If two projective lines both intersect four given projective lines, must the two lines be parallel on an affine open? L.S.,
I am trying to solve an exercise of my algebraic geometry course, which is as follows. 
Given four projective lines in $\mathbb{P}^3$, show that the number of lines intersecting them all is either one, two or infinite. 
I managed to show that there must be a line intersecting them all! Using plucker coordinates and a dimension argument. But now I am stuck in showing that if there aren't infinitely many lines intersecting them all, there can be at most two. Since I am stuck I thought that it could be helpful to maybe show first that two lines both intersecting the given four lines must be parallel on some affine open. But I am also stuck on that, and I don't even now if it is true! Does anyone maybe know whether this is true or not, and maybe give me a starting point to showing that? Then hopefully that'll help me show that the number of lines intersecting all four lines is either one, two or infinite. 
Many thanks!
Willem
 A: If I remember correctly, this classical fact can be analyzed as follows. Recall that $2$ lines $L_1$ and $L_2$ in $\mathbf P^3(\mathbf C)$ are skew if they are not contained in a projective plane in $\mathbf P^3(\mathbf C)$. If they are not skew then they intersect.
Let $L_1,L_2,L_3,L_4$ be $4$ distinct lines in $\mathbf P^3(\mathbf C)$. The easy case to analyze is the case where at least $2$ of the $4$ lines are not skew. In that case the number of lines intersecting the lines $L_1,L_2,L_3,L_4$ is either equal to $1$ or is infinity. 
The interesting case is the case where all lines $L_1,L_2,L_3,L_4$ are pairwise skew. For any $3$ pairwise skew lines in $\mathbf P^3(\mathbf C)$ there is a unique quadric surface containing them. Let $Q$ be the quadratic surface containing $L_1,L_2,L_3$. This quadric surface is necessarily non degenerate, otherwise the $3$ lines would intersect. The surface $Q$ has, therefore, two rulings $\rho_1$ and $\rho_2$. The lines $L_1,L_2,L_3$ are all lines in the same ruling since, again, otherwise they would intersect. We may assume that $L_1,L_2,L_3\in\rho_1$. If the line $L_4$ is contained in the surface $Q$ then it belongs to one of the $2$ rulings. If $L_4\in\rho_1$ then there are infinitely many lines intersecting the $4$ lines, namely all the lines of the other ruling $\rho_2$. If $L_4\in\rho_2$ then $L_4$ would intersect the lines $L_1,L_2,L_3$, which would contradict the fact that $L_4$ is skew with the lines $L_1,L_2,L_3$. The only case that is left, is the case where $L_4$ is not contained in $Q$. Then it intersects $Q$ in two points $P_1$ and $P_2$. The two lines $M_1,M_2$ of the ruling $\rho_2$ that contain $P_1,P_2$, respectively, intersect the lines $L_1,L_2,L_3,L_4$. Any other line $M$ that intersects $L_1,L_2,L_3,L_4$ is equal to $M_1$ or $M_2$. Indeed, since $M$ intersects $L_1,L_2,L_3$, it intersects the quadric $Q$ in at least $3$ points. By Bezout, $M$ is contained in $Q$. Since $M$ intersects $L_1,L_2,L_3$, the line $M$ belongs to the ruling $\rho_2$. Since it intersects $L_4$, it passes through $P_1$ or $P_2$. Hence, $M=M_1$ or $M=M_2$. Therefore, in this case we have $1$ or $2$ lines intersecting $L_1,L_2,L_3,L_4$ depending on whether $M_1=M_2$ or not.
