##
###
Note on valuative dimension in power series rings

Mohamed Khalifa

2014
*
Arabian Journal of Mathematics
*

We give necessary and sufficient conditions for the power series ring R [[x 1 , . . . , x n ]] to be a Jaffard domain, where R is an almost pseudo-valuation domain. Mathematics Subject Classification All rings considered below are (commutative with identity) integral domains. The dimension of a ring R, denoted by dim R, means its Krull dimension. In [10], Jaffard defines the valuative dimension, denoted by dim v R, of an integral domain R to be the maximal rank of the valuation overrings of R.
## more »

... ollowing [1], an integral domain R is said to be a Jaffard ring if dim v R = dim R < ∞. Most standard examples of Jaffard domains are finite-dimensional Noetherian domains and finite-dimensional Prüfer domains [10] . In this paper, we are interested in when the power series ring R [[x]] is a Jaffard domain. It was shown by Arnold in [2] that if a ring R fails to have the SFT(strong finite type)-property, then R [[x]] has infinite dimension, and so has infinite valuative dimension. Recall that an ideal I of a ring R is an SFT-ideal if there exist a positive integer k and a finitely generated ideal F of R such that F ⊆ I and x k ∈ F for each x ∈ I ; moreover, if each ideal of R is an SFT-ideal, then we say that R is an SFT-ring. Let R be an integral domain. When is R [[x]] a Jaffard domain? The answer is known in two cases. First, if R is a Noetherian ring with finite dimension, then R [[x]] is also Noetherian with finite dimension [7, Lemma 2.6], and so a Jaffard domain. The second case was shown by Kang and Park. In [11], they computed the dimension of mixed extension R[x 1 ]] . . . [x n ]] where R is a finite dimensional SFT-Prüfer domain. In [15], Park generalized the result to the case, where R is a finite dimensional SFT-globalized pseudo-valuation domain (for short, GPVD) as shown in the following: Theorem 1 [15, Theorem 2.15] Let R be an m-dimensional SFT-GPVD with associated Prüfer domain T and let I = (R : T ). For each maximal ideal M of R, let N M be the maximal ideal of T such that N M ∩ R = M, set k M := R/M and K M := T /N M , and let k 0,M denote the maximal separable extension of k M in K M . Assume that [x i ]] = [[x i ]] for some i. Then, dim R[x 1 ]] . . . [x n ]] = nm + 1 if for each maximal ideal M of R with ht M = m and M ⊇ I , K M has finite exponent over k 0,M , and [k 0,M : k M ] < ∞ and dim R[x 1 ]] . . . [x n ]] = nm + n otherwise.

doi:10.1007/s40065-014-0112-7
fatcat:eydge2zc2vd73bmycsfwhwzone