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).

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: