Why is $\mathbb{R} \otimes \mathbb{S}^{1} $ infinite? Intuitively I see why this is infinite, but there is no obvious way to apply the tensor relations to reduce every element into a finite set.I don't really have an idea on how to approach this problem (here we are looking at the tensor of abelian groups).
 A: A sketch looks as follows.
Let us choose real numbers $(v_i)_{i \in I}$ such that $\{ 1 \} \sqcup \{ v_i | i \in I \}$ is a basis of $\mathbb{R}$ over $\mathbb{Q}$.
Note that as abelian groups $\mathbb{R} \cong \mathbb{Q} \oplus \bigoplus_{i \in I} \mathbb{Q}$, and $S^1 \cong \mathbb{R}/\mathbb{Z} \cong \mathbb{Q}/\mathbb{Z} \oplus \bigoplus_{i \in I} \mathbb{Q}$.
Note that $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q}/\mathbb{Z} \cong 0$. Note that $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q} \cong \mathbb{Q}$. (You only really need $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q} \neq 0$ for this proof, which follows from the existence of the bilinear mapping $(p,q) \mapsto p \cdot q$).
The formula for tensor product of a direct sum implies then, that $\mathbb{R} \otimes_{\mathbb{Z}} S^1$ can be presented as a direct sum of $|(I \sqcup \{1 \}) \times I|$ copies of $\mathbb{Q}$, which is nonzero.
A: Here’s a similar approach, but wrapped in fancier language.
$\mathbb{R}$ is a torsion-free abelian group so $\mathbb{R} \otimes -$ maps injections to injections.
Now, there is an injection $i: \mathbb{Z} \rightarrow \mathbb{R}/\mathbb{Z}$ (given by the class of any irrational number), so $\mathbb{R} \otimes i: \mathbb{R}=\mathbb{R}  \otimes \mathbb{Z} \rightarrow \mathbb{R} \otimes \mathbb{R} /\mathbb{Z}$ is still injective.
As $\mathbb{R}$ is nonzero, $\mathbb{R} \otimes \mathbb{R}/\mathbb{Z}$ is nonzero either.
