Canonical Injection into Cartesian Product of Modules

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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 1

To demonstrate that $\inj_j$ is an epimorphism, we need to show that:

$(1): \quad \inj_j$ is an injection
$(2): \quad \forall x, y \in G_j: \map {\inj_j} {x +_j y} = \map {\inj_j} x + \map {\inj_j} y$
$(3): \quad \forall x_j \in G_j: \forall \lambda \in R: \map {\inj_j} {\lambda \circ_j x_j} = \lambda \circ \map {\inj_j} {x_j}$

Criteria $(1)$ and $(2)$ are a direct application of Canonical Injection is Monomorphism.


Let $x_j \in G_j$ be arbitrary.

Then we have:

\(\ds \forall x \in G_j: \forall \lambda \in R: \, \) \(\ds \map {\inj_j} {\lambda \circ_j x_j}\) \(=\) \(\ds \tuple {e_1, e_2, \ldots, \lambda \circ_j x_j, \ldots, e_n}\) Definition of Canonical Injection
\(\ds \) \(=\) \(\ds \lambda \circ \tuple {e_1, e_2, \ldots, x_j, \ldots, e_n}\) Definition of Cartesian Product
\(\ds \) \(=\) \(\ds \lambda \circ \map {\inj_j} {x_j}\) Definition of Canonical Injection

$\blacksquare$


Proof 2

$G$ 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.

$\blacksquare$


Sources

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