Exercise 7.5.4

Proof. Let $\Phi :\{\begin{array}{ccc}\hfill \mathrm{GL}(2,F)\hfill & \hfill \to \hfill & \hfill \mathrm{Gal}(F(t)∕F)\hfill \\ \hfill \gamma \hfill & \hfill \mapsto \hfill & \hfill {\sigma }_{{\gamma }^{-1}}.\hfill \end{array}$

Let $\delta =\left(\begin{array}{cc}\hfill e\hfill & \hfill f\hfill \\ \hfill g\hfill & \hfill h\hfill \end{array}\right),\gamma =\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)$ in $\mathrm{GL}(2,F)$. Then $\mathit{\delta \gamma }=\left(\begin{array}{cc}\hfill \mathit{ea}+\mathit{fc}\hfill & \hfill \mathit{eb}+\mathit{fd}\hfill \\ \hfill \mathit{ga}+\mathit{hc}\hfill & \hfill \mathit{gb}+\mathit{hd}\hfill \end{array}\right).$For all $\alpha \in F(t)$, define $\beta ={\sigma }_{\delta }(\alpha )=\alpha \left(\frac{\mathit{et}+f}{\mathit{gt}+h}\right)$.

Therefore

Applying this equality to ${\delta }^{-1},{\gamma }^{-1}$, we obtain

For all $\delta ,\gamma \in \mathrm{GL}(2,F)$,

$\Phi$ is so a group homomorphism.

Note: in terms of group actions, if we write ${\alpha }^{\gamma }=\alpha (\frac{\mathit{at}+b}{\mathit{ct}+d})$, the preceding calculation proves that ${({\alpha }^{\delta })}^{\gamma }={\alpha }^{\mathit{\delta \gamma }}$, so $\gamma \mapsto {\alpha }^{\gamma }=\alpha \cdot \gamma$ defines a right action, and this is equivalent to the fact that $\Phi :\mathit{GL}(2,F)\to F(t)$ defined by $\gamma \mapsto {\alpha }^{{\gamma }^{-1}}$ is a group homomorphism :

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