**Proof.**
Let $\mathfrak p \subset A$ be a prime ideal. Let $\mathfrak q \subset B$ be a prime minimal over $\mathfrak pB$. Then $\mathfrak q \cap A = \mathfrak p$ by going down for $A \to B$ (Algebra, Lemma 10.39.19). Hence $A_\mathfrak p \to B_\mathfrak q$ is a flat local ring map with special fibre of dimension $0$ and hence

\[ \dim (A_\mathfrak p) = \dim (B_\mathfrak q) = \dim (B) - \dim (B/\mathfrak q) \]

(Algebra, Lemma 10.112.7). The second equality because $\mathop{\mathrm{Spec}}(B)$ is equidimensional and $B$ is catenary. Thus $\dim (B/\mathfrak q)$ is independent of the choice of $\mathfrak q$ and we conclude that $\mathop{\mathrm{Spec}}(B/\mathfrak p B)$ is equidimensional of dimension $\dim (B) - \dim (A_\mathfrak p)$. On the other hand, we have $\dim (B/\mathfrak p B) = \dim (A/\mathfrak p) + \dim (B/\mathfrak m_ A B)$ and $\dim (B) = \dim (A) + \dim (B/\mathfrak m_ A B)$ by flatness (see lemma cited above) and we get

\[ \dim (A_\mathfrak p) = \dim (A) - \dim (A/\mathfrak p) \]

for all $\mathfrak p$ in $A$. Applying this to all minimal primes in $A$ we see that $A$ is equidimensional. If $\mathfrak p \subset \mathfrak p'$ is a strict inclusion with no primes in between, then we may apply the above to the prime $\mathfrak p'/\mathfrak p$ in $A/\mathfrak p$ because $A/\mathfrak p \to B/\mathfrak p B$ is flat and $\mathop{\mathrm{Spec}}(B/\mathfrak p B)$ is equidimensional, to get

\[ 1 = \dim ((A/\mathfrak p)_{\mathfrak p'}) = \dim (A/\mathfrak p) - \dim (A/\mathfrak p') \]

Thus $\mathfrak p \mapsto \dim (A/\mathfrak p)$ is a dimension function and we conclude that $A$ is catenary.
$\square$

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