Automorphisms of projective line how does one find the set of Automorphisms of the complex projective line?
PS: no scheme theory is assumed.
 A: 0) I'll use coordinates $(t:z)$ on the projective line $\mathbb P^1(\mathbb C)$, with the embedding $\mathbb C\to \mathbb P^1(\mathbb C)$ given by $z\mapsto (1:z)$  and with $\infty = (0:1)$.  
1) Now, given an automorphism $f:\mathbb P^1(\mathbb C)\to \mathbb P^1(\mathbb C)$, we may assume $f(\infty)=\infty$: if this is not the case and if $f(\infty)=a=(1:a)\in \mathbb C$, we consider the automorphism $g:\mathbb P^1(\mathbb C)\to \mathbb P^1(\mathbb C): (t:z)\mapsto (z-at:t)$ (which maps $a=(1:a)$ to $(0:1)=\infty$) and the new automorphism $g\circ f$  will satisfy $(g\circ f)(\infty)=\infty$.
2) If $f(\infty)=\infty$, we may consider  the restriction $f\mid \mathbb C=f_0:\mathbb C\to\mathbb C$.
It is given by a polynomial $f_0(z)=P(z)$ and since it is a bijection it must have degree $1$ (by the fundamental theorem of algebra, say) : $P(z)=bz+c=(1:bz+c)$ .  
3) Taking into  account   the reduction in 1), we see that the original automorphism must have the form $$(t:z) \mapsto (\gamma z+\delta t:\alpha z+\beta t)   $$ with $\alpha,\beta,\gamma,\delta \in \mathbb C$ and $\alpha\delta-\beta\gamma\neq0$.
Conversely, it is clear that such a formula defines an automorphism of   $\mathbb P^1(\mathbb C)$.
With the obvious traditional abuse of notation we just write this as the Möbius transformation $$f(z)=\frac {\alpha z+\beta}{\gamma z+\delta}     $$ 
5) Summary $$Aut(P^1(\mathbb C)) =PGl_2(\mathbb C)=Gl_2(\mathbb C)/ \mathbb C ^*            $$
