Proper submodules of Noetherian modules are intersections of indecomposable modules 
If $M$ is Noetherian, any submodule $N \neq M$ can be written as a finite intersection of intersection-indecomposable ones.

I just don't know how to begin with the proof of this proposition. 
Very kind of you for reading or answering.
 A: The three equivalent definitions of "noetherian module" (in the presence of the Axiom of Choice) are:


*

*$M$ is noetherian if and only if every submodule of $M$ is finitely generated.

*$M$ is noetherian if and only if every ascending chain of submodules of $M$ stabilizes: $N_1\subseteq N_2\subseteq\cdots$ implies that there exists $m$ such that for all $k\geq m$, $N_m=N_k$.

*$M$ is noetherian if and only if every nonempty collection of submodules of $M$ has maximal elements.
So look at the collection of all proper submodules that are not finite intersections of intersection-indecomposable submodules. If this collection is empty, you are done. If it is not empty, then by (3) above it has a maximal element $N$. Note that $N$ cannot be intersection-indecomposable, nor can $N$ be maximal in $M$ (if $N$ is maximal in $M$, then it is intersection indecomposable). So there must exist two proper submodules $N_1$ and $N_2$ of $M$ such that $N=N_1\cap N_2$, and $N\neq N_1$, $N\neq N_2$. Since $N$ is maximal among submodules that are not finite intersections of intersection-indecomposable submodules, then neither $N_1$ nor $N_2$ belong to the collection. So $N_1$ is a finite intersection of intersection-indecomposable submodules, as is $N_2$, $N_1 = \cap_{j=1}^k M_j$, $N_2 = \cap_{j=k+1}^{\ell}M_j$. Then
$$N = N_1\cap N_2 = \cap_{j=1}^{\ell} M_j$$
is a finite intersection of intersection-indecomposable submodules, a contradiction. The contradiction arises from assuming that the collection of submodules that do not satisfy the condition is nonempty, so the collection is empty and we are done.
Note. Just before stating the lemma, Jacobson explicitly defines "indecomposable" as follows: "We shall call a submodule $N$ of $M$ (intersection) indecomposable if we cannot write $N$ as $N_1\cap N_2$ where $N_i\neq N$ for $i=1,2$." This is the definition I am using above, rather than the more common "not expressible as a direct sum of two proper submodules" (or even "subdirectly irreducible"). 
A: This is obviously false; perhaps you meant something else?  Consider a vector space of dimension 3 as the module M.  It is Noetherian, but its directly indecomposable submodules are all dimension 1.  If N has dimension 2, then N is not the intersection of one-dimensional submodules.
I suspect this is just a translation problem.  There is the Lasker–Noether theorem that expresses each proper submodule of a Noetherian module as an intersection of meet-irreducible (or primary) submodules.
See page 84 and page 102 of Lam's Lectures on Modules and Rings for my understanding of it, or the proper chapter and exercises in Atiyah-MacDonald for a more commutative view.
The part you asked about I think just follows from the definition of Noetherian and should be true in any lattice with appropriate finiteness condition.
In section 7.13 of Jacobson's Basic Algebra II (pagination differs a little, page 438 for me), Jacobson uses (intersection) indecomposable to mean what I have called meet-irreducible.  His proof is by Noetherian induction and holds in any Noetherian lattice.
