Local ring of infinite dimension

Question

Short version: Let $R$ be a commutative ring such that all chains of primes of $R$ with the same extremities have the same finite cardinality. Is $R$ locally finite-dimensional?

Longer version: Let $R$ be a commutative ring. Several slightly different definitions of the notion of catenarity of $R$ are in use, and here I would like to compare two of them.

A chain of primes of $R$ is a totally ordered set of primes of $R$ that has a minimum and a maximum. The minimum and the maximum of such a chain are called its extremities. A chain $C$ of primes of $R$ is called saturated if it is $\subseteq$-maximal among the chains of primes of $R$ with the same extremities as $C$.

Now, we can give the two definitions of catenarity.

(A) All finite chains of primes of $R$ with the same extremities have the same cardinality.

(B) All chains of primes of $R$ with the same extremities have the same finite cardinality.

Clearly, (B) implies (A). Now, we consider the following two statements.

(1) $R$ fulfils (B).

(2) $R$ fulfils (A) and is locally finite-dimensional, i.e., all its primes have finite height.

It is easy to see that (2) implies (1). Moreover, if every prime of $R$ contains only finitely many minimal primes of $R$, then (1) and (2) are equivalent. In particular, they are equivalent if $R$ is noetherian or a domain.

My question is now the following:

Are (1) and (2) equivalent?

Of course, this comes down to asking whether (B) implies that $R$ is locally finite-dimemsional. I guess it does not and hence look for a counterexample. This amounts to finding a local ring $R$ fulfilling (B) with infinitely many minimal primes such that the (finite) dimensions $\dim(R/\mathfrak{p})$ for the minimal primes $\mathfrak{p}$ are unbounded. Does such a ring exist?

Answer 1

Let $k$ be a field. Let $x_{i, j}$, $1 \leq j \leq i \in \mathbf{N}$ be variables. Consider the ring $$ R = k[x_{i, j}]/(x_{i, j} x_{i', j'}, i \not = i') $$ Let $\mathfrak m$ be the maximal ideal generated by all $x_{i, j}$. For any prime ideal $\mathfrak p \not = \mathfrak m$ there exists a unique $i$ such that $x_{i', j'} \in \mathfrak p$ for $i' \not = i$. Hence chains happen in the finite polynomial rings $k[x_{i, 1}, \ldots, x_{i, i}]$. Using this it is easy to see that $R$ is a counter example.