Lemma 10.24.4. 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

is injective.

Lemma 10.24.4. 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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