Exercise 7.5.7

Proof.

(a)

Let $\gamma =\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)\in \mathrm{GL}(2,F)$, and suppose that $[\gamma ]\in \mathrm{PGL}(2,F)$ fixes $\infty$, and also two distinct points $\alpha ,\beta$ of $F$.

If $c\ne 0$, then $[\gamma ]\cdot \infty =a∕c\ne \infty$, which is in contradiction with the fact that $[\gamma ]$ fixes $\infty$. Therefore $c=0$.

If $z\in F$, using $c=0$,

$$\gamma \cdot z=z⟺\frac{\mathit{az}+b}{\mathit{cz}+d}=z⟺c{z}^2+(d-a)z-b=0⟺(d-a)z-b=0.$$

This equation is satisfied by $\alpha$ and $\beta$. The polynomial $(d-a)x-b$ has degree at most 1 and has two distinct roots $\alpha \ne \beta$, thus is the null polynomial. This implies $c=b=0,a=d$, so $\gamma \in F^{\text{∗}}{I}_2$, and $[\gamma ]=e$ is the identity of the group $\mathrm{PGL}(2,F)$.

(b)

Suppose now that $[\gamma ]$ fixes three distinct points $\alpha ,\beta ,\xi$ de $F$.

Let $\delta =\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill \alpha \hfill \end{array}\right)$. Then $\det <!--\; FUNCTION\; APPLICATION\; -->(\delta )=1:[\delta ]\in \mathrm{PGL}(2,F)$, and

$$[\delta ]\cdot \alpha =\infty ,\forall <!--\; FUNCTION\; APPLICATION\; -->z\ne \alpha ,[\delta ]\cdot z=\frac{1}{\alpha -z}.$$

As $\beta ,\xi$ are two distinct elements of $F$, then $\delta (\beta ),\delta (\xi )$ are two distinct points of $F$ since $\delta$ is a bijection of $\hat{F}$.

Moreover $\eta =\mathit{\delta \gamma }{\delta }^{-1}$ satisfies

$$\begin{array}{llll}\hfill [\eta ]\cdot \infty & =[\mathit{\delta \gamma }{\delta }^{-1}]\cdot \infty =[\mathit{\delta \gamma }]\cdot \alpha =[\delta ]\cdot \alpha =\infty & \hfill & \\ \hfill [\eta ]\cdot ([\delta ]\cdot \beta )& =[\mathit{\delta \gamma }{\delta }^{-1}]\cdot ([\delta ]\cdot \beta )=[\mathit{\delta \gamma }]\cdot \beta =[\delta ]\cdot \beta & \hfill & \\ \hfill [\eta ]\cdot ([\delta ]\cdot \xi )& =[\mathit{\delta \gamma }{\delta }^{-1}]\cdot ([\delta ]\cdot \xi )=[\mathit{\delta \gamma }]\cdot \xi =[\delta ]\cdot \xi & \hfill & \\ \hfill & & \hfill \end{array}$$

So $\eta$ fixes the three points $\infty ,[\delta ]\cdot \beta ,[\delta ]\cdot \xi$, where $[\delta ]\cdot \beta ,[\delta ]\cdot \xi$ are two distinct points of $F$. By part (a), $\eta =\lambda I_2,\lambda \in F^{\text{∗}}$. Therefore $\gamma ={\delta }^{-1}\mathit{\eta \delta }=\lambda {\delta }^{-1}\delta =\lambda I_2$, so $[\gamma ]=e$ is the identity of $\mathrm{PGL}(2,F)$.

(c)

By parts (a) and (b), if $[\gamma ]$ fixes three points of $\hat{F}$, then $[\gamma ]=e$.

If $\gamma ,{\gamma }^{\prime }$ satisfy $[\gamma ]\cdot {\alpha }_i=[{\gamma }^{\prime }]\cdot {\alpha }_i,i=1,2,3$, then $[{\gamma }^{\prime }{\gamma }^{-1}]$ fixes three points of $\hat{F}$. Therefore $[{\gamma }^{\prime }{\gamma }^{-1}]=[{\gamma }^{\prime }]{[\gamma ]}^{-1}=e$, so $[{\gamma }^{\prime }]=[\gamma ]$: the uniqueness is proved.