The tensor algebra of a cyclic module is commutative

Let $R$ be a commutative ring with 1, and let $M$ be an $(R,R)$ -bimodule in the usual way. (I.e. $rm = mr$ .) Prove that if $M$ is cyclic as an $R$ -module, then the tensor algebra $\mathcal{T}(M)$ is commutative.

Note that as a ring, $\mathcal{T}(M)$ is generated by 0- and 1-tensors. Thus to show that $\mathcal{T}(M)$ is commutative, it suffices to show that these generators commute pairwise.

Certainly the 0-tensors commute with each other, since $R$ is commutative. Similarly, 0- and 1-tensors commute pairwise using the “commutativity condition” $rm = mr$ . It remains to be seen that 1-tensors commute with each other. To that end, suppose $M = Ra$ , and let $ra,sa \in M$ . Note that $ra \otimes sa = ar \otimes sa$ $= as \otimes ra$ $= sa \otimes ra$ , and indeed the 1-tensors commute. Thus $\mathcal{T}(M)$ is a commutative $R$ -algebra.