Proposition 4

For the a-module M to be of finite length, it is necessary and sufficient that D(M) be of finite length. One then has $z_0(D(M)) = z_0(M)$, $\operatorname{long}_A(M) = \operatorname{long}_A D(M)$, $\operatorname{Ann}_A(M) = \operatorname{Ann}_A(D(M))$, and the a-linear mapping $\alpha_M$ is bijective.

For every $m \in \Omega$, the $A_m$-module $D(M)_m$ is identified with $\operatorname{Hom}_{A_m}(M_m, J_m)$ (II, § 2, No. 7, prop. 19); the first assertion of the proposition then follows from th. 2, c) and from the characterization of modules of finite length given above. Henceforth suppose M to be of finite length; one has $\operatorname{long}_{A_m}(D(M)_m) = \operatorname{long}_{A_m}(M_m)$ for every $m \in \Omega$ (loc. cit.), whence $z_0(D(M)) = z_0(M)$ and $\operatorname{long}_A(D(M)) = \operatorname{long}_A(M)$. Moreover the mapping $(\alpha_M)_m : M_m \to D(D(M))_m$ is identified with the canonical homomorphism $\alpha_{M_m}$, which is bijective (loc. cit.); consequently $\alpha_M$ is bijective.

For every $a \in A$ one has $D(a_M) = a_{D(M)}$ and consequently $\operatorname{Ann}_A(M) \subset \operatorname{Ann}_A(D(M))$. Applying this to the a-module D(M), one deduces the opposite inclusion, whence the equality $\operatorname{Ann}_A(M) = \operatorname{Ann}_A(D(M))$.