Direct Product of Modules is Module

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Theorem

Let $R$ be a ring.

Let $\left\{\left\langle {M_i,+_i,\circ_i} \right\rangle\right\}_{i \in I}$ be a family of $R$-modules.

Let $\left\langle{M, +, \circ}\right\rangle$ be their direct product.


Then $\left\langle{M, +, \circ}\right\rangle$ is a module.


Proof

From External Direct Product of Abelian Groups is Abelian Group it follows that $(M,+)$ is an abelian group.

We need to show that:

$\forall x, y, \in M, \forall \lambda, \mu \in R$:

$(1): \quad \lambda \circ \left({x + y}\right) = \left({\lambda \circ x}\right) + \left({\lambda \circ y}\right)$
$(2): \quad \left({\lambda +_R \mu}\right) \circ x = \left({\lambda \circ x}\right) + \left({\mu \circ x}\right)$
$(3): \quad \left({\lambda \times_R \mu}\right) \circ x = \lambda \circ \left({\mu \circ x}\right)$


Checking the criteria in order:


Criterion 1

$(1): \quad \lambda \circ \left({x + y}\right) = \left({\lambda \circ x}\right) + \left({\lambda \circ y}\right)$

Let $x = \left({x_i}\right)_{i\in I}, y = \left({y_i}\right)_{i\in I} \in M$.

\(\displaystyle \lambda \circ \left({x + y}\right)\) \(=\) \(\displaystyle \lambda \circ \left({ \left({x_i}\right)_{i\in I} + \left({y_i}\right)_{i\in I} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda \circ \left({x_i+_iy_i}\right)_{i\in I}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\lambda\circ_i x_i+_i \lambda\circ_i y_i}\right)_{i\in I}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\lambda\circ_i x_i}\right)_{i\in I}+\left({\lambda\circ_i y_i}\right)_{i\in I}\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda\circ\left({x_i}\right)_{i\in I}+\lambda\circ\left({y_i}\right)_{i\in I}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\lambda \circ x}\right) + \left({\lambda \circ y}\right)\)

So $(1)$ holds.

$\Box$


Criterion 2

$(2): \quad \left({\lambda +_R \mu}\right) \circ x = \left({\lambda \circ x}\right) + \left({\mu \circ x}\right)$

Let $x = \left({x_i}\right)_{i\in I} \in M$.

\(\displaystyle \left({\lambda +_R \mu}\right) \circ x\) \(=\) \(\displaystyle \left({\lambda +_R \mu}\right) \circ \left({x_i}\right)_{i\in I}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({\lambda +_R \mu}\right) \circ_i x_i}\right)_{i\in I}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\lambda \circ_i x_i +_i \mu\circ_i x_i}\right)_{i\in I}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\lambda \circ_i x_i}\right)_{i\in I} + \left({\mu\circ_i x_i}\right)_{i\in I}\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda \circ \left({x_i}\right)_{i\in I} + \lambda \circ \left({y_i}\right)_{i\in I}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\lambda \circ x}\right) + \left({\mu \circ x}\right)\)

So $(2)$ holds.

$\Box$


Criterion 3

$(3): \quad \left({\lambda \times_R \mu}\right) \circ x = \lambda \circ \left({\mu \circ x}\right)$

Let $x = \left({x_i}\right)_{i\in I} \in M$.

\(\displaystyle \left({\lambda \times_R \mu}\right) \circ x\) \(=\) \(\displaystyle \left({\lambda \times_R \mu}\right) \circ \left({x_i}\right)_{i\in I}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({\lambda \times_R \mu}\right)\circ_i x_i}\right)_{i\in I}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\lambda\circ_i\left({\mu\circ_i x_i}\right)}\right)_{i\in I}\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda \circ \left({\mu\circ_i x_i}\right)_{i\in I}\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda \circ \left({ \mu \circ \left({x_i}\right)_{i\in I} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda \circ \left({\mu \circ x}\right)\)

So $(3)$ holds.

$\Box$


Thus all criteria are seen to hold.

The result follows.

$\blacksquare$


Also see