How can one express $\mathbb{Q} / \mathbb{Z}$ as a direct sum of $\mathbb{Q}$ and $\mathbb{Z} / p$ for $p$ prime?

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?

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

2 Answers

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

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 $$\mathbb{Q}/\mathbb{Z}=\bigoplus_{p\text{ prime}} \mathbb{Z}(p^\infty)$$