Lemma 12.9.6. Let $\mathcal{A}$ be an abelian category. Let $A$ be an object of $\mathcal{A}$. The following are equivalent
$A$ is Artinian and Noetherian, and
there exists a filtration $0 \subset A_1 \subset A_2 \subset \ldots \subset A_ n = A$ by subobjects such that $A_ i/A_{i - 1}$ is simple for $i = 1, \ldots , n$.
Proof. Assume (1). If $A$ is zero, then (2) holds. If $A$ is not zero, then there exists a smallest nonzero object $A_1 \subset A$ by the Artinian property. Of course $A_1$ is simple. If $A_1 = A$, then we are done. If not, then we can find $A_1 \subset A_2 \subset A$ minimal with $A_2 \not= A_1$. Then $A_2/A_1$ is simple. Continuing in this way, we can find a sequence $0 \subset A_1 \subset A_2 \subset \ldots $ of subobjects of $A$ such that $A_ i/A_{i - 1}$ is simple. Since $A$ is Noetherian, we conclude that the process stops. Hence (2) follows.
Assume (2). We will prove (1) by induction on $n$. If $n = 1$, then $A$ is simple and clearly Noetherian and Artinian. If the result holds for $n - 1$, then we use the short exact sequence $0 \to A_{n - 1} \to A_ n \to A_ n/A_{n - 1} \to 0$ and Lemmas 12.9.4 and 12.9.5 to conclude for $n$. $\square$
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