Exercise 7, § 2

Let B be an integrally closed ring, with field of fractions L. Put d = dim(B) and t = dim(B[X]). Let L’ be a nontrivial extension of L in which L is algebraically closed and let V be a discrete valuation ring (which is not a field) having residue field L’. We shall denote by K the field of fractions of V and by A the set of elements of V whose image in L’ under the canonical homomorphism V → L’ belongs to B. The ring A is integrally closed with field of fractions K. Let p denote the kernel of the surjective homomorphism A → B. Now prove that every nonzero prime ideal of A contains p, whence dim(A) = dim(B) + 1. Considering an element x of V whose image in L’ does not belong to L, show that the ring of fractions A_p is not a valuation ring, and that pA_p[x] is not minimal among the nonzero prime ideals of A_p[x]. Using Exerc. 6, deduce that dim(A[x]) ≥ t + 2. Conclude that dim(A[x]) = t + 2 by a direct argument. Finally, show that for every pair (d, t ) of integers with d + 1 ≤ t ≤ 2d + 1, there exists an integral domain A of dimension d such that dim(A[X]) = t.