Tensor Product is Module

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Theorem

Let $R$ be a ring.

Let $M$ be a $R$-right module.

Let $N$ be a $R$-left module.

Then:

$\ds T = \bigoplus_{s \mathop \in M \times N} R s$

is a left module.

Proof

Axiom 1

Let $x, y \in T$ with $x = \family {s_i}_{i \mathop \in I}$ and $y = (t_i)_{i \mathop \in I}$.

Let $\lambda\in R$.

Then:

\(\ds \lambda \circ \paren {x + y}\)

\(=\)

\(\ds \lambda \circ (\family {s_i}_{i \mathop \in I} + \family {t_i}_{i \mathop \in I})\)

By definition of elements in direct sum

\(\ds \)

\(=\)

\(\ds \lambda \circ \family {s_i + t_i}_{i \mathop \in I}\)

By addition in direct sum

\(\ds \)

\(=\)

\(\ds \family {\lambda \circ s_i + \lambda \circ t_i}_{i \mathop \in I}\)

By $R$-action in direct sum

\(\ds \)

\(=\)

\(\ds \family {\lambda \circ s_i}_{i \mathop \in I} + \family {\lambda \circ t_i}_{i \mathop \in I}\)

By addition in direct sum

\(\ds \)

\(=\)

\(\ds \lambda \circ \family {s_i}_{i \mathop \in I} + \lambda \circ \family {t_i}_{i \mathop \in I}\)

By $R$-action in direct sum

\(\ds \)

\(=\)

\(\ds \lambda \circ x + \lambda \circ y\)

By definition of elements in direct sum

$\Box$

Axiom 2

Let $x \in T$ with $x = \family {s_i}_{i \mathop \in I}$

Let $\lambda, \mu \in R$.

Then:

\(\ds \paren {\lambda + \mu} \circ x\)

\(=\)

\(\ds \paren {\lambda + \mu} \circ \family {s_i}_{i \mathop \in I}\)

By definition of elements in direct sum

\(\ds \)

\(=\)

\(\ds \family {\paren {\lambda + \mu} \circ s_i}_{i \mathop \in I}\)

By definition or $R$-action in direct sum

\(\ds \)

\(=\)

\(\ds \family {\lambda \circ s_i + \mu \circ s_i}_{i \mathop \in I}\)

By definition or $R$-action in modules

\(\ds \)

\(=\)

\(\ds \family {\lambda \circ s_i}_{i \mathop \in I} + \family {\mu \circ s_i}_{i \mathop \in I}\)

By definition or sum in direct sum

\(\ds \)

\(=\)

\(\ds \lambda \circ \family {s_i}_{i \mathop \in I} + \mu \circ \family {s_i}_{i \mathop \in I}\)

By definition of $R$-action on direct sum

\(\ds \)

\(=\)

\(\ds \lambda \circ x + \mu \circ x\)

By original equality

$\Box$

Axiom 3

Let $x\in T$ with $x = \family {s_i}_{i \mathop \in I}$.

Let $\lambda, \mu \in R$.

Then:

\(\ds \paren {\lambda \times \mu} \circ x\)

\(=\)

\(\ds \paren {\lambda \times \mu} \circ \family {s_i}_{i \mathop \in I}\)

By original equality

\(\ds \)

\(=\)

\(\ds \family {\paren {\lambda \times \mu} \circ s_i}_{i \mathop \in I}\)

By Definition of $R$-action on direct sum

\(\ds \)

\(=\)

\(\ds \family {\lambda \circ \paren {\mu \circ s_i} }_{i \mathop \in I}\)

Definition of Module

\(\ds \)

\(=\)

\(\ds \lambda \circ \family {\mu \circ s_i}_{i \mathop \in I}\)

Definition of $R$-action on direct sum

\(\ds \)

\(=\)

\(\ds \lambda \circ \paren {\mu \circ x}\)

By original equality

$\blacksquare$