A counterexample for a conjecture about the catenarity of polynomial rings. (Q1599096)

It is known that a Noetherian ring \(R\) is universally catenary if \(R[X]\), the polynomial ring in one indeterminate over \(R\), is catenary [\textit{L. J. Ratliff jun.}, Am. J. Math. 92, 99--144 (1970; Zbl 0198.06003)]. The authors show that this is not true in non-Noetherian case. The main result is as follows: Let \(T\) be a Noetherian normal domain with \(\text{height}\,n = 2\) for any maximal ideal \(n\) of \(T\), \(I\) a height one prime ideal of \(T\), \(K\) a field contained in \(T/I\), and \(R\) is the pull-back of \(T \rightarrow T/I \leftarrow K\), that is \(R = \pi^{-1}(K)\) in this case where \(\pi\) is the natural map \(T \to T/I\). Assume that (i) \(\text{trans.deg}\,[T/I:K] \geq 2\) and (ii) \(\text{trans.deg}\,[T/n:K] = 0\) for any maximal ideal \(n\) containing \(I\). Then \(\text{height}\,m = 2\) for any maximal ideal \(m\) of \(R\), \(R[X]\) is catenary and \(R[X,Y]\) is not catenary. For example, let \(R = \mathbb{Z} + p\mathbb{Z}[[t]]\) where \(\mathbb{Z}\) is the ring of integers, \(p\) is a prime number and \(t\) is an indeterminate. Then \(R[X]\) is catenary and \(R[X,Y]\) is not catenary.