# Tensor Product is Module

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## Contents

## 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 |

$\blacksquare$