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*}

Lemma 10.83.2. Let $M$ be an $R$-module. If $(M_{\alpha })_{\alpha \in S}$ is a direct sum dévissage of $M$, then $M \cong \bigoplus _{\alpha + 1 \in S} M_{\alpha + 1}/M_{\alpha }$.

Proof. By property (3) of a direct sum dévissage, there is an inclusion $M_{\alpha + 1}/M_{\alpha } \to M$ for each $\alpha \in S$. Consider the map

\[ f : \bigoplus \nolimits _{\alpha + 1\in S} M_{\alpha + 1}/M_{\alpha } \to M \]

given by the sum of these inclusions. Further consider the restrictions

\[ f_{\beta } : \bigoplus \nolimits _{\alpha + 1 \leq \beta } M_{\alpha + 1}/M_{\alpha } \longrightarrow M \]

for $\beta \in S$. Transfinite induction on $S$ shows that the image of $f_{\beta }$ is $M_{\beta }$. For $\beta =0$ this is true by $(0)$. If $\beta +1$ is a successor ordinal and it is true for $\beta $, then it is true for $\beta + 1$ by (3). And if $\beta $ is a limit ordinal and it is true for $\alpha < \beta $, then it is true for $\beta $ by (2). Hence $f$ is surjective by (1).

Transfinite induction on $S$ also shows that the restrictions $f_{\beta }$ are injective. For $\beta = 0$ it is true. If $\beta +1$ is a successor ordinal and $f_{\beta }$ is injective, then let $x$ be in the kernel and write $x = (x_{\alpha + 1})_{\alpha + 1 \leq \beta + 1}$ in terms of its components $x_{\alpha + 1} \in M_{\alpha + 1}/M_{\alpha }$. By property (3) and the fact that the image of $f_{\beta }$ is $M_{\beta }$ both $(x_{\alpha + 1})_{\alpha + 1 \leq \beta }$ and $x_{\beta + 1}$ map to $0$. Hence $x_{\beta +1} = 0$ and, by the assumption that the restriction $f_{\beta }$ is injective also $x_{\alpha + 1} = 0$ for every $\alpha + 1 \leq \beta $. So $x = 0$ and $f_{\beta +1}$ is injective. If $\beta $ is a limit ordinal consider an element $x$ of the kernel. Then $x$ is already contained in the domain of $f_{\alpha }$ for some $\alpha <\beta $. Thus $x=0$ which finishes the induction. We conclude that $f$ is injective since $f_{\beta }$ is for each $\beta \in S$. $\square$


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