**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?



Local ring of infinite dimension
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.


