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.


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
  • 1
    $\begingroup$ @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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