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?
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2Where the product is taken over all the positive integers... the sum $\oplus_N Z / N Z$ does become zero when you tensor with $\mathbb{Q}$ (since tensor product commutes with colimits) - but I guess you know this, and this is why you are asking the question). – Elle Najt Jan 03 '16 at 03:44
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3I believe it's true in general that for any abelian group $A$, $\mathbb{Q}\otimes_{\mathbb{Z}} A$ is trivial iff $A$ is a torsion group (i.e. if $A$ has no elements of infinite order). If you can prove this, then since $\prod_{n\in N}\mathbb{Z}/n\mathbb{Z}$ has elements of infinite order, and is an abelian group, you would be done. – catfish Jan 03 '16 at 03:45
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@catfish Right. We have $\ker(A\to\mathbb Q\otimes_{\mathbb Z} A)=t(A)$, where $t(A)$ is the torsion of $A$. – user26857 Jan 03 '16 at 10:15
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Solved here. – user26857 Jan 03 '16 at 13:00
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1And here. – user26857 Jan 03 '16 at 15:48
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@user26857 OP hasn't been seen in weeks and I thought the question was interesting enough to be kept, so I put some context in. I hadn't seen that you had linked a duplicate in the comments, though... sorry. – Najib Idrissi Feb 12 '16 at 17:01
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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.
