Prove that this element is nonzero in a tensor product I want to solve the following problem: show that the element $1\otimes (1,1,....)$ is not the zero element in $$\mathbb{Q}\otimes_{\mathbb{Z}} \prod^{\infty}_{n\geq 2}\mathbb{Z}/n\mathbb{Z}$$.
My approach would be to try to define a map from this tensor to a $\mathbb{Z}$-module such that the element in question is not mapped to zero. I tried to start defining a map from $\mathbb{Q}\times \prod^{\infty}_{n\geq 2}\mathbb{Z}/n\mathbb{Z}$. An idea I had was to send an element $(p/q,(x_1,x_2,...))$ to $(p/q)\sum (x_i/2^i)$. This is not well defined though. Is there a way to fix this? Or a better approach to the problem?
 A: If you consider any $\mathbb{Z}$-module $M$ and its torsion part $t(M)$, we have the exact sequence
$$
0\to t(M)\to M\to M/t(M)\to0
$$
that, tensored with $\mathbb{Q}$, says
$$
\mathbb{Q}\otimes_{\mathbb{Z}}M\cong\mathbb{Q}\otimes_{\mathbb{Z}}M/t(M)
$$
So, if $1\otimes x=0$ in $\mathbb{Q}\otimes_{\mathbb{Z}}M$ (for some $x\in M$), then also $1\otimes\pi(x)=0$ in $\mathbb{Q}\otimes_{\mathbb{Z}}M/t(M)$ (for $\pi\colon M\to M/t(M)$ the canonical map).
Now, if $N$ is torsion-free and $y\in N$, $y\ne0$, then $1\otimes y\ne0$ in $\mathbb{Q}\otimes_{\mathbb{Z}}N$, because the map
$$
N\to\mathbb{Q}\otimes_{\mathbb{Z}}N,\qquad y\mapsto 1\otimes y
$$
is injective.
Note that the first part uses the fact that $\mathbb{Q}$ is divisible (so tensoring with it kills the torsion part); the second part uses that $N$ is torsionfree, so flat over $\mathbb{Z}$.
In your case, the element $(1,1,\dotsc)$ has infinite order in $M=\prod_{n>0}\mathbb{Z}/n\mathbb{Z}$, so it goes to a nonzero element in the quotient $M/t(M)$.
A: Here is an alternative approach which however only works for tensor products that can be considered as localizations: If $R$ is a commutative ring and $S\subset R$ is a multiplicative subset of $R$, then given any $R$-module $M$ the $R_S$-module $R_S\otimes_R M$ together with the map $M\to R_S\otimes_R M$, $m\mapsto 1\otimes m$, is a localization of $M$ at $S$. However, you know that the localization can also be defined as $S^{-1} M$ by taking as elements the equivalence classes of formal fractions $\frac{m}{s}$ with $s\in S$ and $m\in M$ under the relation $\frac{m_0}{s_0} = \frac{m_1}{s_1}:\Leftrightarrow t s_1 m_0 = t s_0 m_1$ for some $t\in S$, together with the map $M\to S^{-1}M$ sending $m$ to $\frac{m}{1}$. By uniqueness of localization, these two approaches are uniquely isomorphic over $R_S$ in a way compatible with the morphisms from $M$ - in particular, given $m\in M$, we have $1\otimes m=0$ in $R_S\otimes_R M$ if and only if there exists some $t\in S$ such that $tm=0$.
This applies to $R := {\mathbb Z}$, $S := {\mathbb Z}\setminus\{0\}$ (so that $R_S = {\mathbb Q}$), proving that the kernel of $M\to M\otimes_{\mathbb Z}{\mathbb Q}$ is precisely the torsion subgroup of $M$. Since in your example the element $(1,1,...)\in\prod\limits_{n\geq 2}{\mathbb Z}/n{\mathbb Z}$ is not torsion, it does therefore not vanish under the localization map to ${\mathbb Q}\otimes_{\mathbb Z}\prod\limits_{n\geq 2}{\mathbb Z}/n{\mathbb Z}$. 
A: (This answer might be difficult.)
Let $U$ be a nonprincipal ultrafilter over the set of natural numbers which contains subsets of $\Bbb{N}$ of the form $n\Bbb{N}+1$ for each $n\in \Bbb{N}$. Consider the ultraproduct $A=\prod_{n\in \Bbb{N}} (\Bbb{Z}/n\Bbb{Z})/U$. You can check that $A$ is a $\Bbb{Z}$-module under pointwise operation. We will represent the elements of $A$ of the form $[x]_U$. More precisely, $[x]_U$ is defined as
$$[x]_U=\left\{x\in\prod_{n\in \Bbb{N}} (\Bbb{Z}/n\Bbb{Z}) : \{k\in \Bbb{N}: x_k=y_k\}\in U\right\}$$
where $x=(x_1,x_2,\cdots)$ and $y=(y_1,y_2,\cdots)$
We will define a function $\phi:\Bbb{Q}\times \prod_{n\in \Bbb{N}} (\Bbb{Z}/n\Bbb{Z})\to A$, $\phi(a/b,(x_n)_{n\in\Bbb{N}})=[(ab^{-1}x_n)_{n\in \Bbb{N}}]_U$. If it is well-defined, then it gives the bilinear function and it sends $(1,(1,1,1,\cdots))$ to nonzero element of $A$. 
We must check that $\phi$ is well defined since $b^{-1}$ is not well-defined unless $\gcd(b,n)=1$. However for given $b$, the set of all $n$ such that $\gcd(b,n)>1$ can be ignored because $b\Bbb{N}+1\in U$ and $\gcd(bk+1,b)=1$ for every $k$ so $b^{-1}$ is well-defined for almost all $n$ (i.e. set of all $n$ with $(b,n)=1$ is an element of $U$.) 
A: I come a bit late but wouldn't it be easier to just say that the map :
$$ \mathbb{Q} \times \prod_{n \in \mathbb{N}} \frac{\mathbb{Z}}{n\mathbb{Z}}  \longrightarrow \prod_{n \in \mathbb{N}} \mathbb{Q}$$
and say that it is a $\mathbb{Z}$-bilinear map between $\mathbb{Z}$ modules that send the wanted element to a non zero element?
