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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!!!

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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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  • $\begingroup$ Thank you very much!!! $\endgroup$ Jun 9, 2023 at 7:11

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