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.

  • $\begingroup$ This $Q_p/\mathbb{Z}$ isn't just the Prufer group? $\endgroup$ – user321268 Apr 7 '18 at 19:38
  • $\begingroup$ Here is a proof that $Q_p/\Bbb{Z}$ is divisible. This also implies that that it cannot be a free $\Bbb{Z}$-module. Of course, the simplest way is to use the observation from @mayer_vietoris' answer: Every element of this group is torsion, so even all singleton sets are linearly dependent over $\Bbb{Z}$! As the group is non-trivial it is then automatically non-free. $\endgroup$ – Jyrki Lahtonen Apr 11 '18 at 10:00

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...

  • $\begingroup$ In your first answer @DonAntonio how can you say that $\frac {\frac {k} {m}} {p^{n}} \in Q_{p}$. How do you assure that $\frac {k} {m}$ is an integer i.e. whether or not $m \mid k$. For instance if I take $m:=2$ and $g:=\frac {1} {3}$ then what is your choice $y \in Q_{3}$ for which $g=my$? $\endgroup$ – Dbchatto67 Apr 8 '18 at 5:02
  • $\begingroup$ I have accepted your answer for your alternative method as I still find difficulty to understand your first answer. $\endgroup$ – Dbchatto67 Apr 8 '18 at 5:16
  • 1
    $\begingroup$ @DebabrataChattopadhyay. Yes, the first method is wrong as I misunderstood the definition of $\;Q_p\;$ : I thought $\;k\in\Bbb Q\;$ . Good the second method, which is the usual method to prove $\;\Bbb Q\;$ is not a free abelian group, holds here in spite of my misunderstanding...I shall edit the answer. $\endgroup$ – DonAntonio Apr 8 '18 at 7:19

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.


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