scalar map

Given a ring $R$, a left $R$-module $U$, a right $R$-module $V$ and a two-sided$R$-module $W$then a map $b:U\times V\to W$ is an $R$-scalar map if

1. $b$ is biadditive, that is $b(u+u^{\prime },v)=b(u,v)+b(u^{\prime },v)$ and $b(u,v+v^{\prime })=b(u,v)+b(u,v^{\prime })$for all $u,u^{\prime }\in U$ and $v,v^{\prime }\in V$;

2. $b(ru,v)=rb(u,v)$ and $b(u,vr)=b(u,v)r$ for all $u\in U$, $v\in V$ and $r\in R$.

Such maps can also be called outer linear.

Unlike bilinear maps, scalar maps do not force a commutative multiplicationon $R$ even when the map is non-degenerate and the modules are faithful.For example, if $A$ is an associative ring then the multiplication of $A$,$b:A\times A\to A$ is a $A$-outer linear:

$$b(xy,z)=(xy)z=x(yz)=xb(y,z)$$

and likewise $b(x,yz)=b(x,y)z$. Using a non-commutative ring $A$ confirmsthe claim.

It is immediate however that $⟨b(U,V)⟩$ is in fact an $R$-bimodule.This is because:

$$s(b(u,v)r)=sb(u,vr)=b(su,vr)=sb(u,vr)=(sb(u,v))r$$

for all $u\in U$, $v\in V$ and $s,r\in R$. Therefore it is not uncommon torequire that indeed all of $W$ be an $R$-bimodule.