Exercise 6, § 4

A module M is called semi-Artinian if every nonzero quotient of M contains a simple submodule. Every Artinian module and every vector space over a field is semi-Artinian.

a) Let $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$ be an exact sequence. Then M is semi-Artinian if and only if the same holds for $M'$ and $M''$ (if M is semi-Artinian, consider a maximal element in the set of submodules N of M such that $N\cap M'= 0$).

b) The direct sum of a family of semi-Artinian modules is a semi-Artinian module. c) Let K be a field and A the product ring $K^{\mathbf{N}}$. Prove that the A-module $A_s$ is not semi-Artinian even though it is a product of simple A-modules (if $\mathfrak{s}$ is the socle of $A_s$, prove that the A-module $A_s/\mathfrak{s}$ does not contain any simple modules).

d) Let A be a left Noetherian ring and M a semi-Artinian A-module. Prove that M is the sum of its Artinian submodules (let S be the sum of the Artinian submodules of M; observe that if N is a finitely generated submodule of M with simple image in $M/S$, then $N\cap S$ is Artinian). In particular, if M is finitely generated, then it is Artinian. e) Let D be a field, V an infinite-dimensional D-vector space, and A the subring of End$_D(V)$ generated by the center of D and the endomorphisms of finite rank. Prove that $A_s$ is semi-Artinian (use a) and b)) but is not the sum of its Artinian submodules.