bilinearity and commutative rings

We show that a bilinear map b:U×VW is almost always definable only for commutative rings. The exceptions lie only where non-trivial commutators act trivially on one of the three modules.

Lemma 1.

Let R be a ring and U,V and W be R-modules. If b:U×VW is R-bilinear then b is also R-middle linear.


Given rR, uU and vV then b(ru,v)=rb(u,v) and b(u,rv)=rb(u,v) so b(ru,v)=b(u,rv). ∎

Theorem 2.

Let R be a ring and U,V and W be faithfulPlanetmathPlanetmath R-modules. If b:U×VW is R-bilinear and (left or right) non-degenerate, then R must be commutativePlanetmathPlanetmathPlanetmath.


We may assume that b is left non-degenerate. Let r,sR. Then for all uU and vV it follows that


Therefore b([s,r]u,v)=0, where [s,r]=sr-rs. This makes [s,r]u an element of the left radicalPlanetmathPlanetmath of b as it is true for all vV. However b is non-degenerate so the radical is trivial and so [s,r]u=0 for all uU. Since U is a faithful R-module this makes [s,r]=0 for all s,rR. That is, R is commutative. ∎

Alternatively we can interpret the result in a weaker fashion as:

Corollary 3.

Let R be a ring and U,V and W be R-modules. If b:U×VW is R-bilinear with W=b(U,V) then every element [R,R] acts trivially on one of the three modules U, V or W.


Suppose [r,s][R,R], [r,s]U0 and [r,s]V0. Then we have shown 0=b([r,s]u,v)=[r,s]b(u,v) for all uU and vV. As W=b(U,V) it follows that [r,s]W=0. ∎

Whenever a non-commutative ring is required for a biadditive map U×VW it is therefore often preferable to use a scalar map instead.

Title bilinearity and commutative rings
Canonical name BilinearityAndCommutativeRings
Date of creation 2013-03-22 17:24:19
Last modified on 2013-03-22 17:24:19
Owner Algeboy (12884)
Last modified by Algeboy (12884)
Numerical id 5
Author Algeboy (12884)
Entry type Theorem
Classification msc 13C99