No. Take $R=\mathbb Z$, $M=\mathbb Q$, and
$$
N_i=\mathbb Z/i\mathbb Z\qquad(i\geq2).
$$
Every $N_i$ is a torsion group, so localization at the nonzero integers gives
$$
\mathbb Q\otimes_{\mathbb Z}N_i=0
$$
and hence $\prod_i(\mathbb Q\otimes_{\mathbb Z}N_i)=0$.

By contrast, the element $x=(1\bmod i)_{i\geq2}$ of $\prod_iN_i$ has infinite order: no positive integer is divisible by every $i$. The canonical description
$$
\mathbb Q\otimes_{\mathbb Z}\prod_iN_i
\cong(\mathbb Z\setminus\{0\})^{-1}\left(\prod_iN_i\right)
$$
shows that $1\otimes x$ is nonzero, because an element dies in this localization only if one nonzero integer annihilates it. Thus the left module is nonzero while the right module is zero, giving the <tensor product and infinite direct product> counterexample.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 101 / 1 / iii / Solution

Tensor product need not commute with an infinite direct product. For example,
$$
\mathbb Q\otimes_{\mathbb Z}\prod_{n\geq2}\mathbb Z/n\mathbb Z\ne0,
\qquad
\prod_{n\geq2}\left(\mathbb Q\otimes_{\mathbb Z}\mathbb Z/n\mathbb Z\right)=0.
$$
The element $(1\bmod n)_n$ has infinite order in the product, so it survives rational localization.


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