Exercise 7.5.5

Proof.

If $\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)\notin \mathrm{GL}(2,F)$, then the two rows $(a,b),(c,d)$ are linearly dependent.

Moreover $(c,d)\ne 0$, otherwise $B(t)=\mathit{at}+b$ is zero, in contradiction with $\sigma (t)=A(t)∕B(t)\in F(t)$.

So there exists $\lambda \in F$ such that $(a,b)=\lambda (c,d)$, and then $\sigma (t)=A(t)∕B(t)=\lambda \in F$. As ${\sigma }^{-1}\in \mathrm{Gal}(F(t)∕F)$, $t={\sigma }^{-1}(\lambda )=\lambda \in F$, which is impossible since $t$ is transcendental over $F$.

Conclusion: $\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)\in \mathrm{GL}(2,F)$ □