Tensor product of modules

Remarks

A tensor product exists and is unique up to a unique isomorphism compatible with the maps $\otimes$. One construction is the free module 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 $2 \cdot (1, 1) = (2, 2)$.