I'm reading Klaus Hulek's algebraic geoemtry and there is something that I can't understand.
Here it says that if {p,q} is a counterexample with minimum max{deg p , deg q}, then it can be assumed that the 4 rations are 0, 1, infinite, λ What I understand is that We can apply the linear transformation to {p, q} and change them to {p' , q'} which have the same max{deg p' , deg q'} as p,q. And p' , q' , p'-q', p'-λq' are squares. Is my understanding correct? If so, how to guarantee that max{deg p , deg q} and max{deg p' , deg q'} are same?
Addition) I have one more question. Thanks to the answer below, I now can see that max{deg p , deg q} and max{deg p' , deg q'} are same. But one thing remains that how I can "find" those p' , q' such that p' , q' , p'-q', p'-λq' are squares? Could anyone explain this also?
Clearly, the degrees of both $p'$ and $q'$ are $\leq \max(\deg p, \deg q)$. If $\deg p \neq \deg q$, then it is also clear that the degree of both $p'$ and $q'$ is equal to $\max(\deg p, \deg q)$. So the case that requires some extra care is when $p$ and $q$ have the same degree, say $p = a t^d + $ lower order terms and $q = b t^d + $ lower order terms. Then $p'$ will have degree $d$ unless $a \alpha + b \beta = 0$, and similarly $q'$ will have degree $d$ unless $a \gamma + b \delta = 0$. So the only way for $\max(\deg p', \deg q') < \max(\deg p, \deg q)$ is for there to be a non-trivial solution (in fact a solution with $a \neq 0, b \neq 0$) to the matrix equation $$ \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$ But this contradicts the fact that $\begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$ belongs to $GL(2,\mathbb{C})$.