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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.

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1 Answer 1

up vote 3 down vote accepted

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 $f(a)=0$, $c = 0$

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@Louis no problem! Yes, sometimes the simplest approach is best! –  Tom Oldfield May 5 '13 at 19:01
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But you can prove this quickly via $R[X]/(X-a)=R$ (which in turn is trivial when working with universal properties). No computation is needed. –  Martin Brandenburg May 5 '13 at 19:10

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