# Nonsingular curves on a quadric cone in $\mathbb{P}^3$.

This is Exercise V.2.9 in Hartshorne's Algebraic Geometry:

Exercise V.2.9. Let $$Y$$ be a nonsingular curve on a quadric cone $$X_0\subset\mathbb{P}^3$$. Show that either $$Y$$ is a complete intersection of $$X_0$$ with a surface of degree $$a\geq 1$$, in which case $$\deg Y =2a, g(Y) = (a -1)^2$$, or, $$\deg Y$$ is odd, say $$2a + 1$$, and $$g(Y) = a^2 -a$$.

Now I have showed that if $$\deg Y =2a$$, then $$Y$$ is a complete intersection of $$X_0$$ with a surface of degree $$a$$, hence $$g(Y) = (a -1)^2$$. But I got stuck at the case when $$\deg Y$$ is odd:

Let $$\deg Y=2a+1$$. After blowing up the vertex $$v$$ of $$X_0$$, we get a ruled surface $$\pi:X\to X_0$$ such that $$C_0=\pi^{-1}(v)$$. Our aim is to use adjunction formula to calculate $$g(Y)$$. Hence we need to express $$Y':=\pi^{-1}(Y)$$ in terms of $$C_0$$ and $$f$$, up to numerical equivalence. So we let $$Y'\sim_{num}a'C_0+bf$$. Let $$H$$ be the hyperplane section of $$X_0$$, we can easy to see that $$H':=\pi^{-1}H\sim_{num}C_0+2f$$. Hence $$2a+1=H'\cdot Y'=b$$. Hence $$Y'\sim_{num}a'C_0+(2a+1)f$$. I claim that $$a'=a+1$$, then as $$K\sim_{num}-2C_0-4f$$ we get $$g(Y)=a^2-a$$ by adjunction formula.

My question: But why $$a'=a+1$$? Actually I don't know how to compute $$a'$$!

(How can I get $$a'=a+1$$? Using the method of undetermined coefficients, we get $$a'=a$$ or $$a+1$$. And consider the ruled line of $$X_0$$ we get $$a'=a+1$$.)

Thank you for your any help!!!

The key is to realize how you can compute this from the situation in $$\Bbb P^3$$. Let $$P_0$$ be the cone point and let $$H_0$$ be a hyperplane missing the cone point. Then the projection from $$P_0$$ to $$H_0\cap X_0\cong\Bbb P^1$$ gives a map $$Y\to\Bbb P^1$$, which if $$\deg Y>1$$ is a nonconstant map of curves, hence finite of some degree. Considering a hyperplane $$H$$ through the cone point which isn't tangent to $$Y$$ at the cone point, we see that we must have that $$H\cap Y$$ is $$2a+1$$ points counted with multiplicity, and that there must be the same number of points counted with multiplicity on the two lines $$L_1,L_2$$ which make up $$H\cap X_0$$. So $$Y$$ must pass through the vertex and there must be $$a$$ other points on each of $$L_1$$ and $$L_2$$ counted with multiplicity. This tells us that we must have $$Y'.f=a$$ and $$Y'.C_0=1$$, so we can use the intersection theory on $$X$$ (recall $$C_0^2=-2$$, $$C_0.f=1$$, and $$f^2=0$$) to see that $$Y'\equiv aC_0+(2a+1)f$$ by writing $$Y'\equiv tC_0+uf$$ and then solving for $$t,u$$ based on those equations.
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