TITLE: Tensor product of $\mathbb{Q}$ with an infinite product
QUESTION [6 upvotes]: How can I prove that the tensor $\mathbb{Q} \otimes \left( \prod_n \mathbb{Z}/n\mathbb{Z} \right)$, where the product is taken over all the positive
integers $n$, is not trivial?

REPLY [9 votes]: Proposition: $\mathbb{Q}$ is flat as a $\mathbb{Z}$-module. (Which is to say, the functor $ \_ \otimes_{\mathbb{Z}} \mathbb{Q}$ is exact.)
Pf: The easiest is to observe that tensoring with $\mathbb Q$ is the same as localizing at $\mathbb{Z} \setminus \{0\}$, and localization is exact.
Fancier proof: $\mathbb{Q}$ is the filtered colimit of the free modules $\mathbb{Z}[1/n]$, for $n = 1,2,3 \ldots$. A free module is flat. Now use that $\text{Tor}$ commutes with filtered colimits (since tensoring commutes with colimits, and taking cohomology commutes with filtered colimits).
Suppose that $x \in A$ is an element of infinite order. This gives some $0 \to \mathbb{Z} \to A$. Tensor this exact sequence with $\mathbb{Q}$ to get $0 \to \mathbb{Q} \to A \otimes \mathbb{Q}$, using $\mathbb{Q}$ flat to keep injectivity.
Now, the element $(1,1,\ldots) \in \Pi_n \mathbb{Z} / n \mathbb{Z}$ has infinite order.