Showing a particular quotient is noetherian/artinian

Question

I'm trying to show and answer the following:

Let $M$ be an $R$-module and $L,N$ submodules of $M$ such that $M/L$ and $M/N$ are noetherian/artianian. Then $\frac{M}{L \cap N}$ is noetherian/artinian. If both former quotients are of finite length, what is the length of the latter?

Now, as it stands, for the noetherian case for instance, I imagine that given a chain $$\frac{0+(L \cap N)}{L \cap N} \subseteq \frac{M_1+(L \cap N)}{L \cap N} \subseteq ... \subseteq \frac{M_n+(L \cap N)}{L \cap N} \subseteq ...$$ of submodules of the $\frac{M}{L \cap N}$, if I assume that for some positive integer $k$ both $$\frac{M_k+L}{L}=\frac{M_{k+1}+L}{L}=...$$ and $$\frac{M_k+N}{N}=\frac{M_{k+1}+N}{N}=...,$$ I'll be able to show that for integers $n \geq k$, we'll have $$\frac{M_n+(L \cap N)}{L \cap N} = \frac{M_{n+1}+(L \cap N)}{L \cap N}.$$

But my calculations for showing the nontrivial inclusion do not seem to be working. Aditionally, I imagine there is some better way to show this which does not involve such calculations (using the Isomorphism Theorems, I guess?), but it seems I'm not able to see what to do.

Any help is appreciated, thanks in advance.

Answer 1

I'll just leave an answer here which was suggested to me in the comments by another user. The comments are now gone.

Anyway, the sugestion was to look at the the short exact sequence $$0 \to \frac{M}{L \cap N} \to \frac{M}{L} \oplus \frac{M}{N} \to \frac{M}{L+N} \to 0,$$ which arises as the corkernel part of the Snake Lemma applied to the short exact sequences $$0 \to L \cap N \to L \oplus N \to L+N \to 0$$ (given by $x \to (x,x)$ and $(x,y) \to x-y$) and $$0 \to M \to M \oplus M \to M \to 0.$$ This gives both the result that $\frac{M}{L \cap N}$ is noetherian/artinian under the suitable hypothesis, and its lenght in terms of $\frac{M}{L} \oplus \frac{M}{N}$ and $\frac{M}{L+N}$.

(The following link came with said comments: Some exact sequence of ideals and quotients).

Answer 2

First, I want to say I was just interested in this question and thought I would give it a try, so it might be wrong, but I didn't want to start a new question just to ask for a proof check, so I'll post my attempt here.

Observe the lattice of submodules of $M$ - clearly $N+L$, $N$, $L$ and $0$ form a rhombus in the lattice.

We use an isomorphism theorem to show that $\frac{N+L}{N} \simeq \frac{L}{L\cap N}$, and since $\frac{N+L}{N}$ is a submodule of $\frac{M}{N}$ which is Noetherian, $\frac{L}{L\cap N}$ is also Noetherian.

Using an isomorphism theorem again, we get $\frac {\frac{M}{L\cap N}} {\frac{L}{L\cap N}} \simeq \frac{M}{L}$, and since $\frac{M}{L}$ is Noetherian, so is $\frac {\frac{M}{L\cap N}}{\frac{L}{L\cap N}} $.

Now setting $M_0=0$, $M_1=\frac{L}{L\cap N}$, $M_2= {\frac{M}{L\cap N}} $, and using the theorem:

Theorem: If a non-decreasing series $M_i$, $i \in \{0,..,n\}$ of submodules of $M$ satisfies $M_0 = 0$, $M_n=M$, and $\frac{M_{i+1}} {M_i}$ is Noetherian (Artinian) for $i\in\{0,..,n-1\}$, then $M$ is Noetherian (Artinian).

Therefore $\frac{M}{L\cap N}$ is Noetherian.

If both $\frac{M}{L}$ and $\frac{M}{N}$ are of finite length, looking at the lattice of submodules of $\frac{M}{L\cap N}$, we can find a composition series of $\frac{M}{L\cap N}$ passing through $\frac{L}{L\cap N}$, and so it's length is $l[0,\frac{L}{L\cap N}] + l[\frac{L}{L\cap N},\frac{M}{L\cap N}]$ ($l$ being the length of the respective lattice interval - length of some composition series inside of it, when applied to a lattice interval, and $l$ being the length of module, when applied on a finite length module).

We can find appropriate isomorphisms (using, the isomorphism theorem again), such that these isomorphisms induce us the lattice isomorphisms (isomorphisms between certain lattices and intervals in certain lattices) that we need, and so we can see, that

$l[0,\frac{L}{L\cap N}] = l(\frac{L}{L\cap N})=l(\frac{N+L}{N})$, and $l[\frac{L}{L\cap N},\frac{M}{L\cap N}] = l(\frac {\frac{M}{L\cap N}} {\frac{L}{L\cap N}})=l(\frac{M}{L})$.

Because length of a module is equal to the lenght of any of it's composition series, length of $\frac{M}{L\cap N}$ is $l(\frac{N+L}{N}) + l(\frac{M}{L})$.