# The Prüfer group $\mathbb{Z}(p^{\infty})$ as a $\mathbb{Z}$ module is not projective.

The Prüfer group $$\mathbb{Z}(p^{\infty})$$ as a $$\mathbb{Z}$$ module is not projective. Suppose that $$\mathbb{Z}(p^{\infty})$$ is a $$\mathbb{Z}$$ projective module, then $$\mathbb{Z}(p^{\infty})$$ is isomorphic to $$\mathbb{Z}$$ module $$K$$ such that $$K \oplus K' = F$$, where $$F$$ is a free $$\mathbb{Z}$$ module.So, on the one hand since $$\mathbb{Z}$$ is an integer domain every submodule of $$F$$ (a free module) should be torsion-free.See

A submodule of a free module is torsion-free?

On the other hand, is not difficult to show that $$\mathbb{Q} / \mathbb{Z}$$ is a torsion, module, see

$\mathbb{Q}/\mathbb{Z}$ is a torsion $\mathbb{Z}$-module. But $$\mathbb{Z}(p^{\infty})$$ as a $$\mathbb{Z}$$ is submodule of $$\mathbb{Q} / \mathbb{Z}$$, therefore $$\mathbb{Z}(p^{\infty})$$ is a torsion module contained in a free module over an integer domain. It yields that $$\mathbb{Z}(p^{\infty})$$ as a $$\mathbb{Z}$$ module is not projective. Is my proof correct? $$\mathbb{Z}(p^{\infty})$$ as a group or module is still a little bit alien to me, is there good reference on how to see this group as the $$p$$ primary components of $$\mathbb{Q} / \mathbb{Z}$$ or as a Sylow $$p$$-subrgroup of $$\mathbb{Q} / \mathbb{Z}$$ ??

A projective module over an integral domain is torsion-free, so a nonzero module in which every element is torsion is about the worst thing you could even imagine might be projective. Your proof is correct.

You say $$\mathbf Z(p^\infty)$$ is a bit alien to you and you seek a reference that explains why $$\mathbf Z(p^\infty)$$ is (isomorphic to) the $$p$$-primary subgroup of $$\mathbf Q/\mathbf Z$$. It'd help if you told us how you define $$\mathbf Z(p^\infty)$$, since maybe you're just using a definition that makes things harder to see than they have to be.

The roots of unity in the unit circle are not defined as $$\mathbf Q/\mathbf Z$$, but they are isomorphic to it: a root of unity is $$e^{2\pi i r}$$ where $$r \in \mathbf Q$$, and $$r$$ is only determined by $$e^{2\pi ir}$$ up to adding an integer (because $$e^{2\pi ix} = 1$$ only when $$x \in \mathbf Z$$). Or you could think about the map $$\mathbf Q \to S^1$$ where $$r \mapsto e^{2\pi ir}$$: it is a homomorphism whose image is the roots of unity and its kernel is $$\mathbf Z$$, so $$\mathbf Q/\mathbf Z$$ is isomorphic to the roots of unity.

The $$p$$-power roots of unity are all $$e^{2\pi ir}$$ where $$r$$ has a $$p$$-power denominator. These $$r$$ are the fractions $$a/b$$ where $$b$$ is a pure $$p$$-power and we only care about these numbers up to addition by integers. So it is $$\mathbf Z[1/p]/\mathbf Z$$ as a quotient group, where $$\mathbf Z[1/p] = \{a/p^n : a \in \mathbf Z, n \geq 1\}$$.

• Thanks so much!! Yes, what troubles me about $\mathbb{Z}(p^{\infty})$ is that it is defined most of the time as the subgroup of$p^{n}$-th rooths of the unity of the circle group which I think is defined as $\mathbb{Q} / \mathbb{Z}$ right ? But it is also defined as a the Sylow $p$ subgroup of the quotient group $\mathbb{Q} / \mathbb{Z}$ which makes sense. However I don't know why this both definitions are equivalent.
– Sok
Commented May 21 at 21:27
• Thank you again for the explanation about the different characterization of the Prüfer group.
– Sok
Commented May 23 at 18:53