Theorem 2

Let $M$ be an $A$-module.

a) In order that $M$ be artinian, it is necessary and sufficient that the $\widehat{A}$-module $D_A(M)$ be of finite type. When this is so, the homomorphism $\alpha_M$ is bijective.

b) For $M$ to be of finite type, it is necessary and sufficient that $D_A(M)$ be artinian (as an $A$-module or as a $\widehat{A}$-module). In this case the homomorphism $\widehat{\alpha}_M$ is an isomorphism.

c) For $M$ to be of finite length, it is necessary and sufficient that $D_A(M)$ be of finite length (as an $A$-module or as a $\widehat{A}$-module). In this case $\alpha_M$ is an isomorphism of $M$ onto $D_A(D_A(M))$, and one has $\operatorname{long}_A(D_A(M)) = \operatorname{long}_A(M)$.

Let us first prove that the homomorphism $\alpha_M$ is injective for every $A$-module $M$. Let $m$ be a nonzero element of $M$; its annihilator is contained in $m_A$. There therefore exists a surjective $A$-homomorphism of $Am$ onto $\kappa_A$, and consequently a nonzero homomorphism of $Am$ into $I$. Since $I$ is injective, this extends to a homomorphism $f : M \to I$ such that $f(m) \neq 0$. This proves the injectivity of $\alpha_M$.

Suppose that the $A$-module $M$ is artinian. By lemma 4, there exist an integer $r$ and an injective $A$-linear mapping $f : M \to I^r$. The homomorphism $D_A(f) : D_A(I^r) \to D_A(M)$ is then surjective; since $D_A(I^r)$ is identified with $\widehat{A}^r$, this proves that the $\widehat{A}$-module $D_A(M)$ is of finite type. Similarly, if $M$ is of finite type, there exist an integer $n$ and a surjective homomorphism $u : A^n \to M$; the homomorphism $D_A(u) : D_A(M) \to I^n$ is injective, so that $D_A(M)$ is artinian (as an $A$-module or as a $\widehat{A}$-module).

Suppose that $M$ is artinian. There exist an integer $r$ and an injective $A$-linear mapping $f : M \to I^r$; since $I$ is artinian (lemma 3), the $A$-module $\operatorname{Coker}(f)$ is also so, and one can find an integer $s$ and an exact sequence of $A$-modules

Hence one deduces a commutative diagram with exact rows

The $\widehat{A}$-module $D_{\widehat{A}}(D_A(I))$ is identified with $I$ and $\alpha_I$ with the identity mapping; hence $\alpha_{I^r}$ and $\alpha_{I^s}$ are bijective, and analogously so is $\alpha_M$ (A, X, p. 7, Corollary 3).

If the $A$-module $M$ is finitely generated, there exist integers $m$ and $n$ and an exact sequence of $A$-modules

from which one deduces a commutative diagram with exact rows

Since $\widehat{\alpha}_A$ is equal to $1_{\widehat{A}}$, it follows that $\widehat{\alpha}_M$ is an isomorphism.

It remains to prove the equality $\operatorname{long}_A(M) = \operatorname{long}_A(D_A(M))$ when $M$ is of finite length. We may suppose $M \neq 0$; there then exists an exact sequence

whence one deduces an exact sequence

One has $\operatorname{long}_A(M) = \operatorname{long}_A(N) + 1$ and

one concludes by induction on the integer $\operatorname{long}_A(M)$.