A module has a composition series if and only if it is Noetherian and Artinian, the converse using dependent choice

Let $0=M_0<\cdots <M_n=M$ be a composition series [L1] and induct on $i$ that $M_i$ is Noetherian and Artinian. The zero module $M_0$ satisfies both conditions vacuously. A simple factor $M_i\mathrm{/}{M}_{i-1}$ has only the submodules $0$ and itself, so every chain of its submodules stabilizes and it too satisfies both conditions. Applying [L4] to the short exact sequence $0\to M_{i-1}\to M_i\to M_i\mathrm{/}{M}_{i-1}\to 0$ carries both conditions from $M_{i-1}$ and the simple factor to $M_i$. At $i=n$ this gives the forward implication, which uses no choice principle.

Conversely, assume dependent choice and let $M$ be Noetherian and Artinian. Every submodule of $M$ is Noetherian by [L4], so a nonzero submodule $N$ has a nonempty family of proper submodules, which by the maximal condition of [L2] has a maximal member — a maximal proper submodule of $N$. Dependent choice applied to this relation, starting at $M$, yields a chain $M=N_0>N_1>\cdots$ in which $N_{k+1}$ is a maximal proper submodule of $N_k$ for as long as $N_k\mathrm{\ne }0$.

The chain of step 2.1 is strictly descending while its terms are nonzero, so the descending chain condition forces some $N_r=0$. Since $N_{k+1}$ is maximal proper in $N_k$, the quotient $N_k\mathrm{/}{N}_{k+1}$ is nonzero and has no proper nonzero submodule, hence is simple. Reversing the chain gives $0=N_r<\cdots <N_0=M$, a composition series. For $M=0$ the empty chain is already the required series, so no choice is consumed in that case. This proves the stated claim.