3
$\begingroup$

Consider the $\mathbb{Z}$ modules $\mathbb{Q}$ and $\mathbb{Z} / p$ for $p$ prime. I have a result that says that every injective $\mathbb{Z}$ module is a direct sum of these modules. I also know that $\mathbb{Q} / \mathbb{Z}$ is an injective $\mathbb{Z}$ module, being divisible. How can I reconcile these two notions and express $\mathbb{Q} / \mathbb{Z}$ as a direct sum of $\mathbb{Q}$ and $\mathbb{Z} / p$ for $p$ prime? This doesn't seem possible to me. Am I interpreting one of my stated results in the wrong way?

$\endgroup$
  • 2
    $\begingroup$ $\newcommand{\Z}{\mathbb{Z}}$It's not $\Z/p$, but rather the Prüfer group $\Z(p^{\infty})$. $\endgroup$ – Andreas Caranti May 23 at 10:39
  • $\begingroup$ @AndreasCaranti Isn't that an answer? Why are you putting it in a comment? $\endgroup$ – Arthur May 23 at 10:59
  • $\begingroup$ @Arthur, right, will move it to an answer. $\endgroup$ – Andreas Caranti May 23 at 11:02
6
$\begingroup$

$\newcommand{\Z}{\mathbb{Z}}$It's not $\Z / p$, but rather the Prüfer group $\Z(p^{\infty})$.

$\endgroup$
2
$\begingroup$

Expanding on what Andreas has said.

$\mathbb{Z}/p$ is not injective (tensor it in $0\to\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/p\to 0$). What you should have gotten instead is $\mathbb{Z}(p^\infty)=\injlim_{n\to\infty}\mathbb{Z}/p^n$ being divisible (i.e. injective). Then $$ \mathbb{Q}/\mathbb{Z}=\bigoplus_{p\text{ prime}} \mathbb{Z}(p^\infty) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.