Lemma 10.84.2. 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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