Lemma 4

Let I be a Matlis A-module and M a $A$-module. The following conditions are equivalent :

(i) $M$ is artinian; (ii) every element of $M$ is annihilated by a power of $m_A$, and the socle of $M$ is of finite dimension over $\kappa_A$; (iii) there exist an integer $n \geq 0$ and an injective a-linear mapping of $M$ into $I^n$.

When these conditions are satisfied, every injective envelope of $M$ is isomorphic to $I^s$, where $s$ is the dimension over $\kappa_A$ of the socle of $M$.

(iii) $\Rightarrow$ (i): this is clear since the a-module $I$ is artinian (Lemma 3).

(i) $\Rightarrow$ (ii): suppose that $M$ is artinian. Let $x \in M$; the decreasing sequence of submodules $m_A^n x$ of $M$ is stationary. Let $n$ be an integer such that $m_A^{n+1} x = m_A^n x$; Nakayama's lemma implies that $m_A^n x = 0$. On the other hand, the socle of $M$ is artinian as an a-module, hence also as a $\kappa_A$-vector space, which means that it is of finite dimension.

(ii) $\Rightarrow$ (iii): suppose condition (ii) satisfied; let $e : M \to J$ be an injective envelope of $M$. We have $\mathrm{Ass}(M) \subset \{m_A\}$, hence $\mathrm{Ass}(J) \subset \{m_A\}$ (No. 1, Remark 2), and $J$ is isomorphic to $I^{(c)}$ for a cardinal $c$ (No. 1, Theorem 1). Let $x$ be a nonzero element of $J$ annihilated by $m_A$; since the a-module $Ax$ is simple and its intersection with $e(M)$ is not reduced to 0, $x$ belongs to $e(M)$. Thus $e$ induces an isomorphism of the socle of $M$ onto that of $J$; consequently the socle of $M$ is of dimension $c$, which proves (iii) as well as the last assertion.