# How do we know that automorphisms on polynomials have a polynomial like form?

Let $F$ be a field and $\sigma:F[x]\to F[x]$ be automorphism, $\sigma(a) = a$ for all $a\in F$. I'm supposed to show that $\sigma(f(x)) = f(ax+b)$ for some $a\not = 0$ and $b$ in $F$.

Now I've got a solution that my professor gave me that seems to assume that the automorphism must have the form $\sigma(f(x)) = f(p(x))$ for some $p(x)\in F[x]$. So my question is how does $\sigma$ being an automorphism on $F[x]$ and $\sigma(a) = a$ for all $a\in F$ give us that $\sigma(f(x) = f(p(x))$ for some $p(x)\in F[x]$, why can't there be some weirder looking automorphism?

I've looked at Automorphisms of $F[x]$, however the only solution seems to make the same assumption that my professor makes.

Let $\sigma(x) = p(x)$.

If $f(x) = \sum_{i=0}^n a_i x^i$, then using the fact that $\sigma$ is a ring homomorphism, $$\sigma(f(x)) = \sigma \left( \sum_{i=0}^n a_i x^i \right) = \sum_{i=0}^n \sigma(a_i) \sigma(x)^i = \sum_{i=0}^n a_i p(x)^i = f(p(x)).$$

More generally, if $A$ is any $F$-algebra, then every $F$-algebra homomorphism $F[x] \to A$ takes the form $f(x) \mapsto f(a)$ for some $a \in A$.

• I'm not sure if I'm missing something, but how do we know from this that $p(x)\in F[x]$? – Benji Altman Jun 26 '18 at 5:27
• $\sigma$ maps to $F[x]$, so of course $\sigma(x)\in F[x]$. – Christoph Jun 26 '18 at 5:28
• @Christoph Thank you! This makes so much sense now! – Benji Altman Jun 26 '18 at 5:29
• Another take in the same vein is that $F[x]$ is the free commutative ring generated from the singleton set $\{x\}$. A commutative ring homomorphism from $F[x]$ is thus completely determined by what it does on $x$. In particular, its action on $x$ is extended to an action on arbitrary polynomials $f(x)\in F[x]$ via the equation given. This is true for any commutative ring homomorphism from $F[x]$. The interesting part about of the exercise the OP was working on was then why $x$ gets mapped to $ax+b$ given $\sigma$ fixes $F$. – Derek Elkins left SE Jun 26 '18 at 7:19

Since $\sigma$ fixes $F$, that is,

$\sigma(a) = a, \; a \in F, \tag 1$

we have, for any

$r(x) = \displaystyle \sum_0^n r_i x^i \in F[x], \tag 2$

$\sigma(r(x)) = \sigma \left ( \displaystyle \sum_0^n r_i x^i \right ) = \displaystyle \sum_0^n \sigma(r_i) \sigma(x^i) = \sum_0^n r_i (\sigma(x))^i; \tag 3$

thus, $\sigma$ is determined by its value on $x$, $\sigma(x)$; since by definition,

$\sigma:F[x] \to F[x], \tag 4$

we must have

$\sigma(x) = p(x) \in F[x]; \tag 5$

thus, from (3), we see that

$\sigma(r(x)) = r(\sigma(x)) = r(p(x)). \tag 6$

Hint: If $\sigma \in$ Aut (F[$x$]) such that $\sigma$ (a)=a, for all $\in$ F, deg$\sigma(f(x))$= deg$f(x)$ , for all $f(x)\in$ F[$(x)$]. In particular $\sigma(x)= ax + b$, where $a \neq0$.