# The Stacks Project

## Tag 0AUN

Lemma 10.104.5. 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. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 24147–24163 (see updates for more information).

\begin{lemma}
\label{lemma-catenary-check-local}
Let $R$ be a ring. The following are equivalent
\begin{enumerate}
\item $R$ is catenary,
\item $R_\mathfrak p$ is catenary for all prime ideals $\mathfrak p$,
\item $R_\mathfrak m$ is catenary for all maximal ideals $\mathfrak m$.
\end{enumerate}
Assume $R$ is Noetherian. The following are equivalent
\begin{enumerate}
\item $R$ is universally catenary,
\item $R_\mathfrak p$ is universally catenary for all prime ideals
$\mathfrak p$,
\item $R_\mathfrak m$ is universally catenary for all maximal ideals
$\mathfrak m$.
\end{enumerate}
\end{lemma}

\begin{proof}
The implication (1) $\Rightarrow$ (2) follows from
Lemma \ref{lemma-localization-catenary} 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 \ref{lemma-localization-catenary}.
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.
\end{proof}

Comment #2374 by Matthew Emerton on February 11, 2017 a 8:13 pm UTC

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

Comment #2430 by Johan (site) on February 17, 2017 a 2:58 pm UTC

THanks, fixed here.

## Add a comment on tag 0AUN

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).