Lemma 1   Let $R$ be a ring and $U,V$ and $W$ be $R$ -modules. If $b:U\times V\to W$ 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 $U,V$ and $W$ be faithful $R$ -modules. If $b:U\times V\to W$ 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
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 $[s,r]u=0$ for all $u\in U$ . Since $U$ is a faithful $R$ -module this makes $[s,r]=0$ for all $s,r\in R$ . That is, $R$ is commutative.

Corollary 3   Let $R$ be a ring and $U,V$ and $W$ be $R$ -modules. If $b:U\times V\to W$ is $R$ -bilinear with $W=\langle b(U,V)\rangle$ then every element $[R,R]$ acts trivially on one of the three modules $U$ , $V$ or $W$ .

Proof. Suppose $[r,s]\in [R,R]$ , $[r,s]U\neq 0$ and $[r,s]V\neq 0$ . 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 $W=\langle b(U,V)\rangle$ it follows that $[r,s]W=0$ .