Intersection of twisted cubics in $\mathbb{P}^3$ Suppose we have two twisted cubics $C_1$, $C_2$ in $\mathbb{P}^3$ such that both of them lie in some cubic surface, which means that $h^0(\mathbb{P}^3, I_{C_1\cup C_2}(3))>0$. I want to show that in this case they intersect.
Suppose that they do not intersect. Then $\mathcal{O}_{C_1\cup C_2}=\mathcal{O}_{C_1}\oplus\mathcal{O}_{C_2}$. Twisting by 3 the exact sequence
$$0\to I_{C_1\cup C_2}\to \mathcal{O}_{\mathbb{P}^3}\to\mathcal{O}_{C_1\cup C_2}\to0$$
and taking cohomologies we obtain
$$0\to H^0(\mathbb{P}^3, I_{C_1\cup C_2}(3))\to H^0(\mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3}(3))\stackrel{f}\to H^0(\mathbb{P}^3, \mathcal{O}_{C_1\cup C_2}(3))\to H^1(\mathbb{P}^3, I_{C_1\cup C_2}(3))\to0.$$
But $h^0(\mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3}(3))=20$ and $h^0(\mathbb{P}^3, \mathcal{O}_{C_1\cup C_2}(3))=2h^0(\mathbb{P}^3, \mathcal{O}_{C_1}(3))=2h^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(9))=20$, so it seems that the map $f$ is an isomorphism, which contradicts the assumption $h^0(\mathbb{P}^3, I_{C_1\cup C_2}(3))>0$. 
Is it true or not that $f$ is an isomorphism? If yes, then how to prove this?
$\textbf{Edit}$
It seems that the approach I gave in the post is not the best possible.
If someone could give a canonical answer based on the different argument it would be welcomed! 
$\textbf{Edit 2}$
I received the following suggestion.
Since $C_1$ and $C_2$ do not intersect, $I_{C_1\cup C_2}=I_{C_1}\otimes I_{C_2}$. For $I_{C_i}$ there is a resolution of the form
$$0\to\mathcal{O}(-3)^{\oplus2}\to\mathcal{O}(-2)^{\oplus3}\to I_{C_i}\to0.$$
Tensoring these two resolution and twisting by 3 we obtain the exact sequence
$$0\to\mathcal{O}(-3)^{\oplus4}\to\mathcal{O}(-2)^{\oplus6}\to\mathcal{O}(-1)^{\oplus9}\to I_{C_1}\otimes I_{C_2}(3)\to0.$$
The first three terms do not have cohomologies, so it seems that the fourth doesn't have too. How to prove this? Should I use some spectral sequence?
 A: It is probably true that your map $f$ is an isomorphism if the curves are chosen "sufficiently generally". But that would be a circular argument. It may well fail in general. The best I can come up with at present is the following suggestion.
Suppose that $C_1$ and $C_2$ have $2$ common bisecant lines $L$ and $M$ that meet in a point $P = L \cap M$. (I don't see any particular reason why that should not happen—maybe you can look for an example? The bisecant lines to $C_1$ cover $\mathbb{P}^3$, and similarly $C_2$.)
Then the condition for cubics to contain $C_1$ and $C_2$ would be linearly dependent: argue first on the $4$ points $L \cap C_1$ and $L \cap C_2$ ($4$ conditions on cubics, as should be). Then argue on $4$ points $P$ and $2$ points of $M \cap C_1$ and one of the points of $M \cap C_2$; cubics through these necessarily pass through the second point of $M \cap C_2$. So the two curves $C_1$ and $C_2$ each imposes $10$ conditions on cubics, but these conditions are not linearly independent.
It is true that twisted cubics $C_1$ and $C_2$ cannot be contained in a cubic surface $S$ that is nonsingular along both. I hope you can follow intersection numbers of curves on a surface. Because (after you resolve singularities if necessary), $C_1^2 = +1$ and $C_2^2 = +1$, so that $C_1 \cdot C_2 = 0$ would contradict the algebraic index theorem. But that argument does not work if $S$ has any singularities along the $C_i$.
