Here is a solution, but unfortunately it invokes several deep results of dimension theory.

**Theorem.** (Krull's Hauptidealsatz)

Let $A$ be a ring, and let $a \in A$. If $a$ is not invertible, then every prime of $A$ minimal over $(a)$ has height at most $1$.

Let $A$ be a ring, and let $a_1, \ldots, a_r$ be elements of $A$. If $(a_1, \ldots, a_r) \ne A$, then every prime of $A$ minimal over $(a)$ has height at most $r$.

**Proposition.** A noetherian ring $R$ has finitely many minimal primes.

*Proof.* Let $\Sigma$ be the set of all ideals $I$ of $R$ such that there are infinitely many primes minimal over $I$. (Recall that every ring has at least one minimal prime, so there is always at least one prime minimal over $I$.) Partially order $\Sigma$ by inclusion; suppose, for a contradiction, that $\Sigma$ is non-empty – then $\Sigma$ has a maximal element $I$, because $R$ is noetherian. Obviously, $I$ is not prime, so there are $f$ and $g$ in $R$ such that $f g \in I$ with $f \notin I$ and $g \notin I$. Yet, any prime minimal over $I$ must also be minimal over $I + (f)$ or $I + (g)$, so $I$ has only finitely many minimal primes – a contradiction. $\qquad \blacksquare$

**Corollary.** Let $R$ be a noetherian ring, and let $\mathfrak{n}_1, \ldots, \mathfrak{n}_c$ be the minimal primes of $R$. If $\dim R < \infty$, then
$$\dim R = \max_i \dim R / \mathfrak{n}_i$$

*Proof.* If $\dim R < \infty$, every maximal chain of primes of $R$ must start at a minimal prime and end at a maximal ideal. $\qquad \blacksquare$

Geometrically, what we're saying is that $\operatorname{Spec} R$ can be decomposed into finitely many irreducible components, and the dimension of $\operatorname{Spec} R$ is the maximum of the dimensions of those irreducible components. Thus, when doing dimension theory, we can sometimes get away with assuming that $R$ is an integral domain.

**Fact.** If $A$ is a local ring with maximal ideal $\mathfrak{m}$ and $\hat{A}$ is its $\mathfrak{m}$-adic completion, then $\dim A = \dim \hat{A}$.

**Fact.** A complete noetherian local ring is universally catenary. In particular, in a complete noetherian local *domain*, every saturated chain of primes has the same length.

**Proposition.** Let $A$ be a noetherian local ring with maximal ideal $\mathfrak{m}$ and residue field $k$. If $a \in \mathfrak{m}$, then $\dim A/(a) \ge \dim A - 1$.

*Proof.* We start by reducing to more tractable cases. First, observe that the completion of $A / (a)$ as a $A$-module is the same as the completion of $A / (a)$ as a local ring, so we will assume that $A$ is a complete local ring. Notice that it is enough to show that, for each minimal prime $\mathfrak{n}$ of $A$, there exists a prime $\mathfrak{c}$ of $A$ with $\mathfrak{n} \subseteq \mathfrak{n} + (a) \subseteq \mathfrak{c}$ and $\dim A/\mathfrak{n} - \dim A/\mathfrak{c} \le 1$. Moreover, it is enough to do this for a minimal prime $\mathfrak{n}$ such that $\dim A / \mathfrak{n} = \dim A$. Thus we may assume without loss of generality that $A$ is complete noetherian local domain.

Let $\mathfrak{c}_1, \ldots, \mathfrak{c}_n$ be the primes of $A$ minimal over $(a)$. The case $a = 0$ is uninteresting, so we assume $a \ne 0$. Then, our hypotheses together with Krull's Hauptidealsatz implies the height of each $\mathfrak{c}_i$ is exactly $1$. So $\dim A / (a) + 1 \le \dim A$ – therefore we are looking to prove that $\dim A / (a) = \dim A - 1$ exactly. Any maximal chain of primes containing $(a)$ extended to the left by $(0)$ yields a saturated chain of primes, but $A$ is catenary, so this is also a maximal chain of primes. Therefore $\dim A / (a) = \dim A - 1$, as required. $\qquad \blacksquare$.