Exercise 5, § 1

If A is a ring (not necessarily commutative), M a left A-module, denote by Kdev(M) the deviation of the set of submodules of M, ordered by inclusion. Put Kdev(A) = Kdev(A_s). a) If N is a submodule of an A-module M, one has

b) Suppose A is commutative. If p and q are two prime ideals of A such that $p \subset q$ and $p \neq q$, one has $\operatorname{dev}(A/p) > \operatorname{dev}(A/q)$. (Use the ideals $p + Ax$, where x is an element of $q - p$.) Deduce $\dim(A) \leq \operatorname{Kdev}(A)$. c) Suppose A is commutative and noetherian. Exercise prove that one has $\dim(A) = \operatorname{Kdev}(A)$ (reason by induction on $\dim(A)$, assuming that $\operatorname{Kdev}(B) \leq \dim(B)$ for every commutative noetherian ring B such that $\dim(B) < \dim(A)$; let S be the set of $s \in A$ which do not belong to any of the prime ideals $p$ of A such that $\dim(A) = \dim(A/p)$; now prove that, for every finitely generated A-module M such that $S^{-1}M = 0$, one has Kdev(M) < dim(A); using the fact that $S^{-1}A$ is artinian, deduce that Kdev(A) \leq dim(A)). Proving that one has dim(M) = Kdev(M) for every finitely generated A-module M.