8. Show that the automorphism of $k(t)$ fixing $k$ are precisely the fractional linear transformations defined by $t ↦ \dfrac{at+b}{ct+d}$ for $ad-bc \neq 0$.
9. Determine the fixed field of the automorphism $t ↦ t+1$ of $k(t)$.
Proof: (6) Let $φ$ be such a mapping. It is seen to be a ring homomorphism as evaluation at any polynomial is seen to be a ring homomorphism. Moreover, letting $k[x]_i$ denote the subspace spanned by $1,t,...,t^i$ over $k$, when $a \neq 0$ we see the basis elements of $k[x]_i$ are mapped to linear combinations of its preimage basis, showing $φ$ is bijective on $k[x]_i$ and by extension on $k[x]$ since $∪k[x]_i=k[x]$.
Conversely, any automorphism fixing $k$ is uniquely defined by its action on $t$, so observe the polynomial $f(t)=φ(t)$. We have $φ(g(t))=g(f(t))=t$ for some $g(t)∈k[x]$ by surjectivity, and since $\text{deg }g(f(t)) = \text{deg }f(t)~\text{deg }g(t)$, necessarily $f(t)=at+b$ for some $a \neq 0$.
(8) As before, evaluation is an endomorphism. When $ad-bc \neq 0$, $at+b$ and $ct+d$ are relatively prime (one of them may be in $k$) and by 13.2.18 $[k(t)~:~\text{img }φ]=1$ so $φ$ is surjective. Now, assume $f(\dfrac{at+b}{ct+d})=0$ for $f(t) = ∑a_kt^k$ ($f(t)$ being fractional implies the existence of a nonfractional polynomial satisfying such). Letting $n=\text{deg }f(t)$ and observing $(ct+d)^nf(\dfrac{at+b}{ct+d})= a_n(at+b)^n + (ct+d)g(t) = 0$ for some polynomial $g(t)$, we see by the fact that $k[t]$ is a UFD that $ct+d$ is a unit and injectivity follows by (6).
As before, all endomorphisms are evaluations, and writing them in the form $φ~:~t↦p(t)/q(t)$ for relatively prime $p(t),q(t)$ by 13.2.18 for it to be surjective necessarily the greatest degree is $1$. Clearly $ad-bc \neq 0$ as otherwise $\text{img }φ=k$.
(9) By the previous this is indeed an automorphism. Now, let $f(t)=\dfrac{pt)}{q(t)}$ for relatively prime $p(t),q(t)$ and monic $p(t)$ be a typical element of the fixed field, i.e. $f(t)=f(t+1)$. Then $\dfrac{p(t)}{q(t)}=\dfrac{p(t+1)}{q(t+1)}$ and $p(t)q(t+1)=p(t+1)q(t)$. Assuming $p(t) \neq p(t+1)$ implies $p(t) \not \mid p(t+1)$ since they are monic of the same degree, so there is some irreducible factor on the left not present on the right, a contradiction. Now $p(t)q(t+1)=p(t)q(t)$ so $q(t)=q(t+1)$ and it suffices to find the collection of polynomials in $k[t]$ fixed by $t ↦ t+1$.
If $k$ is of characteristic $0$ then $f(t)=f(t+1)$ implies $f(α)=0$ implies $f(α+1)=0$, so $f(t)$ has no zeros in any field and $f(t)∈k$. In this case the fixed field is merely $k$. Now consider the polynomial $λ(t)=t(t-1)...(t-(p-1))$ in $k[t]$, where $p$ is the characteristic of $k$. Clearly $λ(t)=λ(t+1)$, so all polynomials generated as a ring by $λ(t)$ and $k$ are fixed by $t↦t+1$. Conversely, if $f(t)=f(t+1)$ and $f(0)=a_0$, then for the polynomial $F(t)=f(t)-f(0)$ we have $F(t)=F(t+1)$ and $F(0)=0$ so also $F(1),...,F(p-1)=0$ and $λ(t)~|~F(t)$. By induction on degree $F(t)/λ(t)$ is in the ring generated by $λ(t)$ and $k$ and so too is $f(t)$, so this ring in $k[t]$ (also known as the image of $φ~:~f(t)↦f(λ(t))$ on $k[t]$, to provide a way of efficiently determining whether a polynomial is fixed by $t↦t+1$) extended to a field in $k(t)$ is precisely the field fixed by $t↦t+1$.$~\square$
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