I'm doing Exercise 4 in textbook Algebra by Saunders MacLane and Garrett Birkhoff.
Show that, if $F$ is a field, the group of all those automorphisms of $F[x]$ which leave all elements of $F$ fixed, consists of substitutions given by $x \mapsto a x+b, a \neq 0$ and $b$ in $F$.
Could you please verify if my understanding is correct? Thank you so much for your help!
My attempt:
Consider a map $f: \sum a_n x^n \mapsto \sum a_n (ax+b)^n$. It suffices to show that $f$ is an automorphism. It's trivial to show that it is a homomorphism. Hence it remains to show that it is bijective.
Let $p = \sum a_n x^n\in F[x]$. By polynomial division, there are unique polynomials $q_1,r_1$ such that $p = (ax+b)q_1+r_1$ and $\deg r_1 < \deg q_1$. Inductively, $p = \sum b_n (ax+b)^n$ for some $b_n$'s. The surjectivity then follows. Because such $b_n$'s are unique, the injectivity then follows.
Update: I add the proof for "If $f$ is an automorphism on $F[x]$ such that $f(c)=c$ for all $c \in F$, then $f(x)=ax+b$ for some $a \neq 0$ and $b$ in $F$" here.
If $\deg f(x) < 1$, then $\operatorname{im} f \subseteq F$. If $\deg f(x) > 1$, then $\operatorname{im} f$ does not contain such polynomials whose degrees are $1$. In both cases, $f$ is not surjective. As such, $\deg f(x) = 1$.
Best Answer
Hint: If $\sigma$ is such an automorphism, then $f = \sum_i a_ix^i$ maps to $f^\sigma = \sum_i a_i^\sigma (x^i)^\sigma = \sum_i a_i (x^\sigma)^i =\sum_i a_i g^i$, where $x^\sigma = g$ is a polynomial in $x$.
Since $f$ is an automorphism, $g$ must be a linear polynomial which you can show by comparison of degrees.