# Canonical Injection into Cartesian Product of Modules

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## Contents

## Theorem

Let $G$ be the cartesian product of a sequence $\sequence {G_n}$ of $R$-modules.

Then for each $j \in \closedint 1 n$, the canonical injection $\inj_j$ from $G_j$ into $G$ is a monomorphism.

## Proof

$G$ can be seen as functions:

- $\displaystyle f: A \to \bigcup_{a \mathop \in A} G_a$

Let $a \in A$.

Let $x, y \in G_a$.

Let $r \in R$.

So both $x + y \in G_a$ and $r x \in G_a$.

Let $b \in A$.

### Case 1

Let $b = a$.

Then:

- $\map {\map {\inj_a} {x + y} } b = x + y = \map {\map {\inj_a} x} b + \map {\map {\inj_a} y} b$

and:

- $\map {\map {\inj_a} {r x} } b = r x = r \, \map {\map {\inj_a} x} b$

$\Box$

### Case 2

Let $b \ne a$.

Then:

- $\map {\map {\inj_a} {x + y} } b = 0 + 0 = \map {\map {\inj_a} x} b + \map {\map {\inj_a} y} b$

and:

- $\map {\map {\inj_a} {r x} } b = 0 = \map r 0 = r \, \map {\map {\inj_a} x} b$

Therefore:

- $\map {\inj_a} {x + y} = \map {\inj_a} x + \map {\inj_a} y$

and:

- $\map {\inj_a} {r x} = r \, \map {\inj_a} x$

$\Box$

So $\inj_a$ is a homomorphism.

Combined with Canonical Injection is Injection gives that $\inj_a$ is a monomorphism.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 28$: Example $28.7$