Exercise 16, § 2

Let $A$ be an integral ring and $K$ its field of fractions. The valuative dimension of $A$ is called and denoted by $\dim_v(A)$ the real number $\sup_{A \subset V \subset K} \dim(V)$, where $V$ runs through the set of valuation rings of $K$ containing $A$. a) Now prove that one has $\dim(A) \leq \dim_v(A)$ (exerc. 11). b) If $A$ is prüferian (VII, § 2, exerc. 12), prove that one has $\dim(A) = \dim_v(A)$. c) Prove that, for every $m \in \mathbf{N}$, one has

(One will remark that a valuation ring is prüferian and will use b) and exerc. 6, p. 84.)

d) Prove that if one has $\dim_v(A) = s < \infty$, one has for every integer $m \geqslant s - 1$ the equality

e) Let $s \in \mathbf{N}$. Prove that one has $\dim_v(A) = s$ if and only if $\dim(A[X_1, ..., X_s]) = 2s$ (use $d$) and exerc. 13 and 14). Deduce that, for every integral and noetherian ring $A$, one has $\dim(A) = \dim_v(A)$.