One counterexample for two open questions about the rings \(R(X)\) and \(R\langle X\rangle\) (Q1969363)

Recall that a domain \(R\) is strong \(S\) (resp., catenarian) if, for each consecutive pair \(q\subset q\) of primes in \(R\), the extended primes \(p[X] \subset q[X]\) are consecutive in \(R[X]\) (resp., \(\text{ht}(q)= \text{ht}(p)+1)\). The author constructs a commutative domain \(R\) such that \(R(X)\) is catenarian, but \(R\langle X \rangle\) is not a strong \(S\)-ring. Actually, the construction is done with \(X\) replaced by any finite number of indeterminates.