Lemma 27.11.4. Let $X$ be a scheme. The following are equivalent

$X$ is catenary, and

for any $x \in X$ the local ring $\mathcal{O}_{X, x}$ is catenary.

Lemma 27.11.4. Let $X$ be a scheme. The following are equivalent

$X$ is catenary, and

for any $x \in X$ the local ring $\mathcal{O}_{X, x}$ is catenary.

**Proof.**
Assume $X$ is catenary. Let $x \in X$. By Lemma 27.11.2 we may replace $X$ by an affine open neighbourhood of $x$, and then $\Gamma (X, \mathcal{O}_ X)$ is a catenary ring. By Algebra, Lemma 10.104.4 any localization of a catenary ring is catenary. Whence $\mathcal{O}_{X, x}$ is catenary.

Conversely assume all local rings of $X$ are catenary. Let $Y \subset Y'$ be an inclusion of irreducible closed subsets of $X$. Let $\xi \in Y$ be the generic point. Let $\mathfrak p \subset \mathcal{O}_{X, \xi }$ be the prime corresponding to the generic point of $Y'$, see Schemes, Lemma 25.13.2. By that same lemma the irreducible closed subsets of $X$ in between $Y$ and $Y'$ correspond to primes $\mathfrak q \subset \mathcal{O}_{X, \xi }$ with $\mathfrak p \subset \mathfrak q \subset \mathfrak m_{\xi }$. Hence we see all maximal chains of these are finite and have the same length as $\mathcal{O}_{X, \xi }$ is a catenary ring. $\square$

Your email address will not be published. Required fields are marked.

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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)