Inclusion $\mathbb{Q} \hookrightarrow \mathbb{R}$ induces an injection in scalar extensions

Question

Question

Given a torsion-free $\mathbb{Z}$-module (aka. abelian group) $G$, let $i: \mathbb{Q} \hookrightarrow \mathbb{R}$ be the inclusion. I want to show that $$ i \otimes \mathrm{id}: \mathbb{Q} \mathbin{\otimes_\mathbb{Z}} G \hookrightarrow \mathbb{R} \mathbin{\otimes_\mathbb{Z}} G $$ is injective.

I feel it should really work because $\mathbb{R}$ is "sufficiently independent" of $\mathbb{Q}$, but I haven't been able to show it.

Context

I'm trying to embed $G$ into a $\mathbb{R}$ vector space. I already arrived at $\mathbb{Q} \mathbin{\otimes_\mathbb{Z}} G$ by the proof of theorem 4.26.2 in this note by Keith Conrad, but it does not generalize to $\mathbb{R}$. Also, all spaces are in fact partially ordered in my case, but I noticed that most proofs can be extended to support the ordering.

Answer 1

Torsion-free modules over Dedekind domains are flat, in particular torsion-free abelian groups are flat.

Answer 2

It has nothing to see with that. Quite simply, a torsion-free $\mathbf Z$-module is flat, as is any torsion-free $A$-module over a PID or more generally over a Dedekind ring.