This result appears as Proposition II.3 in

Hamet Seydi, *Anneaux henséliens et conditions de chaînes. III*, Bull. Soc. Math. France **98** (1970), 329–336. Numdam: BSMF_1970__98__329_0. DOI: 10.24033/bsmf.1706. MR: 276222.

Namely, Seydi proves that "every (Noetherian) UFD of dimension three is catenary." The Noetherian assumption is necessary due to a counterexample of Fujita.

There is no proof provided, but I think that Seydi is pointing out that this result is a consequence of the proofs of previous results, namely, Théorème II.2 and Corollaire II.2.4. We give a version of Seydi's proof here, by showing that Noetherian UFD's of dimension at most three are catenary.

*Proof.* Let $A$ be a Noetherian UFD of dimension at most three. Since the property of being catenary can be checked after localizing at every maximal ideal, it suffices to consider the case when $A$ is local. We recall that Ratliff's criterion [Matsumura, Theorem 31.4] says that a Noetherian local domain $B$ is catenary if and only if
$$\operatorname{ht} \mathfrak{p} + \dim(B/\mathfrak{p}) = \dim B\tag{$*$}\label{eq:ratliff}$$
for every prime ideal $\mathfrak{p} \subseteq B$. We also recall that Noetherian domains of dimension $\le 2$ are catenary [Matsumura, Corollary 2 to Theorem 31.7], and so it suffices to consider the case when $\dim A = 3$.

Consider a prime ideal $\mathfrak{p} \subseteq A$. If $\operatorname{ht} \mathfrak{p} = 0$, then $\mathfrak{p} = 0$, in which case \eqref{eq:ratliff} trivially holds for $B$ replaced by $A$. Otherwise, suppose that $\operatorname{ht} \mathfrak{p} \ge 1$. Then, there exists a prime ideal $\mathfrak{q} \subseteq \mathfrak{p}$ such that $\operatorname{ht} \mathfrak{q} = 1$ and such that $\operatorname{ht}(\mathfrak{p} \cdot A/\mathfrak{q}) + 1 = \operatorname{ht}\mathfrak{p}$. Since $A$ is a local UFD, the ideal $\mathfrak{q}$ is principal, and we have $\dim(A/\mathfrak{q}) = 2$. We then have
$$\operatorname{ht}(\mathfrak{p} \cdot A/\mathfrak{q}) + \dim(A/\mathfrak{p}) = \dim(A/\mathfrak{q}) = 2$$
by Ratliff's criterion, since $A/\mathfrak{q}$ is a Noetherian domain of dimension $2$. But $\operatorname{ht}(\mathfrak{p}\cdot A/\mathfrak{q}) + 1 = \operatorname{ht}\mathfrak{p}$, and hence by adding $1$ to both sides of the equation above, we obtain
$$\operatorname{ht}\mathfrak{p} + \dim(A/\mathfrak{p}) = \dim(A/\mathfrak{q}) + 1 = 3 = \dim A.\tag*{$\blacksquare$}$$