# Tensor Product is Module It has been suggested that this article or section be renamed: Free Module on Set is Module One may discuss this suggestion on the talk page.

## Theorem

Let $R$ be a ring.

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

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

Then:

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

is a left module.

## Proof

### Axiom 1

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

Let $\lambda\in R$.

Then:

 $\displaystyle \lambda \circ (x + y)$ $=$ $\displaystyle \lambda \circ ((s_i)_{i\in I} + (t_i)_{i\in I})$ By definition of elements in direct sum $\displaystyle$ $=$ $\displaystyle \lambda \circ (s_i + t_i)_{i\in I}$ By addition in direct sum $\displaystyle$ $=$ $\displaystyle (\lambda \circ s_i + \lambda \circ t_i)_{i\in I}$ By $R$-action in direct sum $\displaystyle$ $=$ $\displaystyle (\lambda \circ s_i)_{i\in I} + (\lambda \circ t_i)_{i\in I}$ By addition in direct sum $\displaystyle$ $=$ $\displaystyle \lambda \circ (s_i)_{i\in I} + \lambda \circ (t_i)_{i\in I}$ By $R$-action in direct sum $\displaystyle$ $=$ $\displaystyle \lambda \circ x + \lambda \circ y$ By definition of elements in direct sum

$\Box$

### Axiom 2

Let $x \in T$ with $x = (s_i)_{i\in I}$

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

Then:

 $\displaystyle \left({\lambda + \mu}\right) \circ x$ $=$ $\displaystyle (\lambda+\mu) \circ (s_i)_{i\in I}$ By definition of elements in direct sum $\displaystyle$ $=$ $\displaystyle ((\lambda + \mu) \circ s_i)_{i\in I}$ By definition or $R$-action in direct sum $\displaystyle$ $=$ $\displaystyle (\lambda \circ s_i + \mu\circ s_i)_{i\in I}$ By definition or $R$-action in modules $\displaystyle$ $=$ $\displaystyle (\lambda \circ s_i)_{i\in I} + (\mu\circ s_i)_{i\in I}$ By definition or sum in direct sum $\displaystyle$ $=$ $\displaystyle \lambda \circ (s_i)_{i\in I} + \mu\circ (s_i)_{i\in I}$ By definition of $R$-action on direct sum $\displaystyle$ $=$ $\displaystyle \lambda \circ x + \mu\circ x$ By original equality

$\Box$

### Axiom 3

Let $x\in T$ with $x = (s_i)_{i\in I}$.

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

Then:

 $\displaystyle (\lambda \times \mu) \circ x$ $=$ $\displaystyle (\lambda \times \mu) \circ (s_i)_{i\in I}$ By original equality $\displaystyle$ $=$ $\displaystyle ((\lambda \times\mu) \circ s_i)_{i\in I}$ By Definition of $R$-action on direct sum $\displaystyle$ $=$ $\displaystyle (\lambda \circ (\mu\circ s_i))_{i\in I}$ By definition of modules $\displaystyle$ $=$ $\displaystyle \lambda \circ (\mu \circ s_i)_{i\in I}$ Definition of $R$-action on direct sum $\displaystyle$ $=$ $\displaystyle \lambda \circ \left({\mu \circ x}\right)$ By original equality

$\blacksquare$