let M and Bi be R-MODULE for all i in I Show that $M \bigotimes_R (\Pi_{i\in I} B_i) \ncong \Pi_{i \in I} (M \bigotimes B_i)$ Take $R=\mathbb{Z}$, $M=\mathbb{Q}$ and $B_n =\frac{\mathbb{Z}}{P^n \mathbb{Z}}$ , n > 0
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3Nice exercise! Where does it come from? I can see one side is zero, and the other isn't. – Angina Seng Nov 27 '18 at 19:01