I'm trying to prove that,
For a commutative ring $R$ with identity, the assignment $\phi: f(x) \to f(cx+b)$ on $R[x]$, where $c,b\in R$ and $c$ is a unit, is a one-to-one map.
To show this, I tried to show that the kernel is zero, so given $f = \sum_k^n a_k x^k \in Ker \phi$, $$f \to \sum_{k=0}^n \sum_{t=0}^k \binom{k}{t} a_k (cx)^t b^{k-t} = 0,$$
but this implies $$a_r (1+b+b^2+...+b^{n-r}) = 0 \quad \forall r=0,1,..,n.$$
However, I stuck at this point; since $R$ is not a integral domain in general, I don't know how to proceed.
I would appreciate any help or hint.
If we let $\psi:g(x) \mapsto g(c^{-1}(x-b))$, it follows that $$ (\psi\phi)(f)(x)=\psi(\phi(f))(x)=\phi(f)(c^{-1}(x-b))=f(cc^{-1}(x-b)+b)=f(x), $$ and $$(\phi\psi)(f)(x)=\phi(\psi(f))(x) = \psi(f)(cx+b)=f(c^{-1}((cx+b)-b))=f(x). $$ Hence, $\psi = \phi^{-1}$ and the bijectivity of $\phi$ follows.
But I couldn't see how you derived the condition $$ a_r(1+b+\cdots b^{n-r})=0. $$ Shouldn't it be $\forall j=0,\ldots,n$ $$ \sum_{k=j}^n a_k\binom{k}{j}b^{k-j}=0? $$ If it is the case, we can show inductively that $a_n=0$, $a_{n-1}=0$, ..., etc.