# Definition:Tensor Product of Modules

## Contents

## Definition

### Commutative ring

Let $R$ be a commutative ring with unity.

Let $M$ and $N$ be $R$-modules.

### Definition 1

Their **tensor product** is a pair $(M \otimes_R N, \theta)$ where:

- $M \otimes_R N$ is an $R$-module
- $\theta : M \times N \to M \otimes_R N$ is an $R$-bilinear mapping

satisfying the following **universal property**:

- For every pair $(P, \omega)$ of an $R$-module and an $R$-bilinear mapping $\omega : M \times N \to P$, there exists a unique $R$-module homomorphism $f : M \otimes_R N \to P$ with $\omega = f \circ \theta$.

### Definition 2

Their **tensor product** is the pair $\struct {M \otimes_R N, \theta}$, where:

- $M \otimes_R N$ is the quotient of the free $R$-module $R^{\paren {M \times N} }$ on the direct product $M \times N$, by the submodule generated by the set of elements of the form:
- $\tuple {\lambda m_1 + m_2, n} - \lambda \tuple {m_1, n} - \tuple {m_2, n}$
- $\tuple {m, \lambda n_1 + n_2} - \lambda \tuple {m, n_1} - \tuple {m, n_2}$
- for $m, m_1, m_2 \in M$, $n, n_1, n_2 \in N$ and $\lambda \in R$, where we denote $tuple {m, n}$ for its image under the canonical mapping $M \times N \to R^{\paren {M \times N} }$.

- $\theta : M \times N \to M \otimes_R N$ is the composition of the canonical mapping $M \times N \to R^{\paren {M \times N} }$ with the quotient module homomorphism $R^{\paren {M \times N} } \to M \otimes_R N$.

### Noncommutative ring

Let $R$ be a ring.

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

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

First construct a left module as a direct sum of all free left modules with a basis that is a single ordered pair in $M \times N$ which is denoted $\map R {m, n}$.

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

That this is indeed a module is demonstrated in Tensor Product is Module.

Next for all $m, m' \in M$, $n, n' \in N$ and $r \in R$ we construct the following free left modules.

- $L_{m, m', n}$ with a basis of $\tuple {m + m', n}$, $\tuple {m, n}$ and $\tuple {m', n}$
- $R_{m, n, n'}$ with a basis of $\tuple {m, n + n'}$, $\tuple {m, n}$ and $\tuple {m, n'}$
- $A_{r, m, n}$ with a basis of $r \tuple {m, n}$ and $\tuple {m r, n}$
- $B_{r, m, n}$ with a basis of $r \tuple {m, n}$ and $\tuple {m, r n}$

Let:

- $D = \displaystyle \map {\bigoplus_{r \in R, n, n' \in N, m, m' \in M} } {L_{m, m', n} \oplus R_{m, n, n'} \oplus A_{r, m, n} \oplus B_{r, m, n} }$

The **tensor product** $M \otimes_R N$ is then our quotient module $T / D$.

## Also denoted as

Elements in $M \otimes N$ are commonly written as $a \otimes b$ for $a \in M$ and $b \in N$.

## Notes

This notation, and definition of quotient module, gives us the following identities for $m, m' \in M$, $n, n' \in N$ and $r \in R$:

- $\paren {m + m'} \otimes n = m \otimes n + m' \otimes n$
- $m \otimes \paren {n + n'} = m \otimes n + m \otimes n'$
- $\paren {m r} \otimes n = m \otimes \paren {r n} = r \paren {m \otimes n}$

Note should be taken that not all elements in $M \otimes N$ comes in the simple form of $a \otimes b$, as an example in $\Z \otimes \Z$ we have $2 \otimes 2 + 3 \otimes 5$ being an element in it but cannot be simplified further using the previous identities.

## Also see

## Sources

- 1989: Nicolas Bourbaki:
*Algebra I* - 2002: Serge Lang:
*Algebra*