The Stacks project

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10.104 Catenary rings

Definition 10.104.1. A ring $R$ is said to be catenary if for any pair of prime ideals $\mathfrak p \subset \mathfrak q$, all maximal chains of primes $\mathfrak p = \mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ e = \mathfrak q$ have the same (finite) length.

Lemma 10.104.2. A ring $R$ is catenary if and only if the topological space $\mathop{\mathrm{Spec}}(R)$ is catenary (see Topology, Definition 5.11.4).

Proof. Immediate from the definition and the characterization of irreducible closed subsets in Lemma 10.25.1. $\square$

In general it is not the case that a finitely generated $R$-algebra is catenary if $R$ is. Thus we make the following definition.

Definition 10.104.3. A Noetherian ring $R$ is said to be universally catenary if every $R$ algebra of finite type is catenary.

We restrict to Noetherian rings as it is not clear this definition is the right one for non-Noetherian rings. By Lemma 10.104.7 to check a Noetherian ring $R$ is universally catenary, it suffices to check each polynomial algebra $R[x_1, \ldots , x_ n]$ is catenary.

Lemma 10.104.4. Any localization of a catenary ring is catenary. Any localization of a Noetherian universally catenary ring is universally catenary.

Proof. Let $A$ be a ring and let $S \subset A$ be a multiplicative subset. The description of $\mathop{\mathrm{Spec}}(S^{-1}A)$ in Lemma 10.16.5 shows that if $A$ is catenary, then so is $S^{-1}A$. If $S^{-1}A \to C$ is of finite type, then $C = S^{-1}B$ for some finite type ring map $A \to B$. Hence if $A$ is Noetherian and universally catenary, then $B$ is catenary and we see that $C$ is catenary too. This proves the lemma. $\square$

Lemma 10.104.5. Let $A$ be a Noetherian universally catenary ring. Any $A$-algebra essentially of finite type over $A$ is universally catenary.

Proof. If $B$ is a finite type $A$-algebra, then $B$ is Noetherian by Lemma 10.30.1. Any finite type $B$-algebra is a finite type $A$-algebra and hence catenary by our assumption that $A$ is universally catenary. Thus $B$ is universally catenary. Any localization of $B$ is universally catenary by Lemma 10.104.4 and this finishes the proof. $\square$

Lemma 10.104.6. Let $R$ be a ring. The following are equivalent

  1. $R$ is catenary,

  2. $R_\mathfrak p$ is catenary for all prime ideals $\mathfrak p$,

  3. $R_\mathfrak m$ is catenary for all maximal ideals $\mathfrak m$.

Assume $R$ is Noetherian. The following are equivalent

  1. $R$ is universally catenary,

  2. $R_\mathfrak p$ is universally catenary for all prime ideals $\mathfrak p$,

  3. $R_\mathfrak m$ is universally catenary for all maximal ideals $\mathfrak m$.

Proof. The implication (1) $\Rightarrow $ (2) follows from Lemma 10.104.4 in both cases. The implication (2) $\Rightarrow $ (3) is immediate in both cases. Assume $R_\mathfrak m$ is catenary for all maximal ideals $\mathfrak m$ of $R$. If $\mathfrak p \subset \mathfrak q$ are primes in $R$, then choose a maximal ideal $\mathfrak q \subset \mathfrak m$. Chains of primes ideals between $\mathfrak p$ and $\mathfrak q$ are in 1-to-1 correspondence with chains of prime ideals between $\mathfrak pR_\mathfrak m$ and $\mathfrak qR_\mathfrak m$ hence we see $R$ is catenary. Assume $R$ is Noetherian and $R_\mathfrak m$ is universally catenary for all maximal ideals $\mathfrak m$ of $R$. Let $R \to S$ be a finite type ring map. Let $\mathfrak q$ be a prime ideal of $S$ lying over the prime $\mathfrak p \subset R$. Choose a maximal ideal $\mathfrak p \subset \mathfrak m$ in $R$. Then $R_\mathfrak p$ is a localization of $R_\mathfrak m$ hence universally catenary by Lemma 10.104.4. Then $S_\mathfrak p$ is catenary as a finite type ring over $R_\mathfrak p$. Hence $S_\mathfrak q$ is catenary as a localization. Thus $S$ is catenary by the first case treated above. $\square$

Lemma 10.104.7. Any quotient of a catenary ring is catenary. Any quotient of a Noetherian universally catenary ring is universally catenary.

Proof. Let $A$ be a ring and let $I \subset A$ be an ideal. The description of $\mathop{\mathrm{Spec}}(A/I)$ in Lemma 10.16.7 shows that if $A$ is catenary, then so is $A/I$. The second statement is a special case of Lemma 10.104.5. $\square$

Lemma 10.104.8. Let $R$ be a Noetherian ring.

  1. $R$ is catenary if and only if $R/\mathfrak p$ is catenary for every minimal prime $\mathfrak p$.

  2. $R$ is universally catenary if and only if $R/\mathfrak p$ is universally catenary for every minimal prime $\mathfrak p$.

Proof. If $\mathfrak a \subset \mathfrak b$ is an inclusion of primes of $R$, then we can find a minimal prime $\mathfrak p \subset \mathfrak a$ and the first assertion is clear. We omit the proof of the second. $\square$

Lemma 10.104.9. A Noetherian Cohen-Macaulay ring is universally catenary. More generally, if $R$ is a Noetherian ring and $M$ is a Cohen-Macaulay $R$-module with $\text{Supp}(M) = \mathop{\mathrm{Spec}}(R)$, then $R$ is universally catenary.

Proof. Since a polynomial algebra over $R$ is Cohen-Macaulay, by Lemma 10.103.7, it suffices to show that a Cohen-Macaulay ring is catenary. Let $R$ be Cohen-Macaulay and $\mathfrak p \subset \mathfrak q$ primes of $R$. By definition $R_{\mathfrak q}$ and $R_{\mathfrak p}$ are Cohen-Macaulay. Take a maximal chain of primes $\mathfrak p = \mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ n = \mathfrak q$. Next choose a maximal chain of primes $\mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset \mathfrak q_ m = \mathfrak p$. By Lemma 10.103.3 we have $n + m = \dim (R_{\mathfrak q})$. And we have $m = \dim (R_{\mathfrak p})$ by the same lemma. Hence $n = \dim (R_{\mathfrak q}) - \dim (R_{\mathfrak p})$ is independent of choices.

To prove the more general statement, argue exactly as above but using Lemmas 10.102.13 and 10.102.9. $\square$

Lemma 10.104.10. Let $(A, \mathfrak m)$ be a Noetherian local ring. The following are equivalent

  1. $A$ is catenary, and

  2. $\mathfrak p \mapsto \dim (A/\mathfrak p)$ is a dimension function on $\mathop{\mathrm{Spec}}(A)$.

Proof. If $A$ is catenary, then $\mathop{\mathrm{Spec}}(A)$ has a dimension function $\delta $ by Topology, Lemma 5.20.4 (and Lemma 10.104.2). We may assume $\delta (\mathfrak m) = 0$. Then we see that

\[ \delta (\mathfrak p) = \text{codim}(V(\mathfrak m), V(\mathfrak p)) = \dim (A/\mathfrak p) \]

by Topology, Lemma 5.20.2. In this way we see that (1) implies (2). The reverse implication follows from Topology, Lemma 5.20.2 as well. $\square$


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