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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
- $b$ is biadditive, that is $b(u+u',v)=b(u,v)+b(u',v)$ and $b(u,v+v')=b(u,v)+b(u,v')$ for all $u,u'\in U$ and $v,v'\in V$ ;
- $b(ru,v)=rb(u,v)$ 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 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 $A$ confirms the claim.
It is immediate however that $\langle b(U,V)\rangle$ is in fact an $R$ -bimodule. This is because:$$s(b(u,v)r)=sb(u,vr)=b(su,vr)=sb(u,vr)=(sb(u,v))$$ 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.
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