translation automorphism of a polynomial ring

Let $R$ be a commutative ring, let $R[X]$ be the polynomial ring over $R$, and let $a$ be an element of $R$. Then we can define a homomorphism ${\tau }_a$ of $R[X]$ by constructing the evaluation homomorphism from $R[X]$ to $R[X]$ taking $r\in R$ to itself and taking $X$ to $X+a$.

To see that ${\tau }_a$ is an automorphism, observe that ${\tau }_{-a}\circ {\tau }_a$ is the identity on $R\subset R[X]$ and takes $X$ to $X$, so by the uniqueness of the evaluation homomorphism, ${\tau }_{-a}\circ {\tau }_a$ is the identity.

Title	translation automorphism of a polynomial ring
Canonical name	TranslationAutomorphismOfAPolynomialRing
Date of creation	2013-03-22 14:16:13
Last modified on	2013-03-22 14:16:13
Owner	archibal (4430)
Last modified by	archibal (4430)
Numerical id	4
Author	archibal (4430)
Entry type	Example
Classification	msc 12E05
Classification	msc 11C08
Classification	msc 13P05
Related topic	IsomorphismSwappingZeroAndUnity