Field automorphism of $F(t)$

Question

Let $t$ be transcendental over $F$. Let $a,b,c,d \in F$ such that $ad-bc \ne 0$. Prove that there is a field automorphism of $F(t)$ given by $\sigma (t) = \frac{at+b}{ct+d}$ and $\sigma(\alpha)=\alpha$, for all $\alpha \in F$.

Here is my attempt:

Define $\sigma:F(t) \rightarrow F(t)$ by $\frac{f(t)}{g(t)} \mapsto \frac{f(\sigma (t))}{g(\sigma(t))}$, where $t \mapsto \frac{at+b}{ct+d}$, where $a,b,c,d \in F$ such that $ad-bc \ne 0$ and $\alpha \mapsto \alpha$, for all $\alpha \in F$.

It's easy to check that $\sigma$ is well-defined and that it is a homomorphism.

Note that $\sigma(F(t))=F(\sigma(t))$. Clearly, $F(\sigma(t)) \subseteq F(t)$.

$\underline{Theorem}$: Let $t$ be transcendental over $F$. If $P=P(t)$ and $Q=Q(t)$ are nonzero relatively prime polynomials in $F[t]$ which are not both constant, then $[F(t):F(P/Q)]$ equals the maximum of $\{deg(P),deg(Q)\}$.

$\underline{Remark}$: As suggested in the comments, it's necessary to check that $(at+b,ct+d)=1$ to make use of the above theorem. In order to do so, observe that $(at+b,ct+d)=1$ iff $ad-bc \neq0$.

Observe that $a=c=0$ would imply that $ad-bc=0$, a contradiction. So, at least one of $c$ or $a$ must be non-zero.

Then, by the above theorem, we have:$[F(t):F(\sigma(t))]=1 \Longrightarrow F(t)=F(\sigma(t)) \Longrightarrow \sigma(F(t))=F(t) \Longrightarrow \sigma$ is surjective.

Clearly, $\sigma$ is a non-zero field homomorphism and therefore must be injective as well. Therefore, it follows then that $\sigma \in Aut(F(t)/F)$, as desired.

Is this a fair proof?

Answer 1

What are $(\sigma f)(t)$ and $(\sigma g)(t)$ when $f(t),g(t) \in F[t]$? Note that if you define \begin{align} \sigma \Big( \frac{a_0+a_1t+\cdots+a_nt^n}{b_0 + b_1t + \cdots + b_mt^m} \Big) &:= \frac{a_0+a_1\Big( \dfrac{at+b}{ct+d} \Big)+\cdots+a_n \Big( \dfrac{at+b}{ct+d} \Big)^n}{b_0 + b_1\Big( \dfrac{at+b}{ct+d} \Big) + \cdots + b_m \Big( \dfrac{at+b}{ct+d} \Big)^m} \\ &= \frac{a_0 + a_1(at+b)(ct+d)^{n-1} + \cdots + a_n(at+b)^n}{b_0 + b_1(at+b)(ct+d)^{n-1} + \cdots + b_m(at+b)^m} \end{align} for any positive integers $n,m$ and $a_i,b_i \in F$, then:

• $\sigma$ is well-defined ($t$ is trascendental over $F$ implies that the denominator of the above fraction doesn't vanish);
• $\sigma$ fixes every element of $F$ (take $b_0=1$ and $a_k=b_k=0$ for $k \geq 1$); and
• $\sigma(t) = \dfrac{at+b}{ct+d}$ (take $a_1=b_0=1$ and $a_k=b_j=0$ for $k \neq 1$ and $j \neq 0$).

In other words, just write $$\sigma \Big( \frac{f(t)}{g(t)} \Big) := \frac{f(\sigma(t))}{g(\sigma(t))}; \quad \sigma(t) := \frac{at+b}{ct+d}$$ and do the proper observations.

Moreover, you forgot to prove that $at+b$ and $ct+d$ are relatively prime! This is easy, since any common divisor of $at+b$ and $ct+d$ must divide $$-\frac c{ad-bc}(at+b) + \frac a{ad-bc}(ct+d) = 1.$$