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 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:

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

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	2013-03-22 17:24:22
Last modified on	2013-03-22 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