Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 6 down vote accepted

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.

share|improve this answer
    
Thank you. ${}{}$ –  Clara Nov 22 '11 at 20:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.