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Let $R$ be a commutative ring with identity. Then $R$ is $\textit{catenary}$ if for each pair of prime ideal $p \subsetneq q$, all maximal chains of prime ideals $p = p_0 \subsetneq p_1 \subsetneq \dots \subsetneq p_n = q$ have the same length.

In some (informal) texts the author conclude that (without further explanation) the ring is catenary since it is a local UFD and its dimension is less than or equal 3. Is this considered trivial?

thank you.

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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 $$\operatorname{ht}\mathfrak{p} + \dim(A/\mathfrak{p}) = \dim(A/\mathfrak{q}) + 1 = 3 = \dim A.\tag*{$\blacksquare$}$$

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  • $\begingroup$ thank you very much Takumi $\endgroup$ – user 1 Jan 8 at 10:40
  • $\begingroup$ I suppose $\mathfrak pA$ is actually $\mathfrak p$. $\endgroup$ – user26857 Jan 9 at 15:27
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    $\begingroup$ @user26857 That's right, thank you for the clarification! I changed the notation to "$\mathfrak{p}\cdot A/\mathfrak{q}$" since I wanted it to be clear that I was considering the height of the extension of $\mathfrak{p}$ to $A/\mathfrak{q}$. $\endgroup$ – Takumi Murayama Jan 9 at 16:55

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