Exercise 6, § 2

a) Using the preceding exercises, now prove that if A is an integral ring of dimension 1, one has dim(A[X]) = 2 if and only if all the localizations at a prime ideal of the integral closure of A are valuation rings. b) Prove that if one has dim(A) = n and dim(A[X]) > n + 1, there exists a prime ideal p of A such that ht(p) = 1 and either dim(A_p[X]) > 2, or dim((A/p)[X]) > dim(A/p) + 1. c) Deduce that if A is Prüferian (VII, § 2, exerc. 12), one has dim(A[X₁, ..., Xₙ]) = dim(A) + n, for every integer n ≥ 0.