Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 9E / ii / Solution

One direction follows because finitely many nilpotent coefficients generate a nilpotent ideal. Conversely, reducing a nilpotent polynomial modulo $N$ gives a nilpotent polynomial over the reduced ring $R/N$; comparing its highest nonzero coefficient shows it must be zero, so every coefficient lay in $N$.