Finite modules over Noetherian rings admit prime filtrations

If $M=0$, then the empty chain $0=M$ is already a prime filtration.

Assume $M\mathrm{\ne }0$. Let $\mathrm{\Sigma }=\{N\le M:N\text{ admits a prime filtration}\}\mathrm{.}$ By step 1.1, the zero submodule belongs to $\mathrm{\Sigma }$. Since $M$ is Noetherian by [L1], [L2] gives a maximal member $N$ of $\mathrm{\Sigma }$. This use of the maximal condition spends DC, supplied by the assumed AC; AC also supplies the choice assumptions of the associated-prime input in the next step.

If $N\mathrm{\ne }M$, then $M\mathrm{/}N$ is a nonzero quotient module by [L5]. Fact [L3] gives an associated prime $\mathfrak{p}$ of $M\mathrm{/}N$, and [L4] yields an embedded copy of $R\mathrm{/}\mathfrak{p}$ in $M\mathrm{/}N$. Let $N^{\mathrm{\prime }}$ be its preimage in $M$. Then $N⊊N^{\mathrm{\prime }}\le M$ and $N^{\mathrm{\prime }}\mathrm{/}N\cong R\mathrm{/}\mathfrak{p}$. Appending $N^{\mathrm{\prime }}$ to a prime filtration of $N$ gives a prime filtration of $N^{\mathrm{\prime }}$, contradicting the maximality of $N$ in step 2.1. Therefore $N=M$.

When $M\mathrm{\ne }0$, step 3.1 shows that $M$ itself has a prime filtration; the zero case was handled in step 1.1.