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[parent] translation automorphism of a polynomial ring (Example)

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.

"translation automorphism of a polynomial ring" is owned by archibal.
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Cross-references: identity, automorphism, evaluation homomorphism, polynomial ring, commutative ring

This is version 1 of translation automorphism of a polynomial ring, born on 2004-03-19.
Object id is 5720, canonical name is TranslationAutomorphismOfAPolynomialRing.
Accessed 967 times total.

AMS MSC12E05 (Field theory and polynomials :: General field theory :: Polynomials )
 11C08 (Number theory :: Polynomials and matrices :: Polynomials)
 13P05 (Commutative rings and algebras :: Computational aspects of commutative algebra :: Polynomials, factorization)

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