# Thread: Commutative ring

1. ## Commutative ring

Let $\displaystyle R$ be a commutative ring with $\displaystyle r \in R$.
Let $\displaystyle f(x)$ be an element of $\displaystyle R[x]$.
I want to show that $\displaystyle f(x)$ is irreducible over $\displaystyle R$ if and only if $\displaystyle f(x + r)$ irreducible over $\displaystyle R$.

2. Originally Posted by bram kierkels
Let $\displaystyle R$ be a commutative ring with $\displaystyle r \in R$.
Let $\displaystyle f(x)$ be an element of $\displaystyle R[x]$.
I want to show that $\displaystyle f(x)$ is irreducible over $\displaystyle R$ if and only if $\displaystyle f(x + r)$ irreducible over $\displaystyle R$.

The map $\displaystyle p(x)\mapsto p(x+r),\,r\in R$ is an automorphism of the ring $\displaystyle R[x]$, and then:

$\displaystyle f(x)$reducible $\displaystyle \Longleftrightarrow f(x)=g(x)q(x),\,g(x),\,q(x)\in R[x]\Longleftrightarrow f(x+r)=g(x+r)q(x+r)$ via the above automorphism $\displaystyle \Longleftrightarrow f(x+r)$ reducible .

Tonio