---
tag: SMFZN9E2ZB
kind: definition
name: Tensor product of modules
msc: [13C05, 15A69, 16D10]
defines: F0HQXQ0HCD
uses: [0310REJJY8, 0NZ8JA25WK, 0NRYHJPQG2, 0443XNARDV, 0DZ9DKAKZR]
sources:
  - url: https://en.wikipedia.org/wiki/Tensor_product_of_modules
---

## Definition

Let $R$ be a [[commutative ring|0310REJJY8]] and let $M$, $N$ be [[$R$-modules|0NZ8JA25WK]].
A map $\beta \colon M \times N \to P$ to an $R$-module $P$ is *$R$-bilinear* if it is $R$-[[linear|0NRYHJPQG2]] in each variable when the other is fixed.
A *tensor product* of $M$ and $N$ is an $R$-module $M \otimes_R N$ with an $R$-bilinear map $\otimes \colon M \times N \to M \otimes_R N$, written $(m, n) \mapsto m \otimes n$.
It must satisfy this universal property.
For every $R$-module $P$ and every $R$-bilinear map $\beta \colon M \times N \to P$ there is exactly one $R$-linear map $\varphi \colon M \otimes_R N \to P$ with $\varphi(m \otimes n) = \beta(m, n)$ for all $m \in M$ and $n \in N$.

## Remarks

A tensor product exists and is unique up to a unique isomorphism compatible with the maps $\otimes$.
One construction is the [[free module|0DZ9DKAKZR]] on the set $M \times N$ modulo the submodule generated by the elements $(m + m', n) - (m, n) - (m', n)$, $(m, n + n') - (m, n) - (m, n')$, $(rm, n) - r(m, n)$ and $(m, rn) - r(m, n)$.
The elements $m \otimes n$ generate $M \otimes_R N$, but a general element is a finite sum $\sum_i m_i \otimes n_i$.
Standard isomorphisms are $R \otimes_R N \cong N$, $M \otimes_R N \cong N \otimes_R M$, and $R^m \otimes_R R^n \cong R^{mn}$.

Example: for integers $a, b \ge 1$, $\Z/a\Z \otimes_\Z \Z/b\Z \cong \Z/d\Z$ with $d = \gcd(a, b)$.
The map $(x, y) \mapsto xy \bmod d$ is well defined and bilinear, so it gives a surjection onto $\Z/d\Z$; conversely $1 \otimes 1$ generates and is killed by $a$ and $b$, hence by $d$.

Non-example: the module $M \oplus N$ with the map $(m, n) \mapsto (m, n)$ is not a tensor product when $M$ and $N$ are nonzero, because this map is not bilinear.
For instance, with $M = N = \Z$, it sends $(2, 1)$ to $(2, 1)$, while bilinearity would require $2 \cdot (1, 1) = (2, 2)$.