Krull-dimension of the power series ring over a nondiscrete valuation domain is uncountable (Q373856)

Some of the results are as follows:NEWLINENEWLINE Corollaries 17-18. Let \(V\) be a finite-dimensional valuation domain. ThenNEWLINENEWLINE (1) if \(V\) is discrete, \(\dim V[[X]]=\dim V+1\) andNEWLINENEWLINE(2) if \(V\) is non-discrete, then \(\dim V[[X]]\) is uncountable.NEWLINENEWLINE This is true even if \(V\) is not-finite-dimensional.NEWLINENEWLINETheorem 19. Let \(V\) be a valuation domain and denote \(V[[X]]_{V\setminus 0}=D\). ThenNEWLINENEWLINE(1) If \(V\) does not have a minimal prime ideal, \(D\) is a rank-one discrete valuation ring.NEWLINENEWLINE (2) If \(V\) has a minimal prime ideal that is not an idempotent, then \(D\) is a principal ideal domain.NEWLINENEWLINE (3) If \(V\) has a minimal prime ideal that is an idempotent, then \(D\) is uncountably-dimensional.NEWLINENEWLINE Theorem 26. Let \(V\) be a rank one non-discrete valuation domain with maximal ideal \(M\) and the value group \(\mathbb R\). Then, there exists an uncountable chain of prime ideals between the prime ideals \(MV[[x]]\) and \(M[[x]]\). There is also an application to entire functions.NEWLINENEWLINE Theorem 27. If \(V\) is a rank one non-discrete domain and \(\mathbf E=V[[X]]_{v\setminus 0}\), then \(\mathbf E\) is not a Prüfer domain; in fact, one can find two incomparable elements of Spec\(\mathbf E\) that have a common (upper) bound.