1
$\begingroup$

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

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $\mathbf Z[1/p] = \{a/p^n : a \in \mathbf Z, n \geq 1\}$.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Sok
    Commented May 21 at 21:27
  • $\begingroup$ Thank you again for the explanation about the different characterization of the Prüfer group. $\endgroup$
    – Sok
    Commented May 23 at 18:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .