## 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.

In particular, we have $m_1 \otimes m_2 - m_2 \otimes m_1 = 0$ for all $m_1$ and $m_2$, so that $\mathcal{S}(M) \cong_R \mathcal{T}(M)$.