Let $K=k(T)$ be the field of rational functions; a k-automorphism of $K$ is a ring homomorphism $\phi: K\rightarrow K$ that is the identity on $k$ and is an automorphism of $K$. Describe the group $\text{Aut}_k(K)$ of $k$-automorphisms of $K$.
I've asked a different question about the same exercise here. I'm looking for a proof verification for this elementary proof.
Let $\phi:K\rightarrow K$ be an automorphism, then $\phi$ is determined by $\phi(T)$ since, for $f,g\in k[T]$, $$\phi\left(\frac{f}{g}\right)=\frac{f(\phi(T))}{g(\phi(T))}\tag{1}$$ Let $\phi(T)=\frac{f}{g}$ with $f,g\in k[T]$, and let $\phi^{-1}(T)=\frac{r}{s}$ with $r,s\in K$. Assume that $f,g$ have no common factor, and the same for $r,s$. Then $$f\left(\frac{r}{s}\right)=Tg\left(\frac{r}{s}\right)\tag{2}$$
Expanding $(2)$ yields,
$$a_n\left(\frac{r}{s}\right)^n+\ldots+a_0=T\left(b_m\left(\frac{r}{s}\right)^m+\ldots+b_0\right)\tag{3}$$
There are 3 cases to consider:
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If $n>m$ then multiplying $(3)$ by $s^n$ yields $$a_nr^n+\ldots+a_0s^n=T(b_mr^ms^{n-m}+\ldots+b_0s^n)$$ Every term except $a_0s^n, Tb_0s^n$ is divisible by $r$, therefore $r\mid (b_0T-a_0)s^n\Rightarrow r\mid b_0T-a_0$ since $r,s$ have no common factor. Therefore $$r=r_1T+r_0$$ Also, $s\mid a_nr^n\Rightarrow s\mid a_n\Rightarrow s=s_0$. We can swap the role of $\frac{f}{g}$ and $\frac{r}{s}$ since $\phi\circ\phi^{-1}=\phi^{-1}\circ\phi$. Running over the same argument as above, we obtain: $$\boxed{f=f_1T+f_0\\g=g_0\\r=r_1T+r_0\\s=s_0}$$
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If $n=m$, then arguing the same as is the 1. case yields $$\boxed{f=f_1T+f_0\\g=g_1T+g_0\\r=r_1T+r_0\\s=s_1T+s_0}$$
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If $n<m$, then arguing the same as is the 1. case yields $$\boxed{f=f_1T+f_0\\g=g_0\\r=r_1T+r_0\\s=s_0}$$
All in all, every automorphism has the form $$\phi(T)=\frac{a+bT}{c+dT}\mid ad-bc\neq 0$$ $ad-bc\neq 0$ since $a+bT\neq k(c+dT)$
Best Answer
Your proof is correct, as far as I can tell. Some minor comments: