Krull dimension of quotient by principal ideal

Solution 1:

This is not true, even if $R$ is Noetherian! It is possible to have a Noetherian ring $R$ and $x \in R$ a nonunit, such that $\dim R/(x) = \dim R - 2 < \infty$. The easiest counterexample I know is the following: if $S$ is a Noetherian local ring, $f \in S$ is a non-nilpotent nonunit, then $\dim S_f \le \dim S - 1$ (since the only maximal ideal of $S$ is no longer prime in $S_f$). But $S_f = S[\frac{1}{f}] \cong S[t]/(tf - 1)$, and $\dim S[t] = \dim S + 1$ (since $S$ is Noetherian), so taking $R = S[t]$, $x = tf - 1$ gives a Noetherian ring with $\dim R/(x) \le \dim R - 2$ (and equality holds if $f$ avoids all minimal primes of $S$).

As an explicit example: the ring $R = \mathbb{Z}_{(2)}[t]$ is Noetherian of dimension $2$, and the element $x = 2t - 1$ is not a unit, but $R/(x) = \mathbb{Z}_{(2)}[t]/(2t-1) \cong \mathbb{Z}_{(2)}[\frac{1}{2}] \cong \mathbb{Q}$ has dimension $0$.