Lemma 10.105.6. 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 10.105.4 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 10.105.4. 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. $\square$

Comment #2374 by Matthew Emerton on

Very minor: there is a typo on line 2 of the proof: $R\mathfrak m$ should be $R_{\mathfrak m}$.

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