TITLE: Unique line in $\mathbb{P}^3$ through a point $p$ and intersecting two disjoint lines
QUESTION [7 upvotes]: I'm a bit stuck with this exercise from a script I'm reading, and I'm not very familiar with projective $n$-space yet. The problem:

Let $L_1$ and $L_2$ be two disjoint lines in $\mathbb{P}^3$, and let $p\in\mathbb{P}^3\smallsetminus(L_1\cup L_2)$. Show that there is a unique line $L\subseteq\mathbb{P}^3$ meeting $L_1$, $L_2$, and $p$ (i.e. such that $p\in L$ and $L\cap L_i\neq\varnothing$ for $i=1,2$).

To be honest, I already have a problem with the term 'line'. As I take it, a line in $\mathbb{P}^3$ should be something cut out by two degree-1 polynomials (homogeneous, since it wouldn't be well defined otherwise, right?). But what are disjoint lines in projective space? As far as I understood it, two distinct lines should always intersect in exactly one point there, so how can they be disjoint? Can it be that these two statements mean a different kind of 'line'?
Any explanation of this, hints, or even a solution would be very appreciated. Thanks in advance!

REPLY [4 votes]: Let $L_1$ and $L_2$ be two disjoint lines in $\mathbb{P}^3$, and let $p\in\mathbb{P}^3\setminus(L_1\cup L_2)$. Show that there is a unique line $L\subseteq\mathbb{P}^3$ such that $p\in L$ and $L\cap L_i\neq\varnothing$ for $i=1,2$.

One possibility is to view this as a linear algebra problem in $k^4$.
Recall (cf. Lemma 6.17 in Gathmann's notes) that we have a one-to-one correspondence
\begin{align}
\{\text{Cones in $\mathbb{A}^{n+1}$}\}&\longleftrightarrow \{\text{Projective varieties in $\mathbb{P}^n$}\}\\
Y&\longmapsto \mathbb{P}(Y):=\{(a_0:\cdots:a_n) : (a_0,\ldots,a_n)\in Y\setminus\{\mathbf{0}\}\}
\end{align}
with inverse given by $X\mapsto C(X):=\{\mathbf{0}\}\cup\{(a_0,\ldots, a_n) : (a_0:\cdots:a_n)\in X\}$ for $X\subseteq\mathbb{P}^n$.
Using this correspondence, the problem can be translated from the language of projective geometry to the language of linear algebra as follows:

The lines $L_1$ and $L_2$ correspond to two-dimensional linear subspaces $U=\mathrm{span}_k\{\mathbf{u}_1,\mathbf{u}_2\}=C(L_1)$ and $V=\mathrm{span}_k\{\mathbf{v}_1,\mathbf{v}_2\}=C(L_2)$ of $k^4$, respectively.

The point $p$ corresponds to a one-dimensional linear subspace $W=\mathrm{span}_k\{\mathbf{w}\}=C(\{p\})$.

$L_1\cap L_2=\varnothing$ corresponds to $U\cap V=\{\mathbf{0}\}$.

$p\not\in L_1\cup L_2$ corresponds to $U\cap W=V\cap W=\{\mathbf{0}\}$.

We want to show that there is a unique two-dimensional subspace $Z$ of $k^4$, such that $W\subseteq Z$, $U\cap Z\neq \{\mathbf{0}\}$ and $V\cap Z\neq \{\mathbf{0}\}$; the desired claim then follows with $L=\mathbb{P}(Z)$.


One way to do this (I leave it for you to check the details!) is to form the hyperplanes $H_1=\mathrm{span}_k\{\mathbf{u}_1,\mathbf{u}_2,\mathbf{w}\}$ and $H_2=\mathrm{span}_k\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{w}\}$, and then set $Z=H_1\cap H_2$.