Proposition 12

Let M be an A-module of finite length. Then the map which sends every submodule N of M to $N^0$ is a bijection from $\mathfrak{S}(M)$ onto $\mathfrak{S}(D(M))$, and the inverse bijection sends every submodule N' of D(M) to the submodule ${N'}^0$ of M; the module $D(N)$ is naturally identified with $D(M)/N^0$ and $D(M/N)$ with $N^0$. Moreover, we have

(2)

for all submodules $N_1, N_2$ of M.

For each submodule N of M, there is an exact sequence

and hence (Prop. 11) an exact sequence

and since the image of $D(M/N)$ in $D(M)$ is obviously $N^0$, we see (Prop. 10) that $\operatorname{long}(N^0) = \operatorname{long}(M) - \operatorname{long}(N)$; since $M$ is identified with $D(D(M))$ by Prop. 10, we have similarly

in addition it is clear that $N \subset N^{00}$, so $N^{00} = N$. Furthermore the first relation in (2) is obvious, and by applying it to the submodules $N_1^0$ and $N_2^0$ of $D(M)$, we have $(N_1^0 + N_2^0)^0 = N_1 \cap N_2$, so $N_1^0 + N_2^0 = (N_1^0 + N_2^0)^{00} = (N_1 \cap N_2)^0$. This completes the proof of the proposition.