## Tag `00NJ`

Chapter 10: Commutative Algebra > Section 10.104: Catenary rings

Lemma 10.104.4. Any localization of a (universally) catenary ring is (universally) catenary.

Proof.Let $A$ be a ring and let $S \subset A$ be a multiplicative subset. The description of $\mathop{\rm Spec}(S^{-1}A)$ 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 universally catenary, then $B$ is catenary and we see that $C$ is catenary too. Combined with Lemma 10.30.1 this proves the lemma. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 24207–24210 (see updates for more information).

```
\begin{lemma}
\label{lemma-localization-catenary}
Any localization of a (universally) catenary ring is (universally) catenary.
\end{lemma}
\begin{proof}
Let $A$ be a ring and let $S \subset A$ be a multiplicative subset.
The description of $\Spec(S^{-1}A)$ in Lemma \ref{lemma-spec-localization}
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 universally catenary, then $B$ is catenary
and we see that $C$ is catenary too. Combined with
Lemma \ref{lemma-Noetherian-permanence} this proves the lemma.
\end{proof}
```

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