How to find $\mathbb{Q} / \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Q}$

Question

I just started to study tensor product and i dont know how to calculate

$\mathbb{Q} / \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Q}$

I assumed that it is $\mathbb{Q}$ and built a bilinear mapping

$f: \mathbb{Q} / \mathbb{Z} \rightarrow \mathbb{Q}$

$f(q_{1}*q_{2}) = q_{1}*q_{2}$

But it didnt help, because i cant build $f^{-1}$

$g: \mathbb{Q} \rightarrow \mathbb{Q} / \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Q}$

$g(ab) = ab \otimes 1 \neq a \otimes b$

Can you help me please?

Answer 1

The general mantra is that tensoring with a field kills torsion. Indeed, your entire object is torsion, so we should expect to see $0$, and we do see this, by just manipulating the elements.

$$\frac{a}{b} \otimes_{\mathbb{Z}} \frac{c}{d} = \frac{a}{b} \otimes_{\mathbb{Z}}\frac{bc}{bd} = \frac{a}{b} \cdot b \otimes_{\mathbb{Z}} \frac{c}{bd}= a \otimes_{\mathbb{Z}}\frac{c}{bd}= 0 \otimes_{\mathbb{Z}} \frac{c}{bd} = 0$$

Answer 2

Hint: In $A\otimes_B C$ we have $ab\otimes c= a\otimes bc$ for $a\in A, b\in B, c\in C$). In your case, let $b$ be the denominator of $a$.