I am trying to make sense of the tensor algebra product, as given as an example in Dummit and Foote. Let $R=\mathbb{Z}$ and let $M=\mathbb{Q}/\mathbb{Z}$ and consider $M$ as an $R$-module. Since $M\otimes_R M=0$, we have that $\mathcal{T}(M)=\mathbb{Z}\oplus(\mathbb{Q}/\mathbb{Z})$.
The authors claim that the multiplication is given by $(r,\overline{p})(s,\overline{q})=(rs,\overline{rq+sp})$, but I do not see why this is true. If we consider these elements as sums, by distributing we should have $(r+\overline{p})(s+\overline{q})=rs+r\overline{q}+\overline{p}s+\overline{pq}=(rs+\overline{rq+sp+pq})$.
My question is what happens to the $\overline{pq}$ in the multiplication given by the book? Since multiplying two fractions doesn't necessarily give an integer, I'm not sure why $pq$ would be in $\mathbb{Z}$ and so would vanish in the quotient. What am I missing?
As pointed out in a comment above, the tensor algebra is graded: $$T(M)=T_0M\oplus T_1M\oplus T_2M\oplus\cdots$$ with $T_nM=M\otimes_RM\otimes_R\cdots\otimes_RM$ ($n$ copies of $M$). The multiplication goes $$T_nM\otimes_RT_mM\longrightarrow T_{n+m}M$$ and is given by concatenation, and distribution over the direct sum. In your case, when you write $$(r,\bar{p})\in T(M)$$ you mean the element $r+\bar{p}$ with $r\in R=T_0M$ and $\bar{p}\in M=T_1M$. Notice that then $$r\cdot s = r\otimes_Rs=rs\in\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}\cong\mathbb{Z}$$ and $$r\cdot\bar{q} = r\otimes_\mathbb{Z}\bar{q}=\overline{rq}\in\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q}/\mathbb{Z}\cong\mathbb{Q}/\mathbb{Z},$$ similarly for $\bar{p}\cdot s$, and $$\bar{p}\cdot\bar{q}=0$$ as it lives in $T_2M=0$.