If $0 \rightarrow L \xrightarrow{f} M \xrightarrow{g} N \rightarrow 0$ is a short exact sequence of $A$-modules, I want to show that $\ell_A(M) = \ell_A(L) + \ell_A(N)$. It can be shown that $N$ is isomorphic to $M/L$ (this may be abuse of notation, I'm just using shorthand since $L \cong f(L)$), by the definition of a short exact sequence. So any chain of submodules in $N$ is a chain of submodules in $M/L$.
First, $M$ has finite length iff it is Artinian, iff $L$ and $N$ are Artinian, iff they also have finite length. So we can assume they all have finite length $\ell_A(L) = s$, $\ell_A(N) = t$, and $\ell_A(M) = n$.
Let $0 \subset L_1 \subset \ldots \subset L_t = L$ and $0 \subset M_1/L \subset \ldots \subset M_s/L = M/L \cong N$. I think I want to show that I can combine the sequences $L_i$ and $M_i$ to make a chain of submodules of $M$ of maximal length. Clearly $L \subset M_1$. I have a hint that I can use Schreier's Theorem on composition series of groups, but I'm not sure which two composition series to compare to get equivalent refinements.
I'm new to commutative algebra and I'm sure I've made a notational mistake in here somewhere. Any hints are appreciated. Thank you.
One direction is easy: From $\ldots\subset L_i\subset\ldots$ and $\ldots\subset N_j\subset\ldots$, we obtain $\ldots\subset f(L_i)\subset\ldots\subset g^{-1}(N_j)\subset\ldots$ and conclude $\ell_A(M)\ge \ell_A(L)+\ell_A(N)$.
Assume we have $0\subset M_1\subset\ldots\subset M_n$. The sequences $g(M_i)$ and $f^{-1}(M_i)$ are also ascending, but not necessarily strictly ascending. In the sequence of the $g(M_i)$, at most $\ell_A(N)+1$ different submoduls (including $0$) of $N$ occur. But every time we have $g(M_i)=g(M_{i-1})$, we must have $M_{i-1}\cap \ker g\subset M_{i}\cap \ker g$, i.e., $f^{-1}(M_{i-1})\subset f^{-1}(M_i)$. Thus after removing duplicates, we obtain a chain of length $n-\ell_A(N)$ in $L$. We conclude $n-\ell_A(N)\le \ell_A(L)$, $\ell_A(M)\le \ell_A(L)+\ell_A(N)$.