Finding Tensor Product

Question

Let $\Pi_{n\in \mathbb{N}}\mathbb{Z}:= M$

Is $ M \otimes_{\mathbb{Z}} \mathbb{Q} \cong \Pi_{n\in \mathbb{N}}\mathbb{Q}$ ? I believe this is true but I don't know how to prove this.

Kindly help me with an idea/ hint.

Thanks in advance.

Answer 1

It is false. There's a natural map

$$\left( \prod_{\mathbb{N}} \mathbb{Z} \right) \otimes \mathbb{Q} \to \prod_{\mathbb{N}} \mathbb{Q}$$

which is injective but not surjective. Its image consists of the subspace of $\prod_{\mathbb{N}} \mathbb{Q}$ consisting of sequences whose denominators are bounded, or equivalently which can be put under a common denominator (basically because tensoring by $\mathbb{Q}$ only allows you to divide an entire integer sequence by some common denominator) and so does not contain, for example, the sequence $n \mapsto \frac{1}{n}$.

(On the other hand these groups are abstractly isomorphic because they're both vector spaces over $\mathbb{Q}$ of continuum dimension. See this math.SE answer which says basically the same thing.)

In general, tensor product is only guaranteed to preserve finite products. You can show that tensoring with a module preserves infinite products iff it's finitely presented (which $\mathbb{Q}$ is not); see this math.SE answer.