Proving that the length of the sum and intersection of two finite length modules is finite.

Question

I am trying to solve the following question:

Let $M_1, M_2 \subset M$ be finite length submodules of a module $M$ (where $M$ need not be of finite length). Show that $ \ M_1 + M_2 = \{x_1 + x_2 \in M \, | \, x_i \in M_i\}$ and $M_1 \cap M_2$ have finite length and that $l(M_1) +l(M_2) = l(M_1 \cap M_2) + l(M_1 + M_2).$

I am really not sure how to prove this and was hoping someone could help me?

Answer 1

The main tool for this is the following:

If $0\to L\to M\to N\to 0$ is a short exact sequence of modules, then $M$ has finite length if and only if both $L$ and $N$ have finite length.

(The proof is similar to the one where you replace "has finite length" by "is Noetherian/Artinian"; see Pete Clark's answer.) Moreover, you can show that in that case, the length of $M$ is equal to the sum of the lengths of $L$ and $N$.

Now, for your situation, use the following short exact sequences: $$0\to M_1\to M_1\oplus M_2\to M_2\to 0,$$ and $$ 0\to M_1\cap M_2\to M_1\oplus M_2\to M_1+M_2\to 0. $$

Answer 2

Suggestion: use the result that lengths are additive in short exact sequences: if

$0 \rightarrow M_1 \rightarrow M \rightarrow M_2 \rightarrow 0$,

then $\ell(M) = \ell(M_1) + \ell(M_2)$: see e.g. $\S 8.4$ of my commutative algebra notes. Now the proof of this result is left as an exercise in my notes, but it is meant to be a rather straightforward exercise: you should get some clues at least by looking at what occurs just before this statement.