Injectivity of map between tensor products with $\mathbb{Q}$ and $\mathbb{R}$ Let $G$ be an abelian group. We have a natural map $i\colon G\otimes\mathbb{Q}\to G\otimes\mathbb{R}$ given on simple tensors by
$$i(g\otimes a)=g\otimes a.$$
Question: Is $i$ injective?
Comment: I understand that for example that the analogous map $G\otimes\mathbb{Z}\to G\otimes\mathbb{Q}$ is not injective in general, since $G\otimes\mathbb{Q}$ is torsion-free. But can there be any other reason that the map $i$ above is not injective?
 A: You have the injective map of $\mathbb{Q}$ vector spaces
$$\mathbb{Q}\to \mathbb{R}$$
Tensor it with the $\mathbb{Q}$-vector space
$G\otimes_{\mathbb{Z}}\mathbb{Q}$ and get a map
$$(G\otimes_{\mathbb{Z}}\mathbb{Q})\otimes_{\mathbb{Q}} \mathbb{Q} \to (G\otimes_{\mathbb{Z}}\mathbb{Q})\otimes_{\mathbb{Q}}\mathbb{R}$$ which "is" the map
$$G\otimes_{\mathbb{Z}}\mathbb{Q}\to G\otimes_{\mathbb{Z}}\mathbb{R}$$
So now we have to show that if we tensor an injective map of $\mathbb{Q}$-vector spaces with another $\mathbb{Q}$-vector space we still get an injective map.  Reduce it to showing this: if $v_i\in V$, $w_j$ in $W$ are linearly independent in $V$, respectively $W$, then $v_i \otimes w_j$ are linearly independent in $V\otimes W$. Now, we can reduce to the case $V$, $W$ finite dimensional (avoiding Hamel bases). Indeed, if a tensor $\sum m_k\otimes n_k$ is $0$ in $M\otimes N$, then it is zero in some finitely generated submodules $M'$, $N'$, of $M$, respectively $N$ ( look at the definition-construction of the tensor product).
