# 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 |
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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 |