The Stacks project

Example 10.89.1. Take $R = \mathbf{Z}$, $M = \mathbf{Q}$, and consider the family $Q_ n = \mathbf{Z}/n$ for $n \geq 1$. Then $\prod _ n (M \otimes Q_ n) = 0$. However there is an injection $\mathbf{Q} \to M \otimes (\prod _ n Q_ n)$ obtained by tensoring the injection $\mathbf{Z} \to \prod _ n Q_ n$ by $M$, so $M \otimes (\prod _ n Q_ n)$ is nonzero. Thus $M \otimes (\prod _ n Q_ n) \to \prod _ n (M \otimes Q_ n)$ is not injective.

On the other hand, take again $R = \mathbf{Z}$, $M = \mathbf{Q}$, and let $Q_ n = \mathbf{Z}$ for $n \geq 1$. The image of $M \otimes (\prod _ n Q_ n) \to \prod _ n (M \otimes Q_ n) = \prod _ n M$ consists precisely of sequences of the form $(a_ n/m)_{n \geq 1}$ with $a_ n \in \mathbf{Z}$ and $m$ some nonzero integer. Hence the map is not surjective.