Hint for a question about automorphisms

Question

let $F$ a field and $f$ an automorphism of the polynomials ring $F[x]$. show that exists $g $automorphism of $F$ such that $f(c) = g(c)$ for all $c \in F$ and exists $a,b \in F, a \neq 0$ such that $f(x) = ax + b. $

can anyone send me a hint to solve this?

Answer 1

$f$ must map $0\mapsto 0$ and $1\mapsto 1$ and then also must map $F[x]^\times \to F[x]^\times$, therefore induces an automorphism of $F = F[x]^\times\cup \{0\}$. So let us call this $g$. Then clearly $g(c)=f(c)$ for all $c\in F$. Now $f(x)$ must be some polynomial in $x$ of some degree $d$. Observe that the degree of $a_nx^n+\cdots +a_1x+a_0$ (with $a_n\ne 0$) then turns out to be $nd$. By surjectivity of $f$, we must have $d=1$.

Answer 2

Hints: if $p, q \in F[x]$ and $pq = 1$, then$p, q \in F$. If $p \in F$ has $\deg(p) > 1$ or $\deg(p) = 0$, then $x$ does not belong to the ideal generated by $p$.