## x being reducible in R[x]

Posted: December 1, 2009 in Elementary Algebra; Problems & Solutions, Rings and Modules
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There are many commutative rings $R$ satisfying this property that the indeterminate $x$ is reducible in the polynomial ring $R[x]$. Here are two examples: in $(\mathbb{Z}/6\mathbb{Z})[x]$ we have $x=(3x+4)(4x+3)$ and in $(\mathbb{Z}/10\mathbb{Z})[x]$ we have $x=(5x+4)(4x+5).$ In general, if a commutative ring $R$  has an idempotent $e \neq 0,1,$ then

$x=(ex + 1-e)((1-e)x + e).$

So if an integer $n > 1$ has at least two distinct prime divisors, then $x$ will be reducible in $(\mathbb{Z}/n\mathbb{Z})[x].$