Commutative ring

Quote:

Originally Posted by bram kierkels

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

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

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

Tonio