Is the projective line minus one point always isomorphic to the affine space?

Question

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.

Answer 1

Yes, the projective line minus any point is the affine line.

I'll suppose the underlying field is $\mathbb F$, and that we are defining $\mathbb P$ to be the set of one dimensional subspaces of $\mathbb F^2$. The affine line is just the field $\mathbb A=\mathbb F$. Let me know if your definitions are substantially different.

For any point $p'\in\mathbb P$ let $p$ be some non-zero point on the corresponding line in $\mathbb F^2$, and let $q\in\mathbb F^2$ be some linearly independent point. By linear independence, the map $\mathbb A\rightarrow\mathbb F^2$ by $x\mapsto xp+q$ never hits zero. In fact it's easy to prove that it hits every one dimensional subspace exactly once, except for the one through $p$. So it gives an injection $\mathbb A\rightarrow \mathbb P$ hitting every point except $p'$.

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In higher dimensions you get the affine space from the projective space by taking away any subspace of dimension one less:

$$\mathbb P^n-\mathbb P^{n-1}=\mathbb A^n$$

(In particular geometers sometimes think of the projective plane, $\mathbb P^2$, as being the usual plane along with the "line at infinity":

$$\mathbb P^2=\mathbb A^2+\mathbb P^1)$$

Of course we can iterate this to express any projective space as a union of affine spaces:

$$\mathbb P^n=\mathbb A^n + \mathbb A^{n-1} + \dots + \mathbb A^0$$

Answer 2

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 $x/y$.