Like the free group and the free module, the tensor product of two $R$-modules is another construction satisfying an interesting universal property, but with a richer structure than these two. Namely, it is (when possible) an $R$-module, denoted $M \otimes _R N,$ such that every bilinear map from $M \times N$ to another $R$-module uniquely factors through $M \otimes _R N.$

Given a unitary ring $R,$ a right $R$-module $M$ and a left $R$-module $N,$ we will first construct the tensor products of $M$ and $N$ as an abelian group. We can then, when possible, give it a module structure.

Definition: The tensor product of $M$ and $N$ over the ring $R$, noted $M \otimes _R N$ or $M \otimes N,$ is the quotient of the free $\mathbf{Z}$-module (or equivalently the free abelian group) over $M \times N$ by the subgroup generated by all elements of the form ( $\dagger$) $(m_1+m_2,n) - (m_1,n) - (m_2,n)$; $(m,n_1+n_2)-(m,n_1)-(m,n_2)$; $(mr,n) - (m,rn).$

for any $m, m_1, m_2 \in M, n,n_1, n_2 \in N$ and $r \in R.$ Elements of $M \otimes N$ are called tensors and the cosets, denoted $m \otimes n$, represented by $(m,n),$ are called simple tensors.

Note the similarity between this definition and that of the free group or free module: we took all couples in $M \times N$ and quotiented out by the minimal relations giving us the most general module possessing the properties we wanted, namely $(m_1 + m_2) \otimes n = m_1 \otimes n + m_2 \otimes n$ ; $m \otimes (n_1 + n_2) = m \otimes n_1 + m \otimes n_2$ ; $mr \otimes n = m \otimes rn.$

for all $m, m_1, m_2 \in M, n, n_1, n_2 \in N, r \in R.$ (Note the similarity between these and the definition of a bilinear map.) It is easy to see that every tensor is a finite sum of simple tensors and that $M \otimes N$ is an abelian group (for addition of tensors).

Definition: For $S$ and $R$ two rings, an $(S,R)$bimodule is a module which is a left $S$-module and a right $R$-module.

Suppose $M$ is an $(S,R)$-bimodule, then $M \otimes _R N$ can be given a left $S$-module structure by letting $s.\left(\displaystyle\sum_{finite}m_i \otimes n_i \right) = \displaystyle\sum_{finite}(s.m_i) \otimes n_i.$

An important remark is that for a commutative ring $R$ and a $R$-module $M,$ letting the left and right actions of $R$ on $M$ be equal by $r.m := m.r$ gives $M$ a natural $(R,R)$-bimodule structure and thus gives $M \otimes _R N$ a left $R$-module structure. To simplify matters, I will therefore assume in the rest of this post that $R$ is commutative and that $M$ has this natural bimodule structure.

Theorem (Universal property): Let $R$ be a commutative and unitary ring, $M, N$ and $L$ three $R$-modules and $\varphi : M \times N \to L$ a bilinear map. Then there exists a unique $R$-module homomorphism $\phi : M \otimes _R N \to L$ such that $\phi(m\otimes n) = \varphi (m,n).$ ie such that this diagram commutes : (Where $\iota(m,n) = m \otimes n.$)

Proof: By the universal property of free abelian groups (identifying abelian groups with $\textbf{Z}$-modules) the map $\varphi$ defines a unique $\textbf{Z}$-module homomorphism $\tilde\varphi : F(M \times N) \to L$ from the free $\textbf{Z}$-module on $M \times N$ to $L$ such that $\tilde\varphi(m,n) = \varphi(m,n).$ Since $\varphi$ is bilinear, $\tilde\varphi$ sends every element of the form ( $\dagger$) to $0 \in L$ so $\ker \tilde\varphi$ contains the subgroup by which we quotiented $F(M \times N)$ to obtain $M \otimes N$ at the beginning of this post. Letting $\phi : M \otimes N \to L$ be $\phi(m\otimes n) = \tilde\varphi(m,n)$ thus gives us a well-defined group homomorphism $\phi : M \otimes N \to L$ such that $\varphi = \phi \circ \iota.$ Moreover, we have $\phi(r(m\otimes n)) = \varphi(rm,n) = r \varphi(m,n) = r\phi(m \otimes n)$

so $\phi$ is also an $R$-module homomorphism. Since $\iota(M \times N)$ generates $M \otimes N,$ (every tensor is a finite sum of simple tensors) we can conclude by linearity that $\phi$ is the unique $R$-module homomorphism satisfying these properties.

QED

Note that for every $R$-module homomorphism $\phi : M \otimes _R N \to L,$ the map $\varphi(m,n) := \phi(m\otimes n)$ is an $R$-bilinear map. So by the last theorem, there is a bijection between the sets $\textbf{Bilin}_R(M \times N, L)$ and $\textbf{Hom}(M \otimes _R N, L).$

Reference: D. S. Dummit, R. M. Foote, Abstract Algebra, 3rd ed.