Let $X$ be a scheme over an algebraically closed field $k$ and let $\operatorname{Aut}(X)$ denote the functor sending a $k$-scheme $T$ to the group $\operatorname{Aut}\_T(X \times\_k T)$ of automorphisms of $X \times\_k T$ over $T$. 


My goal is to have a better grasp of the equality $\operatorname{Lie}(\operatorname{Aut}(X))= H^0(X, \mathcal{T} X)$.  

Therefore I am trying to work through the example where $X = \mathbb{P}\_k^1$ so that $\operatorname{Aut}(X)= PGL(2,k)$. 


The global sections of $\mathcal{T} X$ are are of the form $a\_0 \partial\_z + a\_1 z \partial\_z+ a\_2 z^2 \partial\_z$ where $z=v/u$ is a choice of homogeneous coordinates on $X$. On the other hand, I know every $\phi \in \operatorname{Aut}\_k(X)$ is given by the following map of $k$-algebras. 
$$ z \mapsto \frac{az+b}{cz + d}.$$


What is the identification between the global sections of $\mathcal{T} X$ and the $k$-algebra maps $z \mapsto \frac{az+b}{cz + d}$? 


Solving for the integral curve I end up with the equation $z'(t) = a\_0 + a\_1 z(t) + a\_2z^2(t)$. If $a\_0=0$, this would be a Bernoulli differential equation and I can solve it to find $z(t)= \frac{a\_1 z\_0 e^{a\_1 t}}{a\_1 -a\_2 z\_0 e^{a\_1 t}}$. I think that this corresponds to the $k$-algebra map $z \mapsto (a\_1 a^\*) z /(a\_1 - a\_2 a^\* z )$ where $a^\* \in k^\*$. This is close but not exactly right. 


However, the affine subset itself has automorphisms given by $z \mapsto \alpha + \beta z$. If I compose these maps with the maps $z \mapsto (a\_1 a^\*) z /(a\_1 - a\_2 a^\* z )$ I get from integrating the tangent space I do get the Mobius tranformation. Is this the correct approach? 



Lie Algebra of Automorphism Group of $\mathbb{P}_k^1$
Elements of $\operatorname{Lie}(\operatorname{Aut}(X))$ are not $k$-algebra maps, but rather maps over the ring $k[\epsilon]/(\epsilon^2)$ that reduce to the identity $k$-algebra map under $\epsilon\mapsto 0$. For $X=\mathbb{P}^1$, such maps can be identified with elements of the kernel of $PGL\_2(k[\epsilon]/(\epsilon^2))\to PGL\_2(k)$, or in other words maps
$$
z\mapsto \frac{(1+a\epsilon)z+b\epsilon}{c\epsilon z + (1+d\epsilon)},
$$
with $a$, $b$, $c$, $d\in k$. We can compute
$$
\frac{(1+a\epsilon)z+b\epsilon}{c\epsilon z + (1+d\epsilon)}=\big((1+a\epsilon)z+b\epsilon\big)\big(1-d\epsilon-c\epsilon z\big)=z+\epsilon\big( b+(a-d)z-cz^2 \big),
$$
and the vector field corresponding to the map above is $b\partial\_z+(a-d)z\partial\_z-cz^2\partial\_z$.


