How to show that $\Bbb{P}^n$ is birational to $\Bbb{A}^n$ but they are not isomorphic?

Question

Let $\Bbb{P}^n$ be the projective $n$-space and $\Bbb{A}^n$ the affine $n$-space. It is said that $\Bbb{P}^n$ is birational to $\Bbb{A}^n$ but they are not isomorphic. In the case of $\Bbb{P}^1$ and $\Bbb{A}^1$. I think that we can define rational maps $\varphi: \Bbb{P}^1 \to \Bbb{A}^1$ by $\varphi(x_0 : x_1) = x_1/x_0$ if $x_0 \neq 0$ $\varphi(x_0:x_1)= 0$ if $x_0=0$. Define $\psi: \Bbb{A}^1 \to \Bbb{P}^1$ by $\psi(x) = (1:x)$ if $x\neq 0$ and $\psi(x)=(0:1)$ if $x = 0$. Then $\psi$ and $\varphi$ are inverse of each other. Therefore $\Bbb{P}^1$ and $\Bbb{A}^1$ are birational. Is this true? How to show other cases and the fact that they are not isomorphic? Thank you very much.

Answer 1

There is something wrong with your attempt in trying to show $\Bbb{P}^1$ is birational to $\Bbb{A}^1$. Remember a birational map is only defined on an open subset of your space and not necessarily on all of $\Bbb{P}^1$. There are issues with the map $\Bbb{P}^1 \to \Bbb{A}^1$ you defined above. First it is not injective because $(0,1)$ and $(1,0)$ are both mapped to the same point (so how can there be an " inverse"?).

In fact your map $\Bbb{P}^1 \to \Bbb{A}^1$ cannot possibly be a morphism! Let's assume we are working over $\Bbb{C}$ and argue using complex analysis. If we had a morphism $\varphi : \Bbb{P}^1_{\Bbb{C}} \to \Bbb{C}$ this has to be bounded (because $\Bbb{P}^1_{\Bbb{C}}$ is compact in the analytic topology) and restricting to a morphism $$\varphi|_{\Bbb{P}^1\setminus \{\infty\} = \Bbb{C}} : \Bbb{C} \to \Bbb{C}$$

we get a bounded entire function on $\Bbb{C}$ which must be constant by Liouville's Theorem. The restriction of $\varphi$ to the open dense subset $\Bbb{P}^1 \setminus \{\infty\} = \Bbb{C}$ is constant thus is globally constant. But your map $\varphi$ is not constant!

Here are three proofs that $\Bbb{P}^1$ cannot be isomorphic to $\Bbb{A}^1$.

Proof 1: On $\Bbb{A}^1$ the global sections are $k[x]$ while on $\Bbb{P}^1$ the global sections are the constants. To see this let us recall that the scheme $\Bbb{P}^1$ is obtained by gluing two copies of $\operatorname{Spec} k[x]$ and $\operatorname{Spec} k[y]$ via the isomorphism $x \mapsto 1/y$ on $\Bbb{A}^1 \setminus \{0\}$. Any global section on $\Bbb{P}^1$ restricts to local sections $f(x)$ and $g(y)$ which must agree on the intersection. Viz, $f(1/y) = g(y)$. Immediately this implies that $g,f$ have to be constant so that $H^0(\Bbb{P}^1_k,\mathcal{O}_{\Bbb{P}^1_k}) = k$. In fact this generalizes to show that $H^0(\Bbb{P}^n_k, \mathcal{O}_{\Bbb{P}^n_k} ) = k$ so $\Bbb{P}^n \not\cong \Bbb{A}^n$ for any $n\in \Bbb{N}$.

<p><strong>Proof 2:</strong> On $\Bbb{P}^1$ there are an integer's worth of line bundles  given by all the $\mathcal{O}(n)$'s while on $\Bbb{A}^1$ every line bundle is trivial.</p>

<p><strong>Proof 3:</strong> The $\Bbb{P}^n_k$ is proper over $\operatorname{Spec} k$ while $\Bbb{A}^n_k$ is never proper over $\operatorname{Spec} k$. I believe this is true with $k$ replaced by any ring $A$ because properness is stable under base change.</p>