I'm trying to show
$\mathbb{R} \otimes_\mathbb{Z} (\mathbb{R} / \mathbb{Z})$ is nontrivial,
where my tensor product is defined using the universal property.
This problem can be easily reduced to
Construct a nontrivial $\mathbb{Z}$-bilinear map from $\mathbb{R} \times (\mathbb{R} / \mathbb{Z})$ to some $\mathbb{Z}$-module $M$.
I solved this problem by taking a bypath:
Sketch of Proof. Notice that $\mathbb{R}$ and $\mathbb{R} / \mathbb{Q}$ are $\mathbb{Q}$-vector spaces, so $\mathbb{R} \otimes_\mathbb{Q} (\mathbb{R} / \mathbb{Q})$ is nontrivial. In particular, we have a nontrivial $\mathbb{Q}$-bilinear map $f: \mathbb{R} \times (\mathbb{R} / \mathbb{Q}) \to \mathbb{R} \otimes_\mathbb{Q} (\mathbb{R} / \mathbb{Q})$. Then we can construct the desired map basically by projecting. More precisely, let $\tilde{f}: \mathbb{R} \times (\mathbb{R} / \mathbb{Z}) \to \mathbb{R} \otimes_\mathbb{Q} (\mathbb{R} / \mathbb{Q})$ be $$ \tilde{f}(r, s + \mathbb{Z}) = f(r, s + \mathbb{Q}). $$ Then it is easy to check $\tilde{f}$ is well-defined, and in fact $\mathbb{Z}$-bilinear. It is nontrivial since $f$ is nontrivial. QED.
I'm not satisfied with this proof though. First, it smells somewhat suspicious (I can't find a gap though). Second, it's always better to find a more explicit construction — $\mathbb{R} \otimes_\mathbb{Q} (\mathbb{R} / \mathbb{Q})$ is a mess. (The only explicit thing I could think of is something like $(r, s+\mathbb{Z}) \mapsto rs + \mathbb{Z}$, which is obviously not well defined.) Any good ideas? Thanks in advance.
Consider a direct sum decomposition $$ \def\R{\mathbb{R}}\def\Z{\mathbb{Z}} \R/\Z=\mathbb{Q}/\Z\oplus H $$ where $H\ne\{0\}$ is torsionfree (the torsion part of $\R/\Z$ is indeed $\mathbb{Q}/\Z$). Then $$ \R\otimes\R/\Z\cong \R\otimes H $$ and both $H$ and $\R$ are divisible torsionfree groups, so isomorphic to direct sums of copies of $\mathbb{Q}$. Apply the distributivity of the tensor product with respect to direct sums.