bilinearity and commutative rings


We show that a bilinear map b:U×V→W 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×V→W is R-bilinear then b is also R-middle linear.

Proof.

Given r∈R, u∈U and v∈V then b⁢(r⁢u,v)=r⁢b⁢(u,v) and b⁢(u,r⁢v)=r⁢b⁢(u,v) so b⁢(r⁢u,v)=b⁢(u,r⁢v). ∎

Theorem 2.

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

Proof.

We may assume that b is left non-degenerate. Let r,s∈R. Then for all u∈U and v∈V it follows that

b⁢((s⁢r)⁢u,v)=s⁢b⁢(r⁢u,v)=s⁢b⁢(u,r⁢v)=b⁢(s⁢u,r⁢v)=b⁢((r⁢s)⁢u,v).

Therefore b⁢([s,r]⁢u,v)=0, where [s,r]=s⁢r-r⁢s. This makes [s,r]⁢u an element of the left radicalPlanetmathPlanetmath of b as it is true for all v∈V. However b is non-degenerate so the radical is trivial and so [s,r]⁢u=0 for all u∈U. Since U is a faithful R-module this makes [s,r]=0 for all s,r∈R. 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×V→W 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.

Proof.

Suppose [r,s]∈[R,R], [r,s]⁢U≠0 and [r,s]⁢V≠0. Then we have shown 0=b⁢([r,s]⁢u,v)=[r,s]⁢b⁢(u,v) for all u∈U and v∈V. As W=⟨b⁢(U,V)⟩ it follows that [r,s]⁢W=0. ∎

Whenever a non-commutative ring is required for a biadditive map U×V→W 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