Show that $Q_p / \mathbb Z$ is not free as $\mathbb Z$-module.

Question

Let $p \in \mathbb N$ be a prime. Let

$$Q_p : = \left \{ x \in \mathbb Q : (\exists k \in \mathbb Z)\ \mathrm {and}\ (\exists n \in \mathbb N)\ \mathrm {such}\ \mathrm {that}\ x= \frac {k} {p^n} \right \}.$$

Show that $Q_p / \mathbb Z$ is not free as $\mathbb Z$-module.

How do I proceed? Please tell me some way out here. Then it will be really helpful for me.

Thank you very much.

Answer 1

For any two elements $\;x:=\frac k{p^n}\;,\;\;y:=\frac j{p^m}\in Q_p\;$ , we have that

$$p^n(x+\Bbb Z)+p^m(y+\Bbb Z)=\overline 0\in Q_p/\Bbb Z$$

and thus there can exist at most one single element in a free basis for $\;Q_p/\Bbb Z\;$ . But $\;Q_p/\Bbb Z\;$ is not cyclic...

Answer 2

Another way to see this cannot be a free module, is the torsion elements it contains. Indeed, free abelian group means that there is an index set say $\Delta$ (infinite or finite), such that $Q_p/\mathbb{Z} \cong \mathbb{Z}^{(\Delta)}$. However you can observe that this isomorphism should send a basis to a basis necessarily and in particular elements of the left hand side to elements of the same virtue on the right hand side. So a simple observation gives that on the right hand side any elements has infinite degree, while definitely this is not true for $Q_p/\mathbb{Z}$ (every element has finite order; both arguments come from elementary group theory). Therefore such an isomorphism cannot exist for any $\Delta$.

In summary (and less rigorously) we say that $Q_p/\mathbb{Z}$ has non-trivial torsion. This is one way of proving it. DonAntonio's answer gives you other hints , which are helpful pedagogically to elaborate too. Also since I don't know your background if I put many details skip them, on the other hand if you believe that something is missing you're welcome to ask further questions.