Balanced maps from a right module and a left module, and bilinear maps over a commutative ring

Let $R$ be a unital ring, let $M$ be a right $R$-module, let $N$ be a left $R$-module, and let $A$ be an abelian group, written additively (Unital left and right modules over a ring; unqualified module means left module, Group and abelian group). A map $b:M\times N\to A$ is $R$-balanced if, for all $m,m^{\mathrm{\prime }}\in M$, $n,n^{\mathrm{\prime }}\in N$, and $r\in R$,

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

and

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

Thus a balanced map is additive in each variable and identifies the two ways in which a scalar may cross the pair. Additivity implies $b(0,n)=0=b(m,0)$.

If $R$ is commutative (Commutative ring) and $M,N,P$ are $R$-modules, a map $b:M\times N\to P$ is $R$-bilinear if it is $R$-linear in each variable. Equivalently, it is additive in each variable and satisfies

$b(rm,n)=rb(m,n)=b(m,rn)$

for all $r,m,n$. After a commutative-ring module is regarded as a right module by $mr:=rm$, every bilinear map is balanced.