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A (commutative unitary) Noetherian ring $R$ of finite dimension is said to be catenary if for every prime ideal $\mathfrak{p}$ of $R$ one has $\mathrm{ht}(\mathfrak{p})+\mathrm{dim}(R/\mathfrak{p})=\mathrm{dim}R$.

Let $A$ be a catenary ring and $B$ a finite extension of $A$ (id est $A\subseteq B$ and $B$ is a finite $A$-module). Is $B$ a catenary ring?

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  • $\begingroup$ Is your definition of catenary equivalent to the usual one (that requires that for all primes $p \subsetneq q$ all maximal prime chains $p \subsetneq \cdots \subsetneq q$ have the same length)? $\endgroup$
    – tj_
    Commented Mar 24, 2019 at 10:25

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No. Nagata's famous family of examples of non-catenary rings yields a non-catenary finite extension of a catenary noetherian local domain.

Reference: M. Nagata, On the chain problem of prime ideals, Nagoya Math. J. 10 (1956), 51-64.

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  • $\begingroup$ A simple counterexample is given by $A$ $=$ $\mathbb{Z}$ and $B$ $=$ $\mathbb{Z}[X]/(X^2+X,2X)$. Cf. here. $\endgroup$
    – Matthé van der Lee
    Commented Sep 10, 2022 at 11:31

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