The tensor product $M{\otimes }_{R}N$ from the additive group underlying the free $\mathbb{Z}$-module on $M\times N$, elementary tensors, and finite tensor sums

Let $R$ be a unital ring, $M$ a right $R$-module, and $N$ a left $R$-module. Let

$F:={\mathbb{Z}}^{(M\times N)}$

be the free $\mathbb{Z}$-module on the set $M\times N$ (The free module on a set and its standard basis, Universal property of the free module on a set), and write $e_{(m,n)}$ for its standard basis elements. The additive group of $F$ is abelian. Let $H$ be the subgroup generated (The subgroup $⟨S⟩$ generated by a subset, the cyclic subgroup $⟨g⟩$, and cyclic groups) by all elements

$e_{(m+m^{\mathrm{\prime }},n)}-e_{(m,n)}-e_{(m^{\mathrm{\prime }},n)},$

$e_{(m,n+n^{\mathrm{\prime }})}-e_{(m,n)}-e_{(m,n^{\mathrm{\prime }})},$

and

$e_{(mr,n)}-e_{(m,rn)}$

as $m,m^{\mathrm{\prime }}$ range over $M$, $n,n^{\mathrm{\prime }}$ over $N$, and $r$ over $R$. Since every subgroup of an abelian group is normal, the quotient group $F\mathrm{/}H$ is defined (The quotient group $G\mathrm{/}N$ and coset product $(gN)(hN)=ghN$). The tensor product of $M$ and $N$ over $R$ is

$M{\otimes }_{R}N:=F\mathrm{/}H\mathrm{.}$

The coset of $e_{(m,n)}$ is the elementary tensor $m\otimes n$. Every tensor is a finite sum of elementary tensors, because every element of $F$ is a finite $\mathbb{Z}$-linear combination of basis elements and integer coefficients may be absorbed into either additive variable. The defining relations give

$(m+m^{\mathrm{\prime }})\otimes n=m\otimes n+m^{\mathrm{\prime }}\otimes n,m\otimes (n+n^{\mathrm{\prime }})=m\otimes n+m\otimes n^{\mathrm{\prime }},$

and

$(mr)\otimes n=m\otimes (rn)\mathrm{.}$

In particular $0\otimes n=0=m\otimes 0$. No $R$-module structure on $M{\otimes }_{R}N$ is part of this arbitrary-ring definition; at this stage it is an abelian group.