I am a recent graduate studying a bit of algebra in my free time. I've stumbled across a problem that I haven't been able to write or find a proof for.
Let $R$ be a commutative ring with unity, and let $a,b\in R$, where $a$ is a unit of $R$. If $R[x]$ is the polynomial ring with coefficients from $R$, then the mapping given by $f(x) \rightarrow f(ax+b)$ is an automorphism of $R[x]$.
I can see that if this is a homomorphism, then it's clearly an automorphism because it is invertible; the inverse map being $g(x) \rightarrow g(a^{-1}x-a^{-1}b)$. However, proving it's a homomorphism I've found difficult.
I'm actually not sure it actually is a homomorphism, but it feels like it should be if that makes sense. It feels like if I treated elements of $R[x]$ as functions rather than abstract polynomials it may be easier, but I'm really not sure.
I tried proving it was a homomorphism straight from the definition of addition and multiplication of polynomials, and that was very unpleasant, to say the least (nested sums from the binomial theorem). Also, as far as abstract algebra goes, I haven't really had any experience "plugging things into" polynomials (except for single elements regarding roots, etc.), so $f(ax+b)$ is kind of awkward at this point.
So I guess my question is "is there any slick/non-messy proof that $f(x) \rightarrow f(ax+b)$ is an automorphism of $R[x]$?" Am I missing a really obvious proof?
Thanks!