bilinearity and commutative rings

We show that a bilinear map $b:U\times V\to W$ is almost always definable only for commutative rings.The exceptions lie only where non-trivial commutators act trivially on one of thethree modules.

Lemma 1.

Let $R$ be a ring and $U\mathrm{,}V$ and $W$ be $R$-modules.If $b\mathrm{:}U\mathrm{\times }V\mathrm{\to }W$ is $R$-bilinear then $b$ is also $R$-middle linear.

Proof.

Given $r\in R$, $u\in U$ and $v\in V$ 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 $U\mathrm{,}V$ and $W$ be faithful $R$-modules.If $b\mathrm{:}U\mathrm{\times }V\mathrm{\to }W$ 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 $r,s\in R$. Then for all $u\in U$ and $v\in V$ it follows that

$$\begin{array}{c}b((sr)u,v)=sb(ru,v)=sb(u,rv)=b(su,rv)=b((rs)u,v).\hfill \end{array}$$

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 $v\in V$.However $b$ is non-degenerate so the radical is trivial and so $[s,r]u=0$ forall $u\in U$. Since $U$ is a faithful $R$-module this makes $[s,r]=0$ for all$s,r\in R$. That is, $R$ is commutative.∎

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

Corollary 3.

Let $R$ be a ring and $U\mathrm{,}V$ and $W$ be $R$-modules.If $b\mathrm{:}U\mathrm{\times }V\mathrm{\to }W$ is $R$-bilinear with $W\mathrm{=}\mathrm{⟨}b\mathit{}\mathrm{(}U\mathrm{,}V\mathrm{)}\mathrm{⟩}$ thenevery element $\mathrm{[}R\mathrm{,}R\mathrm{]}$ acts triviallyon one of the three modules $U$, $V$ or $W$.

Proof.

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

Whenever a non-commutative ring is required for a biadditive map $U\times V\to W$it is therefore often preferable to use a scalar map instead.