarrow \OO_{X_0}(a) \longrightarrow0, \] and the standard vanishing \[ H^1(\PP^3,\OO_{\PP^3}(a-2))=0, \] the restriction map \[ H^0(\PP^3,\OO_{\PP^3}(a)) \longrightarrow H^0(X_0,\OO_{X_0}(a)) \] is surjective. Lift the section defining \(Y\) to a degree-\(a\) homogeneous form \(F\). Then \[ Y=X_0\cap V(F) \] scheme-theoretically. Hence \(Y\) is a complete intersection of the quadric cone with a surface of degree \(a\).
7. The two alternatives in the statement are exhaustive.
If \(Y\) avoids the vertex, steps <1>4–<1>6 give \[ \deg Y=2a, \qquad g(Y)=(a-1)^2, \] with \(a\ge1\), and show that \(Y\) is the required complete intersection.
If \(Y\) passes through the vertex, steps <1>4–<1>5 give \[ \deg Y=2a+1, \qquad g(Y)=a^2-a. \] These are exactly the two cases asserted.
8. Q.E.D.
Step 7 proves the required classification.