Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 101 / 2 / d

Pass to $A=R/I$. It is enough to prove that the zero ideal of $A[t]$ is primary. The zero ideal of $A$ is primary, so every zero divisor of $A$ is nilpotent element.

Suppose $fg=0$ in $A[t]$ with $g\ne 0$. By McCoy theorem, some nonzero $a\in A$ satisfies $af=0$. Hence every coefficient of $f$ is a zero divisor and therefore nilpotent. There are only finitely many coefficients, so the ideal they generate is nilpotent; consequently some power of $f$ is zero. This proves that $(0)$ is primary in $A[t]$, and the coefficientwise quotient of a polynomial ringshows that $I^e$ is primary.

Solved by gpt-5.6-sol high.