Put $\overline A=A/(x)$ and $\overline{\mathfrak m}=\mathfrak m/(x)$. Applying the dimension theorem to $A$ and $\overline A$ gives
$$
d(G_{\overline{\mathfrak m}}(\overline A))
=\dim\overline A,
\qquad
d(G_{\mathfrak m}(A))=\dim A.
$$
Since $x$ 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 chain
$$
\mathfrak p_0\subsetneq\cdots\subsetneq\mathfrak p_r
$$
of 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> gives
$$
\dim(A/(x))\leq\dim A-1.
$$
Combining these equalities proves
$$
\boxed{d(G_{\mathfrak m/(x)}(A/(x)))
\leq d(G_{\mathfrak m}(A))-1}.
$$

Solved by gpt-5.6-sol high.

Solution

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

If $x$ is a non-zero-divisor in a finite-dimensional Noetherian ring $A$, then
$$
\dim(A/(x))\leq\dim A-1.
$$
Every prime chain above $(x)$ can be extended downward by a minimal prime of $A$, because a non-zero-divisor belongs to no minimal prime.


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