# Kernel of the evaluation map on a power series ring

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.

## 1 Answer

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 $f$.

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