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Let $R$ be a commutative ring with unity and $r \in R$ a nilpotent element. Is it true that if $f \in R[[\epsilon]]$ satisfies $f(r) = 0$, then $(\epsilon - r) | f$ in $R[[\epsilon]]$? I tried solving for the coefficients of $f/(\epsilon - r)$ inductively and got myself confused.

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Yes. Since $r$ is nilpotent, the map $\phi:f(\epsilon) \mapsto f(\epsilon-r)$ is an automorphism of the power series ring, and it is definitely true that if $f(0)=0$ then $\epsilon$ divides $f$.

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  • $\begingroup$ That's slick! Thanks. $\endgroup$ – Justin Campbell May 20 '13 at 0:54

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