The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Length is additive in short exact sequences.

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 $M''$.

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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