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I am trying to solve the exercise IV.6.1 from Hartshorne's "Algebraic geometry":

A smooth rational curve of degree 4 in $\mathbb{P}^3$ is contained in a unique quadric surface $Q$, and $Q$ is necessarily nonsingular.

Let $X$ be such a curve. We have the following exact sequence $$0\to\mathcal{I}_X\to\mathcal{O}_{\mathbb{P}^3}\to\mathcal{O}_X\to0.$$

Analysing the proof of the Theorem 6.4 from Hartshorne's book it is not hard to see that $\text{h}^0(\mathcal{I}_X(2))\geqslant1$. Thus $X$ is contained in at least one quadric surface. If $X$ is an intersection of two quadrics then we have the following exact sequence $$0\to\mathcal{O}_{\mathbb{P}^3}(-4)\to\mathcal{O}_{\mathbb{P}^3}(-2)\oplus\mathcal{O}_{\mathbb{P}^3}(-2)\to\mathcal{I}_X\to0.$$ The first exact sequence gives $\text{h}^1(\mathcal{O}_X)=\text{h}^2(\mathcal{I}_X)$ and the second one gives $\text{h}^2(\mathcal{I}_X)=\text{h}^3(\mathcal{O}_{\mathbb{P}^3}(-4))=1$. Thus $\text{h}^1(\mathcal{O}_X)=1$, which contradicts the assumption that $X$ is rational. Thus $X$ is contained in a unique quadric surface $Q$.

Now I don't know how to prove that $Q$ is smooth. What is the idea of the proof?

vitaliy
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  • Use: Chapter $I,$ Ex. $5.12$ and Chapter $IV, 6.4.1 (d).$ – Krish May 07 '15 at 20:56
  • Okay. Suppose that $Q$ is a quadric cone in $\mathbb{P}^3$ that contains $X$. I don't understand why $X$ is a complete intersection of $Q$ with a surface of degree 2. – vitaliy May 07 '15 at 22:39
  • You have already proved that $X$ is contained in a unique quadric $Q.$ By a change of variables, we can assume $Q$ is of the form $z_0^2=0, z_0^2+z_1^2=0, z_0^2+z_1^2+z_2^2=0, z_0^2+z_1^2+z_2^2+z_3^2=0.$ In the first two cases $X$ will be of degree one. In the third case, the genus of $X$ can be computed as 1. That leaves the only possibility, the fourth one, which is non-singular. – Krish May 09 '15 at 14:25

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