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According to Hartshorne (exercise IV.6.1), a rational curve of degree 4 in $\mathbb{P}^3$ is contained in a unique smooth quadric surface. If this is the case, then it must define a divisor on it. My question is, what is the type of this divisor? It looks like it must be of type $(1,3)$ (by the following example) but I do not know how to show this in general.

One can work out an example where the quadric is $\{xw-yz=0\}$ and the quartic is $$\{xw-yz=0,wy^2-xz^2=0,z^3-w^2y=0,y^3-x^2z=0\}.$$

This intersects $\{x=0,z=0\}$ with multiplicity $1$ and $\{x=0,y=0\}$ with multiplicity $3$, so we get type $(1,3)$ (which is rather surprising, because I thought we must get $(2,2)$ by symmetry). In general, I can see that it can only be of type $(2,2)$ or $(1,3)$, but do not see how to rule out $(2,2)$.

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2 Answers 2

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Divisors of bi-degree (2,2) come from restricting generic quadrics in $\mathbb{P}^3$ to the quadric containing the curve. So you would need at least two quadrics in $\mathbb{P}^3$ that contain the curve to realize it as a (2,2) divisor, and there is only one as you showed in the exercise. In fact, you can apply adjunction to show that a (non-singular) complete intersection of two quadrics in $\mathbb{P}^3$ has a trivial canonical bundle, hence, it cannot be rational.

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Apply adjunction as in Hartshorne's example V.1.5.2 to obtain $g = 2\cdot 2-2-2+1 = 1 \neq 0$ for a curve of type $(2,2)$ on $Q$.

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  • $\begingroup$ Thanks! I'll wait a bit more to see if someone can offer a more elementary proof, and if not I'll accept this. $\endgroup$
    – adrido
    Commented Nov 5, 2014 at 0:11

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