# Show that the group of all those automorphisms of $F[x]$ which leave all elements of $F$ fixed, consists of substitutions given by $x \mapsto a x+b$

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

• 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. – Robert Shore Aug 28 '20 at 9:05
• @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 $F$"? – Akira Aug 28 '20 at 9:10
• I'd use $\sigma$ rather than $f$ to denote the automorphism (reserving $f$ for elements of $F[x]$), but yes. – Robert Shore Aug 28 '20 at 9:12

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.

• I've just figured out a the missing part. Could you please have a check on my update? – Akira Aug 28 '20 at 10:07

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 $$\text{deg} \left(\sigma(x)\right) = 1.$$