Help filling in gaps for a lemma in the introduction to Hulek's elementary algebraic geometry

Question

This is a step in Hulek's Elementary Algebraic geometry that I'm trying to fill in the gaps for. Any help is appreciated. This is Lemma 0.11 in the book.

Let $p,q\in\mathbb{C}[t]$ be coprime. If there are four distinct values $\lambda/\mu\in\mathbb{C}\cup\{\infty\}$ such that $\lambda p+\mu q$ is a square in $\mathbb{C}[t]$, then $p,q\in\mathbb{C}$.

The problem that I am having is with the first part of the proof of this lemma. He says to note that the hypotheses are unchanged by a linear transformation

$$p'=ap+bq,\ \ q'=cp+dq,\ \ \text{where } \begin{pmatrix} a& b\\ c& d\\ \end{pmatrix}\in \text{GL}(2,\mathbb{C})$$

I can see that $p'$ and $q'$ are coprime but I do not see that they satisfy the hypothesis regarding the squares. He then goes on to say that we can assume that the four ratios we have are $0,1,\infty$, and $\lambda$ to get that $$p,q,p-q,p-\lambda q$$ are squares.

Answer 1

I find it easier to think of how $\text{GL}(2,\mathbb{C})$ acts on the four distinct pairs of solutions in the set $(\lambda , \mu) \in \mathbb{C} \cup \{\infty\}$, such that $\lambda p + \mu q$ is equal to a square. For example, if $$s^{2} = \lambda p + \mu q \longmapsto \lambda(ap + b q) + \mu (c p + dq) = (\lambda a + \mu c)p + (\lambda b + \mu d)q = \lambda^{\prime} p + \mu^{\prime} q $$ then $$\begin{pmatrix} a && c \\ b && d \end{pmatrix} \begin{pmatrix} \lambda \\ \mu \end{pmatrix} = \begin{pmatrix} \lambda^{\prime} \\ \mu^{\prime} \end{pmatrix} $$ are now the new values for which a pair $(\lambda_{i}, \mu_{i})$, $i = 1,2,3,4$, determines a square, of which there are still four. One can then play around with the values for $a,\ldots, d$ to get the four ratios that is then assumed in the proof.

So to get the ratios to be $0, 1, \infty$ and $\alpha \in \mathbb{C}$ (here I use $\alpha$ where Hulek has used $\lambda$), one has to solve $$ (\lambda_{1},\mu_{1}) \longmapsto a\lambda_{1} + c\mu_{1} = 0,\\ (\lambda_{2},\mu_{2}) \longmapsto a\lambda_{2} + c\mu_{2} = b\lambda_{2} + d\mu_{2},\\ (\lambda_{3},\mu_{3}) \longmapsto b\lambda_{3} + d\mu_{3} = 0,\\ $$ for the ratios $\lambda^{\prime}/\mu^{\prime}$ to be those assumed, with a one-dimensional degree of freedom for $\lambda_{4}^{\prime}/\mu_{4}^{\prime} = \alpha \in \mathbb{C}$ since we have four unknowns $a,b,c,d$ and three equations.

Answer 2

I see now, thanks partly to BenCWBrown, that we can assume this by essentially a change of basis for the vector space spanned by $(p,q)$. I will post my solution here in the hopes that it will be helpful for others. I find it easier to work with Möbius transformations, so my solution involves knowledge of them:

There is a one to one correspondence between $\mathbb{P}^1(F)$ and $F_\infty=F\cup\{\infty\}$. Thus it suffices to show that we can map any 4 four points in $F_\infty$ to $0,1,\infty$, and $\alpha$. But we know that the Möbius transformations on $F_\infty$ are completely determined by their action on $3$ points and are given by an element $A\in\text{SL}(2,F)$. For $A=\begin{pmatrix}a& b\\ c& d \end{pmatrix}$, the Möbius transformation $f_A$ is given by $f_A(z)=\frac{az+b}{cz+d}$.

The maps $f_A$ and $\phi_A$ preserve the bijection between $\mathbb{P}^1$ and $F_\infty$ by $\frac{az+b}{cz+d}\rightarrow(az+b:cz+d)$ and $\infty\rightarrow (1:0)$. Thus using Möbius transformations we can send any 3 distinct points to $0,1,\infty$ and it must map another distinct point to some point $\alpha$. $0,1,\infty$ and $\alpha$ correspond to the desired points. In the same manner, we can map any 3 points to $0,-1$, and $\infty$. To send the points $z_1,z_2,z_3$ to $0,1,\infty$, let $f(z)=\frac{(z_3-z_2)(z_1-z)}{(z_1-z_2)(z_3-z)}$. Then $f$ is the desired Möbius transformation. To get $z_1,z_2,z_3$ to map to $w_1,w_2,w_3$, simply compose the above transformation $f(z)$ with the inverse of the transformation that maps $w_1,w_2,w_3$ to $0,1,\infty$.

For a matrix $A$ as above, observe that for $(\lambda:\mu)\in\mathbb{P}^1$ we have that $A$ acts on $\lambda p+\mu q $ by mapping it to $\lambda p'+\mu q'=(\lambda a+\mu c)p+(\lambda b+\mu d)q$. For $i=1,..,4$, let $P_i=(\lambda_i:\mu_i)$ be the points for which $\lambda_ip+\mu_iq$ are squares. Let $A$ be an invertible matrix so that $\phi_{A^T}$ sends $P_1,P_2,P_3,P_4$ to $(1:0),(0:1),(1:-1),(1:-\alpha)$ for some $\alpha\neq0,1$, such an $A$ exists by the above discussion. Then the action of $A$ on the polynomials gives us that $$p'=\lambda_1p+\mu_1q,\ \ q'=\lambda_2p+\mu_2q, \ \ p'-q'=\lambda_3p+\mu_3q,\ \ p'-\alpha q'=\lambda_4p+\mu_4q$$ Hence $p',q',p'-q'$, and $p'-\alpha q'$ are all squares.