# $\mathrm{Aut}(\mathbb P)$ is isomorphic to $\mathrm{ Aut} (k(t) )$

$\def\Aut{\mathrm{Aut}}$I want to prove that the automorphism group of $\mathbb P^1$ it's isomorphic with the Moebius transformation with coefficients over the obvious field. I proved that the automorphism of $k(t)$ are of this form, If I prove that: $$\Aut(\mathbb P)\text{ is isomorphic to }\Aut (k(t) )$$ I'll be done. How can I do it, only using basic facts?

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What's your definition of $\mathbb{P}^1$ and what's your definition of a morphism between projective varieties? –  Qiaochu Yuan Jun 18 '12 at 2:40
If $f:\mathbb P^1\to \mathbb P^1$ is your automorphism and if $f({\infty})=b\neq \infty$, consider $g(z)=\frac {1}{z-b}$.
Then $g\circ f$ sends $\infty$ to $\infty$. Thus you may assume (replacing $f$ by $g\circ f$ if necessary) that $f({\infty})={\infty}$.
You are then reduced to showing that the restricted automorphism $f\mid \mathbb A^1: \mathbb A^1 \to \mathbb A^1$ is given by $z\mapsto az+b$, which result you can probably handle.