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I'm doing Exercise 4 in textbook Algebra by Saunders MacLane and Garrett Birkhoff.

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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, $\deg f(x) = 1$.

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  • $\begingroup$ That just proves that a map of the requested form is an automorphism. It doesn't prove that all such automorphisms are of the requested form. $\endgroup$ – Robert Shore Aug 28 '20 at 9:05
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    $\begingroup$ @RobertShore You meant I remain to show that "If $f: F[x] \to F[x]$ is an automorphism such that $f(c)=c$ for all $c \in F$, then $f(x)=ax+b$ for some $a \neq 0$ and $b$ in $F$"? $\endgroup$ – Akira Aug 28 '20 at 9:10
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    $\begingroup$ I'd use $\sigma$ rather than $f$ to denote the automorphism (reserving $f$ for elements of $F[x]$), but yes. $\endgroup$ – Robert Shore Aug 28 '20 at 9:12
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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.

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    $\begingroup$ I've just figured out a the missing part. Could you please have a check on my update? $\endgroup$ – Akira Aug 28 '20 at 10:07
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As stated in a comment by OP, it remains to show that for all automorphisms $\sigma: F[x] \to F[x]$ that leave every element of $F$ fixed, we have that $\sigma(x) = ax + b$ for some $a, b \in F$ with $a \neq 0$, or equivalently, $\text{deg}\left(\sigma(x)\right) = 1$.

Suppose that $\sigma$ is an automorphism of $F[x]$ that leaves every element of $F$ fixed. Then certainly $\sigma(x) \notin F$ (otherwise we would have $x \in F$). Hence, the degree of $\sigma(x)$ is at least $1$. Letting $k$ be the degree of $\sigma(x)$, there exist scalars $a_0, a_1, \dots, a_k$ in $F$ with $a_k \neq 0$ such that

$$\sigma(x) = \sum_{i = 0}^k a_i x^i.$$ Noting that $\sigma^{-1}$ is an automorphism of $F[x]$ that leaves every element of $F$ fixed, by an identical argument used to show that the degree of $\sigma(x)$ is larger than or equal to $1$, we get that $\text{deg}\left(\sigma^{-1}(x)\right) \geq 1$. Letting $l$ be the degree of $\sigma^{-1} (x)$, there exist scalars $b_0, b_1, \dots, b_l$ in $F$ with $b_l \neq 0$ such that $$\sigma^{-1}(x) = \sum_{j = 0}^l b_j x^j.$$ Moreover

\begin{align} x &= (\sigma^{-1} \sigma)(x) \\ &= \sum_{i = 0}^k a_i \left(\sigma^{-1} (x)\right)^i \\ &= \sum_{i = 0}^k a_i \left( \sum_{j = 0}^l b_j x^j \right)^i. \end{align}

The coefficient of $x^{kl}$ in the above is $a_k (b_l)^k$, which is nonzero. We know that $l \geq 1$, so if $k \geq 2$, then $kl \geq 2$, implying that the degree of the polynomial $x$ exceeds $1$. Since the latter is not true, we have that $k \leq 1$. We showed earlier that $k \geq 1$, and so $$ \text{deg} \left(\sigma(x)\right) = 1.$$

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