Canonical Injection into Cartesian Product of Modules/Proof 2
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Theorem
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +, \circ}_R$ be the cartesian product of a sequence $\sequence {\struct {G_n, +_n, \circ_n}_R}$ of $R$-modules.
Then for each $j \in \closedint 1 n$, the canonical injection $\inj_j$ from $\struct {G_j, +_j, \circ_j}_R$ into $\struct {G, +, \circ}_R$ is a monomorphism.
Proof
$G$ can be seen as functions:
- $\ds f: A \to \bigcup_{a \mathop \in A} G_a$
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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
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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.
$\blacksquare$