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 $g\ne0$. 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 ring>
$$
R[t]/I^e\cong(R/I)[t]
$$
shows that $I^e$ is primary.

Solved by gpt-5.6-sol high.

Solution

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

{c}
{wiki=McCoy's_theorem}

A polynomial $f\in R[t]$ is a zero divisor exactly when some nonzero scalar $a\in R$ annihilates it. One proof chooses a nonzero polynomial $g$ of minimal degree with $fg=0$ and successively kills its leading coefficient.


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 Incoming links: McCoy theorem