Nilpotent elements of polynomial quotient ring

Question

Let $F$ be a field and let $f\in F[x]$ be an irreducible polynomial. Are the non-units of $F[x]/(f^n)$ nilpotent elements?

Answer 1

If $\overline{g}$ is not a unit, it means that $f$ and $g$ are not coprime, otherwise Bézout's theorem would imply $\overline{g}$ is invertible. Since $f$ is irreducible, it means that $f|g$.

Then $f^n|g^n$ so $\overline{g}^n=0$ and $\overline{g}$ is nilpotent.

Answer 2

This quotient ring has only one prime ideal, generated by $\bar f$, hence this prime ideal is the nilradical of the ring, and all other elements are units.