scalar map
Given a ring $R$, a left $R$module $U$, a right $R$module $V$ and a twosided $R$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^{} multiplication on $R$ even when the map is nondegenerate 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 noncommutative ring $A$ confirms the claim.
It is immediate however that $\u27e8b(U,V)\u27e9$ 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 to require that indeed all of $W$ be an $R$bimodule.
Title  scalar map 

Canonical name  ScalarMap 
Date of creation  20130322 17:24:22 
Last modified on  20130322 17:24:22 
Owner  Algeboy (12884) 
Last modified by  Algeboy (12884) 
Numerical id  7 
Author  Algeboy (12884) 
Entry type  Definition 
Classification  msc 13C99 
Synonym  outer linear 
Related topic  BilinearMap 
Defines  scalar map 