Lemma 5

Every artinian $\widehat{A}$-module is artinian as a $A$-module.

Let $M$ be an artinian $\widehat{A}$-module; every element of $M$ is annihilated by a power of $m_{\widehat{A}}$, hence by a power of $m_A$. By Lemma 2 of No. 2, the sub-A-modules of $M$ are its sub-$\widehat{A}$-modules, hence $M$ is artinian as an A-module.

Let us now fix a Matlis $A$-module I. For every $A$-module $M$, let us denote by $D_A(M)$ the $\widehat{A}$-module

The $\widehat{A}$-module $D_A(A)$ is canonically identified with $I$, the $\widehat{A}$-module $D_A(I)$ with $\widehat{A}$ ($n^\circ 2$, prop. 3), and the $\widehat{A}$-module $D_A(\kappa_A)$ with $I_1$ (loc. cit.).

For every $A$-linear mapping $f : M \to N$, we shall denote by

the $\widehat{A}$-linear mapping $\operatorname{Hom}_A(f, 1_I)$. Since the $A$-module $I$ is injective, the sequence $(D_A(g), D_A(f))$ is exact for every exact sequence $(f, g)$ of $A$-linear mappings.

We shall apply these definitions to the ring $\widehat{A}$ endowed with the Matlis module I (lemma 3, a)) ; for every $\widehat{A}$-module $P$, $D_{\widehat{A}}(P)$ is therefore the $\widehat{A}$-submodule $\operatorname{Hom}_{\widehat{A}}(P, I)$ of $D_A(P)$. It amounts to the same thing to say that $P$ is artinian as an $A$-module or as a $\widehat{A}$-module (lemma 5) ; if this is the case, one has $D_{\widehat{A}}(P) = D_A(P)$ (loc. cit.).

Let $M$ be an $A$-module. For $m \in M$, the mapping $f \mapsto f(m)$ of $D_A(M)$ into $I$ is $\widehat{A}$-linear ; let us denote it by $\alpha_M(m)$. Thus one defines an $A$-linear homomorphism

We denote by $\widehat{\alpha}_M : \widehat{A} \otimes_A M \longrightarrow D_{\widehat{A}}(D_A(M))$ the $\widehat{A}$-linear mapping deduced from $\alpha_M$.