I’ve just started studying projective geometry and this is the first result we are proving. By the way, $V$ is a finite-dimensional vector space, and I write $\mathbb{P}(V)$ for the projective space associated with $V$. I also define a projective line in $\mathbb{P}(V)$ as $\mathbb{P}(U)$ for any subspace $U$ of $V$ with dimension $2$.
Theorem I’m trying to show:
Through any two distinct points $P$ and $Q$ in $\mathbb{P}(V)$, there is a unique projective line $L$.
My attempt
Let $P=[p]\neq Q=[q]$ in $\mathbb{P}(V)$. Then the vectors $p,q$ are linearly independent. So $\langle p,q \rangle$ is a 2-dimensional subspace of $V$, hence $\mathbb{P}\langle p,q \rangle$ is a projective line in $\mathbb{P}(V)$.
Now I’m struggling to show explicitly that this projective line, call it $L$, contains $P,Q$ and that it is unique in doing so.
Well, $\langle p,q \rangle$ has 1-dimensional subspaces $\langle p\rangle$ and $\langle q \rangle$, so $\mathbb{P}\langle p,q\rangle = \{\langle p\rangle, \langle q\rangle \}$.
Am I right in saying that $\langle p\rangle=[p]$, and therefore $L$ contains $P,Q$.
I have no idea how to prove uniqueness.
Thank you
From what I understand, I believe you are defining $\mathbb{P}(V)=V\setminus\{0\}/k^\times$.
The answer $\mathbb{P}(\langle p,q\rangle)$ is correct.
The notation $[p]$ denotes the equivalence class associated to $p$ in $\mathbb{P}(V)$. So, I would not say that $[p]=\langle p\rangle$, but rather that $[p]$ is the image of $\langle p\rangle\setminus\{0\}$ in the natural map $V\setminus \{0\}\rightarrow \mathbb{P}(V)$.
Edit: The natural map is the one that takes a non-zero vector $v\in V$ to its equivalence class $[v]\in\mathbb{P}(V)$.
From this it should be clear to see that $[p],[q]\in \mathbb{P}(\langle p, q\rangle)$ becasue $\langle p\rangle\subset \langle p, q\rangle$ and $\langle q\rangle\subset \langle p, q\rangle$.
Next, the statement $\mathbb{P}(\langle p, q\rangle)=\{[p],[q]\}$ is not exactly correct. For instance $[p+q]\in \mathbb{P}(\langle p, q\rangle)$ (which is not equal to the other two points).
The last thing to show is the uniqueness. Suppose $W\subset V$ is also a 2-dimensional vector space such that $[p],[q]\in \mathbb{P}(W)$. Then, $p,q\in W$ and hence $\langle p, q\rangle\subset W$. Comparing dimensions tells you that $W=\langle p,q \rangle$.
Hope this clears any confusion.