Q:

Show that there exists a unique ring homomorphism such that

Hi my question is as follows:
Let A be a domain, and let $A[T]$ be the ring polynomials over A.
Show that for any elements $a,b{\in}A$, there exists a unique ring homomorphism ${\varphi}_{a,b}:A[T]{\to}A[T]$ such that:
(i) ${\varphi}_{a,b}(T)=aT+b$ in $A[T]$ and
(ii) ${\varphi}_{a,b}$ induces the identity map on $A$.
I have thought of using the universal property of polynomials ring by first invoking a ring homomorphism ${\varphi}:A{\to}A[T]$ but this is where I am stuck. I have no idea how to idea how to construct such a map ${\varphi}$ that will allow me to have map ${\varphi}_{a,b}$ satisfying the mentioned hypothesis.
Any hints/help would be welcomed. Thanks!

A:

Your ring homomorphism is an $A$-algebra homomorphism from $A[T]$ to itself. Now for any  $A$-algebra $B$, an algebra homomorphism from $A[T]$ to $B$ is uniquely determined by the image of the indeterminate $T$ in $B$. In other words:
$$\DeclareMathOperator{\Hom}{Hom}\Hom_{A\text{-alg}}(A[T],B)\simeq B,$$
and more  generally, for any set $I$
$$\Hom_{A\text{-alg}}(A[T_i]_{i\in I},B)\simeq B^I$$
($\simeq\;$ denotes here a bijection).

