Meaning of the statement that "$\mathcal{T}^{0}(M) = R$ and $\mathcal{T}^1(M) = M$ commute modulo the symmetric algebra $\mathcal{S}(M)$."

Question

In D & F, chap. $11.5$ page $444$, it is said that

"The tensor algebra $\mathcal{T}(M)$ is generated as a ring by $R = \mathcal{T}^{0}(M)$ and $\mathcal{T}^1(M) = M$, and these elements commute in the quotient ring $\mathcal{S}(M)$ by definition. It follows that the symmetric algebra $\mathcal{S}(M)$ is a commutative ring.

I am not sure exactly what they mean here?

Is it not generally true that if $M$ is an $R$-module, we have $$R \otimes_{R} M \cong M \otimes_{R} R \cong M$$ without regarding any quotient-structure?

I am not sure how to interpret this statement, any clarification appreciated.

Edit: And the isomorphism above is explicitly defined by $$r \otimes m \mapsto rm$$ and since $R$ is commutative, I believe we give it what is called the "standard structure" so that $$r \cdot m = m \cdot r.$$

Is that what is meant, or do they mean something else?

Answer 1

This is confusingly worded. I believe the intended meaning is that the elements of $T^0(M) = R$ and $T^1(M) = M$ commute with themselves and each other in the symmetric algebra, meaning explicitly that

• $\forall r \in R, m \in M : rm = mr \in M$
• $\forall r, s \in R : rs = sr \in R$
• $\forall m, n \in M : mn = nm \in S^2(M)$.

The first two conditions, of course, also hold in the tensor algebra. It's the third condition that specifically holds in the symmetric algebra and not the tensor algebra. Then the argument is that the symmetric algebra is generated by $T^0(M)$ and $T^1(M)$, and since all of these generators commute, the whole algebra is commutative.