Lemma 3

Let A be a ring and $n$ a natural number. An Artinian A-module M that is the sum of a family of submodules of length $\leqslant n$ has finite length.

We use induction on $n$. First, suppose $n= 1$. If M were not of finite length, then we could construct a sequence $(M_m)_{m\in\mathbf{N}}$ of submodules of M of length 1 with $M_m\not\subset (\sum_{i<m}M_i)$ for every $m\in \mathbf{N}$. We would then have $M_m\cap \sum_{i<m}M_i= 0$ for every $m$, and the sum of the family $(M_m)_{m\in\mathbf{N}}$ would be direct. However, this contradicts the fact that the A-module M is Artinian (VIII, p. 2, Example 2).

Now, suppose $n\geqslant 2$. Let $(M_i)_{i\in I}$ be a family of submodules of M of length $\leqslant n$ with sum M. For every $i\in I$, choose a submodule $M'_i$ of $M_i$ of length $\leqslant n-1$ such that $M_i/M'_i$ has length $\leqslant 1$. Set $M'=\sum M'_i$, and denote the image of $M_i$ in $M''= M/M'$ by $M''_i$. The modules $M''_i$ are of length $\leqslant 1$, and their sum is $M''$. The modules $M'$ and $M''$ are Artinian (VIII, p. 3, Proposition 3); by the induction hypothesis, they are of finite length. Hence, M has finite length (II, §1, No. 10, p. 212, Proposition 16).

Let us now prove Theorem 1. Let $\mathscr{S}$ denote the set of left ideals $\mathfrak{a}$ of A such that the module $A_s/\mathfrak{a}$ has finite length. Let $(\mathfrak{a}_i)_{i\in I}$ be a finite family of elements of $\mathscr{S}$. By Proposition 1 of VIII, p. 2, the A-module $A_s/\mathfrak{a}_i$ is Artinian and Noetherian for every $i\in I$. Consequently, $A_s/\bigcap_{i\in I}\mathfrak{a}_i$ is Artinian and Noetherian (VIII, p. 4, Corollary of Proposition 3), hence of finite length (VIII, p. 2, Proposition 1). This proves that $\mathscr{S}$ is a left directed set for the inclusion. Since the ring A is left Artinian, the set $\mathscr{S}$ has a least element $\mathfrak{b}$. We denote the length of the A-module $A_s/\mathfrak{b}$ by $n$.

Let $x$ be an element of $A_s$ and $\mathfrak{a}$ its annihilator (II, §1, No. 12, p. 219). The A-module $Ax$ is isomorphic to $A_s/\mathfrak{a}$. If $Ax$ has finite length, then $\mathfrak{a}$ belongs to $\mathscr{S}$, so $\mathfrak{a}$ contains $\mathfrak{b}$ and $Ax$ has length $\leqslant n$. Thus, every monogenous left ideal of A of finite length has length $\leqslant n$. Let $\mathfrak{c}$ be the sum of these ideals; this is a left ideal of A, of finite length by Lemma 3. Every left ideal of A of finite length is a sum of monogenous left ideals of finite length and is therefore contained in $\mathfrak{c}$. Hence, $\mathfrak{c}$ is the largest left ideal of A of finite length.

If $\mathfrak{c}$ were distinct from A, then the set of left ideals of A containing $\mathfrak{c}$ and distinct from $\mathfrak{c}$ would have a minimal element $\mathfrak{c}'$. The A-module $\mathfrak{c}'/\mathfrak{c}$ would then have length 1, and $\mathfrak{c}'$ would have finite length, which contradicts the fact that $\mathfrak{c}$ is maximal. We therefore have $\mathfrak{c}= A$; the A-module $A_s$ has finite length.