I'm thinking about the following problem: If I take a general point $p \in \mathbb{P}^1$ out of the projective line, is $\mathbb{P}^1 – \{ p \}$ isomorphic to the affine space $\mathbb{A}^1$?
I ask this because if $p = [1, 0] \in \mathbb{P}^1$, the map $[x, 1] \mapsto x$ gives an isomorphism $\mathbb{P}^1 – \{[1, 0] \} \cong \mathbb{A}^1$. I guess that this should be true somehow for a general point $p$, but I can't quite get my head around the map that I need to define.
Best Answer
If you have a 2 by 2 invertible matrix $M$ over $k$, then $M$ induces an isomorphism of $\mathbb P^1$ with itself. If you have a general point $[a:b]$ in $\mathbb P^1$, write down a matrix that sends $[a:b]$ to $[1:0]$ so that you have $\mathbb P^1 - [a:b] \cong \mathbb P^1 - [1:0]$. Then apply your isomorphism above.
By the way, your isomorphism above is not quite right because it's not independent of representative. For example $[2x:2]$ would go to $2x$, not $x$. To get around this, define your isomorphism $\mathbb{P}^1 - \{[1: 0] \} \to \mathbb{A}^1$ by sending $[x:y]$ to $x/y$.