Let $C_1$ and $C_2$ be twisted cubic curves in $\mathbb P^3$. I want to prove that they intersect if and only if they lie in common cubic surface, perhaps singular. The second condition can be reformulated as $h^0(\mathbb P^3, I_{C_1 \cup C_2}(3)) > 0$.
Consider a short exact sequence $$0 \to I_{C_1 \cup C_2}(3) \to \mathcal O_{\mathbb P^3}(3) \to \mathcal O_{C_1 \cup C_2}(3) \to 0,$$ and the corresponding long exact sequence of cohomology (all cohomologies are over $\mathbb P^3$): $$0 \to h^0(\mathbb P^3, I_{C_1 \cup C_2}(3)) \to h^0(\mathcal O_{\mathbb P^3}(3)) \to h^0(\mathcal O_{C_1 \cup C_2}(3)) \to h^1(I_{C_1 \cup C_2}(3)) \to 0.$$
We also know that $h^0(\mathcal O_{\mathbb P^3}(3))=C(6, 3)=20$ and $$h^0(\mathcal O_{C_1 \cup C_2}(3)) \leq 2 h^0(\mathbb P^3, \mathcal O_C(3)) = h^0(\mathbb P^1, \mathcal O_{\mathbb P^1}(9))=20.$$ If cubic curves intersect, then $h^0(\mathcal O_{C_1 \cup C_2}(3)) < 20$, so indeed $h^0(I_{C_1 \cup C_2}(3)) > 0$ and that's all. On the contrary, if they do not intersect, $h^0(\mathcal O_{C_1 \cup C_2}(3)) = 20$, but there still may be $h^0(I_{C_1 \cup C_2}(3))$. Could you help me with it?