Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 101 / 4 / ii / b

Put $\overline{A}=A/(x)$ and $\overline{\mathfrak{m}}=\mathfrak{m}/(x)$. Applying the dimension theorem to $A$ and $\overline{A}$ givesSince $x$ is a non-zero-divisor, it belongs to no minimal prime of the Noetherian ring $A$. Any chain of primes in $A/(x)$ lifts to a chainof primes of $A$ containing $x$. A minimal prime $\mathfrak{q}\subseteq {\mathfrak{p}}_0$ cannot contain $x$, so the inclusion is strict. Prepending $\mathfrak{q}$ gives a chain of length $r+1$ in $A$. Thus the dimension drop by a non-zero-divisor givesCombining these equalities proves

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