Cohomology of rational quartic in $\mathbb{P}^3$ I have to do this exercise. 
Let $X\subseteq\mathbb{P}^3$ a rational curve of degree $4$. Show that
$$H^1(\mathcal{O}_X(1))=0=H^1(\mathcal{I}_X(2))$$
I tried to look at $X$ as closed immersion by linear system, but Hartshorne says that the system is not complete so I'm not sure how to proceed. 
 A: I am going to assume that $X$ is smooth: then it is also nondegenerate, and it is given by the embedding of $\mathbb{P}^1$ via a base point free linear system $V\subseteq H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(4))$ of dimension $\dim V =4$. In particular, this tells us that $H^1(\mathcal{O}_X(1))=H^1(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(4))=0$. For the ideal sheaf cohomology, consider the exact sequence
$$ 0 \to \mathcal{I}_X(2) \to \mathcal{O}_{\mathbb{P}^3}(2) \to \mathcal{O}_X(2) \to 0 $$ 
then this induces an exact sequence in cohomology
$$ 0 \to H^0(\mathcal{I}_X(2)) \to H^0(\mathcal{O}_{\mathbb{P}^3}(2)) \to H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(8)) \to H^1(\mathcal{I}_X(2)) \to 0 $$
so that we need to show that the map $H^0(\mathcal{O}_{\mathbb{P}^3}(2)) \to H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(8))$ is surjective. Since $h^0(\mathcal{O}_{\mathbb{P}^2}(2))=10$ and $h^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(8))=9$, this is equivalent to proving that $h^0(\mathcal{I}_X(2))=1$. Suppose that $X\subseteq Q_1\cap Q_2$, where $Q_1$ and $Q_2$ are two distinct quadrics (note that they must be smooth since $X$ is nondegenerate). Then, comparing the Hilbert polynomials, we have $\chi(\mathcal{O}_X(n)) \leq \chi(\mathcal{O}_{Q_1\cap Q_2}(n))$ for $n>>0$. However we can compute $\chi(\mathcal{O}_X(n))=4n+1$ and $\chi(\mathcal{O}_{Q_1\cap Q_2}(n))=4n$ (for this one we can use the Koszul complex or the adjunction formula) which gives a contradiction.  
