Exercise 23, § 2

Let A be a ring. An A-module $E$ is called Laskerian if it finitely generated and if every submodule of E has a primary decomposition in E. (a) For a finitely generated A-module E to be Laskerian, it is necessary and sufficient that it satisfy the two following axioms: (LA,) For every submodule F of E and every prime ideal p of A, the saturation of F with respect to p in E is of the form $F : (a)$ for some $a \notin p$. (LA,) For every submodule F of E, every decreasing sequence $(\mathrm{sat}_{S_n}(F))$ (where $S_n$ is any decreasing sequence of multiplicative subsets of $\mathbf{A}$) is stationary. (To show that the conditions are sufficient, prove first, using (LA,) and Exercises 18 (a) and 22 (a), that, for every submodule F of E, there exists a submodule Q of E which is primary for some ideal $p \supset F : E$ and a submodule $G = F + aE$, where $a \notin p$, such that $F = Q \cap G$. Then argue by reductio ad absurdum: show that there would exist an infinite sequence $(Q_n)$ of $p_n$-primary submodules of E and a strictly increasing sequence $(G_n)$ of submodules of E such that, if we write $H_n = Q_1 \cap Q_2 \cap \ldots \cap Q_n$: (1) $F = H_n \cap G_n$; (2) $G_n$ is maximal among the submodules G containing $G_{n-1}$ and such that

F = H_n \cap G; (3) there exists $a, \notin p_n$ such that $a_n E \subset G_n$. Show then that, if $S_n = \bigcap_{1 \leq i \leq n} (A - p_i)$, S, meets $G_n : E$ and deduce that $\operatorname{sat}_{S_n}(F) = H_n$. Conclude with the aid of (LA_{II}).)

(b) Give an example of a ring A such that the A-module A does not satisfy (LA,) but where every ideal $a \subset A$ is irreducible and the set of saturations $\operatorname{sat}_S(a)$ for all the multiplicative subsets S of A is finite (cf. Exercise 19).*

(c) Let K be a field, $B = K[[X]]$ the algebra of formal power series over K and m the maximal ideal of B consisting of the formal power series without constant term. In the product algebra $B^\mathbf{N}$, consider the subalgebra A generated by 1 and the ideal $n = m^{(\mathbf{N})}$. Show that in A the only prime ideals distinct from n are the ideals the direct sums in n of all the component ideals except one; every strictly increasing sequence of prime ideals of A then has at most two elements. Let $a \neq A$ be an ideal of A and let S be a multiplicative subset of A not meeting a; if a_n (resp. S_n) is the projection of a (resp. S) onto the n-th factor B, of $B^\mathbf{N}$, show that $\operatorname{sat}_S(a)$ is the direct sum of the ideals $\operatorname{sat}_{S_n}(a_n)$ (observe that, if $s \in S, s_n \in S,$ is the n-th projection of s and $x_n \in \operatorname{sat}_{S_n}(a_n)$ is such that $s_n x_n \in a,$ then $s^2 x_n \in a$). Deduce that A satisfies axiom (LA$_I$) but not (LA$_{II}$) (use Exercise 20 (b)).

(d) Show that, if an A-module E satisfies axiom (LA$_I$), every submodule of E is the intersection of a family (finite or otherwise) of primary submodules in E (argue as in (a), by transfinite induction).

(e) Let E be a finitely generated A-module and $E' \subset E$ a finitely generated submodule. For E to be Laskerian, it is necessary and sufficient that E' and $E/E'$ be so. In particular, if A is a Laskerian ring, every finitely generated A-module is Laskerian. If E is a Laskerian A-module, $S^{-1}E$ is a Laskerian $S^{-1}A$-module for every multiplicative subset S of A.