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(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 faithful R-modules.
If b:U×V→W is R-bilinear and (left or right) non-degenerate,
then R must be commutative
.
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((sr)u,v)=sb(ru,v)=sb(u,rv)=b(su,rv)=b((rs)u,v). |
Therefore b([s,r]u,v)=0, where [s,r]=sr-rs. This makes
[s,r]u an element of the left radical 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 |