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

1. $X$ is catenary, and

2. 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. $\square$

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).