Length is additive in short exact sequences.

Lemma 10.52.3. 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 $M''$.

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. $\square$

Comment #3587 by slogan_bot on

Suggested slogan: Length is additive in short exact sequences.

There are also:

• 2 comment(s) on Section 10.52: Length

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).