Factorization of polynomial over arbitrary commutative ring I have a somewhat silly sounding question: > Let $R$ be an arbitrary commutative ring with $1$. Let $f \in R[X]$. > > Do > > a) $f(1) = 0 \Rightarrow \exists g \in R[X]: f= (X-1)g$ > > b) for some $a \in R$: $f(a)= 0 \Rightarrow \exists g \in R[X]: f= (X-a)g$ > > hold? It's not difficult to show that it is $f =(h + X - 1)g$ for a nilpotent $h \in R[X]$. From this it also follows quickly that $cf= (X-1)(cg)$ for some $c \in R$. That's everything I can show so far. Question might sound silly, but I'm thankful for any input. Remark: although the division algorithm works in all rings, decomposition in arbitrary rings will not be unique.

Unless I'm missing something, both of the things you suggest are true. This is because the division algorithm works in $R[X]$ for any ring $R$, provided that the lead coefficient of the divisor is a unit.

$1$ is certainly a unit, so we can always divide by a monic polynomial. Thus for any $a \in R$, $f\in R[X]$, $f(X)=(X-a)g(X)+c$ for some constant $c$. Thus if $f(a)=0$, $c = 0$