Let $B=k[T]$ with $k$ a field; a k-automorphism of $B$ is a ring homomorphism $\phi: B\rightarrow B$ that is the identity on $k$ and is an automorphism of $B$. Describe the group $\text{Aut}_k(B)$ of $k$-automorphisms of $B$.
For $f=\sum_{i=0}^na_iT^i\in B$, since $\phi(f)=\sum_{i=0}^na_i\phi(T)^i$, the automorphism is determined by $\phi(T)$. Let therefore $\phi(T)=\sum_{i=0}^na_iT^i$. Let $c\neq 0,1$ (assume $k\neq\Bbb F_2$), since $\phi$ is $k$-automorphism $$c\phi(T)=\phi(cT)\tag{1}$$$$\sum_{i=0}^nca_iT^i=\sum_{i=0}^nc^ia_iT^i\iff\sum_{i=0}^n(c-c^i)a_iT^i=0\tag{2}$$ This gives:
$$\begin{align}(c-1)a_0=0\\(c-c)a_1=0\\(c-c^2)a_2=0\\\vdots\\(c-c^n)a_n=0\end{align}\tag{3}$$
Which gives $\phi(T)=a_1T$. Since $\phi(1)=1$, it follows that $a_1=1$.
Therefore $$\text{Aut}_k(B)\cong \{e\}\tag{4}$$ That is, the only $k$-automorphism is the identity, $\phi(T)=T$. Is this correct?
Edit: I think I got it now,
Let $\phi\in\text{Aut}_k(k[T])$, and $g=\sum_{i=0}^na_iT^i$. As $\phi(g)=\sum_{i=0}^na_i\phi(T)^i=g(\phi(T))$, we have $\phi(g)=g(\phi(T))$, that is, the automorphism $\phi$ "evaluates" any polynomial $g$ at some polynomial $\phi(T)=f$. I therefore define $\phi:=\phi_f$ to emphasize this, where $$\phi_f(g)=g(f(T))\tag{5}$$
It is clear that for $g,h\in k[T]$ (for the same reason that evaluation is a homomorphism):
- $\phi_f(g+h)=\phi_f(g)+\phi_f(h)$
- $\phi_f(gh)=\phi_f(g)\phi_f(h)$
- $\phi_f(c)=c$ for $c\in k$, that is, $\phi_f$ acts as the identity on $k$
Also $$\deg(\phi_f(g))=\deg(f)\deg(g)\tag{6}$$ As $\phi_f$ needs to be surjective we need to have $\deg(\phi_f(g))=1$ for some $g\in k[T]$, this gives $$1=\deg(f)\deg(g)\tag{7}$$ Both $\deg(f),\deg(g)\geq 0$ (integers) for $f,g\neq 0$ which gives $\deg(f)=\deg(g)=1$, thus $$\phi_f(T)=a_0+a_1T\tag{8}$$ We find the inverse: $$\phi_f^{-1}(T)=-a_0a_1^{-1}+a_1^{-1}T\tag{9}$$ which exists if and only if $a_1\neq 0$.
Thus $$\boxed{\text{Aut}_k(k[T])=\langle k\oplus k^\times, *\rangle\mid (a_0,a_1)*(b_0,b_1)=(a_0b_1+b_0,a_1b_1)}\tag{10}$$
Best Answer
What if $k=F_2$?
$$\sum_{i=0}^nca_iT^i=\sum_{i=0}^nc^ia_iT^i "$$
Do you mean that $\phi(cT)=\sum_{i=0}^nc^ia_iT^i$? What if $\phi(\sum_{i=0}^na_iT^i):=\sum_{i=0}^na_i(T+1)^i$? (the inverse of this map is $\psi(\sum_{i=0}^na_iT^i):=\sum_{i=0}^na_i(T-1)^i$).
Why $a_1=1$? (and why $a_0=0$ (by 2)?).
(I understand you are not looking for the solution, but only the correctness of your "proof").