Lemma 10.51.3. If $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of modules over $R$ then the length of $M$ is the sum of the lengths of $M'$ and $M''$.

** Length is additive in short exact sequences. **

**Proof.**
Given filtrations of $M'$ and $M''$ of lengths $n', n''$ it is easy to make a corresponding filtration of $M$ of length $n' + n''$. Thus we see that $\text{length}_ R M \geq \text{length}_ R M' + \text{length}_ R M''$. Conversely, given a filtration $M_0 \subset M_1 \subset \ldots \subset M_ n$ of $M$ consider the induced filtrations $M_ i' = M_ i \cap M'$ and $M_ i'' = \mathop{\mathrm{Im}}(M_ i \to M'')$. Let $n'$ (resp. $n''$) be the number of steps in the filtration $\{ M'_ i\} $ (resp. $\{ M''_ i\} $). If $M_ i' = M_{i + 1}'$ and $M_ i'' = M_{i + 1}''$ then $M_ i = M_{i + 1}$. Hence we conclude that $n' + n'' \geq n$. Combined with the earlier result we win.
$\square$

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