Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 9E (Groups, Rings and Modules)

If $a^m=b^n=0$, then every term in $(a+b{)}^{m+n-1}$ vanishes; also $(ra{)}^m=0$. Thus the nilpotents form an ideal $N$. The quotient $R/N$ has no nonzero nilpotents.

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 $N$.

If $b^k=0$, then $(u+b{)}^{-1}=u^{-1}{\sum }_{j=0}^{k-1}(-b{u}^{-1}{)}^j$. For the polynomial, its positive-degree part is nilpotent by (ii), so adding it to the unit $a_0$ gives a unit.

Choose nonzero $g$ of least degree with $fg=0$. If $b$ is its leading coefficient, then the top coefficient equation gives $a_{n}b=0$; minimality applied after removing the top term forces $bf=0$. In $(\mathbb{Z}/m\mathbb{Z})[X]$, a nonzero annihilator $r$ therefore satisfies $r{a}_i=0$ for every $i$; a prime divisor of $m/\gcd (r,m)$ divides all $a_i$ and $m$, so the displayed gcd exceeds one.