# 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 TranslationAutomorphismOfAPolynomialRing 2013-03-22 14:16:13 2013-03-22 14:16:13 archibal (4430) archibal (4430) 4 archibal (4430) Example msc 12E05 msc 11C08 msc 13P05 IsomorphismSwappingZeroAndUnity