# Does this map extend to a unique automorphism of the polynomial ring of an integral domain?

Suppose that $A$ is an integral domain. If $a,b$ are element of $A$ and $a$ is a unit, then how come the change of variable $X\mapsto aX+b$ extends to a unique automorphism of the polynomial ring $A[X]$, which equals the identity map when restricted to $A$?

Thanks for any ideas or ways to get started.

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Well, to start with the polynomial Ring $A[X]$ has the universal property, that for every morphism $f : A \rightarrow B$ of Rings and every $c \in B$ there is a unique homomorphism $\overline{f}: A[X] \rightarrow B$ which extends $f$ and maps $X$ to $c$. That is, if $p = \sum_{i = 1}^n a_i X^i$ is given, we have $$\overline{f}(p) = \sum_{i = 1}^n f(a_i) c^i.$$ In your case, you can take $B = A[X]$, $f: A \rightarrow A[X]$ the canonical inclusion and $c = a X + b$. Now the only thing left to be proved is that this is indeed a bijection. This can be done by explicitly stating the inverse and here you will need that $a$ is a unit.
Thank you. ${}{}$ –  Clara Nov 22 '11 at 20:24