Automorphisms of a rational function field (D&F 14.1.8)

Prove that the automorphisms of the rational function field \(k(t)\) which fix \(k\) are precisely the fractional linear transformations determined by \(t \rightarrow \frac{at+b}{ct+d}\) for \(a,b,c,d \in k\), \(ad-bc \neq 0\).

It is easy to verify that fractional linear transformations are automorphisms of \(k(t)\) fixing \(k\). Conversely, let \(\varphi \in \text{Aut}(k(t)/k)\). Then \(k(\varphi(t)) = \varphi(k(t)) = k(t)\) (1st equality follows from the definition of a homomorphism, 2nd equality follows from surjectivity), so that \([k(t) : k(\varphi(t))] = 1\). If \(\varphi(t) = p(t)/q(t)\) for \(p(t),q(t) \in k[t]\), where \(p(t),q(t)\) are relatively prime and \(q(t) \neq 0\), then by 13.2.18 the LHS is equal to \(\max (\deg p, \deg q)\). Thus \(p(t) = at+b\) and \(q(t) =  ct+d\) for \(a,b,c,d \in k\). Lastly, if \(ad-bc = 0\), then \(a/c = x = b/d\) for some \(x \in k\) and so \(p(t) = xq(t)\), contradicting the fact that they are relatively prime. Notice in particular that we cannot have \(a = c = 0\).