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 found that every homomorphism looks like an evaluation $T\mapsto \phi(T)$, since for any $f,g\in k[T]$ we have
$$\phi\left(\frac{f}{g}\right)=\phi\left(\frac{\sum_{i=0}^na_iT^i}{\sum_{j=0}^mb_jT^j}\right)=\frac{\sum_{i=0}^na_i\phi(T)^i}{\sum_{j=0}^mb_j\phi(T)^j}=\frac{\phi(f)}{\phi(g)}=\frac{f(\phi(T))}{g(\phi(T))}\tag{1}$$ and that this is a homomorphism, for precisely the same reason that any evaluation is a homomorphism.
I then define $\phi:=\phi_{\frac{f}{g}}$ to map $T\mapsto\frac{f}{g}$, where $\frac{f}{g}\in K$. Since $\phi_{\frac{f}{g}}$ is an automorphism I should have an inverse $\phi^{-1}_{\frac{f}{g}}=\phi_{\frac{r}{s}}$ which maps $T\mapsto\frac{r}{s}$. This means that $$\phi_{\frac{r}{s}}\left(\phi_{\frac{f}{g}}(T)\right)=\frac{f\left(\frac{r}{s}\right)}{g\left(\frac{r}{s}\right)}=T\tag{2}$$
$$ f\left(\frac{r}{s}\right)=Tg\left(\frac{r}{s}\right)\tag{3}$$
My strategy is now to make eq. $(3)$ a polynomial equation in $k[T]$, and then compute the degree. (Similar to the first part of the exercise) I multiply both sides of eq. $(3)$ by $s^{\deg(f)+\deg(g)}$.
$$f\left(\frac{r}{s}\right)s^{\deg(f)+\deg(g)}=s^{\deg(g)}\sum_{i=0}^{\deg(f)}a_is^{\deg(f)-i}r^i\tag{4}$$ The degree of the polynomial in $(4)$ is then $$\deg(g)\deg(s)+\max_{0\leq i\leq\deg(f)}\{\deg(s)(\deg(f)-i)+i\deg(r)\}\\=\deg(g)\deg(s)+\max_{0\leq i\leq\deg(f)}\{\deg(f)\deg(s)+i(\deg(r)-\deg(s))\}\tag{5}$$
Similarly for $g$,
$$Tg\left(\frac{r}{s}\right)s^{\deg(f)+\deg(g)}=Ts^{\deg(f)}\sum_{j=0}^{\deg(g)}b_js^{\deg(g)-j}r^j\tag{6}$$ with degree:
$$1+\deg(f)\deg(s)+\max_{0\leq j\leq\deg(g)}\{\deg(s)(\deg(g)-j)+j\deg(r)\}\\=1+\deg(f)\deg(s)+\max_{0\leq j\leq\deg(g)}\{\deg(g)\deg(s)+j(\deg(r)-\deg(s))\}\tag{7}$$
All in all, we have $$1+\deg(f)\deg(s)+\max_{0\leq j\leq\deg(g)}\{\deg(g)\deg(s)+j(\deg(r)-\deg(s))\}\\=\deg(g)\deg(s)+\max_{0\leq i\leq\deg(f)}\{\deg(f)\deg(s)+i(\deg(r)-\deg(s))\}\tag{8}$$ which simplifies to $$1=\max_{0\leq i\leq\deg(f)}\{i(\deg(r)-\deg(s))\}-\max_{0\leq j\leq\deg(g)}\{j(\deg(r)-\deg(s))\}\tag{9}$$
There are 3 cases to consider in eq. $(9)$:
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$\deg(r)>\deg(s)$, in this case the maximum degree occurs when $i=\deg(f), j=\deg(g)$. Rearranging eq. $(9)$ then gives $$1=(\deg(f)-\deg(g))(\deg(r)-\deg(s))\tag{10}$$ This forces $\deg(f)=\deg(g)+1$, $\deg(r)=\deg(s)+1$ and also tells us that it is impossible to have $\deg(f)=\deg(g)$, which sounds wrong.
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$\deg(r)=\deg(s)$, this case simplifies to $$1=0\tag{11}$$ which is also a contradiction.
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$\deg(r)<\deg(s)$, this gives exactly the same result as the 2nd. case, since the maximum degree occurs at $i=j=0$.
In particular, eq. $(10)$ does not allow any linear transformation as endomorphisms. This is wrong since any $$\psi(T)=\frac{a_0+a_1T}{b_0+b_1T}\mid a_0b_1-a_1b_0\neq 0\tag{12}$$ has the inverse $$\psi^{-1}(T)=\frac{a_0-b_0T}{-a_1+b_1T}\tag{13}$$ and are therefore endomorphisms with $\deg(f)=\deg(g)$, and $\deg(f)=\deg(g)=\deg(r)=\deg(s)=1$.
I seem to get contradictions in all 3 cases above. I would be very happy if someone could point out my mistakes! I'm very unsure about eq. $(10)$, because it intuitively looks like something one would get, blindly calculating the degrees of the polynomial-eq. $(3)$, that is $\deg(f)(\deg(r)-\deg(s))=1+\deg(g)(\deg(r)-\deg(s))$.
Best Answer
Equality (5) breaks when $\deg (r)=\deg (s)$ and the leading coefficients $l_r$ and $l_s$ of $r$ and $s$ are such that $\frac{l_r}{l_s}$ is a root of $f$:
With $d=\deg (r)=\deg (s)$ the leading term of $\sum_{i=0}^{\deg(f)}a_is^{\deg(f)-i}r^i$ becomes $$T^{d\deg(f)}\sum_{i=0}^{\deg(f)}a_i{l_s}^{\deg(f)-i}{l_r}^i={l_s}^{\deg(f)}\,T^{d\deg(f)}\sum_{i=0}^{\deg(f)}a_i{(\frac{l_r}{l_s})^i}={l_s}^{\deg(f)}\,T^{d\deg(f)}f(\frac{l_r}{l_s})=0$$