Write $\mathbb{P}^3_{\mathbb{C}}$ as a union of disjoint lines

Question

Is there a set $\Gamma=\{L \subseteq \mathbb{P}^3_{\mathbb{C}}: L \textrm{ is a projective line}\}$ such that every point $p \in \mathbb{P}^3_{\mathbb{C}}$ lies on exactly one line $L_p \in \Gamma$?

I know that this is possible for the real numbers, so I wonder wonder if it is correct for the complex numbers too.

Answer 1

After thinking a little bit more, I found the answer by myself. Sorry for asking this question too soon. The answer is yes: Let $H$ be the division ring of Quaternions. Then $H^2$ is a four dimensional $\mathbb{C}$-vector space. For each point $0 \neq p=(p_1,p_2) \in H^2$ consider $L_p=\{(z \cdot p_1, z \cdot p_2) \in H^2: \, z \in H\}$. It is straight forward to check that the $L_p$ are $\mathbb{C}$-linear subspaces of dimension two and $L_p \cap L_q \neq 0$ implies $L_p=L_q$.