Show that it cannot be an isomorphism I want to show that $\psi: \mathbb{Q} \otimes_{\mathbb{Z}} \left( \displaystyle \prod_{n \in \mathbb{N}} \mathbb{Z}_{2^n} \right) \to \displaystyle \prod_{n \in \mathbb{N}}(\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z}_{2^n})$ cannot be an isomorphism.
My attempt consists in showing $\displaystyle \prod_{n \in \mathbb{N}}(\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z}_{2^n})=0$ and $ \mathbb{Q} \otimes_{\mathbb{Z}} \left( \displaystyle \prod_{n \in \mathbb{N}} \mathbb{Z}_{2^n} \right) \neq 0$, but maybe it is not true.
Let $q\otimes a \in \mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z}_{2^k}$, for some $k \in \mathbb{N}$. Then $$q \otimes_{\mathbb{Z}}a = \frac{q 2^k}{2^k} \otimes_{\mathbb{Z}}a = \frac{q}{2^k} \otimes_{\mathbb{Z}} 2^ka = \frac{q}{2^k} \otimes_{\mathbb{Z}} 0 = 0.$$
So I guess the product is zero. But I don't know to show that the otherside is nonzero. Can you help me? Also, if my attempt is not correct, can you give me a hint?
 A: You have successfully show that for each $n \in \mathbb{N}$, $\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Z}_{n} \simeq 0$ as rings (and in particular, $\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Z}_{2^n} \simeq 0$. Thus, we see that $\prod\limits_{n \in \mathbb{N}} \mathbb{Q} \otimes_\mathbb{Z} \mathbb{Z}_{2^n} \simeq 0$.
It suffices to show that there is a nonzero element of $\mathbb{Q} \otimes_\mathbb{Z} \prod\limits_{n \in \mathbb{N}} \mathbb{Z}_{2^n}$. In particular, the element $1 \otimes (1, 1, ...)$ is nonzero. The proof (of a slightly altered version) of that is here.
To recap that proof, consider that $S^{-1}M \simeq S^{-1}R \otimes_R M$ for every multiplicative set $S \subseteq R$. Here, $S^{-1}M = \{(m, s) | m \in M, s \in S\} / \sim$, where $\sim$ is defined by $(m, s) \sim (m', s')$ iff there exists some $k \in S$ such that $kms' = km's$. This is the localization of a module.
In particular, consider $S = \mathbb{Z} \setminus \{0\}$. Then $\mathbb{Q} = S^{-1} \mathbb{Z}$, so we have $\mathbb{Q} \otimes \prod\limits_{n \in \mathbb{N}} \mathbb{Z}_{2^n} = S^{-1} \mathbb{Z} \otimes \prod\limits_{n \in \mathbb{N}} \mathbb{Z}_{2^n} \simeq S^{-1} \prod\limits_{n \in \mathbb{N}} \mathbb{Z}_{2^n}$.
Now suppose that $(1, 1, ...) = 0$ in $S^{-1} \prod\limits_{n \in \mathbb{N}} \mathbb{Z}_{2^n}$. Then there must exist some $k \in S$ such that $k (1, 1, ...) = 0$ in $\prod\limits_{n \in \mathbb{N}} \mathbb{Z}_{2^n}$. That is, we have $(k, k, ...) = 0$ in $\prod\limits_{n \in \mathbb{N}} \mathbb{Z}_{2^n}$. That is, for all $n$, $k \equiv 0 \mod 2^n$. But clearly, this can only hold for $k = 0$, and $0 \notin S$. Thus, we see that $(1, 1, ...) \neq 0$ in $S^{-1} \prod\limits_{n \in \mathbb{N}} \mathbb{Z}_{2^n}$.
