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 10.23.4. Let $R$ be a ring. Let $f_1, \ldots , f_ n \in R$. Let $M$ be an $R$-module. Then $M \to \bigoplus M_{f_ i}$ is injective if and only if

\[ M \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} M, \quad m \longmapsto (f_1m, \ldots , f_ nm) \]

is injective.

Proof. The map $M \to \bigoplus M_{f_ i}$ is injective if and only if for all $m \in M$ and $e_1, \ldots , e_ n \geq 1$ such that $f_ i^{e_ i}m = 0$, $i = 1, \ldots , n$ we have $m = 0$. This clearly implies the displayed map is injective. Conversely, suppose the displayed map is injective and $m \in M$ and $e_1, \ldots , e_ n \geq 1$ are such that $f_ i^{e_ i}m = 0$, $i = 1, \ldots , n$. If $e_ i = 1$ for all $i$, then we immediately conclude that $m = 0$ from the injectivity of the displayed map. Next, we prove this holds for any such data by induction on $e = \sum e_ i$. The base case is $e = n$, and we have just dealt with this. If some $e_ i > 1$, then set $m' = f_ im$. By induction we see that $m' = 0$. Hence we see that $f_ i m = 0$, i.e., we may take $e_ i = 1$ which decreases $e$ and we win. $\square$


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